PHYSICAL CHEMISTRY

ATKINS’ PHYSICAL CHEMISTRY This page intentionally left blank ATKINS’ PHYSICAL CHEMISTRY Eighth Edition Peter Atkins Professor of Chemistry, University of Oxford, and Fellow of Lincoln College, Oxford Julio de Paula Professor and Dean of the College of Arts and Sciences Lewis and Clark College, Portland, Oregon W. H. Freeman and Company New York Library of Congress Control Number: 2005936591 Physical Chemistry, Eighth Edition © 2006 by Peter Atkins and Julio de Paula All rights reserved ISBN: 0-7167-8759-8 EAN: 9780716787594 Published in Great Britain by Oxford University Press This edition has been authorized by Oxford University Press for sale in the United States and Canada only and not for export therefrom. First printing W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 www.whfreeman.com Preface We have taken the opportunity to refresh both the content and presentation of this text while—as for all its editions—keeping it flexible to use, accessible to students, broad in scope, and authoritative. The bulk of textbooks is a perennial concern: we have sought to tighten the presentation in this edition. However, it should always be borne in mind that much of the bulk arises from the numerous pedagogical features that we include (such as Worked examples and the Data section), not necessarily from density of information. The most striking change in presentation is the use of colour. We have made every effort to use colour systematically and pedagogically, not gratuitously, seeing as a medium for making the text more attractive but using it to convey concepts and data more clearly. The text is still divided into three parts, but material has been moved between chapters and the chapters have been reorganized. We have responded to the shift in emphasis away from classical thermodynamics by combining several chapters in Part 1 (Equilibrium), bearing in mind that some of the material will already have been covered in earlier courses. We no longer make a distinction between ‘concepts’ and ‘machinery’, and as a result have provided a more compact presentation of thermodynamics with less artificial divisions between the approaches. Similarly, equilibrium electrochemistry now finds a home within the chapter on chemical equilibrium, where space has been made by reducing the discussion of acids and bases. In Part 2 (Structure) the principal changes are within the chapters, where we have sought to bring into the discussion contemporary techniques of spectroscopy and approaches to computational chemistry. In recognition of the major role that physical chemistry plays in materials science, we have a short sequence of chapters on materials, which deal respectively with hard and soft matter. Moreover, we have introduced concepts of nanoscience throughout much of Part 2. Part 3 has lost its chapter on dynamic electrochemistry, but not the material. We regard this material as highly important in a contemporary context, but as a final chapter it rarely received the attention it deserves. To make it more readily accessible within the context of courses and to acknowledge that the material it covers is at home intellectually with other material in the book, the description of electron transfer reactions is now a part of the sequence on chemical kinetics and the description of processes at electrodes is now a part of the general discussion of solid surfaces. We have discarded the Boxes of earlier editions. They have been replaced by more fully integrated and extensive Impact sections, which show how physical chemistry is applied to biology, materials, and the environment. By liberating these topics from their boxes, we believe they are more likely to be used and read; there are end-ofchapter problems on most of the material in these sections. In the preface to the seventh edition we wrote that there was vigorous discussion in the physical chemistry community about the choice of a ‘quantum first’ or a ‘thermodynamics first’ approach. That discussion continues. In response we have paid particular attention to making the organization flexible. The strategic aim of this revision is to make it possible to work through the text in a variety of orders and at the end of this Preface we once again include two suggested road maps. The concern expressed in the seventh edition about the level of mathematical ability has not evaporated, of course, and we have developed further our strategies for showing the absolute centrality of mathematics to physical chemistry and to make it accessible. Thus, we give more help with the development of equations, motivate vi PREFACE them, justify them, and comment on the steps. We have kept in mind the struggling student, and have tried to provide help at every turn. We are, of course, alert to the developments in electronic resources and have made a special effort in this edition to encourage the use of the resources on our Web site (at www.whfreeman.com/pchem8) where you can also access the eBook. In particular, we think it important to encourage students to use the Living graphs and their considerable extension as Explorations in Physical Chemistry. To do so, wherever we call out a Living graph (by an icon attached to a graph in the text), we include an Exploration in the figure legend, suggesting how to explore the consequences of changing parameters. Overall, we have taken this opportunity to refresh the text thoroughly, to integrate applications, to encourage the use of electronic resources, and to make the text even more flexible and up to date. Oxford Portland P.W.A. J.de P. PREFACE vii About the book There are numerous features in this edition that are designed to make learning physical chemistry more effective and more enjoyable. One of the problems that make the subject daunting is the sheer amount of information: we have introduced several devices for organizing the material: see Organizing the information. We appreciate that mathematics is often troublesome, and therefore have taken care to give help with this enormously important aspect of physical chemistry: see Mathematics and Physics support. Problem solving—especially, ‘where do I start?’—is often a challenge, and we have done our best to help overcome this first hurdle: see Problem solving. Finally, the web is an extraordinary resource, but it is necessary to know where to start, or where to go for a particular piece of information; we have tried to indicate the right direction: see About the Web site. The following paragraphs explain the features in more detail. Organizing the information Checklist of key ideas Checklist of key ideas 1. A gas is a form of matter that fills any container it occupies. 2. An equation of state interrelates pressure, volume, temperature, and amount of substance: p = f(T,V,n). 3. The pressure is the force divided by the area to which the force is applied. The standard pressure is p7 = 1 bar (105 Pa). 4. Mechanical equilibrium is the condition of equality of pressure on either side of a movable wall. 5. Temperature is the property that indicates the direction of the flow of energy through a thermally conducting, rigid wall. 6. A diathermic boundary is a boundary that permits the passage of energy as heat. An adiabatic boundary is a boundary that prevents the passage of energy as heat. 7. Thermal equilibrium is a condition in which no change of state occurs when two objects A and B are in contact through a diathermic boundary. 8. The Zeroth Law of thermodynamics states that, if A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then C is also in thermal equilibrium with A. 9. The Celsius and thermodynamic temperature scales are related by T/K = θ/°C + 273.15. 10. A perfect gas obeys the perfect gas equation, pV = nRT, exactly 12. The partial pressure of any gas i xJ = nJ/n is its mole fraction in a pressure. 13. In real gases, molecular interact state; the true equation of state i coefficients B, C, . . . : pVm = RT Here we collect together the major concepts introduced in the chapter. We suggest checking off the box that precedes each entry when you feel confident about the topic. 14. The vapour pressure is the press with its condensed phase. 15. The critical point is the point at end of the horizontal part of the a single point. The critical const pressure, molar volume, and tem critical point. 16. A supercritical fluid is a dense fl temperature and pressure. 17. The van der Waals equation of s the true equation of state in whi by a parameter a and repulsions parameter b: p = nRT/(V − nb) − 18. A reduced variable is the actual corresponding critical constant IMPACT ON NANOSCIENCE I20.2 Nanowires We have already remarked (Impacts I9.1, I9.2, and I19.3) that research on nanometre-sized materials is motivated by the possibility that they will form the basis for cheaper and smaller electronic devices. The synthesis of nanowires, nanometre-sized atomic assemblies that conduct electricity, is a major step in the fabrication of nanodevices. An important type of nanowire is based on carbon nanotubes, which, like graphite, can conduct electrons through delocalized π molecular orbitals that form from unhybridized 2p orbitals on carbon. Recent studies have shown a correlation between structure and conductivity in single-walled nanotubes (SWNTs) that does not occur in graphite. The SWNT in Fig. 20.45 is a semiconductor. If the hexagons are rotated by 60° about their sixfold axis, the resulting SWNT is a metallic conductor. Carbon nanotubes are promising building blocks not only because they have useful electrical properties but also because they have unusual mechanical properties. For example, an SWNT has a Young’s modulus that is approximately five times larger and a tensile strength that is approximately 375 times larger than that of steel. Silicon nanowires can be made by focusing a pulsed laser beam on to a solid target composed of silicon and iron. The laser ejects Fe and Si atoms from the surface of the Impact sections Where appropriate, we have separated the principles from their applications: the principles are constant and straightforward; the applications come and go as the subject progresses. The Impact sections show how the principles developed in the chapter are currently being applied in a variety of modern contexts. ABOUT THE BOOK q Notes on good practice A note on good practice We write T = 0, not T = 0 K for the zero temperature on the thermodynamic temperature scale. This scale is absolute, and the lowest temperature is 0 regardless of the size of the divisions on the scale (just as we write p = 0 for zero pressure, regardless of the size of the units we adopt, such as bar or pascal). However, we write 0°C because the Celsius scale is not absolute. The material on regular solutions presented in Section 5.4 gives further insight into the origin of deviations from Raoult’s law and its relation to activity coefficients. The starting point is the expression for the Gibbs energy of mixing for a regular solution (eqn 5.31). We show in the following Justification that eqn 5.31 implies that the activity coefficients are given by expressions of the form ln γB = βxA2 Science is a precise activity and its language should be used accurately. We have used this feature to help encourage the use of the language and procedures of science in conformity to international practice and to help avoid common mistakes. Justifications 5.8 The activities of regular solutions ln γA = βxB2 ix (5.57) These relations are called the Margules equations. Justification 5.4 The Margules equations The Gibbs energy of mixing to form a nonideal solution is On first reading it might be sufficient to appreciate the ‘bottom line’ rather than work through detailed development of a mathematical expression. However, mathematical development is an intrinsic part of physical chemistry, and it is important to see how a particular expression is obtained. The Justifications let you adjust the level of detail that you require to your current needs, and make it easier to review material. ∆mixG = nRT{xA ln aA + xB ln aB} This relation follows from the derivation of eqn 5.31 with activities in place of mole fractions. If each activity is replaced by γ x, this expression becomes ∆mixG = nRT{xA ln xA + xB ln xB + xA ln γA + xB ln γB} Now we introduce the two expressions in eqn 5.57, and use xA + xB = 1, which gives ∆mixG = nRT{xA ln xA + xB ln xB + βxAx B2 + βxBxA2} = nRT{xA ln xA + xB ln xB + βxAxB(xA + xB)} = nRT{xA ln xA + xB ln xB + βxAxB} as required by eqn 5.31. Note, moreover, that the activity coefficients behave correctly for dilute solutions: γA → 1 as xB → 0 and γB → 1 as xA → 0. Molecular interpretation 5.2 The lowering of vapour pressure of a solvent in a mixture The molecular origin of the lowering of the chemical potential is not the energy of interaction of the solute and solvent particles, because the lowering occurs even in an ideal solution (for which the enthalpy of mixing is zero). If it is not an enthalpy effect, it must be an entropy effect. The pure liquid solvent has an entropy that reflects the number of microstates available to its molecules. Its vapour pressure reflects the tendency of the solution towards greater entropy, which can be achieved if the liquid vaporizes to form a gas. When a solute is present, there is an additional contribution to the entropy of the liquid, even in an ideal solution. Because the entropy of the liquid is already higher than that of the pure liquid, there is a weaker tendency to form the gas (Fig. 5.22). The effect of the solute appears as a lowered vapour pressure, and hence a higher boiling point. Similarly, the enhanced molecular randomness of the solution opposes the tendency to freeze. Consequently, a lower temperature must be reached before equilibrium between solid and solution is achieved. Hence, the freezing point is lowered. Molecular interpretation sections Historically, much of the material in the first part of the text was developed before the emergence of detailed models of atoms, molecules, and molecular assemblies. The Molecular interpretation sections enhance and enrich coverage of that material by explaining how it can be understood in terms of the behaviour of atoms and molecules. x ABOUT THE BOOK Further information Further information Further information 5.1 The Debye–Hückel theory of ionic solutions Imagine a solution in which all the ions have their actual positions, but in which their Coulombic interactions have been turned off. The difference in molar Gibbs energy between the ideal and real solutions is equal to we, the electrical work of charging the system in this arrangement. For a salt M p Xq, we write where rD is called the Debye length. Wh potential is virtually the same as the uns small, the shielded potential is much sm potential, even for short distances (Fig. 1.0 ideal Gm 5 4 4 6 4 4 7 5 4 6 4 7 Gm we = (pµ+ + qµ −) − (pµ +ideal + qµ −ideal) 0.8 Potential, !/(Z /rD) = p(µ+ − µ +ideal) + q(µ− − µ −ideal) From eqn 5.64 we write µ+ − µ +ideal = µ− − µ −ideal = RT ln γ± So it follows that ln γ± = we s=p+q sRT (5.73) Zi Zi = r zte φi = 966 r 0.4 0.3 0 0 4πε −r/rD e (5.75) 1 3 0.5 Distan (5.74) The ionic atmosphere causes the potential to decay with distance more sharply than this expression implies. Such shielding is a familiar problem in electrostatics, and its effect is taken into account by replacing the Coulomb potential by the shielded Coulomb potential, an expression of the form Zi 0.6 0.2 This equation tells us that we must first find the final distribution of the ions and then the work of charging them in that distribution. The Coulomb potential at a distance r from an isolated ion of charge zie in a medium of permittivity ε is φi = In some cases, we have judged that a derivation is too long, too detailed, or too different in level for it to be included in the text. In these cases, the derivations will be found less obtrusively at the end of the chapter. The variation of the shielded C distance for different values of the Deby Debye length, the more sharply the pote case, a is an arbitrary unit of length. Fig. 5.36 Exploration Write an expression f unshielded and shielded Coulom Then plot this expression against rD and interpretation for the shape of the plot. Appendices Appendix 2 MATHEMATICAL TECHNIQUES A2.6 Partial derivatives A partial derivative of a function of more than one variable of the function with respect to one of the variables, all the constant (see Fig. 2.*). Although a partial derivative show when one variable changes, it may be used to determine when more than one variable changes by an infinitesimal a tion of x and y, then when x and y change by dx and dy, res df = Physical chemistry draws on a lot of background material, especially in mathematics and physics. We have included a set of Appendices to provide a quick survey of some of the information relating to units, physics, and mathematics that we draw on in the text. A ∂f D A ∂f D dx + dy C ∂x F y C ∂y F x where the symbol ∂ is used (instead of d) to denote a parti df is also called the differential of f. For example, if f = ax 3y A ∂f D = 3ax 2y C ∂x F y 1000 A ∂f D = ax 3 + 2by C ∂y F x Synoptic tables and the Data section DATA SECTION Table 2.8 Expansion coefficients, α, and isothermal compressibilities, κT a/(10 − 4 K−1 ) Table 2.9 Inversion temperatures, no points, and Joule–Thomson coefficient kT /(10 −6 atm−1 ) Liquids TI /K Air Benzene 12.4 92.1 Argon Carbon tetrachloride 12.4 90.5 Carbon dioxide Ethanol 11.2 76.8 Helium Mercury 1.82 38.7 Hydrogen Water 2.1 49.6 Krypton Solids Copper 0.501 0.735 Diamond 0.030 0.187 Iron 0.354 0.589 Lead 0.861 2.21 The values refer to 20°C. Data: AIP(α), KL(κT). Tf /K 603 723 83.8 1500 194.7s 40 202 14.0 1090 116.6 Methane 968 90.6 Neon 231 24.5 Nitrogen 621 63.3 Oxygen 764 54.8 s: sublimes. Data: AIP, JL, and M.W. Zemansky, Heat and New York (1957). Long tables of data are helpful for assembling and solving exercises and problems, but can break up the flow of the text. We provide a lot of data in the Data section at the end of the text and short extracts in the Synoptic tables in the text itself to give an idea of the typical values of the physical quantities we are introducing. ABOUT THE BOOK xi Mathematics and Physics support e n s r e , Comment 2.5 Comment 1.2 A hyperbola is a curve obtained by plotting y against x with xy = constant. e e The partial-differential operation (∂z/∂x)y consists of taking the first derivative of z(x,y) with respect to x, treating y as a constant. For example, if z(x,y) = x 2y, then A ∂z D A ∂[x 2y] D dx 2 B E =B E =y = 2yx C ∂x F y C ∂x F y dx Comments A topic often needs to draw on a mathematical procedure or a concept of physics; a Comment is a quick reminder of the procedure or concept. Partial derivatives are reviewed in Appendix 2. 978 Appendices Appendix 3 ESSENTIAL CONCEPTS OF PHYSICS Classical mechanics Classical mechanics describes the behaviour of objects in t expresses the fact that the total energy is constant in the ab other expresses the response of particles to the forces acti pz p There is further information on mathematics and physics in Appendices 2 and 3, respectively. These appendices do not go into great detail, but should be enough to act as reminders of topics learned in other courses. A3.3 The trajectory in terms of the energy The velocity, V, of a particle is the rate of change of its po V= py dr dt The velocity is a vector, with both direction and magnit velocity is the speed, v. The linear momentum, p, of a pa its velocity, V, by px p = mV The linear momentum of a particle is a vector property and points in the direction of motion. A3.1 Like the velocity vector, the linear momentum vector poi of the particle (Fig. A3.1). In terms of the linear momentu ticle is 2 Problem solving Illustrations Illustration 5.2 Using Henry’s law To estimate the molar solubility of oxygen in water at 25°C and a partial pressure of 21 kPa, its partial pressure in the atmosphere at sea level, we write bO2 = pO2 KO2 = 21 kPa 7.9 × 104 kPa kg mol−1 = 2.9 × 10−4 mol kg−1 The molality of the saturated solution is therefore 0.29 mmol kg−1. To convert this quantity to a molar concentration, we assume that the mass density of this dilute solution is essentially that of pure water at 25°C, or ρH2O = 0.99709 kg dm−3. It follows that the molar concentration of oxygen is [O2] = bO2 × ρH2O = 0.29 mmol kg−1 × 0.99709 kg dm−3 = 0.29 mmol dm−3 A note on good practice The number of significant figures in the result of a calcu- lation should not exceed the number in the data (only two in this case). Self-test 5.5 Calculate the molar solubility of nitrogen in water exposed to air at 25°C; partial pressures were calculated in Example 1.3. [0.51 mmol dm−3] An Illustration (don’t confuse this with a diagram!) is a short example of how to use an equation that has just been introduced in the text. In particular, we show how to use data and how to manipulate units correctly. xii ABOUT THE BOOK Worked examples Example 8.1 Calculating the number of photons Calculate the number of photons emitted by a 100 W yellow lamp in 1.0 s. Take the wavelength of yellow light as 560 nm and assume 100 per cent efficiency. Method Each photon has an energy hν, so the total number of photons needed to produce an energy E is E/hν. To use this equation, we need to know the frequency of the radiation (from ν = c/λ) and the total energy emitted by the lamp. The latter is given by the product of the power (P, in watts) and the time interval for which the lamp is turned on (E = P∆t). Answer The number of photons is N= E hν = P∆t h(c/λ) = A Worked example is a much more structured form of Illustration, often involving a more elaborate procedure. Every Worked example has a Method section to suggest how to set up the problem (another way might seem more natural: setting up problems is a highly personal business). Then there is the worked-out Answer. λP∆t hc Substitution of the data gives N= (5.60 × 10−7 m) × (100 J s−1) × (1.0 s) (6.626 × 10−34 J s) × (2.998 × 108 m s−1) = 2.8 × 1020 Note that it would take nearly 40 min to produce 1 mol of these photons. A note on good practice To avoid rounding and other numerical errors, it is best to carry out algebraic mainpulations first, and to substitute numerical values into a single, final formula. Moreover, an analytical result may be used for other data without having to repeat the entire calculation. Self-test 8.1 How many photons does a monochromatic (single frequency) infrared rangefinder of power 1 mW and wavelength 1000 nm emit in 0.1 s? [5 × 1014] Self-tests Self-test 3.12 Calculate the change in Gm for ice at −10°C, with density 917 kg m−3, [+2.0 J mol−1] when the pressure is increased from 1.0 bar to 2.0 bar. Discussion questions Discussion questions 1.1 Explain how the perfect gas equation of state arises by combination of Boyle’s law, Charles’s law, and Avogadro’s principle. 1.2 Explain the term ‘partial pressure’ and explain why Dalton’s law is a limiting law. 1.3 Explain how the compression factor varies with pressure and temperature and describe how it reveals information about intermolecular interactions in real gases. Each Worked example, and many of the Illustrations, has a Selftest, with the answer provided as a check that the procedure has been mastered. There are also free-standing Self-tests where we thought it a good idea to provide a question to check understanding. Think of Self-tests as in-chapter Exercises designed to help monitor your progress. 1.4 What is the significance of the critical co 1.5 Describe the formulation of the van der rationale for one other equation of state in T 1.6 Explain how the van der Waals equation behaviour. The end-of-chapter material starts with a short set of questions that are intended to encourage reflection on the material and to view it in a broader context than is obtained by solving numerical problems. ABOUT THE BOOK Exercises and Problems Exercises Molar absorption coefficient, & 14.1a The term symbol for the ground state of N 2+ is 2 Σ g. What is the total spin and total orbital angular momentum of the molecule? Show that the term symbol agrees with the electron configuration that would be predicted using the building-up principle. 14.1b One of the excited states of the C2 molecule has the valence electron configuration 1σ g21σ u21π u31π 1g. Give the multiplicity and parity of the term. 14.2a The molar absorption coefficient of a substance dissolved in hexane is known to be 855 dm3 mol−1 cm−1 at 270 nm. Calculate the percentage reduction in intensity when light of that wavelength passes through 2.5 mm of a solution of concentration 3.25 mmol dm−3. 14.2b The molar absorption coefficient of a substance dissolved in hexane is known to be 327 dm3 mol−1 cm−1 at 300 nm. Calculate the percentage reduction in intensity when light of that wavelength passes through 1.50 mm of a solution of concentration 2.22 mmol dm−3. 14.3a A solution of an unknown component of a biological sample when placed in an absorption cell of path length 1.00 cm transmits 20.1 per cent of light of 340 nm incident upon it. If the concentration of the component is 0.111 mmol dm−3, what is the molar absorption coefficient? 14.3b When light of wavelength 400 nm passes through 3.5 mm of a solution of an absorbing substance at a concentration 0.667 mmol dm−3, the transmission is 65.5 per cent. Calculate the molar absorption coefficient of the solute at this wavelength and express the answer in cm2 mol−1. & (%$) " &max{1 # &max %~max Wavenumb Fig. 14.49 14.7b The following data were obtained for th in methylbenzene using a 2.50 mm cell. Calcu coefficient of the dye at the wavelength emplo [dye]/(mol dm−3) 0.0010 0.0050 0.0 T/(per cent) 73 21 4.2 ll fill d h l Problems Assume all gases are perfect unless stated otherwise. Note that 1 atm = 1.013 25 bar. Unless otherwise stated, thermochemical data are for 298.15 K. 2.1 A sample consisting of 1 mol of perfect gas atoms (for which CV,m = –32 R) is taken through the cycle shown in Fig. 2.34. (a) Determine the temperature at the points 1, 2, and 3. (b) Calculate q, w, ∆U, and ∆H for each step and for the overall cycle. If a numerical answer cannot be obtained from the information given, then write in +, −, 0, or ? as appropriate. 1.00 Table 2.2. Calculate the standard enthalpy of from its value at 298 K. 2.8 A sample of the sugar d-ribose (C5H10O Numerical problems Pressure, p/atm xiii 2.9 The standard enthalpy of formation of t bis(benzene)chromium was measured in a c reaction Cr(C6H6)2(s) → Cr(s) + 2 C6H6(g) t Find the corresponding reaction enthalpy an of formation of the compound at 583 K. The heat capacity of benzene is 136.1 J K−1 mol−1 81.67 J K−1 mol−1 as a gas. 2 1 in a calorimeter and then ignited in the prese temperature rose by 0.910 K. In a separate ex the combustion of 0.825 g of benzoic acid, fo combustion is −3251 kJ mol−1, gave a temper the internal energy of combustion of d-ribos Isotherm 2.10‡ From the enthalpy of combustion dat 3 0.50 22.44 44.88 3 Volume, V/dm Fig. 2.34 2.2 A sample consisting of 1.0 mol CaCO3(s) was heated to 800°C, when it decomposed. The heating was carried out in a container fitted with a piston that was initially resting on the solid. Calculate the work done during complete decomposition at 1.0 atm. What work would be done if instead of having a piston the container was open to the atmosphere? alkanes methane through octane, test the ext ∆cH 7 = k{(M/(g mol−1)}n holds and find the Predict ∆cH 7 for decane and compare to the 2.11 It is possible to investigate the thermoc hydrocarbons with molecular modelling me software to predict ∆cH 7 values for the alkan calculate ∆cH 7 values, estimate the standard CnH2(n+1)(g) by performing semi-empirical c or PM3 methods) and use experimental stan values for CO2(g) and H2O(l). (b) Compare experimental values of ∆cH 7 (Table 2.5) and the molecular modelling method. (c) Test th ∆cH 7 = k{(M/(g mol−1)}n holds and find the 2 12‡ When 1 3584 g of sodium acetate trih The real core of testing understanding is the collection of endof-chapter Exercises and Problems. The Exercises are straightforward numerical tests that give practice with manipulating numerical data. The Problems are more searching. They are divided into ‘numerical’, where the emphasis is on the manipulation of data, and ‘theoretical’, where the emphasis is on the manipulation of equations before (in some cases) using numerical data. At the end of the Problems are collections of problems that focus on practical applications of various kinds, including the material covered in the Impact sections. About the Web site The Web site to accompany Physical Chemistry is available at: www.whfreeman.com/pchem8 It includes the following features: Living graphs A Living graph is indicated in the text by the icon attached to a graph. This feature can be used to explore how a property changes as a variety of parameters are changed. To encourage the use of this resource (and the more extensive Explorations in Physical Chemistry) we have added a question to each figure where a Living graph is called out. z 10.16 The boundary surfaces of d orbitals. Two nodal planes in each orbital intersect at the nucleus and separate the lobes of each orbital. The dark and light areas denote regions of opposite sign of the wavefunction. Exploration To gain insight into the shapes of the f orbitals, use mathematical software to plot the boundary surfaces of the spherical harmonics Y3,m (θ,ϕ). l y x dx 2# y 2 dz 2 dyz dxy dz ABOUT THE WEB SITE Artwork An instructor may wish to use the illustrations from this text in a lecture. Almost all the illustrations are available and can be used for lectures without charge (but not for commercial purposes without specific permission). This edition is in full colour: we have aimed to use colour systematically and helpfully, not just to make the page prettier. Tables of data All the tables of data that appear in the chapter text are available and may be used under the same conditions as the figures. Web links There is a huge network of information available about physical chemistry, and it can be bewildering to find your way to it. Also, a piece of information may be needed that we have not included in the text. The web site might suggest where to find the specific data or indicate where additional data can be found. xv integrating all student media resources and adds features unique to the eBook. The eBook also offers instructors unparalleled flexibility and customization options not previously possible with any printed textbook. Access to the eBook is included with purchase of the special package of the text (0-7167-85862), through use of an activation code card. Individual eBook copies can be purchased on-line at www.whfreeman.com. Key features of the eBook include: • Easy access from any Internet-connected computer via a standard Web browser. • Quick, intuitive navigation to any section or subsection, as well as any printed book page number. • Integration of all Living Graph animations. • Text highlighting, down to the level of individual phrases. • A book marking feature that allows for quick reference to any page. • A powerful Notes feature that allows students or instructors to add notes to any page. Tools • A full index. Interactive calculators, plotters and a periodic table for the study of chemistry. • Full-text search, including an option to also search the glossary and index. • Automatic saving of all notes, highlighting, and bookmarks. Group theory tables Comprehensive group theory tables are available for downloading. Explorations in Physical Chemistry Now from W.H. Freeman & Company, the new edition of the popular Explorations in Physical Chemistry is available on-line at www.whfreeman.com/explorations, using the activation code card included with Physical Chemistry 8e. The new edition consists of interactive Mathcad® worksheets and, for the first time, interactive Excel® workbooks. They motivate students to simulate physical, chemical, and biochemical phenomena with their personal computers. Harnessing the computational power of Mathcad® by Mathsoft, Inc. and Excel® by Microsoft Corporation, students can manipulate over 75 graphics, alter simulation parameters, and solve equations to gain deeper insight into physical chemistry. Complete with thought-stimulating exercises, Explorations in Physical Chemistry is a perfect addition to any physical chemistry course, using any physical chemistry text book. The Physical Chemistry, Eighth Edition eBook A complete online version of the textbook. The eBook offers students substantial savings and provides a rich learning experience by taking full advantage of the electronic medium Additional features for lecturers: • Custom chapter selection: Lecturers can choose the chapters that correspond with their syllabus, and students will get a custom version of the eBook with the selected chapters only. • Instructor notes: Lecturers can choose to create an annotated version of the eBook with their notes on any page. When students in their course log in, they will see the lecturer’s version. • Custom content: Lecturer notes can include text, web links, and even images, allowing lecturers to place any content they choose exactly where they want it. Physical Chemistry is now available in two volumes! For maximum flexibility in your physical chemistry course, this text is now offered as a traditional, full text or in two volumes. The chapters from Physical Chemistry, 8e that appear in each volume are as follows: Volume 1: Thermodynamics and Kinetics (0-7167-8567-6) 1. The properties of gases 2. The first law xvi 3. 4. 5. 6. 7. 21. 22. 23. 24. ABOUT THE WEB SITE The second law Physical transformations of pure substances Simple mixtures Phase diagrams Chemical equilibrium Molecules in motion The rates of chemical reactions The kinetics of complex reactions Molecular reaction dynamics Data section Answers to exercises Answers to problems Index Volume 2: Quantum Chemistry, Spectroscopy, and Statistical Thermodynamics (0-7167-8569-2) 8. Quantum theory: introduction and principles 9. Quantum theory: techniques and applications 10. 11. 12. 13. 14. 15. 16. 17. Atomic structure and atomic spectra Molecular structure Molecular symmetry Spectroscopy 1: rotational and vibrational spectra Spectroscopy 2: electronic transitions Spectroscopy 3: magnetic resonance Statistical thermodynamics: the concepts Statistical thermodynamics: the machinery Data section Answers to exercises Answers to problems Index Solutions manuals As with previous editions Charles Trapp, Carmen Giunta, and Marshall Cady have produced the solutions manuals to accompany this book. A Student’s Solutions Manual (0-71676206-4) provides full solutions to the ‘a’ exercises and the odd-numbered problems. An Instructor’s Solutions Manual (0-7167-2566-5) provides full solutions to the ‘b’ exercises and the even-numbered problems. About the authors Julio de Paula is Professor of Chemistry and Dean of the College of Arts & Sciences at Lewis & Clark College. A native of Brazil, Professor de Paula received a B.A. degree in chemistry from Rutgers, The State University of New Jersey, and a Ph.D. in biophysical chemistry from Yale University. His research activities encompass the areas of molecular spectroscopy, biophysical chemistry, and nanoscience. He has taught courses in general chemistry, physical chemistry, biophysical chemistry, instrumental analysis, and writing. Peter Atkins is Professor of Chemistry at Oxford University, a fellow of Lincoln College, and the author of more than fifty books for students and a general audience. His texts are market leaders around the globe. A frequent lecturer in the United States and throughout the world, he has held visiting prefessorships in France, Israel, Japan, China, and New Zealand. He was the founding chairman of the Committee on Chemistry Education of the International Union of Pure and Applied Chemistry and a member of IUPAC’s Physical and Biophysical Chemistry Division. Acknowledgements A book as extensive as this could not have been written without significant input from many individuals. We would like to reiterate our thanks to the hundreds of people who contributed to the first seven editions. Our warm thanks go Charles Trapp, Carmen Giunta, and Marshall Cady who have produced the Solutions manuals that accompany this book. Many people gave their advice based on the seventh edition, and others reviewed the draft chapters for the eighth edition as they emerged. We therefore wish to thank the following colleagues most warmly: Joe Addison, Governors State University Joseph Alia, University of Minnesota Morris David Andrews, University of East Anglia Mike Ashfold, University of Bristol Daniel E. Autrey, Fayetteville State University Jeffrey Bartz, Kalamazoo College Martin Bates, University of Southampton Roger Bickley, University of Bradford E.M. Blokhuis, Leiden University Jim Bowers, University of Exeter Mark S. Braiman, Syracuse University Alex Brown, University of Alberta David E. Budil, Northeastern University Dave Cook, University of Sheffield Ian Cooper, University of Newcastle-upon-Tyne T. Michael Duncan, Cornell University Christer Elvingson, Uppsala University Cherice M. Evans, Queens College—CUNY Stephen Fletcher, Loughborough University Alyx S. Frantzen, Stephen F. Austin State University David Gardner, Lander University Roberto A. Garza-López, Pomona College Robert J. Gordon, University of Illinois at Chicago Pete Griffiths, Cardiff University Robert Haines, University of Prince Edward Island Ron Haines, University of New South Wales Arthur M. Halpern, Indiana State University Tom Halstead, University of York Todd M. Hamilton, Adrian College Gerard S. Harbison, University Nebraska at Lincoln Ulf Henriksson, Royal Institute of Technology, Sweden Mike Hey, University of Nottingham Paul Hodgkinson, University of Durham Robert E. Howard, University of Tulsa Mike Jezercak, University of Central Oklahoma Clarence Josefson, Millikin University Pramesh N. Kapoor, University of Delhi Peter Karadakov, University of York Miklos Kertesz, Georgetown University Neil R. Kestner, Louisiana State University Sanjay Kumar, Indian Institute of Technology Jeffry D. Madura, Duquesne University Andrew Masters, University of Manchester Paul May, University of Bristol Mitchell D. Menzmer, Southwestern Adventist University David A. Micha, University of Florida Sergey Mikhalovsky, University of Brighton Jonathan Mitschele, Saint Joseph’s College Vicki D. Moravec, Tri-State University Gareth Morris, University of Manchester Tony Morton-Blake, Trinity College, Dublin Andy Mount, University of Edinburgh Maureen Kendrick Murphy, Huntingdon College John Parker, Heriot Watt University Jozef Peeters, University of Leuven Michael J. Perona, CSU Stanislaus Nils-Ola Persson, Linköping University Richard Pethrick, University of Strathclyde John A. Pojman, The University of Southern Mississippi Durga M. Prasad, University of Hyderabad Steve Price, University College London S. Rajagopal, Madurai Kamaraj University R. Ramaraj, Madurai Kamaraj University David Ritter, Southeast Missouri State University Bent Ronsholdt, Aalborg University Stephen Roser, University of Bath Kathryn Rowberg, Purdue University Calumet S.A. Safron, Florida State University Kari Salmi, Espoo-Vantaa Institute of Technology Stephan Sauer, University of Copenhagen Nicholas Schlotter, Hamline University Roseanne J. Sension, University of Michigan A.J. Shaka, University of California Joe Shapter, Flinders University of South Australia Paul D. Siders, University of Minnesota, Duluth Harjinder Singh, Panjab University Steen Skaarup, Technical University of Denmark David Smith, University of Exeter Patricia A. Snyder, Florida Atlantic University Olle Söderman, Lund University Peter Stilbs, Royal Institute of Technology, Sweden Svein Stølen, University of Oslo Fu-Ming Tao, California State University, Fullerton Eimer Tuite, University of Newcastle Eric Waclawik, Queensland University of Technology Yan Waguespack, University of Maryland Eastern Shore Terence E. Warner, University of Southern Denmark ACKNOWLEDGEMENTS Richard Wells, University of Aberdeen Ben Whitaker, University of Leeds Christopher Whitehead, University of Manchester Mark Wilson, University College London Kazushige Yokoyama, State University of New York at Geneseo Nigel Young, University of Hull Sidney H. Young, University of South Alabama xix We also thank Fabienne Meyers (of the IUPAC Secretariat) for helping us to bring colour to most of the illustrations and doing so on a very short timescale. We would also like to thank our two publishers, Oxford University Press and W.H. Freeman & Co., for their constant encouragement, advice, and assistance, and in particular our editors Jonathan Crowe, Jessica Fiorillo, and Ruth Hughes. Authors could not wish for a more congenial publishing environment. This page intentionally left blank Summary of contents PART 1 1 2 3 4 5 6 7 PART 2 8 9 10 11 12 13 14 15 16 17 18 19 20 PART 3 21 22 23 24 25 Equilibrium 1 The properties of gases The First Law The Second Law Physical transformations of pure substances Simple mixtures Phase diagrams Chemical equilibrium 3 28 76 117 136 174 200 Structure 241 Quantum theory: introduction and principles Quantum theory: techniques and applications Atomic structure and atomic spectra Molecular structure Molecular symmetry Molecular spectroscopy 1: rotational and vibrational spectra Molecular spectroscopy 2: electronic transitions Molecular spectroscopy 3: magnetic resonance Statistical thermodynamics 1: the concepts Statistical thermodynamics 2: applications Molecular interactions Materials 1: macromolecules and aggregates Materials 2: the solid state 243 277 320 362 404 430 481 513 560 589 620 652 697 Change 745 Molecules in motion The rates of chemical reactions The kinetics of complex reactions Molecular reaction dynamics Processes at solid surfaces 747 791 830 869 909 Appendix 1: Quantities, units and notational conventions Appendix 2: Mathematical techniques Appendix 3: Essential concepts of physics Data section Answers to ‘a’ exercises Answers to selected problems Index 959 963 979 988 1028 1034 1040 This page intentionally left blank Contents PART 1 Equilibrium 1 The properties of gases 1 3 The perfect gas 3 1.1 1.2 I1.1 3 7 The states of gases The gas laws Impact on environmental science: The gas laws and the weather 11 Real gases 14 1.3 1.4 1.5 14 17 21 Molecular interactions The van der Waals equation The principle of corresponding states Checklist of key ideas Further reading Discussion questions Exercises Problems 2 The First Law 23 23 23 24 25 28 The basic concepts 28 2.1 2.2 2.3 2.4 2.5 I2.1 29 30 33 37 40 2.6 Work, heat, and energy The internal energy Expansion work Heat transactions Enthalpy Impact on biochemistry and materials science: Differential scanning calorimetry Adiabatic changes 46 47 Thermochemistry 49 2.7 I2.2 2.8 2.9 49 52 54 56 Standard enthalpy changes Impact on biology: Food and energy reserves Standard enthalpies of formation The temperature-dependence of reaction enthalpies Discussion questions Exercises Problems 70 70 73 3 The Second Law 76 The direction of spontaneous change 77 3.1 3.2 I3.1 3.3 3.4 77 78 85 87 92 The dispersal of energy Entropy Impact on engineering: Refrigeration Entropy changes accompanying specific processes The Third Law of thermodynamics Concentrating on the system 3.5 3.6 The Helmholtz and Gibbs energies Standard reaction Gibbs energies 94 95 100 Combining the First and Second Laws 102 3.7 3.8 3.9 The fundamental equation Properties of the internal energy Properties of the Gibbs energy 102 103 105 Checklist of key ideas Further reading Further information 3.1: The Born equation Further information 3.2: Real gases: the fugacity Discussion questions Exercises Problems 109 110 110 111 112 113 114 4 Physical transformations of pure substances 117 Phase diagrams 117 4.1 4.2 I4.1 117 118 4.3 The stabilities of phases Phase boundaries Impact on engineering and technology: Supercritical fluids Three typical phase diagrams 119 120 State functions and exact differentials 57 Phase stability and phase transitions 122 2.10 2.11 2.12 57 59 63 4.4 4.5 4.6 4.7 122 122 126 129 Exact and inexact differentials Changes in internal energy The Joule–Thomson effect Checklist of key ideas Further reading Further information 2.1: Adiabatic processes Further information 2.2: The relation between heat capacities 67 68 69 69 The thermodynamic criterion of equilibrium The dependence of stability on the conditions The location of phase boundaries The Ehrenfest classification of phase transitions Checklist of key ideas Further reading Discussion questions 131 132 132 xxiv CONTENTS Exercises Problems 132 133 7 Chemical equilibrium 200 Spontaneous chemical reactions 200 136 7.1 7.2 200 202 The thermodynamic description of mixtures 136 5.1 5.2 5.3 I5.1 136 141 143 The response of equilibria to the conditions 210 7.3 7.4 I7.1 210 211 5 Simple mixtures Partial molar quantities The thermodynamics of mixing The chemical potentials of liquids Impact on biology: Gas solubility and breathing 147 The properties of solutions 148 5.4 5.5 I5.2 148 150 Liquid mixtures Colligative properties Impact on biology: Osmosis in physiology and biochemistry 156 Activities 158 5.6 5.7 5.8 5.9 The solvent activity The solute activity The activities of regular solutions The activities of ions in solution 158 159 162 163 Checklist of key ideas Further reading Further information 5.1: The Debye–Hückel theory of ionic solutions Discussion questions Exercises Problems 166 167 167 169 169 171 The Gibbs energy minimum The description of equilibrium How equilibria respond to pressure The response of equilibria to temperature Impact on engineering: The extraction of metals from their oxides Equilibrium electrochemistry 216 7.5 7.6 7.7 7.8 7.9 I7.2 216 217 218 222 224 Half-reactions and electrodes Varieties of cells The electromotive force Standard potentials Applications of standard potentials Impact on biochemistry: Energy conversion in biological cells Checklist of key ideas Further reading Discussion questions Exercises Problems PART 2 Structure 8 Quantum theory: introduction and principles 6 Phase diagrams 215 225 233 234 234 235 236 241 243 174 The origins of quantum mechanics 243 244 249 253 Phases, components, and degrees of freedom 174 6.1 6.2 Definitions The phase rule 174 176 8.1 8.2 I8.1 Two-component systems 179 The dynamics of microscopic systems 254 6.3 6.4 6.5 6.6 I6.1 I6.2 179 182 185 189 191 8.3 8.4 254 256 Vapour pressure diagrams Temperature–composition diagrams Liquid–liquid phase diagrams Liquid–solid phase diagrams Impact on materials science: Liquid crystals Impact on materials science: Ultrapurity and controlled impurity Checklist of key ideas Further reading Discussion questions Exercises Problems The failures of classical physics Wave–particle duality Impact on biology: Electron microscopy The Schrödinger equation The Born interpretation of the wavefunction Quantum mechanical principles 260 192 8.5 8.6 8.7 260 269 272 193 194 194 195 197 Checklist of key ideas Further reading Discussion questions Exercises Problems The information in a wavefunction The uncertainty principle The postulates of quantum mechanics 273 273 274 274 275 CONTENTS 9 Quantum theory: techniques and applications 277 11 Molecular structure xxv 362 Translational motion 277 The Born–Oppenheimer approximation 362 9.1 9.2 9.3 I9.1 278 283 286 Valence-bond theory 363 11.1 11.2 363 365 288 Molecular orbital theory 368 11.3 11.4 11.5 I11.1 368 373 379 A particle in a box Motion in two and more dimensions Tunnelling Impact on nanoscience: Scanning probe microscopy Vibrational motion 290 9.4 9.5 291 292 The energy levels The wavefunctions Rotational motion 297 9.6 9.7 297 I9.2 9.8 Rotation in two dimensions: a particle on a ring Rotation in three dimensions: the particle on a sphere Impact on nanoscience: Quantum dots Spin 301 306 308 Techniques of approximation 310 9.9 9.10 310 311 Time-independent perturbation theory Time-dependent perturbation theory Checklist of key ideas Further reading Further information 9.1: Dirac notation Further information 9.2: Perturbation theory Discussion questions Exercises Problems 10 Atomic structure and atomic spectra 312 313 313 313 316 316 317 320 Homonuclear diatomic molecules Polyatomic molecules The hydrogen molecule-ion Homonuclear diatomic molecules Heteronuclear diatomic molecules Impact on biochemistry: The biochemical reactivity of O2, N2, and NO 385 Molecular orbitals for polyatomic systems 386 11.6 11.7 11.8 387 392 396 The Hückel approximation Computational chemistry The prediction of molecular properties Checklist of key ideas Further reading Discussion questions Exercises Problems 12 Molecular symmetry 398 399 399 399 400 404 The symmetry elements of objects 404 12.1 12.2 12.3 405 406 411 Operations and symmetry elements The symmetry classification of molecules Some immediate consequences of symmetry The structure and spectra of hydrogenic atoms 320 Applications to molecular orbital theory and spectroscopy 413 10.1 10.2 10.3 321 326 335 12.4 12.5 12.6 413 419 423 The structures of many-electron atoms 336 10.4 10.5 336 344 Checklist of key ideas Further reading Discussion questions Exercises Problems The structure of hydrogenic atoms Atomic orbitals and their energies Spectroscopic transitions and selection rules The orbital approximation Self-consistent field orbitals The spectra of complex atoms 345 I10.1 10.6 10.7 10.8 10.9 346 346 347 348 352 Impact on astrophysics: Spectroscopy of stars Quantum defects and ionization limits Singlet and triplet states Spin–orbit coupling Term symbols and selection rules Checklist of key ideas Further reading Further information 10.1: The separation of motion Discussion questions Exercises Problems 356 357 357 358 358 359 Character tables and symmetry labels Vanishing integrals and orbital overlap Vanishing integrals and selection rules 425 426 426 426 427 13 Molecular spectroscopy 1: rotational and vibrational spectra 430 General features of spectroscopy 431 13.1 13.2 13.3 I13.1 431 432 436 Experimental techniques The intensities of spectral lines Linewidths Impact on astrophysics: Rotational and vibrational spectroscopy of interstellar space 438 xxvi CONTENTS Pure rotation spectra 441 13.4 13.5 13.6 13.7 13.8 441 443 446 449 450 Moments of inertia The rotational energy levels Rotational transitions Rotational Raman spectra Nuclear statistics and rotational states Discussion questions Exercises Problems 15 Molecular spectroscopy 3: magnetic resonance 508 509 510 513 The vibrations of diatomic molecules 452 The effect of magnetic fields on electrons and nuclei 513 13.9 13.10 13.11 13.12 13.13 452 454 455 457 15.1 15.2 15.3 513 515 516 Molecular vibrations Selection rules Anharmonicity Vibration–rotation spectra Vibrational Raman spectra of diatomic molecules 459 The vibrations of polyatomic molecules 13.14 Normal modes 13.15 Infrared absorption spectra of polyatomic molecules I13.2 Impact on environmental science: Global warming 13.16 Vibrational Raman spectra of polyatomic molecules I13.3 Impact on biochemistry: Vibrational microscopy 13.17 Symmetry aspects of molecular vibrations 460 460 Checklist of key ideas Further reading Further information 13.1: Spectrometers Further information 13.2: Selection rules for rotational and vibrational spectroscopy Discussion questions Exercises Problems 469 470 470 14 Molecular spectroscopy 2: electronic transitions 464 466 466 473 476 476 478 481 The characteristics of electronic transitions 481 14.1 The electronic spectra of diatomic molecules 14.2 The electronic spectra of polyatomic molecules I14.1 Impact on biochemistry: Vision 482 487 490 The fates of electronically excited states 492 14.3 Fluorescence and phosphorescence I14.2 Impact on biochemistry: Fluorescence 492 microscopy Dissociation and predissociation 494 495 14.4 Lasers 496 14.5 14.6 496 500 General principles of laser action Applications of lasers in chemistry Checklist of key ideas Further reading Further information 14.1: Examples of practical lasers Nuclear magnetic resonance 517 15.4 15.5 15.6 15.7 517 518 524 461 462 505 506 506 The energies of electrons in magnetic fields The energies of nuclei in magnetic fields Magnetic resonance spectroscopy The NMR spectrometer The chemical shift The fine structure Conformational conversion and exchange processes 532 Pulse techniqes in NMR 533 15.8 15.9 I15.1 15.10 15.11 15.12 15.13 533 536 540 541 542 544 548 The magnetization vector Spin relaxation Impact on medicine: Magnetic resonance imaging Spin decoupling The nuclear Overhauser effect Two-dimensional NMR Solid-state NMR Electron paramagnetic resonance 549 15.14 15.15 15.16 I15.2 The EPR spectrometer The g-value Hyperfine structure Impact on biochemistry: Spin probes 549 550 551 553 Checklist of key ideas Further reading Further information 15.1: Fourier transformation of the FID curve Discussion questions Exercises Problems 554 555 16 Statistical thermodynamics 1: the concepts 560 555 556 556 557 The distribution of molecular states 561 16.1 Configurations and weights 16.2 The molecular partition function I16.1 Impact on biochemistry: The helix–coil 561 564 transition in polypeptides 571 The internal energy and the entropy 573 16.3 16.4 573 575 The internal energy The statistical entropy CONTENTS The canonical partition function 577 16.5 16.6 The canonical ensemble The thermodynamic information in the partition function 16.7 Independent molecules 577 Checklist of key ideas Further reading Further information 16.1: The Boltzmann distribution Further information 16.2: The Boltzmann formula Further information 16.3: Temperatures below zero Discussion questions Exercises Problems 581 582 582 583 584 585 586 586 17 Statistical thermodynamics 2: applications 589 578 579 Fundamental relations 589 17.1 17.2 589 591 The thermodynamic functions The molecular partition function Using statistical thermodynamics 599 17.3 17.4 17.5 17.6 17.7 17.8 599 601 604 606 609 610 Mean energies Heat capacities Equations of state Molecular interactions in liquids Residual entropies Equilibrium constants Checklist of key ideas Further reading Discussion questions Exercises Problems 18 Molecular interactions 615 615 617 617 618 620 Electric properties of molecules 620 18.1 18.2 18.3 620 624 627 Electric dipole moments Polarizabilities Relative permittivities Interactions between molecules 629 18.4 Interactions between dipoles 18.5 Repulsive and total interactions I18.1 Impact on medicine: Molecular recognition 629 637 and drug design 638 Gases and liquids 640 18.6 18.7 18.8 640 641 645 Molecular interactions in gases The liquid–vapour interface Condensation Checklist of key ideas Further reading Further information 18.1: The dipole–dipole interaction Further information 18.2: The basic principles of molecular beams Discussion questions Exercises Problems 19 Materials 1: macromolecules and aggregates xxvii 646 646 646 647 648 648 649 652 Determination of size and shape 652 19.1 19.2 19.3 19.4 19.5 I19.1 653 655 657 660 663 Mean molar masses Mass spectrometry Laser light scattering Ultracentrifugation Electrophoresis Impact on biochemistry: Gel electrophoresis in genomics and proteomics 19.6 Viscosity 664 665 Structure and dynamics 667 19.7 19.8 19.9 I19.2 19.10 19.11 19.12 667 668 673 674 675 679 681 The different levels of structure Random coils The structure and stability of synthetic polymers Impact on technology: Conducting polymers The structure of proteins The structure of nucleic acids The stability of proteins and nucleic acids Self-assembly 681 19.13 19.14 19.15 I19.3 682 685 687 Colloids Micelles and biological membranes Surface films Impact on nanoscience: Nanofabrication with self-assembled monolayers Checklist of key ideas Further reading Further information 19.1: The Rayleigh ratio Discussion questions Exercises Problems 20 Materials 2: the solid state 690 690 691 691 692 692 693 697 Crystal lattices 697 20.1 20.2 20.3 I20.1 697 700 702 Lattices and unit cells The identification of lattice planes The investigation of structure Impact on biochemistry: X-ray crystallography of biological macromolecules 20.4 Neutron and electron diffraction 711 713 xxviii CONTENTS Crystal structure 715 20.5 20.6 20.7 715 717 720 Metallic solids Ionic solids Molecular solids and covalent networks The properties of solids 721 20.8 20.9 I20.2 20.10 20.11 20.12 721 723 728 728 733 736 Mechanical properties Electrical properties Impact on nanoscience: Nanowires Optical properties Magnetic properties Superconductors Checklist of key ideas Further reading Discussion questions Exercises Problems 738 739 739 740 741 PART 3 Change 745 21 Molecules in motion 747 Molecular motion in gases 747 21.1 The kinetic model of gases I21.1 Impact on astrophysics: The Sun as a ball of 748 perfect gas 21.2 Collision with walls and surfaces 21.3 The rate of effusion 21.4 Transport properties of a perfect gas 754 755 756 757 Molecular motion in liquids 761 21.5 21.6 21.7 21.8 I21.2 761 761 764 769 Experimental results The conductivities of electrolyte solutions The mobilities of ions Conductivities and ion–ion interactions Impact on biochemistry: Ion channels and ion pumps 770 Diffusion 772 21.9 The thermodynamic view 21.10 The diffusion equation I21.3 Impact on biochemistry: Transport of non- 772 776 electrolytes across biological membranes 21.11 Diffusion probabilities 21.12 The statistical view Checklist of key ideas Further reading Further information 21.1: The transport characteristics of a perfect gas Discussion questions Exercises Problems 22 The rates of chemical reactions Empirical chemical kinetics 791 22.1 22.2 22.3 22.4 22.5 792 794 798 804 807 783 783 784 785 786 788 Experimental techniques The rates of reactions Integrated rate laws Reactions approaching equilibrium The temperature dependence of reaction rates Accounting for the rate laws 809 22.6 Elementary reactions 22.7 Consecutive elementary reactions I22.1 Impact on biochemistry: The kinetics of the 809 811 helix–coil transition in polypeptides Unimolecular reactions 818 820 22.8 Checklist of key ideas Further reading Further information 22.1: The RRK model of unimolecular reactions Discussion questions Exercises Problems 23 The kinetics of complex reactions 823 823 824 825 825 826 830 Chain reactions 830 23.1 23.2 830 833 The rate laws of chain reactions Explosions Polymerization kinetics 835 23.3 23.4 835 836 Stepwise polymerization Chain polymerization Homogeneous catalysis 839 23.5 23.6 839 840 Features of homogeneous catalysis Enzymes Photochemistry 23.7 I23.1 I23.2 779 780 781 791 23.8 I23.3 Kinetics of photophysical and photochemical processes Impact on environmental science: The chemistry of stratospheric ozone Impact on biochemistry: Harvesting of light during plant photosynthesis Complex photochemical processes Impact on medicine: Photodynamic therapy Checklist of key ideas Further reading Further information 23.1: The Förster theory of resonance energy transfer Discussion questions Exercises Problems 845 845 853 856 858 860 861 862 862 863 863 864 CONTENTS 24 Molecular reaction dynamics 869 Reactive encounters 869 24.1 24.2 24.3 870 876 879 Collision theory Diffusion-controlled reactions The material balance equation Transition state theory 880 24.4 24.5 The Eyring equation Thermodynamic aspects 880 883 The dynamics of molecular collisions 885 24.6 24.7 24.8 24.9 886 887 888 Reactive collisions Potential energy surfaces Some results from experiments and calculations The investigation of reaction dynamics with ultrafast laser techniques 892 25.12 I25.3 25.13 I25.4 Working galvanic cells Impact on technology: Fuel cells Corrosion Impact on technology: Protecting materials against corrosion Checklist of key ideas Further reading Further information 25.1: The relation between electrode potential and the Galvani potential Discussion questions Exercises Problems Appendix 1 Quantities, units, and notational conventions xxix 945 947 948 949 951 951 952 952 953 955 959 Electron transfer in homogeneous systems 894 Names of quantities 959 24.10 24.11 24.12 I24.1 894 896 898 Units 960 Notational conventions 961 Further reading 962 The rates of electron transfer processes Theory of electron transfer processes Experimental results Impact on biochemistry: Electron transfer in and between proteins Checklist of key ideas Further reading Further information 24.1: The Gibbs energy of activation of electron transfer and the Marcus cross-relation Discussion questions Exercises Problems 900 902 903 903 904 904 905 Appendix 2 Mathematical techniques 963 Basic procedures 963 A2.1 Logarithms and exponentials A2.2 Complex numbers and complex functions A2.3 Vectors 963 963 964 Calculus 965 965 967 968 969 969 971 The growth and structure of solid surfaces 909 25.1 25.2 910 911 A2.4 A2.5 A2.6 A2.7 A2.8 A2.9 The extent of adsorption 916 Statistics and probability 973 25.3 25.4 25.5 I25.1 916 917 922 925 A2.10 Random selections A2.11 Some results of probability theory 973 974 Matrix algebra 975 A2.12 Matrix addition and multiplication A2.13 Simultaneous equations A2.14 Eigenvalue equations 975 976 977 Further reading 978 25 Processes at solid surfaces Surface growth Surface composition Physisorption and chemisorption Adsorption isotherms The rates of surface processes Impact on biochemistry: Biosensor analysis 909 Heterogeneous catalysis 926 25.6 Mechanisms of heterogeneous catalysis 25.7 Catalytic activity at surfaces I25.2 Impact on technology: Catalysis in the 927 928 chemical industry 929 Processes at electrodes 932 25.8 25.9 25.10 25.11 932 934 940 944 The electrode–solution interface The rate of charge transfer Voltammetry Electrolysis Differentiation and integration Power series and Taylor expansions Partial derivatives Functionals and functional derivatives Undetermined multipliers Differential equations Appendix 3 Essential concepts of physics 979 Energy 979 A3.1 Kinetic and potential energy A3.2 Energy units 979 979 xxx CONTENTS Classical mechanics 980 A3.3 The trajectory in terms of the energy A3.4 Newton’s second law A3.5 Rotational motion A3.6 The harmonic oscillator 980 980 981 982 Waves 983 A3.7 A3.8 A3.9 A3.10 983 983 984 985 The electromagnetic field Features of electromagnetic radiation Refraction Optical activity Electrostatics 985 A3.11 A3.12 A3.13 A3.14 986 986 986 987 The Coulomb interaction The Coulomb potential The strength of the electric field Electric current and power Further reading Data section Answers to ‘b’ exercises Answers to selected problems Index 987 988 1028 1034 1040 List of impact sections I1.1 I2.1 I2.2 I3.1 I4.1 I5.1 I5.2 I6.1 I6.2 I7.1 I7.2 I8.1 I9.1 I9.2 I10.1 I11.1 I13.1 I13.2 I13.3 I14.1 I14.2 I15.1 I15.2 I16.1 I18.1 I19.1 I19.2 I19.3 I20.1 I20.2 I21.1 I21.2 I21.3 I22.1 I23.1 I23.2 I23.3 I24.1 I25.1 I25.2 I25.3 I25.4 Impact on environmental science: The gas laws and the weather Impact on biochemistry and materials science: Differential scanning calorimetry Impact on biology: Food and energy reserves Impact on engineering: Refrigeration Impact on engineering and technology: Supercritical fluids Impact on biology: Gas solubility and breathing Impact on biology: Osmosis in physiology and biochemistry Impact on materials science: Liquid crystals Impact on materials science: Ultrapurity and controlled impurity Impact on engineering: The extraction of metals from their oxides Impact on biochemistry: Energy conversion in biological cells Impact on biology: Electron microscopy Impact on nanoscience: Scanning probe microscopy Impact on nanoscience: Quantum dots Impact on astrophysics: Spectroscopy of stars Impact on biochemistry: The biochemical reactivity of O2, N2, and NO Impact on astrophysics: Rotational and vibrational spectroscopy interstellar space Impact on environmental science: Global warming Impact on biochemistry: Vibrational microscopy Impact on biochemistry: Vision Impact on biochemistry: Fluorescence microscopy Impact on medicine: Magnetic resonance imaging Impact on biochemistry: Spin probes Impact on biochemistry: The helix–coil transition in polypeptides Impact on medicine: Molecular recognition and drug design Impact on biochemistry: Gel electrophoresis in genomics and proteomics Impact on technology: Conducting polymers Impact on nanoscience: Nanofabrication with self-assembled monolayers Impact on biochemistry: X-ray crystallography of biological macromolecules Impact on nanoscience: Nanowires Impact on astrophysics: The Sun as a ball of perfect gas Impact on biochemistry: Ion channels and ion pumps Impact on biochemistry: Transport of non-electrolytes across biological membranes Impact on biochemistry: The kinetics of the helix–coil transition in polypeptides Impact on environmental science: The chemistry of stratospheric ozone Impact on biochemistry: Harvesting of light during plant photosynthesis Impact on medicine: Photodynamic therapy Impact on biochemistry: Electron transfer in and between proteins Impact on biochemistry: Biosensor analysis Impact on technology: Catalysis in the chemical industry Impact on technology: Fuel cells Impact on technology: Protecting materials against corrosion 11 46 52 85 119 147 156 191 192 215 225 253 288 306 346 385 438 462 466 490 494 540 553 571 638 664 674 690 711 728 754 770 779 818 853 856 860 900 925 929 947 949 This page intentionally left blank PART 1 Equilibrium Part 1 of the text develops the concepts that are needed for the discussion of equilibria in chemistry. Equilibria include physical change, such as fusion and vaporization, and chemical change, including electrochemistry. The discussion is in terms of thermodynamics, and particularly in terms of enthalpy and entropy. We see that we can obtain a unified view of equilibrium and the direction of spontaneous change in terms of the chemical potentials of substances. The chapters in Part 1 deal with the bulk properties of matter; those of Part 2 will show how these properties stem from the behaviour of individual atoms. 1 The properties of gases 2 The First Law 3 The Second Law 4 Physical transformations of pure substances 5 Simple mixtures 6 Phase diagrams 7 Chemical equilibrium This page intentionally left blank 1 The properties of gases This chapter establishes the properties of gases that will be used throughout the text. It begins with an account of an idealized version of a gas, a perfect gas, and shows how its equation of state may be assembled experimentally. We then see how the properties of real gases differ from those of a perfect gas, and construct an equation of state that describes their properties. The simplest state of matter is a gas, a form of matter that fills any container it occupies. Initially we consider only pure gases, but later in the chapter we see that the same ideas and equations apply to mixtures of gases too. The perfect gas 1.1 The states of gases 1.2 The gas laws I1.1 Impact on environmental science: The gas laws and the weather Real gases 1.3 Molecular interactions 1.4 The van der Waals equation The perfect gas 1.5 The principle of corresponding We shall find it helpful to picture a gas as a collection of molecules (or atoms) in continuous random motion, with average speeds that increase as the temperature is raised. A gas differs from a liquid in that, except during collisions, the molecules of a gas are widely separated from one another and move in paths that are largely unaffected by intermolecular forces. states Checklist of key ideas Further reading Discussion questions Exercises Problems 1.1 The states of gases The physical state of a sample of a substance, its physical condition, is defined by its physical properties. Two samples of a substance that have the same physical properties are in the same state. The state of a pure gas, for example, is specified by giving its volume, V, amount of substance (number of moles), n, pressure, p, and temperature, T. However, it has been established experimentally that it is sufficient to specify only three of these variables, for then the fourth variable is fixed. That is, it is an experimental fact that each substance is described by an equation of state, an equation that interrelates these four variables. The general form of an equation of state is p = f(T,V,n) (1.1) This equation tells us that, if we know the values of T, V, and n for a particular substance, then the pressure has a fixed value. Each substance is described by its own equation of state, but we know the explicit form of the equation in only a few special cases. One very important example is the equation of state of a ‘perfect gas’, which has the form p = nRT/V, where R is a constant. Much of the rest of this chapter will examine the origin of this equation of state and its applications. 4 1 THE PROPERTIES OF GASES Table 1.1 Pressure units Name Symbol Value pascal 1 Pa 1 N m−2, 1 kg m−1 s−2 bar 1 bar 105 Pa atmosphere 1 atm 101.325 kPa torr 1 Torr (101 325/760) Pa = 133.32 . . . Pa millimetres of mercury 1 mmHg 133.322 . . . Pa pound per square inch 1 psi 6.894 757 . . . kPa (a) Pressure Comment 1.1 The International System of units (SI, from the French Système International d’Unités) is discussed in Appendix 1. Movable wall High Low pressure pressure Pressure is defined as force divided by the area to which the force is applied. The greater the force acting on a given area, the greater the pressure. The origin of the force exerted by a gas is the incessant battering of the molecules on the walls of its container. The collisions are so numerous that they exert an effectively steady force, which is experienced as a steady pressure. The SI unit of pressure, the pascal (Pa), is defined as 1 newton per metre-squared: 1 Pa = 1 N m−2 [1.2a] In terms of base units, 1 Pa = 1 kg m−1 s−2 [1.2b] Several other units are still widely used (Table 1.1); of these units, the most commonly used are atmosphere (1 atm = 1.013 25 × 105 Pa exactly) and bar (1 bar = 105 Pa). A pressure of 1 bar is the standard pressure for reporting data; we denote it p7. (a) Motion Equal pressures Self-test 1.1 Calculate the pressure (in pascals and atmospheres) exerted by a mass of 1.0 kg pressing through the point of a pin of area 1.0 × 10−2 mm2 at the surface of the Earth. Hint. The force exerted by a mass m due to gravity at the surface of the Earth is mg, where g is the acceleration of free fall (see endpaper 2 for its standard value). [0.98 GPa, 9.7 × 103 atm] (b) Low pressure High pressure (c) When a region of high pressure is separated from a region of low pressure by a movable wall, the wall will be pushed into one region or the other, as in (a) and (c). However, if the two pressures are identical, the wall will not move (b). The latter condition is one of mechanical equilibrium between the two regions. Fig. 1.1 If two gases are in separate containers that share a common movable wall (Fig. 1.1), the gas that has the higher pressure will tend to compress (reduce the volume of) the gas that has lower pressure. The pressure of the high-pressure gas will fall as it expands and that of the low-pressure gas will rise as it is compressed. There will come a stage when the two pressures are equal and the wall has no further tendency to move. This condition of equality of pressure on either side of a movable wall (a ‘piston’) is a state of mechanical equilibrium between the two gases. The pressure of a gas is therefore an indication of whether a container that contains the gas will be in mechanical equilibrium with another gas with which it shares a movable wall. (b) The measurement of pressure The pressure exerted by the atmosphere is measured with a barometer. The original version of a barometer (which was invented by Torricelli, a student of Galileo) was an inverted tube of mercury sealed at the upper end. When the column of mercury is in mechanical equilibrium with the atmosphere, the pressure at its base is equal to that 1.1 THE STATES OF GASES exerted by the atmosphere. It follows that the height of the mercury column is proportional to the external pressure. ' Example 1.1 Calculating the pressure exerted by a column of liquid Derive an equation for the pressure at the base of a column of liquid of mass density ρ (rho) and height h at the surface of the Earth. l Method Pressure is defined as p = F/A where F is the force applied to the area A, and F = mg. To calculate F we need to know the mass m of the column of liquid, which is its mass density, ρ, multiplied by its volume, V: m = ρV. The first step, therefore, is to calculate the volume of a cylindrical column of liquid. 1 Answer Let the column have cross-sectional area A; then its volume is Ah and its mass is m = ρAh. The force the column of this mass exerts at its base is Diathermic wall F = mg = ρAhg High temperature The pressure at the base of the column is therefore p= F ρAhg = = ρgh A A (1.3) Note that the pressure is independent of the shape and cross-sectional area of the column. The mass of the column of a given height increases as the area, but so does the area on which the force acts, so the two cancel. (a) Self-test 1.2 Derive an expression for the pressure at the base of a column of liquid of length l held at an angle θ (theta) to the vertical (1). Low temperature Energy as heat [p = ρgl cos θ] The pressure of a sample of gas inside a container is measured by using a pressure gauge, which is a device with electrical properties that depend on the pressure. For instance, a Bayard–Alpert pressure gauge is based on the ionization of the molecules present in the gas and the resulting current of ions is interpreted in terms of the pressure. In a capacitance manometer, the deflection of a diaphragm relative to a fixed electrode is monitored through its effect on the capacitance of the arrangement. Certain semiconductors also respond to pressure and are used as transducers in solid-state pressure gauges. Equal temperatures (b) Low temperature High temperature (c) Temperature The concept of temperature springs from the observation that a change in physical state (for example, a change of volume) can occur when two objects are in contact with one another, as when a red-hot metal is plunged into water. Later (Section 2.1) we shall see that the change in state can be interpreted as arising from a flow of energy as heat from one object to another. The temperature, T, is the property that indicates the direction of the flow of energy through a thermally conducting, rigid wall. If energy flows from A to B when they are in contact, then we say that A has a higher temperature than B (Fig. 1.2). It will prove useful to distinguish between two types of boundary that can separate the objects. A boundary is diathermic (thermally conducting) if a change of state is observed when two objects at different temperatures are brought into contact.1 A 1 The word dia is from the Greek for ‘through’. (c) Energy flows as heat from a region at a higher temperature to one at a lower temperature if the two are in contact through a diathermic wall, as in (a) and (c). However, if the two regions have identical temperatures, there is no net transfer of energy as heat even though the two regions are separated by a diathermic wall (b). The latter condition corresponds to the two regions being at thermal equilibrium. Fig. 1.2 5 6 1 THE PROPERTIES OF GASES A Equilibrium Equilibrium C B Equilibrium The experience summarized by the Zeroth Law of thermodynamics is that, if an object A is in thermal equilibrium with B and B is in thermal equilibrium with C, then C is in thermal equilibrium with A. Fig. 1.3 metal container has diathermic walls. A boundary is adiabatic (thermally insulating) if no change occurs even though the two objects have different temperatures. A vacuum flask is an approximation to an adiabatic container. The temperature is a property that indicates whether two objects would be in ‘thermal equilibrium’ if they were in contact through a diathermic boundary. Thermal equilibrium is established if no change of state occurs when two objects A to B are in contact through a diathermic boundary. Suppose an object A (which we can think of as a block of iron) is in thermal equilibrium with an object B (a block of copper), and that B is also in thermal equilibrium with another object C (a flask of water). Then it has been found experimentally that A and C will also be in thermal equilibrium when they are put in contact (Fig. 1.3). This observation is summarized by the Zeroth Law of thermodynamics: If A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then C is also in thermal equilibrium with A. The Zeroth Law justifies the concept of temperature and the use of a thermometer, a device for measuring the temperature. Thus, suppose that B is a glass capillary containing a liquid, such as mercury, that expands significantly as the temperature increases. Then, when A is in contact with B, the mercury column in the latter has a certain length. According to the Zeroth Law, if the mercury column in B has the same length when it is placed in thermal contact with another object C, then we can predict that no change of state of A and C will occur when they are in thermal contact. Moreover, we can use the length of the mercury column as a measure of the temperatures of A and C. In the early days of thermometry (and still in laboratory practice today), temperatures were related to the length of a column of liquid, and the difference in lengths shown when the thermometer was first in contact with melting ice and then with boiling water was divided into 100 steps called ‘degrees’, the lower point being labelled 0. This procedure led to the Celsius scale of temperature. In this text, temperatures on the Celsius scale are denoted θ and expressed in degrees Celsius (°C). However, because different liquids expand to different extents, and do not always expand uniformly over a given range, thermometers constructed from different materials showed different numerical values of the temperature between their fixed points. The pressure of a gas, however, can be used to construct a perfect-gas temperature scale that is independent of the identity of the gas. The perfect-gas scale turns out to be identical to the thermodynamic temperature scale to be introduced in Section 3.2c, so we shall use the latter term from now on to avoid a proliferation of names. On the thermodynamic temperature scale, temperatures are denoted T and are normally reported in kelvins, K (not °K). Thermodynamic and Celsius temperatures are related by the exact expression T/K = θ/°C + 273.15 (1.4) This relation, in the form θ/°C = T/K − 273.15, is the current definition of the Celsius scale in terms of the more fundamental Kelvin scale. It implies that a difference in temperature of 1°C is equivalent to a difference of 1 K. A note on good practice We write T = 0, not T = 0 K for the zero temperature on the thermodynamic temperature scale. This scale is absolute, and the lowest temperature is 0 regardless of the size of the divisions on the scale (just as we write p = 0 for zero pressure, regardless of the size of the units we adopt, such as bar or pascal). However, we write 0°C because the Celsius scale is not absolute. 1.2 THE GAS LAWS 7 Illustration 1.1 Converting temperatures To express 25.00°C as a temperature in kelvins, we use eqn 1.4 to write T/K = (25.00°C)/°C + 273.15 = 25.00 + 273.15 = 298.15 Increasing temperature, T Pressure, p Note how the units (in this case, °C) are cancelled like numbers. This is the procedure called ‘quantity calculus’ in which a physical quantity (such as the temperature) is the product of a numerical value (25.00) and a unit (1°C). Multiplication of both sides by the unit K then gives T = 298.15 K. A note on good practice When the units need to be specified in an equation, the approved procedure, which avoids any ambiguity, is to write (physical quantity)/ units, which is a dimensionless number, just as (25.00°C)/°C = 25.00 in this Illustration. Units may be multiplied and cancelled just like numbers. 0 1.2 The gas laws The equation of state of a gas at low pressure was established by combining a series of empirical laws. (a) The perfect gas law (1.5)° Charles’s law: V = constant × T, at constant n, p (1.6a)° p = constant × T, at constant n, V (1.6b)° Avogadro’s principle: V = constant × n at constant p, T 2 Volume, V The pressure–volume dependence of a fixed amount of perfect gas at different temperatures. Each curve is a hyperbola (pV = constant) and is called an isotherm. Fig. 1.4 Exploration3 Explore how the We assume that the following individual gas laws are familiar: Boyle’s law: pV = constant, at constant n, T 0 pressure of 1.5 mol CO2(g) varies with volume as it is compressed at (a) 273 K, (b) 373 K from 30 dm3 to 15 dm3. (1.7)° Boyle’s and Charles’s laws are examples of a limiting law, a law that is strictly true only in a certain limit, in this case p → 0. Equations valid in this limiting sense will be signalled by a ° on the equation number, as in these expressions. Avogadro’s principle is commonly expressed in the form ‘equal volumes of gases at the same temperature and pressure contain the same numbers of molecules’. In this form, it is increasingly true as p → 0. Although these relations are strictly true only at p = 0, they are reasonably reliable at normal pressures (p ≈ 1 bar) and are used widely throughout chemistry. Figure 1.4 depicts the variation of the pressure of a sample of gas as the volume is changed. Each of the curves in the graph corresponds to a single temperature and hence is called an isotherm. According to Boyle’s law, the isotherms of gases are hyperbolas. An alternative depiction, a plot of pressure against 1/volume, is shown in Fig. 1.5. The linear variation of volume with temperature summarized by Charles’s law is illustrated in Fig. 1.6. The lines in this illustration are examples of isobars, or lines showing the variation of properties at constant pressure. Figure 1.7 illustrates the linear variation of pressure with temperature. The lines in this diagram are isochores, or lines showing the variation of properties at constant volume. 2 Avogadro’s principle is a principle rather than a law (a summary of experience) because it depends on the validity of a model, in this case the existence of molecules. Despite there now being no doubt about the existence of molecules, it is still a model-based principle rather than a law. 3 To solve this and other Explorations, use either mathematical software or the Living graphs from the text’s web site. Comment 1.2 A hyperbola is a curve obtained by plotting y against x with xy = constant. Volume, V Pressure, p 1 THE PROPERTIES OF GASES Pressure, p 8 Decreasing pressure, p Increasing temperature, T Extrapolation 0 0 1/V Straight lines are obtained when the pressure is plotted against 1/V at constant temperature. Fig. 1.5 Exploration Repeat Exploration 1.4, but plot the data as p against 1/V. Extrapolation 0 0 Temperature, T The variation of the volume of a fixed amount of gas with the temperature at constant pressure. Note that in each case the isobars extrapolate to zero volume at T = 0, or θ = −273°C. Fig. 1.6 Exploration Explore how the volume of 1.5 mol CO2(g) in a container maintained at (a) 1.00 bar, (b) 0.50 bar varies with temperature as it is cooled from 373 K to 273 K. Decreasing volume, V Extrapolation 0 0 Temperature, T The pressure also varies linearly with the temperature at constant volume, and extrapolates to zero at T = 0 (−273°C). Fig. 1.7 Exploration Explore how the pressure of 1.5 mol CO2(g) in a container of volume (a) 30 dm3, (b) 15 dm3 varies with temperature as it is cooled from 373 K to 273 K. A note on good practice To test the validity of a relation between two quantities, it is best to plot them in such a way that they should give a straight line, for deviations from a straight line are much easier to detect than deviations from a curve. The empirical observations summarized by eqns 1.5–7 can be combined into a single expression: pV = constant × nT This expression is consistent with Boyle’s law (pV = constant) when n and T are constant, with both forms of Charles’s law (p ∝ T, V ∝ T) when n and either V or p are held constant, and with Avogadro’s principle (V ∝ n) when p and T are constant. The constant of proportionality, which is found experimentally to be the same for all gases, is denoted R and called the gas constant. The resulting expression pV = nRT (1.8)° is the perfect gas equation. It is the approximate equation of state of any gas, and becomes increasingly exact as the pressure of the gas approaches zero. A gas that obeys eqn 1.8 exactly under all conditions is called a perfect gas (or ideal gas). A real gas, an actual gas, behaves more like a perfect gas the lower the pressure, and is described exactly by eqn 1.8 in the limit of p → 0. The gas constant R can be determined by evaluating R = pV/nT for a gas in the limit of zero pressure (to guarantee that it is 1.2 THE GAS LAWS behaving perfectly). However, a more accurate value can be obtained by measuring the speed of sound in a low-pressure gas (argon is used in practice) and extrapolating its value to zero pressure. Table 1.2 lists the values of R in a variety of units. Table 1.2 The gas constant R J K−1 mol−1 8.314 47 Molecular interpretation 1.1 The kinetic model of gases The molecular explanation of Boyle’s law is that, if a sample of gas is compressed to half its volume, then twice as many molecules strike the walls in a given period of time than before it was compressed. As a result, the average force exerted on the walls is doubled. Hence, when the volume is halved the pressure of the gas is doubled, and p × V is a constant. Boyle’s law applies to all gases regardless of their chemical identity (provided the pressure is low) because at low pressures the average separation of molecules is so great that they exert no influence on one another and hence travel independently. The molecular explanation of Charles’s law lies in the fact that raising the temperature of a gas increases the average speed of its molecules. The molecules collide with the walls more frequently and with greater impact. Therefore they exert a greater pressure on the walls of the container. These qualitative concepts are expressed quantitatively in terms of the kinetic model of gases, which is described more fully in Chapter 21. Briefly, the kinetic model is based on three assumptions: 9 8.205 74 × 10 −2 dm3 atm K−1 mol−1 8.314 47 × 10 −2 dm3 bar K−1 mol−1 8.314 47 Pa m3 K−1 mol−1 1 62.364 dm3 Torr K−1 mol−1 1.987 21 cal K−1 mol−1 1. The gas consists of molecules of mass m in ceaseless random motion. 2. The size of the molecules is negligible, in the sense that their diameters are much smaller than the average distance travelled between collisions. 3. The molecules interact only through brief, infrequent, and elastic collisions. An elastic collision is a collision in which the total translational kinetic energy of the molecules is conserved. From the very economical assumptions of the kinetic model, it can be deduced (as we shall show in detail in Chapter 21) that the pressure and volume of the gas are related by pV = 13 nMc 2 (1.9)° where M = mNA, the molar mass of the molecules, and c is the root mean square speed of the molecules, the square root of the mean of the squares of the speeds, v, of the molecules: c = !v2"1/2 (1.10) We see that, if the root mean square speed of the molecules depends only on the temperature, then at constant temperature pV = constant which is the content of Boyle’s law. Moreover, for eqn 1.9 to be the equation of state of a perfect gas, its right-hand side must be equal to nRT. It follows that the root mean square speed of the molecules in a gas at a temperature T must be ⎛ 3RT ⎞ c=⎜ ⎟ ⎝ M ⎠ 1/2 (1.11)° We can conclude that the root mean square speed of the molecules of a gas is proportional to the square root of the temperature and inversely proportional to the square root of the molar mass. That is, the higher the temperature, the higher the root mean square speed of the molecules, and, at a given temperature, heavy molecules travel more slowly than light molecules. The root mean square speed of N2 molecules, for instance, is found from eqn 1.11 to be 515 m s−1 at 298 K. Comment 1.3 For an object of mass m moving at a speed 1, the kinetic energy is EK = –12 m12. The potential energy, EP or V, of an object is the energy arising from its position (not speed). No universal expression for the potential energy can be given because it depends on the type of interaction the object experiences. 1 THE PROPERTIES OF GASES p µ 1/V isotherm VµT isobar A region of the p,V,T surface of a fixed amount of perfect gas. The points forming the surface represent the only states of the gas that can exist. Fig. 1.8 T Volu me, V Tem pera ture , Tem pera ture ,T Volu me, V Pressure, p pµT isochore Surface of possible states Pressure, p 10 Sections through the surface shown in Fig. 1.8 at constant temperature give the isotherms shown in Fig. 1.4 and the isobars shown in Fig. 1.6. Fig. 1.9 The surface in Fig. 1.8 is a plot of the pressure of a fixed amount of perfect gas against its volume and thermodynamic temperature as given by eqn 1.8. The surface depicts the only possible states of a perfect gas: the gas cannot exist in states that do not correspond to points on the surface. The graphs in Figs. 1.4 and 1.6 correspond to the sections through the surface (Fig. 1.9). Example 1.2 Using the perfect gas equation In an industrial process, nitrogen is heated to 500 K in a vessel of constant volume. If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the working temperature if it behaved as a perfect gas? Method We expect the pressure to be greater on account of the increase in tem- perature. The perfect gas law in the form PV/nT = R implies that, if the conditions are changed from one set of values to another, then because PV/nT is equal to a constant, the two sets of values are related by the ‘combined gas law’: p1V1 p2V2 = n1T1 n2T2 n p V Initial Same 100 Same Final Same ? Same 2 T 300 500 (1.12)° The known and unknown data are summarized in (2). Answer Cancellation of the volumes (because V1 = V2) and amounts (because n1 = n2) on each side of the combined gas law results in p1 p2 = T1 T2 which can be rearranged into p2 = T2 × p1 T1 1.2 THE GAS LAWS Substitution of the data then gives p2 = 500 K × (100 atm) = 167 atm 300 K Experiment shows that the pressure is actually 183 atm under these conditions, so the assumption that the gas is perfect leads to a 10 per cent error. Self-test 1.3 What temperature would result in the same sample exerting a pressure of 300 atm? [900 K] The perfect gas equation is of the greatest importance in physical chemistry because it is used to derive a wide range of relations that are used throughout thermodynamics. However, it is also of considerable practical utility for calculating the properties of a gas under a variety of conditions. For instance, the molar volume, Vm = V/n, of a perfect gas under the conditions called standard ambient temperature and pressure (SATP), which means 298.15 K and 1 bar (that is, exactly 105 Pa), is easily calculated from Vm = RT/p to be 24.789 dm3 mol−1. An earlier definition, standard temperature and pressure (STP), was 0°C and 1 atm; at STP, the molar volume of a perfect gas is 22.414 dm3 mol−1. Among other applications, eqn 1.8 can be used to discuss processes in the atmosphere that give rise to the weather. IMPACT ON ENVIRONMENTAL SCIENCE I1.1 The gas laws and the weather The biggest sample of gas readily accessible to us is the atmosphere, a mixture of gases with the composition summarized in Table 1.3. The composition is maintained moderately constant by diffusion and convection (winds, particularly the local turbulence called eddies) but the pressure and temperature vary with altitude and with the local conditions, particularly in the troposphere (the ‘sphere of change’), the layer extending up to about 11 km. Table 1.3 The composition of dry air at sea level Percentage Component By volume By mass Nitrogen, N2 78.08 75.53 Oxygen, O2 20.95 23.14 Argon, Ar 0.93 1.28 Carbon dioxide, CO2 0.031 0.047 Hydrogen, H2 5.0 × 10 −3 2.0 × 10 −4 Neon, Ne 1.8 × 10 −3 1.3 × 10 −3 5.2 × 10 −4 7.2 × 10 −5 2.0 × 10 −4 1.1 × 10 −4 1.1 × 10 −4 3.2 × 10 −4 Nitric oxide, NO 5.0 × 10 −5 1.7 × 10 −6 Xenon, Xe 8.7 × 10 −6 1.2 × 10 −5 7.0 × 10 −6 1.2 × 10 −5 2.0 × 10 −6 3.3 × 10 −6 Helium, He Methane, CH4 Krypton, Kr Ozone, O3: summer winter 11 12 1 THE PROPERTIES OF GASES 30 In the troposphere the average temperature is 15°C at sea level, falling to –57°C at the bottom of the tropopause at 11 km. This variation is much less pronounced when expressed on the Kelvin scale, ranging from 288 K to 216 K, an average of 268 K. If we suppose that the temperature has its average value all the way up to the tropopause, then the pressure varies with altitude, h, according to the barometric formula: Altitude, h/km 20 15 p = p0e−h/H 10 6 0 0 Pressure, p p0 Fig. 1.10 The variation of atmospheric pressure with altitude, as predicted by the barometric formula and as suggested by the ‘US Standard Atmosphere’, which takes into account the variation of temperature with altitude. Exploration How would the graph shown in the illustration change if the temperature variation with altitude were taken into account? Construct a graph allowing for a linear decrease in temperature with altitude. H H H L H Fig. 1.11 A typical weather map; in this case, for the United States on 1 January 2000. N Wind L Rotation where p0 is the pressure at sea level and H is a constant approximately equal to 8 km. More specifically, H = RT/Mg, where M is the average molar mass of air and T is the temperature. The barometric formula fits the observed pressure distribution quite well even for regions well above the troposphere (see Fig. 1.10). It implies that the pressure of the air and its density fall to half their sea-level value at h = H ln 2, or 6 km. Local variations of pressure, temperature, and composition in the troposphere are manifest as ‘weather’. A small region of air is termed a parcel. First, we note that a parcel of warm air is less dense than the same parcel of cool air. As a parcel rises, it expands adiabatically (that is, without transfer of heat from its surroundings), so it cools. Cool air can absorb lower concentrations of water vapour than warm air, so the moisture forms clouds. Cloudy skies can therefore be associated with rising air and clear skies are often associated with descending air. The motion of air in the upper altitudes may lead to an accumulation in some regions and a loss of molecules from other regions. The former result in the formation of regions of high pressure (‘highs’ or anticyclones) and the latter result in regions of low pressure (‘lows’, depressions, or cyclones). These regions are shown as H and L on the accompanying weather map (Fig. 1.11). The lines of constant pressure—differing by 4 mbar (400 Pa, about 3 Torr)—marked on it are called isobars. The elongated regions of high and low pressure are known, respectively, as ridges and troughs. In meteorology, large-scale vertical movement is called convection. Horizontal pressure differentials result in the flow of air that we call wind (see Fig.1.12). Winds coming from the north in the Northern hemisphere and from the south in the Southern hemisphere are deflected towards the west as they migrate from a region where the Earth is rotating slowly (at the poles) to where it is rotating most rapidly (at the equator). Winds travel nearly parallel to the isobars, with low pressure to their left in the Northern hemisphere and to the right in the Southern hemisphere. At the surface, where wind speeds are lower, the winds tend to travel perpendicular to the isobars from high to low pressure. This differential motion results in a spiral outward flow of air clockwise in the Northern hemisphere around a high and an inward counterclockwise flow around a low. The air lost from regions of high pressure is restored as an influx of air converges into the region and descends. As we have seen, descending air is associated with clear skies. It also becomes warmer by compression as it descends, so regions of high pressure are associated with high surface temperatures. In winter, the cold surface air may prevent the complete fall of air, and result in a temperature inversion, with a layer of warm air over a layer of cold air. Geographical conditions may also trap cool air, as in Los Angeles, and the photochemical pollutants we know as smog may be trapped under the warm layer. L (b) Mixtures of gases S Fig. 1.12 The flow of air (‘wind’) around regions of high and low pressure in the Northern and Southern hemispheres. When dealing with gaseous mixtures, we often need to know the contribution that each component makes to the total pressure of the sample. The partial pressure, pJ, of a gas J in a mixture (any gas, not just a perfect gas), is defined as pJ = xJ p [1.13] 1.2 THE GAS LAWS where xJ is the mole fraction of the component J, the amount of J expressed as a fraction of the total amount of molecules, n, in the sample: xJ = nJ n n = nA + nB + · · · [1.14] When no J molecules are present, xJ = 0; when only J molecules are present, xJ = 1. It follows from the definition of xJ that, whatever the composition of the mixture, xA + xB + · · · = 1 and therefore that the sum of the partial pressures is equal to the total pressure: pA + pB + · · · = (xA + xB + · · · )p = p (1.15) This relation is true for both real and perfect gases. When all the gases are perfect, the partial pressure as defined in eqn 1.13 is also the pressure that each gas would occupy if it occupied the same container alone at the same temperature. The latter is the original meaning of ‘partial pressure’. That identification was the basis of the original formulation of Dalton’s law: The pressure exerted by a mixture of gases is the sum of the pressures that each one would exist if it occupied the container alone. Now, however, the relation between partial pressure (as defined in eqn 1.13) and total pressure (as given by eqn 1.15) is true for all gases and the identification of partial pressure with the pressure that the gas would exert on its own is valid only for a perfect gas. Example 1.3 Calculating partial pressures The mass percentage composition of dry air at sea level is approximately N2: 75.5; O2: 23.2; Ar: 1.3. What is the partial pressure of each component when the total pressure is 1.00 atm? Method We expect species with a high mole fraction to have a proportionally high partial pressure. Partial pressures are defined by eqn 1.13. To use the equation, we need the mole fractions of the components. To calculate mole fractions, which are defined by eqn 1.14, we use the fact that the amount of molecules J of molar mass MJ in a sample of mass mJ is nJ = mJ/MJ. The mole fractions are independent of the total mass of the sample, so we can choose the latter to be 100 g (which makes the conversion from mass percentages very easy). Thus, the mass of N2 present is 75.5 per cent of 100 g, which is 75.5 g. Answer The amounts of each type of molecule present in 100 g of air, in which the masses of N2, O2, and Ar are 75.5 g, 23.2 g, and 1.3 g, respectively, are n(N 2) = 75.5 g 75.5 mol = − 1 28.02 28.02 g mol n(O2) = 23.2 g 23.2 mol = 32.00 g mol −1 32.00 n(Ar) = 1.3 g 1.3 = mol 39.95 g mol −1 39.95 These three amounts work out as 2.69 mol, 0.725 mol, and 0.033 mol, respectively, for a total of 3.45 mol. The mole fractions are obtained by dividing each of the 13 14 1 THE PROPERTIES OF GASES above amounts by 3.45 mol and the partial pressures are then obtained by multiplying the mole fraction by the total pressure (1.00 atm): Mole fraction: Partial pressure/atm: N2 0.780 0.780 O2 0.210 0.210 Ar 0.0096 0.0096 We have not had to assume that the gases are perfect: partial pressures are defined as pJ = xJ p for any kind of gas. Self-test 1.4 When carbon dioxide is taken into account, the mass percentages are 75.52 (N2), 23.15 (O2), 1.28 (Ar), and 0.046 (CO2). What are the partial pressures when the total pressure is 0.900 atm? [0.703, 0.189, 0.0084, 0.00027 atm] Real gases Real gases do not obey the perfect gas law exactly. Deviations from the law are particularly important at high pressures and low temperatures, especially when a gas is on the point of condensing to liquid. 0 Repulsions dominant Potential energy Contact 1.3 Molecular interactions Separation Attractions dominant Fig. 1.13 The variation of the potential energy of two molecules on their separation. High positive potential energy (at very small separations) indicates that the interactions between them are strongly repulsive at these distances. At intermediate separations, where the potential energy is negative, the attractive interactions dominate. At large separations (on the right) the potential energy is zero and there is no interaction between the molecules. Real gases show deviations from the perfect gas law because molecules interact with one another. Repulsive forces between molecules assist expansion and attractive forces assist compression. Repulsive forces are significant only when molecules are almost in contact: they are short-range interactions, even on a scale measured in molecular diameters (Fig. 1.13). Because they are short-range interactions, repulsions can be expected to be important only when the average separation of the molecules is small. This is the case at high pressure, when many molecules occupy a small volume. On the other hand, attractive intermolecular forces have a relatively long range and are effective over several molecular diameters. They are important when the molecules are fairly close together but not necessarily touching (at the intermediate separations in Fig. 1.13). Attractive forces are ineffective when the molecules are far apart (well to the right in Fig. 1.13). Intermolecular forces are also important when the temperature is so low that the molecules travel with such low mean speeds that they can be captured by one another. At low pressures, when the sample occupies a large volume, the molecules are so far apart for most of the time that the intermolecular forces play no significant role, and the gas behaves virtually perfectly. At moderate pressures, when the average separation of the molecules is only a few molecular diameters, the attractive forces dominate the repulsive forces. In this case, the gas can be expected to be more compressible than a perfect gas because the forces help to draw the molecules together. At high pressures, when the average separation of the molecules is small, the repulsive forces dominate and the gas can be expected to be less compressible because now the forces help to drive the molecules apart. (a) The compression factor The compression factor, Z, of a gas is the ratio of its measured molar volume, Vm = V/n, to the molar volume of a perfect gas, Vmo , at the same pressure and temperature: 1.3 MOLECULAR INTERACTIONS Z= Vm o Vm [1.16] Because the molar volume of a perfect gas is equal to RT/p, an equivalent expression o is Z = RT/pV m , which we can write as pVm = RTZ (1.17) Because for a perfect gas Z = 1 under all conditions, deviation of Z from 1 is a measure of departure from perfect behaviour. Some experimental values of Z are plotted in Fig. 1.14. At very low pressures, all the gases shown have Z ≈ 1 and behave nearly perfectly. At high pressures, all the gases have Z > 1, signifying that they have a larger molar volume than a perfect gas. Repulsive forces are now dominant. At intermediate pressures, most gases have Z < 1, indicating that the attractive forces are reducing the molar volume relative to that of a perfect gas. (b) Virial coefficients Figure 1.15 shows the experimental isotherms for carbon dioxide. At large molar volumes and high temperatures the real-gas isotherms do not differ greatly from perfect-gas isotherms. The small differences suggest that the perfect gas law is in fact the first term in an expression of the form pVm = RT(1 + B′p + C′p2 + · · · ) (1.18) This expression is an example of a common procedure in physical chemistry, in which a simple law that is known to be a good first approximation (in this case pV = nRT) is 2 140 120 80 p/atm Compression factor, Z 40°C Perfect F 31.04°C (Tc) * 60 CH4 20°C E D C B 1.00 H2 C2H4 NH3 0 0 100 H2 1 50°C 0.98 0.96 200 40 p/atm 10 CH4 NH3 400 600 p /atm A 0°C 20 C2H4 0 800 Fig. 1.14 The variation of the compression factor, Z, with pressure for several gases at 0°C. A perfect gas has Z = 1 at all pressures. Notice that, although the curves approach 1 as p → 0, they do so with different slopes. 0 0.2 0.4 Vm /(dm3 mol -1) 0.6 Fig. 1.15 Experimental isotherms of carbon dioxide at several temperatures. The ‘critical isotherm’, the isotherm at the critical temperature, is at 31.04°C. The critical point is marked with a star. 15 16 1 THE PROPERTIES OF GASES treated as the first term in a series in powers of a variable (in this case p). A more convenient expansion for many applications is Comment 1.4 Series expansions are discussed in Appendix 2. ⎛ ⎞ B C pVm = RT ⎜1 + + 2 + ⋅ ⋅ ⋅⎟ Vm V m ⎝ ⎠ Synoptic Table 1.4* Second virial coefficients, B/(cm3 mol−1) Temperature 273 K 600 K −21.7 11.9 −149.7 −12.4 N2 −10.5 21.7 Xe −153.7 −19.6 Ar CO2 (1.19) These two expressions are two versions of the virial equation of state.4 By comparing the expression with eqn 1.17 we see that the term in parentheses can be identified with the compression factor, Z. The coefficients B, C, . . . , which depend on the temperature, are the second, third, . . . virial coefficients (Table 1.4); the first virial coefficient is 1. The third virial coefficient, C, is usually less important than the second coefficient, B, in the sense that 2 at typical molar volumes C/V m << B/Vm. We can use the virial equation to demonstrate the important point that, although the equation of state of a real gas may coincide with the perfect gas law as p → 0, not all its properties necessarily coincide with those of a perfect gas in that limit. Consider, for example, the value of dZ/dp, the slope of the graph of compression factor against pressure. For a perfect gas dZ/dp = 0 (because Z = 1 at all pressures), but for a real gas from eqn 1.18 we obtain dZ = B′ + 2pC′ + · · · → B′ dp * More values are given in the Data section. as p→0 (1.20a) However, B′ is not necessarily zero, so the slope of Z with respect to p does not necessarily approach 0 (the perfect gas value), as we can see in Fig. 1.14. Because several physical properties of gases depend on derivatives, the properties of real gases do not always coincide with the perfect gas values at low pressures. By a similar argument, Compression factor, Z dZ → B as Vm → ∞, d(1/Vm ) Higher temperature Boyle temperature 1 corresponding to p → 0 (1.20b) Because the virial coefficients depend on the temperature, there may be a temperature at which Z → 1 with zero slope at low pressure or high molar volume (Fig. 1.16). At this temperature, which is called the Boyle temperature, TB, the properties of the real gas do coincide with those of a perfect gas as p → 0. According to eqn 1.20b, Z has zero slope as p → 0 if B = 0, so we can conclude that B = 0 at the Boyle temperature. It then follows from eqn 1.19 that pVm ≈ RTB over a more extended range of pressures than at other temperatures because the first term after 1 (that is, B/Vm) in the 2 virial equation is zero and C/V m and higher terms are negligibly small. For helium TB = 22.64 K; for air TB = 346.8 K; more values are given in Table 1.5. Perfect gas Lower temperature 0 Synoptic Table 1.5* Critical constants of gases Pressure, p The compression factor, Z, approaches 1 at low pressures, but does so with different slopes. For a perfect gas, the slope is zero, but real gases may have either positive or negative slopes, and the slope may vary with temperature. At the Boyle temperature, the slope is zero and the gas behaves perfectly over a wider range of conditions than at other temperatures. Fig. 1.16 pc /atm Vc /(cm3 mol−1) Tc /K Zc TB /K Ar 48.0 75.3 150.7 0.292 411.5 CO2 72.9 714.8 94.0 304.2 0.274 He 2.26 57.8 5.2 0.305 O2 50.14 78.0 154.8 0.308 22.64 405.9 * More values are given in the Data section. 4 The name comes from the Latin word for force. The coefficients are sometimes denoted B2, B3, . . . . 1.4 THE VAN DER WAALS EQUATION (c) Condensation Now consider what happens when we compress a sample of gas initially in the state marked A in Fig. 1.15 at constant temperature (by pushing in a piston). Near A, the pressure of the gas rises in approximate agreement with Boyle’s law. Serious deviations from that law begin to appear when the volume has been reduced to B. At C (which corresponds to about 60 atm for carbon dioxide), all similarity to perfect behaviour is lost, for suddenly the piston slides in without any further rise in pressure: this stage is represented by the horizontal line CDE. Examination of the contents of the vessel shows that just to the left of C a liquid appears, and there are two phases separated by a sharply defined surface. As the volume is decreased from C through D to E, the amount of liquid increases. There is no additional resistance to the piston because the gas can respond by condensing. The pressure corresponding to the line CDE, when both liquid and vapour are present in equilibrium, is called the vapour pressure of the liquid at the temperature of the experiment. At E, the sample is entirely liquid and the piston rests on its surface. Any further reduction of volume requires the exertion of considerable pressure, as is indicated by the sharply rising line to the left of E. Even a small reduction of volume from E to F requires a great increase in pressure. (d) Critical constants The isotherm at the temperature Tc (304.19 K, or 31.04°C for CO2) plays a special role in the theory of the states of matter. An isotherm slightly below Tc behaves as we have already described: at a certain pressure, a liquid condenses from the gas and is distinguishable from it by the presence of a visible surface. If, however, the compression takes place at Tc itself, then a surface separating two phases does not appear and the volumes at each end of the horizontal part of the isotherm have merged to a single point, the critical point of the gas. The temperature, pressure, and molar volume at the critical point are called the critical temperature, Tc, critical pressure, pc, and critical molar volume, Vc, of the substance. Collectively, pc, Vc, and Tc are the critical constants of a substance (Table 1.5). At and above Tc, the sample has a single phase that occupies the entire volume of the container. Such a phase is, by definition, a gas. Hence, the liquid phase of a substance does not form above the critical temperature. The critical temperature of oxygen, for instance, signifies that it is impossible to produce liquid oxygen by compression alone if its temperature is greater than 155 K: to liquefy oxygen—to obtain a fluid phase that does not occupy the entire volume—the temperature must first be lowered to below 155 K, and then the gas compressed isothermally. The single phase that fills the entire volume when T > Tc may be much denser than we normally consider typical of gases, and the name supercritical fluid is preferred. 1.4 The van der Waals equation We can draw conclusions from the virial equations of state only by inserting specific values of the coefficients. It is often useful to have a broader, if less precise, view of all gases. Therefore, we introduce the approximate equation of state suggested by J.D. van der Waals in 1873. This equation is an excellent example of an expression that can be obtained by thinking scientifically about a mathematically complicated but physically simple problem, that is, it is a good example of ‘model building’. The van der Waals equation is p= ⎛ n⎞ nRT − a⎜ ⎟ V − nb ⎝V ⎠ 2 (1.21a) Comment 1.5 The web site contains links to online databases of properties of gases. 17 18 1 THE PROPERTIES OF GASES Synoptic Table 1.6* van der Waals coefficients a/(atm dm6 mol−2) b/(10−2 dm3 mol−1) Ar 1.337 3.20 CO2 3.610 4.29 He 0.0341 2.38 Xe 4.137 5.16 * More values are given in the Data section. and a derivation is given in Justification 1.1. The equation is often written in terms of the molar volume Vm = V/n as p= RT a − 2 Vm − b V m (1.21b) The constants a and b are called the van der Waals coefficients. They are characteristic of each gas but independent of the temperature (Table 1.6). Justification 1.1 The van der Waals equation of state The repulsive interactions between molecules are taken into account by supposing that they cause the molecules to behave as small but impenetrable spheres. The nonzero volume of the molecules implies that instead of moving in a volume V they are restricted to a smaller volume V − nb, where nb is approximately the total volume taken up by the molecules themselves. This argument suggests that the perfect gas law p = nRT/V should be replaced by p= nRT V − nb when repulsions are significant. The closest distance of two hard-sphere molecules of radius r, and volume Vmolecule = 34 πr 3, is 2r, so the volume excluded is 34 π(2r)3, or 8Vmolecule. The volume excluded per molecule is one-half this volume, or 4Vmolecule, so b ≈ 4VmoleculeNA. The pressure depends on both the frequency of collisions with the walls and the force of each collision. Both the frequency of the collisions and their force are reduced by the attractive forces, which act with a strength proportional to the molar concentration, n/V, of molecules in the sample. Therefore, because both the frequency and the force of the collisions are reduced by the attractive forces, the pressure is reduced in proportion to the square of this concentration. If the reduction of pressure is written as −a(n/V)2, where a is a positive constant characteristic of each gas, the combined effect of the repulsive and attractive forces is the van der Waals equation of state as expressed in eqn 1.21. In this Justification we have built the van der Waals equation using vague arguments about the volumes of molecules and the effects of forces. The equation can be derived in other ways, but the present method has the advantage that it shows how to derive the form of an equation out of general ideas. The derivation also has the advantage of keeping imprecise the significance of the coefficients a and b: they are much better regarded as empirical parameters than as precisely defined molecular properties. Example 1.4 Using the van der Waals equation to estimate a molar volume Estimate the molar volume of CO2 at 500 K and 100 atm by treating it as a van der Waals gas. Method To express eqn 1.21b as an equation for the molar volume, we multiply 2 both sides by (Vm – b)V m , to obtain 2 2 (Vm – b)V m p = RTV m – (Vm – b)a and, after division by p, collect powers of Vm to obtain ⎛ RT ⎞ 2 ⎛ a ⎞ ab V m3 − ⎜ b + V m + ⎜ ⎟ Vm − =0 ⎟ p ⎠ p ⎝ ⎝ p⎠ 1.4 THE VAN DER WAALS EQUATION 19 Although closed expressions for the roots of a cubic equation can be given, they are very complicated. Unless analytical solutions are essential, it is usually more expedient to solve such equations with commercial software. Answer According to Table 1.6, a = 3.592 dm6 atm mol−2 and b = 4.267 × 10−2 dm3 mol−1. Under the stated conditions, RT/p = 0.410 dm3 mol−1. The coefficients in the equation for Vm are therefore b + RT/p = 0.453 dm3 mol−1 a/p = 3.61 × 10−2 (dm3 mol−1)2 ab/p = 1.55 × 10−3 (dm3 mol−1)3 Therefore, on writing x = Vm/(dm3 mol−1), the equation to solve is x 3 − 0.453x 2 + (3.61 × 10−2)x − (1.55 × 10−3) = 0 The acceptable root is x = 0.366, which implies that Vm = 0.366 dm3 mol−1. For a perfect gas under these conditions, the molar volume is 0.410 dm3 mol−1. Self-test 1.5 Calculate the molar volume of argon at 100°C and 100 atm on the [0.298 dm3 mol−1] assumption that it is a van der Waals gas. (a) The reliability of the equation We now examine to what extent the van der Waals equation predicts the behaviour of real gases. It is too optimistic to expect a single, simple expression to be the true equation of state of all substances, and accurate work on gases must resort to the virial equation, use tabulated values of the coefficients at various temperatures, and analyse the systems numerically. The advantage of the van der Waals equation, however, is that it is analytical (that is, expressed symbolically) and allows us to draw some general conclusions about real gases. When the equation fails we must use one of the other equations of state that have been proposed (some are listed in Table 1.7), invent a new one, or go back to the virial equation. That having been said, we can begin to judge the reliability of the equation by comparing the isotherms it predicts with the experimental isotherms in Fig. 1.15. Some Table 1.7 Selected equations of state Critical constants Equation Reduced form* pc Vc Tc 3b 8a 27bR 3b 2 ⎛ 2a ⎞ ⎜ ⎟ 3 ⎝ 3bR ⎠ 2b a 4bR Perfect gas p= RT Vm van der Waals p= RT a − 2 Vm − b V m p= 8Tr 3 − 3Vr − 1 V 2r a 27b2 Berthelot p= RT a − 2 Vm − b TV m p= 8Tr 3 − 3Vr − 1 TrV 2r 1 ⎛ 2aR ⎞ ⎜ ⎟ 12 ⎝ 3b3 ⎠ Dieterici p= RTe−a/RTVm Vm − b p= e2Tre−2/TrVr 2Vr − 1 a 4e2b2 Virial p= ⎫ RT ⎧ B(T ) C(T ) + 2 + ⋅ ⋅ ⋅⎬ ⎨1 + Vm ⎩⎪ Vm Vm ⎭⎪ * Reduced variables are defined in Section 1.5. 1/2 1/2 20 1 THE PROPERTIES OF GASES 1.5 1.5 Reduced pressure, p/pc Pressure 1.5 1.0 Te m pe ra tu re 0.8 Volum e 1 1.0 0.5 0.8 The surface of possible states allowed by the van der Waals equation. Compare this surface with that shown in Fig. 1.8. Fig. 1.17 0 0.1 1 Reduced volume, V/Vc 10 Van der Waals isotherms at several values of T/Tc. Compare these curves with those in Fig. 1.15. The van der Waals loops are normally replaced by horizontal straight lines. The critical isotherm is the isotherm for T/Tc = 1. Fig. 1.18 Exploration Calculate the molar volume of chlorine gas on the basis of the van der Waals equation of state at 250 K and 150 kPa and calculate the percentage difference from the value predicted by the perfect gas equation. Equal areas 3 calculated isotherms are shown in Figs. 1.17 and 1.18. Apart from the oscillations below the critical temperature, they do resemble experimental isotherms quite well. The oscillations, the van der Waals loops, are unrealistic because they suggest that under some conditions an increase of pressure results in an increase of volume. Therefore they are replaced by horizontal lines drawn so the loops define equal areas above and below the lines: this procedure is called the Maxwell construction (3). The van der Waals coefficients, such as those in Table 1.7, are found by fitting the calculated curves to the experimental curves. (b) The features of the equation The principal features of the van der Waals equation can be summarized as follows. (1) Perfect gas isotherms are obtained at high temperatures and large molar volumes. When the temperature is high, RT may be so large that the first term in eqn 1.21b greatly exceeds the second. Furthermore, if the molar volume is large in the sense Vm >> b, then the denominator Vm − b ≈ Vm. Under these conditions, the equation reduces to p = RT/Vm, the perfect gas equation. (2) Liquids and gases coexist when cohesive and dispersing effects are in balance. The van der Waals loops occur when both terms in eqn 1.21b have similar magnitudes. The first term arises from the kinetic energy of the molecules and their repulsive interactions; the second represents the effect of the attractive interactions. (3) The critical constants are related to the van der Waals coefficients. 1.5 THE PRINCIPLE OF CORRESPONDING STATES For T < Tc, the calculated isotherms oscillate, and each one passes through a minimum followed by a maximum. These extrema converge as T → Tc and coincide at T = Tc; at the critical point the curve has a flat inflexion (4). From the properties of curves, we know that an inflexion of this type occurs when both the first and second derivatives are zero. Hence, we can find the critical constants by calculating these derivatives and setting them equal to zero: dp RT 2a =− + 3 =0 dVm (Vm − b)2 V m d2 p 2RT 6a = − 4 =0 2 3 dV m (Vm − b) Vm at the critical point. The solutions of these two equations (and using eqn 1.21b to calculate pc from Vc and Tc) are Vc = 3b pc = a 27b2 Tc = 8a 27 Rb (1.22) These relations provide an alternative route to the determination of a and b from the values of the critical constants. They can be tested by noting that the critical compression factor, Zc, is predicted to be equal to Zc = pcVc 3 = RTc 8 (1.23) for all gases. We see from Table 1.5 that, although Zc < 38 = 0.375, it is approximately constant (at 0.3) and the discrepancy is reasonably small. 1.5 The principle of corresponding states An important general technique in science for comparing the properties of objects is to choose a related fundamental property of the same kind and to set up a relative scale on that basis. We have seen that the critical constants are characteristic properties of gases, so it may be that a scale can be set up by using them as yardsticks. We therefore introduce the dimensionless reduced variables of a gas by dividing the actual variable by the corresponding critical constant: pr = p pc Vr = Vm Vc Tr = T Tc [1.24] If the reduced pressure of a gas is given, we can easily calculate its actual pressure by using p = pr pc, and likewise for the volume and temperature. Van der Waals, who first tried this procedure, hoped that gases confined to the same reduced volume, Vr, at the same reduced temperature, Tr, would exert the same reduced pressure, pr. The hope was largely fulfilled (Fig. 1.19). The illustration shows the dependence of the compression factor on the reduced pressure for a variety of gases at various reduced temperatures. The success of the procedure is strikingly clear: compare this graph with Fig. 1.14, where similar data are plotted without using reduced variables. The observation that real gases at the same reduced volume and reduced temperature exert the same reduced pressure is called the principle of corresponding states. The principle is only an approximation. It works best for gases composed of spherical molecules; it fails, sometimes badly, when the molecules are non-spherical or polar. The van der Waals equation sheds some light on the principle. First, we express eqn 1.21b in terms of the reduced variables, which gives pr pc = RTrTc a − VrVc − b V 2rV c2 4 21 1 THE PROPERTIES OF GASES 1.0 2.0 0.8 Compression factor, Z 22 1.2 0.6 1.0 0.4 Nitrogen Methane Propane 0.2 Ethene 0 0 1 2 3 4 Reduced pressure, pr 5 6 7 Fig. 1.19 The compression factors of four of the gases shown in Fig. 1.14 plotted using reduced variables. The curves are labelled with the reduced temperature Tr = T/Tc. The use of reduced variables organizes the data on to single curves. Exploration Is there a set of conditions at which the compression factor of a van der Waals gas passes through a minimum? If so, how does the location and value of the minimum value of Z depend on the coefficients a and b? Then we express the critical constants in terms of a and b by using eqn 1.22: apr 8aTr a = − 2 2 2 27b(3bVr − b) 9b V r 27b which can be reorganized into pr = 8Tr 3 − 2 3Vr − 1 V r (1.25) This equation has the same form as the original, but the coefficients a and b, which differ from gas to gas, have disappeared. It follows that if the isotherms are plotted in terms of the reduced variables (as we did in fact in Fig. 1.18 without drawing attention to the fact), then the same curves are obtained whatever the gas. This is precisely the content of the principle of corresponding states, so the van der Waals equation is compatible with it. Looking for too much significance in this apparent triumph is mistaken, because other equations of state also accommodate the principle (Table 1.7). In fact, all we need are two parameters playing the roles of a and b, for then the equation can always be manipulated into reduced form. The observation that real gases obey the principle approximately amounts to saying that the effects of the attractive and repulsive interactions can each be approximated in terms of a single parameter. The importance of the principle is then not so much its theoretical interpretation but the way that it enables the properties of a range of gases to be coordinated on to a single diagram (for example, Fig. 1.19 instead of Fig. 1.14). DISCUSSION QUESTIONS 23 Checklist of key ideas 1. A gas is a form of matter that fills any container it occupies. 2. An equation of state interrelates pressure, volume, temperature, and amount of substance: p = f(T,V,n). 3. The pressure is the force divided by the area to which the force is applied. The standard pressure is p7 = 1 bar (105 Pa). 4. Mechanical equilibrium is the condition of equality of pressure on either side of a movable wall. 5. Temperature is the property that indicates the direction of the flow of energy through a thermally conducting, rigid wall. 6. A diathermic boundary is a boundary that permits the passage of energy as heat. An adiabatic boundary is a boundary that prevents the passage of energy as heat. 7. Thermal equilibrium is a condition in which no change of state occurs when two objects A and B are in contact through a diathermic boundary. 8. The Zeroth Law of thermodynamics states that, if A is in thermal equilibrium with B, and B is in thermal equilibrium with C, then C is also in thermal equilibrium with A. 9. The Celsius and thermodynamic temperature scales are related by T/K = θ/°C + 273.15. 10. A perfect gas obeys the perfect gas equation, pV = nRT, exactly under all conditions. 11. Dalton’s law states that the pressure exerted by a mixture of gases is the sum of the partial pressures of the gases. 12. The partial pressure of any gas is defined as pJ = xJ p, where xJ = nJ/n is its mole fraction in a mixture and p is the total pressure. 13. In real gases, molecular interactions affect the equation of state; the true equation of state is expressed in terms of virial 2 coefficients B, C, . . . : pVm = RT(1 + B/Vm + C/V m + · · · ). 14. The vapour pressure is the pressure of a vapour in equilibrium with its condensed phase. 15. The critical point is the point at which the volumes at each end of the horizontal part of the isotherm have merged to a single point. The critical constants pc, Vc, and Tc are the pressure, molar volume, and temperature, respectively, at the critical point. 16. A supercritical fluid is a dense fluid phase above its critical temperature and pressure. 17. The van der Waals equation of state is an approximation to the true equation of state in which attractions are represented by a parameter a and repulsions are represented by a parameter b: p = nRT/(V − nb) − a(n/V)2. 18. A reduced variable is the actual variable divided by the corresponding critical constant. 19. According to the principle of corresponding states, real gases at the same reduced volume and reduced temperature exert the same reduced pressure. Further reading Articles and texts J.L. Pauley and E.H. Davis, P-V-T isotherms of real gases: Experimental versus calculated values. J. Chem. Educ. 63, 466 (1986). M. Ross, Equations of state. In Encyclopedia of applied physics (ed. G.L. Trigg), 6, 291. VCH, New York (1993). A. J. Walton, Three phases of matter. Oxford University Press (1983). R.P. Wayne, Chemistry of atmospheres, an introduction to the chemistry of atmospheres of earth, the planets, and their satellites. Oxford University Press (2000). Sources of data and information J.H. Dymond and E.B. Smith, The virial coefficients of pure gases and mixtures. Oxford University Press (1980). A.D. McNaught and A. Wilkinson, Compendium of chemical terminology. Blackwell Scientific, Oxford (1997). Discussion questions 1.1 Explain how the perfect gas equation of state arises by combination of Boyle’s law, Charles’s law, and Avogadro’s principle. 1.2 Explain the term ‘partial pressure’ and explain why Dalton’s law is a limiting law. 1.3 Explain how the compression factor varies with pressure and temperature and describe how it reveals information about intermolecular interactions in real gases. 1.4 What is the significance of the critical constants? 1.5 Describe the formulation of the van der Waals equation and suggest a rationale for one other equation of state in Table 1.7. 1.6 Explain how the van der Waals equation accounts for critical behaviour. 24 1 THE PROPERTIES OF GASES Exercises 1.1(a) (a) Could 131 g of xenon gas in a vessel of volume 1.0 dm3 exert a pressure of 20 atm at 25°C if it behaved as a perfect gas? If not, what pressure would it exert? (b) What pressure would it exert if it behaved as a van der Waals gas? 1.7(b) The following data have been obtained for oxygen gas at 273.15 K. 1.1(b) (a) Could 25 g of argon gas in a vessel of volume 1.5 dm3 exert a Vm /(dm3 mol−1) 29.9649 ρ/(g dm−3) 1.07144 pressure of 2.0 bar at 30°C if it behaved as a perfect gas? If not, what pressure would it exert? (b) What pressure would it exert if it behaved as a van der Waals gas? 1.2(a) A perfect gas undergoes isothermal compression, which reduces its volume by 2.20 dm3. The final pressure and volume of the gas are 5.04 bar and 4.65 dm3, respectively. Calculate the original pressure of the gas in (a) bar, (b) atm. Calculate the best value of the gas constant R from them and the best value of the molar mass of O2. p/atm 0.750 000 0.500 000 44.8090 0.714110 0.250 000 89.6384 0.356975 1.8(a) At 500°C and 93.2 kPa, the mass density of sulfur vapour is 3.710 kg m−3. What is the molecular formula of sulfur under these conditions? 1.8(b) At 100°C and 1.60 kPa, the mass density of phosphorus vapour is 0.6388 kg m−3. What is the molecular formula of phosphorus under these conditions? 1.2(b) A perfect gas undergoes isothermal compression, which reduces its 1.9(a) Calculate the mass of water vapour present in a room of volume 400 m3 volume by 1.80 dm3. The final pressure and volume of the gas are 1.97 bar and 2.14 dm3, respectively. Calculate the original pressure of the gas in (a) bar, (b) Torr. 1.9(b) Calculate the mass of water vapour present in a room of volume 250 m3 −2 1.3(a) A car tyre (i.e. an automobile tire) was inflated to a pressure of 24 lb in (1.00 atm = 14.7 lb in−2) on a winter’s day when the temperature was –5°C. What pressure will be found, assuming no leaks have occurred and that the volume is constant, on a subsequent summer’s day when the temperature is 35°C? What complications should be taken into account in practice? 1.3(b) A sample of hydrogen gas was found to have a pressure of 125 kPa when the temperature was 23°C. What can its pressure be expected to be when the temperature is 11°C? that contains air at 27°C on a day when the relative humidity is 60 per cent. that contains air at 23°C on a day when the relative humidity is 53 per cent. 1.10(a) Given that the density of air at 0.987 bar and 27°C is 1.146 kg m−3, calculate the mole fraction and partial pressure of nitrogen and oxygen assuming that (a) air consists only of these two gases, (b) air also contains 1.0 mole per cent Ar. 1.10(b) A gas mixture consists of 320 mg of methane, 175 mg of argon, and 225 mg of neon. The partial pressure of neon at 300 K is 8.87 kPa. Calculate (a) the volume and (b) the total pressure of the mixture. 1.11(a) The density of a gaseous compound was found to be 1.23 kg m−3 at 1.4(a) A sample of 255 mg of neon occupies 3.00 dm at 122 K. Use the perfect 330 K and 20 kPa. What is the molar mass of the compound? gas law to calculate the pressure of the gas. 1.11(b) In an experiment to measure the molar mass of a gas, 250 cm3 of the 1.4(b) A homeowner uses 4.00 × 103 m3 of natural gas in a year to heat a gas was confined in a glass vessel. The pressure was 152 Torr at 298 K and, after correcting for buoyancy effects, the mass of the gas was 33.5 mg. What is the molar mass of the gas? 3 home. Assume that natural gas is all methane, CH4, and that methane is a perfect gas for the conditions of this problem, which are 1.00 atm and 20°C. What is the mass of gas used? 1.5(a) A diving bell has an air space of 3.0 m3 when on the deck of a boat. What is the volume of the air space when the bell has been lowered to a depth of 50 m? Take the mean density of sea water to be 1.025 g cm−3 and assume that the temperature is the same as on the surface. 1.5(b) What pressure difference must be generated across the length of a 15 cm vertical drinking straw in order to drink a water-like liquid of density 1.0 g cm−3? 1.6(a) A manometer consists of a U-shaped tube containing a liquid. One side is connected to the apparatus and the other is open to the atmosphere. The pressure inside the apparatus is then determined from the difference in heights of the liquid. Suppose the liquid is water, the external pressure is 770 Torr, and the open side is 10.0 cm lower than the side connected to the apparatus. What is the pressure in the apparatus? (The density of water at 25°C is 0.997 07 g cm−3.) 1.6(b) A manometer like that described in Exercise 1.6a contained mercury in place of water. Suppose the external pressure is 760 Torr, and the open side is 10.0 cm higher than the side connected to the apparatus. What is the pressure in the apparatus? (The density of mercury at 25°C is 13.55 g cm−3.) 1.7(a) In an attempt to determine an accurate value of the gas constant, R, a 3 student heated a container of volume 20.000 dm filled with 0.251 32 g of helium gas to 500°C and measured the pressure as 206.402 cm of water in a manometer at 25°C. Calculate the value of R from these data. (The density of water at 25°C is 0.997 07 g cm−3; the construction of a manometer is described in Exercise 1.6a.) 1.12(a) The densities of air at −85°C, 0°C, and 100°C are 1.877 g dm−3, 1.294 g dm−3, and 0.946 g dm−3, respectively. From these data, and assuming that air obeys Charles’s law, determine a value for the absolute zero of temperature in degrees Celsius. 1.12(b) A certain sample of a gas has a volume of 20.00 dm3 at 0°C and 1.000 atm. A plot of the experimental data of its volume against the Celsius temperature, θ, at constant p, gives a straight line of slope 0.0741 dm3 (°C)−1. From these data alone (without making use of the perfect gas law), determine the absolute zero of temperature in degrees Celsius. 1.13(a) Calculate the pressure exerted by 1.0 mol C2H6 behaving as (a) a perfect gas, (b) a van der Waals gas when it is confined under the following conditions: (i) at 273.15 K in 22.414 dm3, (ii) at 1000 K in 100 cm3. Use the data in Table 1.6. 1.13(b) Calculate the pressure exerted by 1.0 mol H2S behaving as (a) a perfect gas, (b) a van der Waals gas when it is confined under the following conditions: (i) at 273.15 K in 22.414 dm3, (ii) at 500 K in 150 cm3. Use the data in Table 1.6. 1.14(a) Express the van der Waals parameters a = 0.751 atm dm6 mol−2 and b = 0.0226 dm3 mol−1 in SI base units. 1.14(b) Express the van der Waals parameters a = 1.32 atm dm6 mol−2 and b = 0.0436 dm3 mol−1 in SI base units. 1.15(a) A gas at 250 K and 15 atm has a molar volume 12 per cent smaller than that calculated from the perfect gas law. Calculate (a) the compression factor under these conditions and (b) the molar volume of the gas. Which are dominating in the sample, the attractive or the repulsive forces? PROBLEMS 25 1.15(b) A gas at 350 K and 12 atm has a molar volume 12 per cent larger than that calculated from the perfect gas law. Calculate (a) the compression factor under these conditions and (b) the molar volume of the gas. Which are dominating in the sample, the attractive or the repulsive forces? 1.19(a) The critical constants of methane are pc = 45.6 atm, Vc = 98.7 cm3 mol−1, and Tc = 190.6 K. Calculate the van der Waals parameters of the gas and estimate the radius of the molecules. 1.16(a) In an industrial process, nitrogen is heated to 500 K at a constant and Tc = 305.4 K. Calculate the van der Waals parameters of the gas and estimate the radius of the molecules. volume of 1.000 m3. The gas enters the container at 300 K and 100 atm. The mass of the gas is 92.4 kg. Use the van der Waals equation to determine the approximate pressure of the gas at its working temperature of 500 K. For nitrogen, a = 1.352 dm6 atm mol−2, b = 0.0387 dm3 mol−1. 1.16(b) Cylinders of compressed gas are typically filled to a pressure of 200 bar. For oxygen, what would be the molar volume at this pressure and 25°C based on (a) the perfect gas equation, (b) the van der Waals equation. For oxygen, a = 1.364 dm6 atm mol−2, b = 3.19 × 10−2 dm3 mol−1. 1.17(a) Suppose that 10.0 mol C2H6(g) is confined to 4.860 dm3 at 27°C. Predict the pressure exerted by the ethane from (a) the perfect gas and (b) the van der Waals equations of state. Calculate the compression factor based on these calculations. For ethane, a = 5.507 dm6 atm mol−2, b = 0.0651 dm3 mol−1. 1.17(b) At 300 K and 20 atm, the compression factor of a gas is 0.86. Calculate (a) the volume occupied by 8.2 mmol of the gas under these conditions and (b) an approximate value of the second virial coefficient B at 300 K. 1.18(a) A vessel of volume 22.4 dm3 contains 2.0 mol H2 and 1.0 mol N2 at 273.15 K. Calculate (a) the mole fractions of each component, (b) their partial pressures, and (c) their total pressure. 1.18(b) A vessel of volume 22.4 dm3 contains 1.5 mol H2 and 2.5 mol N2 at 273.15 K. Calculate (a) the mole fractions of each component, (b) their partial pressures, and (c) their total pressure. 1.19(b) The critical constants of ethane are pc = 48.20 atm, Vc = 148 cm3 mol−1, 1.20(a) Use the van der Waals parameters for chlorine to calculate approximate values of (a) the Boyle temperature of chlorine and (b) the radius of a Cl2 molecule regarded as a sphere. 1.20(b) Use the van der Waals parameters for hydrogen sulfide to calculate approximate values of (a) the Boyle temperature of the gas and (b) the radius of a H2S molecule regarded as a sphere (a = 4.484 dm6 atm mol−2, b = 0.0434 dm3 mol−1). 1.21(a) Suggest the pressure and temperature at which 1.0 mol of (a) NH3, (b) Xe, (c) He will be in states that correspond to 1.0 mol H2 at 1.0 atm and 25°C. 1.21(b) Suggest the pressure and temperature at which 1.0 mol of (a) H2S, (b) CO2, (c) Ar will be in states that correspond to 1.0 mol N2 at 1.0 atm and 25°C. 1.22(a) A certain gas obeys the van der Waals equation with a = 0.50 m6 Pa mol−2. Its volume is found to be 5.00 × 10−4 m3 mol−1 at 273 K and 3.0 MPa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure? 1.22(b) A certain gas obeys the van der Waals equation with a = 0.76 m6 Pa mol−2. Its volume is found to be 4.00 × 10−4 m3 mol−1 at 288 K and 4.0 MPa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure? Problems* Numerical problems 1.4 The molar mass of a newly synthesized fluorocarbon was measured in a 1.1 Recent communication with the inhabitants of Neptune have revealed that they have a Celsius-type temperature scale, but based on the melting point (0°N) and boiling point (100°N) of their most common substance, hydrogen. Further communications have revealed that the Neptunians know about perfect gas behaviour and they find that, in the limit of zero pressure, the value of pV is 28 dm3 atm at 0°N and 40 dm3 atm at 100°N. What is the value of the absolute zero of temperature on their temperature scale? 1.2 Deduce the relation between the pressure and mass density, ρ, of a perfect gas of molar mass M. Confirm graphically, using the following data on dimethyl ether at 25°C, that perfect behaviour is reached at low pressures and find the molar mass of the gas. p/kPa 12.223 25.20 36.97 60.37 85.23 101.3 ρ/(kg m−3) 0.225 0.456 0.664 1.062 1.468 1.734 1.3 Charles’s law is sometimes expressed in the form V = V0(1 + αθ), where θ is the Celsius temperature, α is a constant, and V0 is the volume of the sample at 0°C. The following values for α have been reported for nitrogen at 0°C: p/Torr 749.7 599.6 333.1 98.6 103α /(°C)−1 3.6717 3.6697 3.6665 3.6643 For these data calculate the best value for the absolute zero of temperature on the Celsius scale. gas microbalance. This device consists of a glass bulb forming one end of a beam, the whole surrounded by a closed container. The beam is pivoted, and the balance point is attained by raising the pressure of gas in the container, so increasing the buoyancy of the enclosed bulb. In one experiment, the balance point was reached when the fluorocarbon pressure was 327.10 Torr; for the same setting of the pivot, a balance was reached when CHF3 (M = 70.014 g mol−1) was introduced at 423.22 Torr. A repeat of the experiment with a different setting of the pivot required a pressure of 293.22 Torr of the fluorocarbon and 427.22 Torr of the CHF3. What is the molar mass of the fluorocarbon? Suggest a molecular formula. 1.5 A constant-volume perfect gas thermometer indicates a pressure of 6.69 kPa at the triple point temperature of water (273.16 K). (a) What change of pressure indicates a change of 1.00 K at this temperature? (b) What pressure indicates a temperature of 100.00°C? (c) What change of pressure indicates a change of 1.00 K at the latter temperature? 1.6 A vessel of volume 22.4 dm3 contains 2.0 mol H2 and 1.0 mol N2 at 273.15 K initially. All the H2 reacted with sufficient N2 to form NH3. Calculate the partial pressures and the total pressure of the final mixture. 1.7 Calculate the molar volume of chlorine gas at 350 K and 2.30 atm using (a) the perfect gas law and (b) the van der Waals equation. Use the answer to (a) to calculate a first approximation to the correction term for attraction and then use successive approximations to obtain a numerical answer for part (b). * Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady. 26 1 THE PROPERTIES OF GASES 1.8 At 273 K measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2, where B and C are the second and third virial coefficients in the expansion of Z in powers of 1/Vm. Assuming that the perfect gas law holds sufficiently well for the estimation of the second and third terms of the expansion, calculate the compression factor of argon at 100 atm and 273 K. From your result, estimate the molar volume of argon under these conditions. 1.9 Calculate the volume occupied by 1.00 mol N2 using the van der Waals equation in the form of a virial expansion at (a) its critical temperature, (b) its Boyle temperature, and (c) its inversion temperature. Assume that the pressure is 10 atm throughout. At what temperature is the gas most perfect? Use the following data: Tc = 126.3 K, a = 1.352 dm6 atm mol−2, b = 0.0387 dm3 mol−1. 1.10‡ The second virial coefficient of methane can be approximated 2 by the empirical equation B′(T) = a + be−c/T , where a = −0.1993 bar−1, −1 2 b = 0.2002 bar , and c = 1131 K with 300 K < T < 600 K. What is the Boyle temperature of methane? 1.11 The mass density of water vapour at 327.6 atm and 776.4 K is 133.2 kg m−3. Given that for water Tc = 647.4 K, pc = 218.3 atm, a = 5.464 dm6 atm mol−2, b = 0.03049 dm3 mol−1, and M = 18.02 g mol−1, calculate (a) the molar volume. Then calculate the compression factor (b) from the data, (c) from the virial expansion of the van der Waals equation. 1.12 The critical volume and critical pressure of a certain gas are 160 cm3 mol−1 and 40 atm, respectively. Estimate the critical temperature by assuming that the gas obeys the Berthelot equation of state. Estimate the radii of the gas molecules on the assumption that they are spheres. 1.13 Estimate the coefficients a and b in the Dieterici equation of state from the critical constants of xenon. Calculate the pressure exerted by 1.0 mol Xe when it is confined to 1.0 dm3 at 25°C. Theoretical problems 1.14 Show that the van der Waals equation leads to values of Z < 1 and Z > 1, and identify the conditions for which these values are obtained. 1.15 Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. The expansion you will need is (1 − x)−1 = 1 + x + x2 + · · · . Measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K. What are the values of a and b in the corresponding van der Waals equation of state? 1.16‡ Derive the relation between the critical constants and the Dieterici equation parameters. Show that Zc = 2e−2 and derive the reduced form of the Dieterici equation of state. Compare the van der Waals and Dieterici predictions of the critical compression factor. Which is closer to typical experimental values? 1.17 A scientist proposed the following equation of state: p= RT Vm − B 2 Vm + C 3 Vm Show that the equation leads to critical behaviour. Find the critical constants of the gas in terms of B and C and an expression for the critical compression factor. 1.18 Equations 1.18 and 1.19 are expansions in p and 1/Vm, respectively. Find the relation between B, C and B′, C′. 1.19 The second virial coefficient B′ can be obtained from measurements of the density ρ of a gas at a series of pressures. Show that the graph of p/ρ against p should be a straight line with slope proportional to B′. Use the data on dimethyl ether in Problem 1.2 to find the values of B′ and B at 25°C. 1.20 The equation of state of a certain gas is given by p = RT/Vm + 2 (a + bT)/V m , where a and b are constants. Find (∂V/∂T)p. 1.21 The following equations of state are occasionally used for approximate calculations on gases: (gas A) pVm = RT(1 + b/Vm), (gas B) p(Vm – b) = RT. Assuming that there were gases that actually obeyed these equations of state, would it be possible to liquefy either gas A or B? Would they have a critical temperature? Explain your answer. 1.22 Derive an expression for the compression factor of a gas that obeys the equation of state p(V – nb) = nRT, where b and R are constants. If the pressure and temperature are such that Vm = 10b, what is the numerical value of the compression factor? 1.23‡ The discovery of the element argon by Lord Rayleigh and Sir William Ramsay had its origins in Rayleigh’s measurements of the density of nitrogen with an eye toward accurate determination of its molar mass. Rayleigh prepared some samples of nitrogen by chemical reaction of nitrogencontaining compounds; under his standard conditions, a glass globe filled with this ‘chemical nitrogen’ had a mass of 2.2990 g. He prepared other samples by removing oxygen, carbon dioxide, and water vapour from atmospheric air; under the same conditions, this ‘atmospheric nitrogen’ had a mass of 2.3102 g (Lord Rayleigh, Royal Institution Proceedings 14, 524 (1895)). With the hindsight of knowing accurate values for the molar masses of nitrogen and argon, compute the mole fraction of argon in the latter sample on the assumption that the former was pure nitrogen and the latter a mixture of nitrogen and argon. 1.24‡ A substance as elementary and well known as argon still receives research attention. Stewart and Jacobsen have published a review of thermodynamic properties of argon (R.B. Stewart and R.T. Jacobsen, J. Phys. Chem. Ref. Data 18, 639 (1989)) that included the following 300 K isotherm. p/MPa 0.4000 0.5000 0.6000 0.8000 1.000 Vm/(dm3 mol−1) 6.2208 4.9736 4.1423 3.1031 2.4795 p/MPa 1.500 2.000 2.500 3.000 4.000 1.6483 1.2328 0.98357 0.81746 0.60998 3 −1 Vm/(dm mol ) (a) Compute the second virial coefficient, B, at this temperature. (b) Use non-linear curve-fitting software to compute the third virial coefficient, C, at this temperature. Applications: to environmental science 1.25 Atmospheric pollution is a problem that has received much attention. Not all pollution, however, is from industrial sources. Volcanic eruptions can be a significant source of air pollution. The Kilauea volcano in Hawaii emits 200–300 t of SO2 per day. If this gas is emitted at 800°C and 1.0 atm, what volume of gas is emitted? 1.26 Ozone is a trace atmospheric gas that plays an important role in screening the Earth from harmful ultraviolet radiation, and the abundance of ozone is commonly reported in Dobson units. One Dobson unit is the thickness, in thousandths of a centimetre, of a column of gas if it were collected as a pure gas at 1.00 atm and 0°C. What amount of O3 (in moles) is found in a column of atmosphere with a cross-sectional area of 1.00 dm2 if the abundance is 250 Dobson units (a typical mid-latitude value)? In the seasonal Antarctic ozone hole, the column abundance drops below 100 Dobson units; how many moles of ozone are found in such a column of air above a 1.00 dm2 area? Most atmospheric ozone is found between 10 and 50 km above the surface of the earth. If that ozone is spread uniformly through this portion of the atmosphere, what is the average molar concentration corresponding to (a) 250 Dobson units, (b) 100 Dobson units? 1.27 The barometric formula relates the pressure of a gas of molar mass M at an altitude h to its pressure p0 at sea level. Derive this relation by showing that PROBLEMS the change in pressure dp for an infinitesimal change in altitude dh where the density is ρ is dp = −ρgdh. Remember that ρ depends on the pressure. Evaluate (a) the pressure difference between the top and bottom of a laboratory vessel of height 15 cm, and (b) the external atmospheric pressure at a typical cruising altitude of an aircraft (11 km) when the pressure at ground level is 1.0 atm. 1.29‡ The preceding problem is most readily solved (see the Solutions 1.28 Balloons are still used to deploy sensors that monitor meteorological phenomena and the chemistry of the atmosphere. It is possible to investigate some of the technicalities of ballooning by using the perfect gas law. Suppose your balloon has a radius of 3.0 m and that it is spherical. (a) What amount of H2 (in moles) is needed to inflate it to 1.0 atm in an ambient temperature of 25°C at sea level? (b) What mass can the balloon lift at sea level, where the density of air is 1.22 kg m−3? (c) What would be the payload if He were used instead of H2? 1.30 ‡ Chlorofluorocarbons such as CCl3F and CCl2F2 have been linked 27 manual) with the use of the Archimedes principle, which states that the lifting force is equal to the difference between the weight of the displaced air and the weight of the balloon. Prove the Archimedes principle for the atmosphere from the barometric formula. Hint. Assume a simple shape for the balloon, perhaps a right circular cylinder of cross–sectional area A and height h. to ozone depletion in Antarctica. As of 1994, these gases were found in quantities of 261 and 509 parts per trillion (1012) by volume (World Resources Institute, World resources 1996–97). Compute the molar concentration of these gases under conditions typical of (a) the mid-latitude troposphere (10°C and 1.0 atm) and (b) the Antarctic stratosphere (200 K and 0.050 atm). 2 The basic concepts 2.1 Work, heat, and energy 2.2 The internal energy 2.3 Expansion work 2.4 Heat transactions 2.5 Enthalpy I2.1 Impact on biochemistry and materials science: Differential scanning calorimetry 2.6 Adiabatic changes Thermochemistry 2.7 Standard enthalpy changes I2.2 Impact on biology: Food and energy reserves 2.8 Standard enthalpies of formation 2.9 The temperature-dependence of reaction enthalpies The First Law This chapter introduces some of the basic concepts of thermodynamics. It concentrates on the conservation of energy—the experimental observation that energy can be neither created nor destroyed—and shows how the principle of the conservation of energy can be used to assess the energy changes that accompany physical and chemical processes. Much of this chapter examines the means by which a system can exchange energy with its surroundings in terms of the work it may do or the heat that it may produce. The target concept of the chapter is enthalpy, which is a very useful book-keeping property for keeping track of the heat output (or requirements) of physical processes and chemical reactions at constant pressure. We also begin to unfold some of the power of thermodynamics by showing how to establish relations between different properties of a system. We shall see that one very useful aspect of thermodynamics is that a property can be measured indirectly by measuring others and then combining their values. The relations we derive also enable us to discuss the liquefaction of gases and to establish the relation between the heat capacities of a substance under different conditions. The release of energy can be used to provide heat when a fuel burns in a furnace, to produce mechanical work when a fuel burns in an engine, and to generate electrical work when a chemical reaction pumps electrons through a circuit. In chemistry, we encounter reactions that can be harnessed to provide heat and work, reactions that liberate energy which is squandered (often to the detriment of the environment) but which give products we require, and reactions that constitute the processes of life. Thermodynamics, the study of the transformations of energy, enables us to discuss all these matters quantitatively and to make useful predictions. State functions and exact differentials 2.10 Exact and inexact differentials 2.11 Changes in internal energy 2.12 The Joule–Thomson effect Checklist of key ideas Further reading Further information 2.1: Adiabatic processes Further information 2.2: The relation between heat capacities Discussion questions Exercises Problems The basic concepts For the purposes of physical chemistry, the universe is divided into two parts, the system and its surroundings. The system is the part of the world in which we have a special interest. It may be a reaction vessel, an engine, an electrochemical cell, a biological cell, and so on. The surroundings comprise the region outside the system and are where we make our measurements. The type of system depends on the characteristics of the boundary that divides it from the surroundings (Fig. 2.1). If matter can be transferred through the boundary between the system and its surroundings the system is classified as open. If matter cannot pass through the boundary the system is classified as closed. Both open and closed systems can exchange energy with their surroundings. For example, a closed system can expand and thereby raise a weight in the surroundings; it may also transfer energy to them if they are at a lower temperature. 2.1 WORK, HEAT, AND ENERGY An isolated system is a closed system that has neither mechanical nor thermal contact with its surroundings. 2.1 Work, heat, and energy The fundamental physical property in thermodynamics is work: work is motion against an opposing force. Doing work is equivalent to raising a weight somewhere in the surroundings. An example of doing work is the expansion of a gas that pushes out a piston and raises a weight. A chemical reaction that drives an electric current through a resistance also does work, because the same current could be driven through a motor and used to raise a weight. The energy of a system is its capacity to do work. When work is done on an otherwise isolated system (for instance, by compressing a gas or winding a spring), the capacity of the system to do work is increased; in other words, the energy of the system is increased. When the system does work (when the piston moves out or the spring unwinds), the energy of the system is reduced and it can do less work than before. Experiments have shown that the energy of a system may be changed by means other than work itself. When the energy of a system changes as a result of a temperature difference between the system and its surroundings we say that energy has been transferred as heat. When a heater is immersed in a beaker of water (the system), the capacity of the system to do work increases because hot water can be used to do more work than the same amount of cold water. Not all boundaries permit the transfer of energy even though there is a temperature difference between the system and its surroundings. An exothermic process is a process that releases energy as heat into its surroundings. All combustion reactions are exothermic. An endothermic process is a process in which energy is acquired from its surroundings as heat. An example of an endothermic process is the vaporization of water. To avoid a lot of awkward circumlocution, we say that in an exothermic process energy is transferred ‘as heat’ to the surroundings and in an endothermic process energy is transferred ‘as heat’ from the surroundings into the system. However, it must never be forgotten that heat is a process (the transfer of energy as a result of a temperature difference), not an entity. An endothermic process in a diathermic container results in energy flowing into the system as heat. An exothermic process in a similar diathermic container results in a release of energy as heat into the surroundings. When an endothermic process takes place in an adiabatic container, it results in a lowering of temperature of the system; an exothermic process results in a rise of temperature. These features are summarized in Fig. 2.2. Molecular interpretation 2.1 Heat and work In molecular terms, heating is the transfer of energy that makes use of disorderly molecular motion. The disorderly motion of molecules is called thermal motion. The thermal motion of the molecules in the hot surroundings stimulates the molecules in the cooler system to move more vigorously and, as a result, the energy of the system is increased. When a system heats its surroundings, molecules of the system stimulate the thermal motion of the molecules in the surroundings (Fig. 2.3). In contrast, work is the transfer of energy that makes use of organized motion (Fig. 2.4). When a weight is raised or lowered, its atoms move in an organized way (up or down). The atoms in a spring move in an orderly way when it is wound; the (a) An open system can exchange matter and energy with its surroundings. (b) A closed system can exchange energy with its surroundings, but it cannot exchange matter. (c) An isolated system can exchange neither energy nor matter with its surroundings. Fig. 2.1 29 30 2 THE FIRST LAW When energy is transferred to the surroundings as heat, the transfer stimulates random motion of the atoms in the surroundings. Transfer of energy from the surroundings to the system makes use of random motion (thermal motion) in the surroundings. Fig. 2.3 (a) When an endothermic process occurs in an adiabatic system, the temperature falls; (b) if the process is exothermic, then the temperature rises. (c) When an endothermic process occurs in a diathermic container, energy enters as heat from the surroundings, and the system remains at the same temperature. (d) If the process is exothermic, then energy leaves as heat, and the process is isothermal. Fig. 2.2 When a system does work, it stimulates orderly motion in the surroundings. For instance, the atoms shown here may be part of a weight that is being raised. The ordered motion of the atoms in a falling weight does work on the system. Fig. 2.4 electrons in an electric current move in an orderly direction when it flows. When a system does work it causes atoms or electrons in its surroundings to move in an organized way. Likewise, when work is done on a system, molecules in the surroundings are used to transfer energy to it in an organized way, as the atoms in a weight are lowered or a current of electrons is passed. The distinction between work and heat is made in the surroundings. The fact that a falling weight may stimulate thermal motion in the system is irrelevant to the distinction between heat and work: work is identified as energy transfer making use of the organized motion of atoms in the surroundings, and heat is identified as energy transfer making use of thermal motion in the surroundings. In the compression of a gas, for instance, work is done as the atoms of the compressing weight descend in an orderly way, but the effect of the incoming piston is to accelerate the gas molecules to higher average speeds. Because collisions between molecules quickly randomize their directions, the orderly motion of the atoms of the weight is in effect stimulating thermal motion in the gas. We observe the falling weight, the orderly descent of its atoms, and report that work is being done even though it is stimulating thermal motion. 2.2 The internal energy In thermodynamics, the total energy of a system is called its internal energy, U. The internal energy is the total kinetic and potential energy of the molecules in the system (see Comment 1.3 for the definitions of kinetic and potential energy).1 We denote by ∆U the change in internal energy when a system changes from an initial state i with internal energy Ui to a final state f of internal energy Uf : ∆U = Uf − Ui [2.1] 1 The internal energy does not include the kinetic energy arising from the motion of the system as a whole, such as its kinetic energy as it accompanies the Earth on its orbit round the Sun. 2.2 THE INTERNAL ENERGY The internal energy is a state function in the sense that its value depends only on the current state of the system and is independent of how that state has been prepared. In other words, it is a function of the properties that determine the current state of the system. Changing any one of the state variables, such as the pressure, results in a change in internal energy. The internal energy is an extensive property. That the internal energy is a state function has consequences of the greatest importance, as we start to unfold in Section 2.10. Internal energy, heat, and work are all measured in the same units, the joule (J). The joule, which is named after the nineteenth-century scientist J.P. Joule, is defined as 1 J = 1 kg m2 s−2 A joule is quite a small unit of energy: for instance, each beat of the human heart consumes about 1 J. Changes in molar internal energy, ∆Um, are typically expressed in kilojoules per mole (kJ mol−1). Certain other energy units are also used, but are more common in fields other than thermodynamics. Thus, 1 electronvolt (1 eV) is defined as the kinetic energy acquired when an electron is accelerated from rest through a potential difference of 1 V; the relation between electronvolts and joules is 1 eV ≈ 0.16 aJ (where 1 aJ = 10−18 J). Many processes in chemistry have an energy of several electronvolts. Thus, the energy to remove an electron from a sodium atom is close to 5 eV. Calories (cal) and kilocalories (kcal) are still encountered. The current definition of the calorie in terms of joules is 1 cal = 4.184 J exactly An energy of 1 cal is enough to raise the temperature of 1 g of water by 1°C. Molecular interpretation 2.2 The internal energy of a gas A molecule has a certain number of degrees of freedom, such as the ability to translate (the motion of its centre of mass through space), rotate around its centre of mass, or vibrate (as its bond lengths and angles change). Many physical and chemical properties depend on the energy associated with each of these modes of motion. For example, a chemical bond might break if a lot of energy becomes concentrated in it. The equipartition theorem of classical mechanics is a useful guide to the average energy associated with each degree of freedom when the sample is at a temperature T. First, we need to know that a ‘quadratic contribution’ to the energy means a contribution that can be expressed as the square of a variable, such as the position or the velocity. For example, the kinetic energy an atom of mass m as it moves through space is EK = –12 mv x2 + –12 mv y2 + –12 mv z2 and there are three quadratic contributions to its energy. The equipartition theorem then states that, for a collection of particles at thermal equilibrium at a temperature T, the average value of each quadratic contribution to the energy is the same and equal to –12 kT, where k is Boltzmann’s constant (k = 1.381 × 10−23 J K −1). The equipartition theorem is a conclusion from classical mechanics and is applicable only when the effects of quantization can be ignored (see Chapters 16 and 17). In practice, it can be used for molecular translation and rotation but not vibration. At 25°C, –12 kT = 2 zJ (where 1 zJ = 10−21 J), or about 13 meV. According to the equipartition theorem, the average energy of each term in the expression above is –12 kT. Therefore, the mean energy of the atoms is –23 kT and the 31 Comment 2.1 An extensive property is a property that depends on the amount of substance in the sample. An intensive property is a property that is independent of the amount of substance in the sample. Two examples of extensive properties are mass and volume. Examples of intensive properties are temperature, mass density (mass divided by volume), and pressure. 32 2 THE FIRST LAW total energy of the gas (there being no potential energy contribution) is –23 NkT, or –23 nRT (because N = nNA and R = NAk). We can therefore write Um = Um(0) + –23 RT The rotational modes of molecules and the corresponding average energies at a temperature T. (a) A linear molecule can rotate about two axes perpendicular to the line of the atoms. (b) A nonlinear molecule can rotate about three perpendicular axes. Fig. 2.5 where Um(0) is the molar internal energy at T = 0, when all translational motion has ceased and the sole contribution to the internal energy arises from the internal structure of the atoms. This equation shows that the internal energy of a perfect gas increases linearly with temperature. At 25°C, –23 RT = 3.7 kJ mol−1, so translational motion contributes about 4 kJ mol−1 to the molar internal energy of a gaseous sample of atoms or molecules (the remaining contribution arises from the internal structure of the atoms and molecules). When the gas consists of polyatomic molecules, we need to take into account the effect of rotation and vibration. A linear molecule, such as N2 and CO2, can rotate around two axes perpendicular to the line of the atoms (Fig. 2.5), so it has two rotational modes of motion, each contributing a term –12 kT to the internal energy. Therefore, the mean rotational energy is kT and the rotational contribution to the molar internal energy is RT. By adding the translational and rotational contributions, we obtain Um = Um(0) + –52 RT (linear molecule, translation and rotation only) A nonlinear molecule, such as CH4 or water, can rotate around three axes and, again, each mode of motion contributes a term –12 kT to the internal energy. Therefore, the mean rotational energy is –32 kT and there is a rotational contribution of –32 RT to the molar internal energy of the molecule. That is, Um = Um(0) + 3RT (nonlinear molecule, translation and rotation only) The internal energy now increases twice as rapidly with temperature compared with the monatomic gas. The internal energy of interacting molecules in condensed phases also has a contribution from the potential energy of their interaction. However, no simple expressions can be written down in general. Nevertheless, the crucial molecular point is that, as the temperature of a system is raised, the internal energy increases as the various modes of motion become more highly excited. It has been found experimentally that the internal energy of a system may be changed either by doing work on the system or by heating it. Whereas we may know how the energy transfer has occurred (because we can see if a weight has been raised or lowered in the surroundings, indicating transfer of energy by doing work, or if ice has melted in the surroundings, indicating transfer of energy as heat), the system is blind to the mode employed. Heat and work are equivalent ways of changing a system’s internal energy. A system is like a bank: it accepts deposits in either currency, but stores its reserves as internal energy. It is also found experimentally that, if a system is isolated from its surroundings, then no change in internal energy takes place. This summary of observations is now known as the First Law of thermodynamics and expressed as follows: The internal energy of an isolated system is constant. We cannot use a system to do work, leave it isolated for a month, and then come back expecting to find it restored to its original state and ready to do the same work again. The evidence for this property is that no ‘perpetual motion machine’ (a machine that 2.3 EXPANSION WORK does work without consuming fuel or some other source of energy) has ever been built. These remarks may be summarized as follows. If we write w for the work done on a system, q for the energy transferred as heat to a system, and ∆U for the resulting change in internal energy, then it follows that ∆U = q + w (2.2) Equation 2.2 is the mathematical statement of the First Law, for it summarizes the equivalence of heat and work and the fact that the internal energy is constant in an isolated system (for which q = 0 and w = 0). The equation states that the change in internal energy of a closed system is equal to the energy that passes through its boundary as heat or work. It employs the ‘acquisitive convention’, in which w > 0 or q > 0 if energy is transferred to the system as work or heat and w < 0 or q < 0 if energy is lost from the system as work or heat. In other words, we view the flow of energy as work or heat from the system’s perspective. Illustration 2.1 The sign convention in thermodynamics If an electric motor produced 15 kJ of energy each second as mechanical work and lost 2 kJ as heat to the surroundings, then the change in the internal energy of the motor each second is ∆U = −2 kJ − 15 kJ = −17 kJ Suppose that, when a spring was wound, 100 J of work was done on it but 15 J escaped to the surroundings as heat. The change in internal energy of the spring is ∆U = +100 kJ − 15 kJ = +85 kJ 2.3 Expansion work The way can now be opened to powerful methods of calculation by switching attention to infinitesimal changes of state (such as infinitesimal change in temperature) and infinitesimal changes in the internal energy dU. Then, if the work done on a system is dw and the energy supplied to it as heat is dq, in place of eqn 2.2 we have dU = dq + dw (2.3) To use this expression we must be able to relate dq and dw to events taking place in the surroundings. We begin by discussing expansion work, the work arising from a change in volume. This type of work includes the work done by a gas as it expands and drives back the atmosphere. Many chemical reactions result in the generation or consumption of gases (for instance, the thermal decomposition of calcium carbonate or the combustion of octane), and the thermodynamic characteristics of a reaction depend on the work it can do. The term ‘expansion work’ also includes work associated with negative changes of volume, that is, compression. (a) The general expression for work The calculation of expansion work starts from the definition used in physics, which states that the work required to move an object a distance dz against an opposing force of magnitude F is dw = −Fdz [2.4] 33 34 2 THE FIRST LAW The negative sign tells us that, when the system moves an object against an opposing force, the internal energy of the system doing the work will decrease. Now consider the arrangement shown in Fig. 2.6, in which one wall of a system is a massless, frictionless, rigid, perfectly fitting piston of area A. If the external pressure is pex, the magnitude of the force acting on the outer face of the piston is F = pex A. When the system expands through a distance dz against an external pressure pex, it follows that the work done is dw = −pex Adz. But Adz is the change in volume, dV, in the course of the expansion. Therefore, the work done when the system expands by dV against a pressure pex is dw = −pexdV (2.5) To obtain the total work done when the volume changes from Vi to Vf we integrate this expression between the initial and final volumes: When a piston of area A moves out through a distance dz, it sweeps out a volume dV = Adz. The external pressure pex is equivalent to a weight pressing on the piston, and the force opposing expansion is F = pex A. Fig. 2.6 # w=− Vf pexdV (2.6) Vi The force acting on the piston, pex A, is equivalent to a weight that is raised as the system expands. If the system is compressed instead, then the same weight is lowered in the surroundings and eqn 2.6 can still be used, but now Vf < Vi. It is important to note that it is still the external pressure that determines the magnitude of the work. This somewhat perplexing conclusion seems to be inconsistent with the fact that the gas inside the container is opposing the compression. However, when a gas is compressed, the ability of the surroundings to do work is diminished by an amount determined by the weight that is lowered, and it is this energy that is transferred into the system. Other types of work (for example, electrical work), which we shall call either nonexpansion work or additional work, have analogous expressions, with each one the product of an intensive factor (the pressure, for instance) and an extensive factor (the change in volume). Some are collected in Table 2.1. For the present we continue with the work associated with changing the volume, the expansion work, and see what we can extract from eqns 2.5 and 2.6. (b) Free expansion By free expansion we mean expansion against zero opposing force. It occurs when pex = 0. According to eqn 2.5, dw = 0 for each stage of the expansion. Hence, overall: Free expansion: w=0 (2.7) Table 2.1 Varieties of work* Type of work dw Comments Units† Expansion −pexdV pex is the external pressure dV is the change in volume Pa m3 Surface expansion γ dσ γ is the surface tension dσ is the change in area N m−1 m2 Extension fdl f is the tension dl is the change in length N m Electrical φ dQ φ is the electric potential dQ is the change in charge V C * In general, the work done on a system can be expressed in the form dw = −Fdz, where F is a ‘generalized force’ and dz is a ‘generalized displacement’. † For work in joules (J). Note that 1 N m = 1 J and 1 V C = 1 J. 2.3 EXPANSION WORK 35 That is, no work is done when a system expands freely. Expansion of this kind occurs when a system expands into a vacuum. (c) Expansion against constant pressure Now suppose that the external pressure is constant throughout the expansion. For example, the piston may be pressed on by the atmosphere, which exerts the same pressure throughout the expansion. A chemical example of this condition is the expansion of a gas formed in a chemical reaction. We can evaluate eqn 2.6 by taking the constant pex outside the integral: # w = −pex Vf dV = −pex(Vf − Vi) Vi Therefore, if we write the change in volume as ∆V = Vf − Vi, w = −pex ∆V (2.8) This result is illustrated graphically in Fig. 2.7, which makes use of the fact that an integral can be interpreted as an area. The magnitude of w, denoted |w|, is equal to the area beneath the horizontal line at p = pex lying between the initial and final volumes. A p,V-graph used to compute expansion work is called an indicator diagram; James Watt first used one to indicate aspects of the operation of his steam engine. Comment 2.2 (d) Reversible expansion The value of the integral A reversible change in thermodynamics is a change that can be reversed by an infinitesimal modification of a variable. The key word ‘infinitesimal’ sharpens the everyday meaning of the word ‘reversible’ as something that can change direction. We say that a system is in equilibrium with its surroundings if an infinitesimal change in the conditions in opposite directions results in opposite changes in its state. One example of reversibility that we have encountered already is the thermal equilibrium of two systems with the same temperature. The transfer of energy as heat between the two is reversible because, if the temperature of either system is lowered infinitesimally, then energy flows into the system with the lower temperature. If the temperature of either system at thermal equilibrium is raised infinitesimally, then energy flows out of the hotter system. Suppose a gas is confined by a piston and that the external pressure, pex, is set equal to the pressure, p, of the confined gas. Such a system is in mechanical equilibrium with its surroundings (as illustrated in Section 1.1) because an infinitesimal change in the external pressure in either direction causes changes in volume in opposite directions. If the external pressure is reduced infinitesimally, then the gas expands slightly. If the external pressure is increased infinitesimally, then the gas contracts slightly. In either case the change is reversible in the thermodynamic sense. If, on the other hand, the external pressure differs measurably from the internal pressure, then changing pex infinitesimally will not decrease it below the pressure of the gas, so will not change the direction of the process. Such a system is not in mechanical equilibrium with its surroundings and the expansion is thermodynamically irreversible. To achieve reversible expansion we set pex equal to p at each stage of the expansion. In practice, this equalization could be achieved by gradually removing weights from the piston so that the downward force due to the weights always matched the changing upward force due to the pressure of the gas. When we set pex = p, eqn 2.5 becomes dw = −pexdV = −pdV The work done by a gas when it expands against a constant external pressure, pex, is equal to the shaded area in this example of an indicator diagram. Fig. 2.7 (2.9)rev (Equations valid only for reversible processes are labelled with a subscript rev.) Although the pressure inside the system appears in this expression for the work, it # f(x)dx is b a equal to the area under the graph of f(x) between x = a and x = b. For instance, the area under the curve f(x) = x 2 shown in the illustration that lies between x = 1 and 3 is # x dx = (–x + constant) 3 2 1 1 3 3 3 1 = –31 (33 − 13) = –26 3– ≈ 8.67 36 2 THE FIRST LAW does so only because pex has been set equal to p to ensure reversibility. The total work of reversible expansion is therefore # w=− Vf pdV (2.10)rev Vi We can evaluate the integral once we know how the pressure of the confined gas depends on its volume. Equation 2.10 is the link with the material covered in Chapter 1 for, if we know the equation of state of the gas, then we can express p in terms of V and evaluate the integral. (e) Isothermal reversible expansion Comment 2.3 An integral that occurs throughout thermodynamics is # x dx = (lnx + constant) b a 1 b a = ln b a Consider the isothermal, reversible expansion of a perfect gas. The expansion is made isothermal by keeping the system in thermal contact with its surroundings (which may be a constant-temperature bath). Because the equation of state is pV = nRT, we know that at each stage p = nRT/V, with V the volume at that stage of the expansion. The temperature T is constant in an isothermal expansion, so (together with n and R) it may be taken outside the integral. It follows that the work of reversible isothermal expansion of a perfect gas from Vi to Vf at a temperature T is # w = −nRT Vf Vi The work done by a perfect gas when it expands reversibly and isothermally is equal to the area under the isotherm p = nRT/V. The work done during the irreversible expansion against the same final pressure is equal to the rectangular area shown slightly darker. Note that the reversible work is greater than the irreversible work. Fig. 2.8 Exploration Calculate the work of isothermal reversible expansion of 1.0 mol CO2(g) at 298 K from 1.0 m3 to 3.0 m3 on the basis that it obeys the van der Waals equation of state. dV V = −nRT ln Vf Vi (2.11)°rev When the final volume is greater than the initial volume, as in an expansion, the logarithm in eqn 2.11 is positive and hence w < 0. In this case, the system has done work on the surroundings and the internal energy of the system has decreased as a result.2 The equations also show that more work is done for a given change of volume when the temperature is increased. The greater pressure of the confined gas then needs a higher opposing pressure to ensure reversibility. We can express the result of the calculation as an indicator diagram, for the magnitude of the work done is equal to the area under the isotherm p = nRT/V (Fig. 2.8). Superimposed on the diagram is the rectangular area obtained for irreversible expansion against constant external pressure fixed at the same final value as that reached in the reversible expansion. More work is obtained when the expansion is reversible (the area is greater) because matching the external pressure to the internal pressure at each stage of the process ensures that none of the system’s pushing power is wasted. We cannot obtain more work than for the reversible process because increasing the external pressure even infinitesimally at any stage results in compression. We may infer from this discussion that, because some pushing power is wasted when p > pex, the maximum work available from a system operating between specified initial and final states and passing along a specified path is obtained when the change takes place reversibly. We have introduced the connection between reversibility and maximum work for the special case of a perfect gas undergoing expansion. Later (in Section 3.5) we shall see that it applies to all substances and to all kinds of work. Example 2.1 Calculating the work of gas production Calculate the work done when 50 g of iron reacts with hydrochloric acid in (a) a closed vessel of fixed volume, (b) an open beaker at 25°C. 2 We shall see later that there is a compensating influx of energy as heat, so overall the internal energy is constant for the isothermal expansion of a perfect gas. 2.4 HEAT TRANSACTIONS Method We need to judge the magnitude of the volume change and then to decide how the process occurs. If there is no change in volume, there is no expansion work however the process takes place. If the system expands against a constant external pressure, the work can be calculated from eqn 2.8. A general feature of processes in which a condensed phase changes into a gas is that the volume of the former may usually be neglected relative to that of the gas it forms. Answer In (a) the volume cannot change, so no expansion work is done and w = 0. In (b) the gas drives back the atmosphere and therefore w = −pex ∆V. We can neglect the initial volume because the final volume (after the production of gas) is so much larger and ∆V = Vf − Vi ≈ Vf = nRT/pex, where n is the amount of H2 produced. Therefore, w = −pex ∆V ≈ −pex × nRT pex = −nRT Because the reaction is Fe(s) + 2 HCl(aq) → FeCl2(aq) + H2(g), we know that 1 mol H2 is generated when 1 mol Fe is consumed, and n can be taken as the amount of Fe atoms that react. Because the molar mass of Fe is 55.85 g mol−1, it follows that w≈− 50 g 55.85 g mol−1 × (8.3145 J K−1 mol−1) × (298 K) ≈ −2.2 kJ The system (the reaction mixture) does 2.2 kJ of work driving back the atmosphere. Note that (for this perfect gas system) the magnitude of the external pressure does not affect the final result: the lower the pressure, the larger the volume occupied by the gas, so the effects cancel. Self-test 2.1 Calculate the expansion work done when 50 g of water is electrolysed under constant pressure at 25°C. [−10 kJ] 2.4 Heat transactions In general, the change in internal energy of a system is dU = dq + dwexp + dwe (2.12) where dwe is work in addition (e for ‘extra’) to the expansion work, dwexp. For instance, dwe might be the electrical work of driving a current through a circuit. A system kept at constant volume can do no expansion work, so dwexp = 0. If the system is also incapable of doing any other kind of work (if it is not, for instance, an electrochemical cell connected to an electric motor), then dwe = 0 too. Under these circumstances: dU = dq (at constant volume, no additional work) (2.13a) We express this relation by writing dU = dqV , where the subscript implies a change at constant volume. For a measurable change, ∆U = qV (2.13b) It follows that, by measuring the energy supplied to a constant-volume system as heat (q > 0) or obtained from it as heat (q < 0) when it undergoes a change of state, we are in fact measuring the change in its internal energy. 37 38 2 THE FIRST LAW (a) Calorimetry Calorimetry is the study of heat transfer during physical and chemical processes. A calorimeter is a device for measuring energy transferred as heat. The most common device for measuring ∆U is an adiabatic bomb calorimeter (Fig. 2.9). The process we wish to study—which may be a chemical reaction—is initiated inside a constantvolume container, the ‘bomb’. The bomb is immersed in a stirred water bath, and the whole device is the calorimeter. The calorimeter is also immersed in an outer water bath. The water in the calorimeter and of the outer bath are both monitored and adjusted to the same temperature. This arrangement ensures that there is no net loss of heat from the calorimeter to the surroundings (the bath) and hence that the calorimeter is adiabatic. The change in temperature, ∆T, of the calorimeter is proportional to the heat that the reaction releases or absorbs. Therefore, by measuring ∆T we can determine qV and hence find ∆U. The conversion of ∆T to qV is best achieved by calibrating the calorimeter using a process of known energy output and determining the calorimeter constant, the constant C in the relation q = C∆T (2.14a) The calorimeter constant may be measured electrically by passing a constant current, I, from a source of known potential difference, V, through a heater for a known period of time, t, for then A constant-volume bomb calorimeter. The ‘bomb’ is the central vessel, which is strong enough to withstand high pressures. The calorimeter (for which the heat capacity must be known) is the entire assembly shown here. To ensure adiabaticity, the calorimeter is immersed in a water bath with a temperature continuously readjusted to that of the calorimeter at each stage of the combustion. q = IV t Fig. 2.9 (2.14b) Alternatively, C may be determined by burning a known mass of substance (benzoic acid is often used) that has a known heat output. With C known, it is simple to interpret an observed temperature rise as a release of heat. Illustration 2.2 The calibration of a calorimeter If we pass a current of 10.0 A from a 12 V supply for 300 s, then from eqn 2.14b the energy supplied as heat is q = (10.0 A) × (12 V) × (300 s) = 3.6 × 104 A V s = 36 kJ because 1 A V s = 1 J. If the observed rise in temperature is 5.5 K, then the calorimeter constant is C = (36 kJ)/(5.5 K) = 6.5 kJ K−1. Comment 2.4 Electrical charge is measured in coulombs, C. The motion of charge gives rise to an electric current, I, measured in coulombs per second, or amperes, A, where 1 A = 1 C s−1. If a constant current I flows through a potential difference V (measured in volts, V), the total energy supplied in an interval t is Energy supplied = IV t Because 1 A V s = 1 (C s−1) V s = 1 C V = 1 J, the energy is obtained in joules with the current in amperes, the potential difference in volts, and the time in seconds. We write the electrical power, P, as P = (energy supplied)/(time interval) = IV t/t = IV (b) Heat capacity The internal energy of a substance increases when its temperature is raised. The increase depends on the conditions under which the heating takes place and for the present we suppose that the sample is confined to a constant volume. For example, the sample may be a gas in a container of fixed volume. If the internal energy is plotted against temperature, then a curve like that in Fig. 2.10 may be obtained. The slope of the tangent to the curve at any temperature is called the heat capacity of the system at that temperature. The heat capacity at constant volume is denoted CV and is defined formally as3 3 If the system can change its composition, it is necessary to distinguish between equilibrium and fixedcomposition values of CV. All applications in this chapter refer to a single substance, so this complication can be ignored. 2.4 HEAT TRANSACTIONS Fig. 2.10 The internal energy of a system increases as the temperature is raised; this graph shows its variation as the system is heated at constant volume. The slope of the tangent to the curve at any temperature is the heat capacity at constant volume at that temperature. Note that, for the system illustrated, the heat capacity is greater at B than at A. CV = Fig. 2.11 The internal energy of a system varies with volume and temperature, perhaps as shown here by the surface. The variation of the internal energy with temperature at one particular constant volume is illustrated by the curve drawn parallel to T. The slope of this curve at any point is the partial derivative (∂U/∂T)V. A ∂U D C ∂T F V [2.15] In this case, the internal energy varies with the temperature and the volume of the sample, but we are interested only in its variation with the temperature, the volume being held constant (Fig. 2.11). Illustration 2.3 Estimating a constant-volume heat capacity The heat capacity of a monatomic perfect gas can be calculated by inserting the expression for the internal energy derived in Molecular interpretation 2.2. There we saw that Um = Um(0) + –32 RT, so from eqn 2.15 CV,m = ∂ ∂T (Um(0) + –32 RT) = –32 R The numerical value is 12.47 J K−1 mol−1. Heat capacities are extensive properties: 100 g of water, for instance, has 100 times the heat capacity of 1 g of water (and therefore requires 100 times the energy as heat to bring about the same rise in temperature). The molar heat capacity at constant volume, CV,m = CV /n, is the heat capacity per mole of material, and is an intensive property (all molar quantities are intensive). Typical values of CV,m for polyatomic gases are close to 25 J K−1 mol−1. For certain applications it is useful to know the specific heat capacity (more informally, the ‘specific heat’) of a substance, which is the heat capacity of the sample divided by the mass, usually in grams: CV,s = CV /m. The specific heat capacity of water at room temperature is close to 4 J K−1 g−1. In general, Comment 2.5 The partial-differential operation (∂z/∂x)y consists of taking the first derivative of z(x,y) with respect to x, treating y as a constant. For example, if z(x,y) = x 2y, then A ∂z D A ∂[x 2y] D dx 2 B E =B E =y = 2yx C ∂x F y C ∂x F y dx Partial derivatives are reviewed in Appendix 2. 39 40 2 THE FIRST LAW heat capacities depend on the temperature and decrease at low temperatures. However, over small ranges of temperature at and above room temperature, the variation is quite small and for approximate calculations heat capacities can be treated as almost independent of temperature. The heat capacity is used to relate a change in internal energy to a change in temperature of a constant-volume system. It follows from eqn 2.15 that dU = CV dT (at constant volume) (2.16a) That is, at constant volume, an infinitesimal change in temperature brings about an infinitesimal change in internal energy, and the constant of proportionality is CV. If the heat capacity is independent of temperature over the range of temperatures of interest, a measurable change of temperature, ∆T, brings about a measurable increase in internal energy, ∆U, where ∆U = CV ∆T (at constant volume) (2.16b) Because a change in internal energy can be identified with the heat supplied at constant volume (eqn 2.13b), the last equation can be written qV = CV ∆T (2.17) This relation provides a simple way of measuring the heat capacity of a sample: a measured quantity of energy is transferred as heat to the sample (electrically, for example), and the resulting increase in temperature is monitored. The ratio of the energy transferred as heat to the temperature rise it causes (qV /∆T) is the constant-volume heat capacity of the sample. A large heat capacity implies that, for a given quantity of energy transferred as heat, there will be only a small increase in temperature (the sample has a large capacity for heat). An infinite heat capacity implies that there will be no increase in temperature however much energy is supplied as heat. At a phase transition, such as at the boiling point of water, the temperature of a substance does not rise as energy is supplied as heat: the energy is used to drive the endothermic transition, in this case to vaporize the water, rather than to increase its temperature. Therefore, at the temperature of a phase transition, the heat capacity of a sample is infinite. The properties of heat capacities close to phase transitions are treated more fully in Section 4.7. 2.5 Enthalpy The change in internal energy is not equal to the energy transferred as heat when the system is free to change its volume. Under these circumstances some of the energy supplied as heat to the system is returned to the surroundings as expansion work (Fig. 2.12), so dU is less than dq. However, we shall now show that in this case the energy supplied as heat at constant pressure is equal to the change in another thermodynamic property of the system, the enthalpy. (a) The definition of enthalpy The enthalpy, H, is defined as When a system is subjected to constant pressure and is free to change its volume, some of the energy supplied as heat may escape back into the surroundings as work. In such a case, the change in internal energy is smaller than the energy supplied as heat. Fig. 2.12 H = U + pV [2.18] where p is the pressure of the system and V is its volume. Because U, p, and V are all state functions, the enthalpy is a state function too. As is true of any state function, the change in enthalpy, ∆H, between any pair of initial and final states is independent of the path between them. 2.5 ENTHALPY Although the definition of enthalpy may appear arbitrary, it has important implications for thermochemisty. For instance, we show in the following Justification that eqn 2.18 implies that the change in enthalpy is equal to the energy supplied as heat at constant pressure (provided the system does no additional work): dH = dq (at constant pressure, no additional work) (2.19a) For a measurable change, ∆H = qp (2.19b) Justification 2.1 The relation ∆H = qp For a general infinitesimal change in the state of the system, U changes to U + dU, p changes to p + dp, and V changes to V + dV, so from the definition in eqn 2.18, H changes from U + pV to H + dH = (U + dU) + (p + dp)(V + dV) = U + dU + pV + pdV + Vdp + dpdV The last term is the product of two infinitesimally small quantities and can therefore be neglected. As a result, after recognizing U + pV = H on the right, we find that H changes to H + dH = H + dU + pdV + Vdp and hence that dH = dU + pdV + Vdp If we now substitute dU = dq + dw into this expression, we get dH = dq + dw + pdV + Vdp If the system is in mechanical equilibrium with its surroundings at a pressure p and does only expansion work, we can write dw = −pdV and obtain dH = dq + Vdp Now we impose the condition that the heating occurs at constant pressure by writing dp = 0. Then dH = dq (at constant pressure, no additional work) as in eqn 2.19a. The result expressed in eqn 2.19 states that, when a system is subjected to a constant pressure, and only expansion work can occur, the change in enthalpy is equal to the energy supplied as heat. For example, if we supply 36 kJ of energy through an electric heater immersed in an open beaker of water, then the enthalpy of the water increases by 36 kJ and we write ∆H = +36 kJ. (b) The measurement of an enthalpy change An enthalpy change can be measured calorimetrically by monitoring the temperature change that accompanies a physical or chemical change occurring at constant pressure. A calorimeter for studying processes at constant pressure is called an isobaric calorimeter. A simple example is a thermally insulated vessel open to the atmosphere: the heat released in the reaction is monitored by measuring the change in temperature 41 42 2 THE FIRST LAW of the contents. For a combustion reaction an adiabatic flame calorimeter may be used to measure ∆T when a given amount of substance burns in a supply of oxygen (Fig. 2.13). Another route to ∆H is to measure the internal energy change by using a bomb calorimeter, and then to convert ∆U to ∆H. Because solids and liquids have small molar volumes, for them pVm is so small that the molar enthalpy and molar internal energy are almost identical (Hm = Um + pVm ≈ Um). Consequently, if a process involves only solids or liquids, the values of ∆H and ∆U are almost identical. Physically, such processes are accompanied by a very small change in volume, the system does negligible work on the surroundings when the process occurs, so the energy supplied as heat stays entirely within the system. The most sophisticated way to measure enthalpy changes, however, is to use a differential scanning calorimeter (DSC). Changes in enthalpy and internal energy may also be measured by noncalorimetric methods (see Chapter 7). Example 2.2 Relating ∆H and ∆U The internal energy change when 1.0 mol CaCO3 in the form of calcite converts to aragonite is +0.21 kJ. Calculate the difference between the enthalpy change and the change in internal energy when the pressure is 1.0 bar given that the densities of the solids are 2.71 g cm−3 and 2.93 g cm−3, respectively. A constant-pressure flame calorimeter consists of this component immersed in a stirred water bath. Combustion occurs as a known amount of reactant is passed through to fuel the flame, and the rise of temperature is monitored. Fig. 2.13 Method The starting point for the calculation is the relation between the enthalpy of a substance and its internal energy (eqn 2.18). The difference between the two quantities can be expressed in terms of the pressure and the difference of their molar volumes, and the latter can be calculated from their molar masses, M, and their mass densities, ρ, by using ρ = M/Vm. Answer The change in enthalpy when the transition occurs is ∆H = H(aragonite) − H(calcite) = {U(a) + pV(a)} − {U(c) + pV(c)} = ∆U + p{V(a) − V(c)} = ∆U + p∆V The volume of 1.0 mol CaCO3 (100 g) as aragonite is 34 cm3, and that of 1.0 mol CaCO3 as calcite is 37 cm3. Therefore, p∆V = (1.0 × 105 Pa) × (34 − 37) × 10− 6 m3 = −0.3 J (because 1 Pa m3 = 1 J). Hence, ∆H − ∆U = −0.3 J which is only 0.1 per cent of the value of ∆U. We see that it is usually justifiable to ignore the difference between the enthalpy and internal energy of condensed phases, except at very high pressures, when pV is no longer negligible. Self-test 2.2 Calculate the difference between ∆H and ∆U when 1.0 mol Sn(s, grey) of density 5.75 g cm−3 changes to Sn(s, white) of density 7.31 g cm−3 at 10.0 bar. At 298 K, ∆H = +2.1 kJ. [∆H − ∆U = −4.4 J] The enthalpy of a perfect gas is related to its internal energy by using pV = nRT in the definition of H: H = U + pV = U + nRT (2.20)° 2.5 ENTHALPY This relation implies that the change of enthalpy in a reaction that produces or consumes gas is ∆H = ∆U + ∆ng RT (2.21)° where ∆ng is the change in the amount of gas molecules in the reaction. Illustration 2.4 The relation between ∆H and ∆U for gas-phase reactions In the reaction 2 H2(g) + O2(g) → 2 H2O(l), 3 mol of gas-phase molecules is replaced by 2 mol of liquid-phase molecules, so ∆ng = −3 mol. Therefore, at 298 K, when RT = 2.5 kJ mol−1, the enthalpy and internal energy changes taking place in the system are related by ∆H − ∆U = (−3 mol) × RT ≈ −7.4 kJ Note that the difference is expressed in kilojoules, not joules as in Example 2.2. The enthalpy change is smaller (in this case, less negative) than the change in internal energy because, although heat escapes from the system when the reaction occurs, the system contracts when the liquid is formed, so energy is restored to it from the surroundings. Example 2.3 Calculating a change in enthalpy Water is heated to boiling under a pressure of 1.0 atm. When an electric current of 0.50 A from a 12 V supply is passed for 300 s through a resistance in thermal contact with it, it is found that 0.798 g of water is vaporized. Calculate the molar internal energy and enthalpy changes at the boiling point (373.15 K). Method Because the vaporization occurs at constant pressure, the enthalpy change is equal to the heat supplied by the heater. Therefore, the strategy is to calculate the energy supplied as heat (from q = IV t), express that as an enthalpy change, and then convert the result to a molar enthalpy change by division by the amount of H2O molecules vaporized. To convert from enthalpy change to internal energy change, we assume that the vapour is a perfect gas and use eqn 2.21. Answer The enthalpy change is ∆H = qp = (0.50 A) × (12 V) × (300 s) = +(0.50 × 12 × 300) J Here we have used 1 A V s = 1 J (see Comment 2.4). Because 0.798 g of water is (0.798 g)/(18.02 g mol−1) = (0.798/18.02) mol H2O, the enthalpy of vaporization per mole of H2O is ∆Hm = + 0.50 × 12 × 300 J (0.798/18.02) mol = +41 kJ mol−1 In the process H2O(l) → H2O(g) the change in the amount of gas molecules is ∆ng = +1 mol, so ∆Um = ∆Hm − RT = +38 kJ mol−1 The plus sign is added to positive quantities to emphasize that they represent an increase in internal energy or enthalpy. Notice that the internal energy change is smaller than the enthalpy change because energy has been used to drive back the surrounding atmosphere to make room for the vapour. 43 44 2 THE FIRST LAW Self-test 2.3 The molar enthalpy of vaporization of benzene at its boiling point (353.25 K) is 30.8 kJ mol−1. What is the molar internal energy change? For how long would the same 12 V source need to supply a 0.50 A current in order to vaporize a 10 g sample? [+27.9 kJ mol−1, 660 s] (c) The variation of enthalpy with temperature Fig. 2.14 The slope of the tangent to a curve of the enthalpy of a system subjected to a constant pressure plotted against temperature is the constant-pressure heat capacity. The slope may change with temperature, in which case the heat capacity varies with temperature. Thus, the heat capacities at A and B are different. For gases, at a given temperature the slope of enthalpy versus temperature is steeper than that of internal energy versus temperature, and Cp,m is larger than CV,m. The enthalpy of a substance increases as its temperature is raised. The relation between the increase in enthalpy and the increase in temperature depends on the conditions (for example, constant pressure or constant volume). The most important condition is constant pressure, and the slope of the tangent to a plot of enthalpy against temperature at constant pressure is called the heat capacity at constant pressure, Cp, at a given temperature (Fig. 2.14). More formally: Cp = A ∂H D C ∂T F p [2.22] The heat capacity at constant pressure is the analogue of the heat capacity at constant volume, and is an extensive property.4 The molar heat capacity at constant pressure, Cp,m, is the heat capacity per mole of material; it is an intensive property. The heat capacity at constant pressure is used to relate the change in enthalpy to a change in temperature. For infinitesimal changes of temperature, dH = CpdT (at constant pressure) (2.23a) If the heat capacity is constant over the range of temperatures of interest, then for a measurable increase in temperature ∆H = Cp ∆T (at constant pressure) (2.23b) Because an increase in enthalpy can be equated with the energy supplied as heat at constant pressure, the practical form of the latter equation is qp = Cp ∆T (2.24) This expression shows us how to measure the heat capacity of a sample: a measured quantity of energy is supplied as heat under conditions of constant pressure (as in a sample exposed to the atmosphere and free to expand), and the temperature rise is monitored. The variation of heat capacity with temperature can sometimes be ignored if the temperature range is small; this approximation is highly accurate for a monatomic perfect gas (for instance, one of the noble gases at low pressure). However, when it is necessary to take the variation into account, a convenient approximate empirical expression is Cp,m = a + bT + c T2 (2.25) The empirical parameters a, b, and c are independent of temperature (Table 2.2). 4 As in the case of CV , if the system can change its composition it is necessary to distinguish between equilibrium and fixed-composition values. All applications in this chapter refer to pure substances, so this complication can be ignored. 2.5 ENTHALPY 45 Synoptic Table 2.2* Temperature variation of molar heat capacities, Cp,m/(J K−1 mol−1) = a + bT + c/T 2 a b/(10−3 K) c/(105 K2) C(s, graphite) 16.86 4.77 −8.54 CO2(g) 44.22 8.79 −8.62 H2O(l) 75.29 0 N2(g) 28.58 3.77 0 −0.50 * More values are given in the Data section. Example 2.4 Evaluating an increase in enthalpy with temperature What is the change in molar enthalpy of N2 when it is heated from 25°C to 100°C? Use the heat capacity information in Table 2.2. Method The heat capacity of N2 changes with temperature, so we cannot use eqn 2.23b (which assumes that the heat capacity of the substance is constant). Therefore, we must use eqn 2.23a, substitute eqn 2.25 for the temperature dependence of the heat capacity, and integrate the resulting expression from 25°C to 100°C. Answer For convenience, we denote the two temperatures T1 (298 K) and T2 (373 K). The integrals we require are # H(T2) dH = H(T1) # T2 T1 A c D a + bT + 2 dT C T F Notice how the limits of integration correspond on each side of the equation: the integration over H on the left ranges from H(T1), the value of H at T1, up to H(T2), the value of H at T2, while on the right the integration over the temperature ranges from T1 to T2. Now we use the integrals #dx = x + constant #x dx = –x + constant # x 1 2 2 dx 2 =− 1 x + constant to obtain H(T2) − H(T1) = a(T2 − T1) + –12 b(T 22 − T 12) − c A 1 1D − C T2 T1 F Substitution of the numerical data results in H(373 K) = H(298 K) + 2.20 kJ mol−1 If we had assumed a constant heat capacity of 29.14 J K−1 mol−1 (the value given by eqn 2.25 at 25°C), we would have found that the two enthalpies differed by 2.19 kJ mol−1. Self-test 2.4 At very low temperatures the heat capacity of a solid is proportional to T 3, and we can write Cp = aT 3. What is the change in enthalpy of such a substance when it is heated from 0 to a temperature T (with T close to 0)? [∆H = –14 aT 4] Most systems expand when heated at constant pressure. Such systems do work on the surroundings and therefore some of the energy supplied to them as heat escapes Comment 2.6 Integrals commonly encountered in physical chemistry are listed inside the front cover. 46 2 THE FIRST LAW back to the surroundings. As a result, the temperature of the system rises less than when the heating occurs at constant volume. A smaller increase in temperature implies a larger heat capacity, so we conclude that in most cases the heat capacity at constant pressure of a system is larger than its heat capacity at constant volume. We show later (Section 2.11) that there is a simple relation between the two heat capacities of a perfect gas: Cp − CV = nR (2.26)° It follows that the molar heat capacity of a perfect gas is about 8 J K−1 mol−1 larger at constant pressure than at constant volume. Because the heat capacity at constant volume of a monatomic gas is about 12 J K−1 mol−1, the difference is highly significant and must be taken into account. IMPACT ON BIOCHEMISTRY AND MATERIALS SCIENCE I2.1 Differential scanning calorimetry A differential scanning calorimeter. The sample and a reference material are heated in separate but identical metal heat sinks. The output is the difference in power needed to maintain the heat sinks at equal temperatures as the temperature rises. Fig. 2.15 A differential scanning calorimeter (DSC) measures the energy transferred as heat to or from a sample at constant pressure during a physical or chemical change. The term ‘differential’ refers to the fact that the behaviour of the sample is compared to that of a reference material which does not undergo a physical or chemical change during the analysis. The term ‘scanning’ refers to the fact that the temperatures of the sample and reference material are increased, or scanned, during the analysis. A DSC consists of two small compartments that are heated electrically at a constant rate. The temperature, T, at time t during a linear scan is T = T0 + αt, where T0 is the initial temperature and α is the temperature scan rate (in kelvin per second, K s−1). A computer controls the electrical power output in order to maintain the same temperature in the sample and reference compartments throughout the analysis (see Fig. 2.15). The temperature of the sample changes significantly relative to that of the reference material if a chemical or physical process involving the transfer of energy as heat occurs in the sample during the scan. To maintain the same temperature in both compartments, excess energy is transferred as heat to or from the sample during the process. For example, an endothermic process lowers the temperature of the sample relative to that of the reference and, as a result, the sample must be heated more strongly than the reference in order to maintain equal temperatures. If no physical or chemical change occurs in the sample at temperature T, we write the heat transferred to the sample as qp = Cp ∆T, where ∆T = T − T0 and we have assumed that Cp is independent of temperature. The chemical or physical process requires the transfer of qp + qp,ex, where qp,ex is excess energy transferred as heat, to attain the same change in temperature of the sample. We interpret qp,ex in terms of an apparent change in the heat capacity at constant pressure of the sample, Cp, during the temperature scan. Then we write the heat capacity of the sample as Cp + Cp,ex, and qp + qp,ex = (Cp + Cp,ex)∆T It follows that Fig. 2.16 A thermogram for the protein ubiquitin at pH = 2.45. The protein retains its native structure up to about 45oC and then undergoes an endothermic conformational change. (Adapted from B. Chowdhry and S. LeHarne, J. Chem. Educ. 74, 236 (1997).) Cp,ex = qp,ex ∆T = qp,ex αt = Pex α where Pex = qp,ex /t is the excess electrical power necessary to equalize the temperature of the sample and reference compartments. A DSC trace, also called a thermogram, consists of a plot of Pex or Cp,ex against T (see Fig. 2.16). Broad peaks in the thermogram indicate processes requiring transfer of energy as heat. From eqn 2.23a, the enthalpy change associated with the process is 2.6 ADIABATIC CHANGES ∆H = # 47 T2 Cp,exdT T1 where T1 and T2 are, respectively, the temperatures at which the process begins and ends. This relation shows that the enthalpy change is then the area under the curve of Cp,ex against T. With a DSC, enthalpy changes may be determined in samples of masses as low as 0.5 mg, which is a significant advantage over bomb or flame calorimeters, which require several grams of material. Differential scanning calorimetry is used in the chemical industry to characterize polymers and in the biochemistry laboratory to assess the stability of proteins, nucleic acids, and membranes. Large molecules, such as synthetic or biological polymers, attain complex three-dimensional structures due to intra- and intermolecular interactions, such as hydrogen bonding and hydrophobic interactions (Chapter 18). Disruption of these interactions is an endothermic process that can be studied with a DSC. For example, the thermogram shown in the illustration indicated that the protein ubiquitin retains its native structure up to about 45°C. At higher temperatures, the protein undergoes an endothermic conformational change that results in the loss of its three-dimensional structure. The same principles also apply to the study of structural integrity and stability of synthetic polymers, such as plastics. 2.6 Adiabatic changes We are now equipped to deal with the changes that occur when a perfect gas expands adiabatically. A decrease in temperature should be expected: because work is done but no heat enters the system, the internal energy falls, and therefore the temperature of the working gas also falls. In molecular terms, the kinetic energy of the molecules falls as work is done, so their average speed decreases, and hence the temperature falls. The change in internal energy of a perfect gas when the temperature is changed from Ti to Tf and the volume is changed from Vi to Vf can be expressed as the sum of two steps (Fig. 2.17). In the first step, only the volume changes and the temperature is held constant at its initial value. However, because the internal energy of a perfect gas is independent of the volume the molecules occupy, the overall change in internal energy arises solely from the second step, the change in temperature at constant volume. Provided the heat capacity is independent of temperature, this change is ∆U = CV (Tf − Ti) = CV ∆T Because the expansion is adiabatic, we know that q = 0; because ∆U = q + w, it then follows that ∆U = wad. The subscript ‘ad’ denotes an adiabatic process. Therefore, by equating the two values we have obtained for ∆U, we obtain wad = CV ∆T (2.27) That is, the work done during an adiabatic expansion of a perfect gas is proportional to the temperature difference between the initial and final states. That is exactly what we expect on molecular grounds, because the mean kinetic energy is proportional to T, so a change in internal energy arising from temperature alone is also expected to be proportional to ∆T. In Further information 2.1 we show that the initial and final temperatures of a perfect gas that undergoes reversible adiabatic expansion (reversible expansion in a thermally insulated container) can be calculated from A Vi D Tf = Ti C Vf F 1/c (2.28a)°rev Fig. 2.17 To achieve a change of state from one temperature and volume to another temperature and volume, we may consider the overall change as composed of two steps. In the first step, the system expands at constant temperature; there is no change in internal energy if the system consists of a perfect gas. In the second step, the temperature of the system is reduced at constant volume. The overall change in internal energy is the sum of the changes for the two steps. 48 2 THE FIRST LAW where c = CV,m/R, or equivalently ViT ci = VfT fc (2.28b)°rev This result is often summarized in the form VT = constant. c Illustration 2.5 Work of adiabatic expansion Consider the adiabatic, reversible expansion of 0.020 mol Ar, initially at 25°C, from 0.50 dm3 to 1.00 dm3. The molar heat capacity of argon at constant volume is 12.48 J K−1 mol−1, so c = 1.501. Therefore, from eqn 2.28a, A 0.50 dm3 D Tf = (298 K) × C 1.00 dm3 F 1/1.501 = 188 K It follows that ∆T = −110 K, and therefore, from eqn 2.27, that w = {(0.020 mol) × (12.48 J K−1 mol−1)} × (−110 K) = −27 J Note that temperature change is independent of the amount of gas but the work is not. Self-test 2.5 Calculate the final temperature, the work done, and the change of internal energy when ammonia is used in a reversible adiabatic expansion from 0.50 dm3 to 2.00 dm3, the other initial conditions being the same. [195 K, −56 J, −56 J] We also show in Further information 2.1 that the pressure of a perfect gas that undergoes reversible adiabatic expansion from a volume Vi to a volume Vf is related to its initial pressure by pfV γf = piV γi (2.29)°rev γ Fig. 2.18 An adiabat depicts the variation of pressure with volume when a gas expands adiabatically. (a) An adiabat for a perfect gas undergoing reversible expansion. (b) Note that the pressure declines more steeply for an adiabat than it does for an isotherm because the temperature decreases in the former. Exploration Explore how the parameter γ affects the dependence of the pressure on the volume. Does the pressure–volume dependence become stronger or weaker with increasing volume? where γ = Cp,m/CV,m. This result is summarized in the form pV = constant. For a monatomic perfect gas, CV,m = –23 R (see Illustration 2.3), and from eqn 2.26 Cp,m = –52 R; so γ = –35 . For a gas of nonlinear polyatomic molecules (which can rotate as well as translate), CV,m = 3R, so γ = –43 . The curves of pressure versus volume for adiabatic change are known as adiabats, and one for a reversible path is illustrated in Fig. 2.18. Because γ > 1, an adiabat falls more steeply (p ∝ 1/V γ ) than the corresponding isotherm (p ∝ 1/V). The physical reason for the difference is that, in an isothermal expansion, energy flows into the system as heat and maintains the temperature; as a result, the pressure does not fall as much as in an adiabatic expansion. Illustration 2.6 The pressure change accompanying adiabatic expansion When a sample of argon (for which γ = –35 ) at 100 kPa expands reversibly and adiabatically to twice its initial volume the final pressure will be A Vi D A 1D pf = pi = C Vf F C 2F γ 5/3 × (100 kPa) = 32 kPa For an isothermal doubling of volume, the final pressure would be 50 kPa. 2.7 STANDARD ENTHALPY CHANGES Thermochemistry The study of the energy transferred as heat during the course of chemical reactions is called thermochemistry. Thermochemistry is a branch of thermodynamics because a reaction vessel and its contents form a system, and chemical reactions result in the exchange of energy between the system and the surroundings. Thus we can use calorimetry to measure the energy supplied or discarded as heat by a reaction, and can identify q with a change in internal energy (if the reaction occurs at constant volume) or a change in enthalpy (if the reaction occurs at constant pressure). Conversely, if we know ∆U or ∆H for a reaction, we can predict the energy (transferred as heat) the reaction can produce. We have already remarked that a process that releases energy by heating the surroundings is classified as exothermic and one that absorbs energy by cooling the surroundings is classified as endothermic. Because the release of energy by heating the surroundings signifies a decrease in the enthalpy of a system (at constant pressure), we can now see that an exothermic process at constant pressure is one for which ∆H < 0. Conversely, because the absorption of energy by cooling the surroundings results in an increase in enthalpy, an endothermic process at constant pressure has ∆H > 0. 2.7 Standard enthalpy changes Changes in enthalpy are normally reported for processes taking place under a set of standard conditions. In most of our discussions we shall consider the standard enthalpy change, ∆H 7, the change in enthalpy for a process in which the initial and final substances are in their standard states: The standard state of a substance at a specified temperature is its pure form at 1 bar.5 For example, the standard state of liquid ethanol at 298 K is pure liquid ethanol at 298 K and 1 bar; the standard state of solid iron at 500 K is pure iron at 500 K and 1 bar. The standard enthalpy change for a reaction or a physical process is the difference between the products in their standard states and the reactants in their standard states, all at the same specified temperature. As an example of a standard enthalpy change, the standard enthalpy of vaporization, ∆vapH 7, is the enthalpy change per mole when a pure liquid at 1 bar vaporizes to a gas at 1 bar, as in H2O(l) → H2O(g) ∆ vapH 7(373 K) = +40.66 kJ mol−1 As implied by the examples, standard enthalpies may be reported for any temperature. However, the conventional temperature for reporting thermodynamic data is 298.15 K (corresponding to 25.00°C). Unless otherwise mentioned, all thermodynamic data in this text will refer to this conventional temperature. A note on good practice The attachment of the name of the transition to the symbol ∆, as in ∆ vapH, is the modern convention. However, the older convention, ∆Hvap, is still widely used. The new convention is more logical because the subscript identifies the type of change, not the physical observable related to the change. 5 The definition of standard state is more sophisticated for a real gas (Further information 3.2) and for solutions (Sections 5.6 and 5.7). 49 50 2 THE FIRST LAW Synoptic Table 2.3* Standard enthalpies of fusion and vaporization at the transition temperature, ∆trsH 7/(kJ mol−1) Tf /K Ar 83.81 C6H6 278.61 H2O 273.15 3.5 He Fusion 1.188 10.59 Tb/K 87.29 Vaporization 6.506 353.2 30.8 6.008 373.15 40.656 (44.016 at 298 K) 0.021 4.22 0.084 * More values are given in the Data section. (a) Enthalpies of physical change The standard enthalpy change that accompanies a change of physical state is called the standard enthalpy of transition and is denoted ∆ trsH 7 (Table 2.3). The standard enthalpy of vaporization, ∆vapH 7, is one example. Another is the standard enthalpy of fusion, ∆fusH 7, the standard enthalpy change accompanying the conversion of a solid to a liquid, as in H2O(s) → H2O(l) ∆fusH 7(273 K) = +6.01 kJ mol−1 As in this case, it is sometimes convenient to know the standard enthalpy change at the transition temperature as well as at the conventional temperature. Because enthalpy is a state function, a change in enthalpy is independent of the path between the two states. This feature is of great importance in thermochemistry, for it implies that the same value of ∆H 7 will be obtained however the change is brought about between the same initial and final states. For example, we can picture the conversion of a solid to a vapour either as occurring by sublimation (the direct conversion from solid to vapour), H2O(s) → H2O(g) ∆subH 7 or as occurring in two steps, first fusion (melting) and then vaporization of the resulting liquid: H2O(s) → H2O(l) ∆ fusH 7 H2O(l) → H2O(g) ∆ vapH 7 Overall: H2O(s) → H2O(g) ∆fusH 7 + ∆ vapH 7 Because the overall result of the indirect path is the same as that of the direct path, the overall enthalpy change is the same in each case (1), and we can conclude that (for processes occurring at the same temperature) ∆subH 7 = ∆ fusH 7 + ∆ vapH 7 (2.30) An immediate conclusion is that, because all enthalpies of fusion are positive, the enthalpy of sublimation of a substance is greater than its enthalpy of vaporization (at a given temperature). Another consequence of H being a state function is that the standard enthalpy changes of a forward process and its reverse differ in sign (2): ∆H 7(A → B) = −∆H 7(B → A) (2.31) For instance, because the enthalpy of vaporization of water is +44 kJ mol−1 at 298 K, its enthalpy of condensation at that temperature is −44 kJ mol−1. 2.7 STANDARD ENTHALPY CHANGES Table 2.4 Enthalpies of transition Transition Process Symbol* Transition Phase α → phase β ∆trsH Fusion s→l ∆fusH Vaporization l→g ∆vapH Sublimation s→g ∆subH Mixing Pure → mixture ∆mixH ∆solH Solution Solute → solution Hydration X±(g) → X±(aq) ∆hydH Atomization Species(s, l, g) → atoms(g) ∆atH Ionization X(g) → X+(g) + e−(g) ∆ionH − − Electron gain X(g) + e (g) → X (g) Reaction Reactants → products ∆egH ∆rH Combustion Compounds(s, l, g) + O2(g) → CO2(g), H2O(l, g) ∆cH Formation Elements → compound ∆fH Activation Reactants → activated complex ∆‡H * IUPAC recommendations. In common usage, the transition subscript is often attached to ∆H, as in ∆Htrs. The different types of enthalpies encountered in thermochemistry are summarized in Table 2.4. We shall meet them again in various locations throughout the text. (b) Enthalpies of chemical change Now we consider enthalpy changes that accompany chemical reactions. There are two ways of reporting the change in enthalpy that accompanies a chemical reaction. One is to write the thermochemical equation, a combination of a chemical equation and the corresponding change in standard enthalpy: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) ∆H 7 = −890 kJ ∆H 7 is the change in enthalpy when reactants in their standard states change to products in their standard states: Pure, separate reactants in their standard states → pure, separate products in their standard states Except in the case of ionic reactions in solution, the enthalpy changes accompanying mixing and separation are insignificant in comparison with the contribution from the reaction itself. For the combustion of methane, the standard value refers to the reaction in which 1 mol CH4 in the form of pure methane gas at 1 bar reacts completely with 2 mol O2 in the form of pure oxygen gas to produce 1 mol CO2 as pure carbon dioxide at 1 bar and 2 mol H2O as pure liquid water at 1 bar; the numerical value is for the reaction at 298 K. Alternatively, we write the chemical equation and then report the standard reaction enthalpy, ∆r H 7. Thus, for the combustion of reaction, we write CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) For the reaction 2A+B→3C+D ∆ r H 7 = −890 kJ mol−1 51 52 2 THE FIRST LAW Synoptic Table 2.5* Standard enthalpies of formation and combustion of organic compounds at 298 K ∆ f H 7/(kJ mol−1) ∆ c H 7/(kJ mol−1) Benzene, C6H6(l) +49.0 −3268 Ethane, C2H6(g) −84.7 −1560 −1274 Glucose, C6H12O6(s) Methane, CH4(g) Methanol, CH3OH(l) −2808 −74.8 −890 −238.7 −721 * More values are given in the Data section. the standard reaction enthalpy is ∆ r H 7 = {3Hm7 (C) + Hm7 (D)} − {2Hm7 (A) + Hm7 (B)} where Hm7 (J) is the standard molar enthalpy of species J at the temperature of interest. Note how the ‘per mole’ of ∆ r H 7 comes directly from the fact that molar enthalpies appear in this expression. We interpret the ‘per mole’ by noting the stoichiometic coefficients in the chemical equation. In this case ‘per mole’ in ∆ r H 7 means ‘per 2 mol A’, ‘per mole B’, ‘per 3 mol C’, or ‘per mol D’. In general, ∆r H 7 = ∑ νH m7 − ∑ νH m7 Products (2.32) Reactants where in each case the molar enthalpies of the species are multiplied by their stoichiometric coefficients, ν.6 Some standard reaction enthalpies have special names and a particular significance. For instance, the standard enthalpy of combustion, ∆c H 7, is the standard reaction enthalpy for the complete oxidation of an organic compound to CO2 gas and liquid H2O if the compound contains C, H, and O, and to N2 gas if N is also present. An example is the combustion of glucose: C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) ∆c H 7 = −2808 kJ mol−1 The value quoted shows that 2808 kJ of heat is released when 1 mol C6H12O6 burns under standard conditions (at 298 K). Some further values are listed in Table 2.5. IMPACT ON BIOLOGY I2.2 Food and energy reserves The thermochemical properties of fuels Table 2.6 and foods are commonly discussed in terms of their specific enthalpy, the enthalpy of combustion per gram of material. Thus, if the standard enthalpy of combustion is ∆c H 7 and the molar mass of the compound is M, then the specific enthalpy is ∆c H 7/M. Table 2.6 lists the specific enthalpies of several fuels. A typical 18–20 year old man requires a daily input of about 12 MJ; a woman of the same age needs about 9 MJ. If the entire consumption were in the form of glucose (3; which has a specific enthalpy of 16 kJ g−1), that would require the consumption of 750 g of glucose for a man and 560 g for a woman. In fact, digestible carbohydrates have a slightly higher specific enthalpy (17 kJ g−1) than glucose itself, so a carbohydrate 6 In this and similar expressions, all stoichiometric coefficients are positive. For a more sophisticated way of writing eqn 2.32, see Section 7.2. 2.7 STANDARD ENTHALPY CHANGES Table 2.6 Thermochemical properties of some fuels Fuel Combustion equation Hydrogen H2(g) + –12 O2(g) → H2O(l) Methane CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) Octane –2 O2(g) C8H18(l) + –25 → 8 CO2(g) + 9 H2O(l) Methanol CH3OH(l) + –32 O2(g) → CO2(g) + 2 H2O(l) ∆ c H 7/ (kJ mol−1) Specific enthalpy/ (kJ g−1) Enthalpy density/ (kJ dm−3) −286 142 13 −890 55 40 −5471 48 3.8 × 104 −726 23 1.8 × 104 diet is slightly less daunting than a pure glucose diet, as well as being more appropriate in the form of fibre, the indigestible cellulose that helps move digestion products through the intestine. The specific enthalpy of fats, which are long-chain esters like tristearin (beef fat), is much greater than that of carbohydrates, at around 38 kJ g−1, slightly less than the value for the hydrocarbon oils used as fuel (48 kJ g−1). Fats are commonly used as an energy store, to be used only when the more readily accessible carbohydrates have fallen into short supply. In Arctic species, the stored fat also acts as a layer of insulation; in desert species (such as the camel), the fat is also a source of water, one of its oxidation products. Proteins are also used as a source of energy, but their components, the amino acids, are often too valuable to squander in this way, and are used to construct other proteins instead. When proteins are oxidized (to urea, CO(NH2)2), the equivalent enthalpy density is comparable to that of carbohydrates. The heat released by the oxidation of foods needs to be discarded in order to maintain body temperature within its typical range of 35.6–37.8°C. A variety of mechanisms contribute to this aspect of homeostasis, the ability of an organism to counteract environmental changes with physiological responses. The general uniformity of temperature throughout the body is maintained largely by the flow of blood. When heat needs to be dissipated rapidly, warm blood is allowed to flow through the capillaries of the skin, so producing flushing. Radiation is one means of discarding heat; another is evaporation and the energy demands of the enthalpy of vaporization of water. Evaporation removes about 2.4 kJ per gram of water perspired. When vigorous exercise promotes sweating (through the influence of heat selectors on the hypothalamus), 1–2 dm3 of perspired water can be produced per hour, corresponding to a heat loss of 2.4–5.0 MJ h−1. (c) Hess’s law Standard enthalpies of individual reactions can be combined to obtain the enthalpy of another reaction. This application of the First Law is called Hess’s law: The standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided. The individual steps need not be realizable in practice: they may be hypothetical reactions, the only requirement being that their chemical equations should balance. The thermodynamic basis of the law is the path-independence of the value of ∆r H 7 and the implication that we may take the specified reactants, pass through any (possibly hypothetical) set of reactions to the specified products, and overall obtain the same change of enthalpy. The importance of Hess’s law is that information about a 53 54 2 THE FIRST LAW reaction of interest, which may be difficult to determine directly, can be assembled from information on other reactions. Example 2.5 Using Hess’s law The standard reaction enthalpy for the hydrogenation of propene, CH2 =CHCH3(g) + H2(g) → CH3CH2CH3(g) is −124 kJ mol−1. The standard reaction enthalpy for the combustion of propane, CH3CH2CH3(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) is −2220 kJ mol−1. Calculate the standard enthalpy of combustion of propene. Method The skill to develop is the ability to assemble a given thermochemical equation from others. Add or subtract the reactions given, together with any others needed, so as to reproduce the reaction required. Then add or subtract the reaction enthalpies in the same way. Additional data are in Table 2.5. Answer The combustion reaction we require is C3H6(g) + –92 O2(g) → 3 CO2(g) + 3 H2O(l) This reaction can be recreated from the following sum: ∆r H 7/(kJ mol−1) C3H6(g) + H2(g) → C3H8(g) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l) H2O(l) → H2(g) + –12 O2(g) C3H6(g) + –92 O2(g) → 3 CO2(g) + 3 H2O(l) Synoptic Table 2.7* Standard enthalpies of formation of inorganic compounds at 298 K −124 −2220 +286 −2058 Self-test 2.6 Calculate the enthalpy of hydrogenation of benzene from its enthalpy of combustion and the enthalpy of combustion of cyclohexane. [−205 kJ mol−1] ∆f H 7/(kJ mol−1) H2O(l) −285.83 H2O(g) −187.78 NH3(g) −46.11 N2H4(l) +50.63 NO2(g) 33.18 N2O4(g) +9.16 NaCl(s) −411.15 KCl(s) −436.75 * More values are given in the Data section. Comment 2.7 The NIST WebBook listed in the web site for this book links to online databases of thermochemical data. 2.8 Standard enthalpies of formation The standard enthalpy of formation, ∆f H 7, of a substance is the standard reaction enthalpy for the formation of the compound from its elements in their reference states. The reference state of an element is its most stable state at the specified temperature and 1 bar. For example, at 298 K the reference state of nitrogen is a gas of N2 molecules, that of mercury is liquid mercury, that of carbon is graphite, and that of tin is the white (metallic) form. There is one exception to this general prescription of reference states: the reference state of phosphorus is taken to be white phosphorus despite this allotrope not being the most stable form but simply the more reproducible form of the element. Standard enthalpies of formation are expressed as enthalpies per mole of molecules or (for ionic substances) formula units of the compound. The standard enthalpy of formation of liquid benzene at 298 K, for example, refers to the reaction 6 C(s, graphite) + 3 H2(g) → C6H6(l) and is +49.0 kJ mol−1. The standard enthalpies of formation of elements in their reference states are zero at all temperatures because they are the enthalpies of such ‘null’ reactions as N2(g) → N2(g). Some enthalpies of formation are listed in Tables 2.5 and 2.7. 2.8 STANDARD ENTHALPIES OF FORMATION The standard enthalpy of formation of ions in solution poses a special problem because it is impossible to prepare a solution of cations alone or of anions alone. This problem is solved by defining one ion, conventionally the hydrogen ion, to have zero standard enthalpy of formation at all temperatures: ∆ f H 7(H+, aq) = 0 [2.33] Thus, if the enthalpy of formation of HBr(aq) is found to be −122 kJ mol−1, then the whole of that value is ascribed to the formation of Br−(aq), and we write ∆ f H 7(Br−, aq) = −122 kJ mol−1. That value may then be combined with, for instance, the enthalpy formation of AgBr(aq) to determine the value of ∆ f H 7(Ag+, aq), and so on. In essence, this definition adjusts the actual values of the enthalpies of formation of ions by a fixed amount, which is chosen so that the standard value for one of them, H+(aq), has the value zero. (a) The reaction enthalpy in terms of enthalpies of formation Conceptually, we can regard a reaction as proceeding by decomposing the reactants into their elements and then forming those elements into the products. The value of ∆ r H 7 for the overall reaction is the sum of these ‘unforming’ and forming enthalpies. Because ‘unforming’ is the reverse of forming, the enthalpy of an unforming step is the negative of the enthalpy of formation (4). Hence, in the enthalpies of formation of substances, we have enough information to calculate the enthalpy of any reaction by using ∆rH 7 = ∑ν∆f H 7 − ∑ν∆f H 7 Products (2.34) Reactants where in each case the enthalpies of formation of the species that occur are multiplied by their stoichiometric coefficients. Illustration 2.7 Using standard enthalpies of formation The standard reaction enthalpy of 2 HN3(l) + 2 NO(g) → H2O2(l) + 4 N2(g) is calculated as follows: ∆r H 7 = {∆f H 7(H2O2,l) + 4∆f H 7(N2,g)} − {2∆f H 7(HN3,l) + 2∆f H 7(NO,g)} = {−187.78 + 4(0)} kJ mol−1 − {2(264.0) + 2(90.25)} kJ mol−1 = −896.3 kJ mol−1 (b) Enthalpies of formation and molecular modelling We have seen how to construct standard reaction enthalpies by combining standard enthalpies of formation. The question that now arises is whether we can construct standard enthalpies of formation from a knowledge of the chemical constitution of the species. The short answer is that there is no thermodynamically exact way of expressing enthalpies of formation in terms of contributions from individual atoms and bonds. In the past, approximate procedures based on mean bond enthalpies, ∆H(A-B), the average enthalpy change associated with the breaking of a specific A-B bond, A-B(g) → A(g) + B(g) ∆H(A-B) 55 56 2 THE FIRST LAW have been used. However, this procedure is notoriously unreliable, in part because the ∆H(A-B) are average values for a series of related compounds. Nor does the approach distinguish between geometrical isomers, where the same atoms and bonds may be present but experimentally the enthalpies of formation might be significantly different. Computer-aided molecular modelling has largely displaced this more primitive approach. Commercial software packages use the principles developed in Chapter 11 to calculate the standard enthalpy of formation of a molecule drawn on the computer screen. These techniques can be applied to different conformations of the same molecule. In the case of methylcyclohexane, for instance, the calculated conformational energy difference ranges from 5.9 to 7.9 kJ mol−1, with the equatorial conformer having the lower standard enthalpy of formation. These estimates compare favourably with the experimental value of 7.5 kJ mol−1. However, good agreement between calculated and experimental values is relatively rare. Computational methods almost always predict correctly which conformer is more stable but do not always predict the correct magnitude of the conformational energy difference. An illustration of the content of Kirchhoff’s law. When the temperature is increased, the enthalpy of the products and the reactants both increase, but may do so to different extents. In each case, the change in enthalpy depends on the heat capacities of the substances. The change in reaction enthalpy reflects the difference in the changes of the enthalpies. Fig. 2.19 2.9 The temperature-dependence of reaction enthalpies The standard enthalpies of many important reactions have been measured at different temperatures. However, in the absence of this information, standard reaction enthalpies at different temperatures may be calculated from heat capacities and the reaction enthalpy at some other temperature (Fig. 2.19). In many cases heat capacity data are more accurate that reaction enthalpies so, providing the information is available, the procedure we are about to describe is more accurate that a direct measurement of a reaction enthalpy at an elevated temperature. It follows from eqn 2.23a that, when a substance is heated from T1 to T2, its enthalpy changes from H(T1) to H(T2) = H(T1) + # T2 CpdT (2.35) T1 (We have assumed that no phase transition takes place in the temperature range of interest.) Because this equation applies to each substance in the reaction, the standard reaction enthalpy changes from ∆rH 7(T1) to ∆ r H 7(T2) = ∆r H 7(T1) + # T2 ∆rC 7pdT (2.36) T1 where ∆rC p7 is the difference of the molar heat capacities of products and reactants under standard conditions weighted by the stoichiometric coefficients that appear in the chemical equation: ∆ rC 7p = ∑ν C 7p,m − ∑ν C 7p,m Products [2.37] Reactants Equation 2.36 is known as Kirchhoff’s law. It is normally a good approximation to assume that ∆rCp is independent of the temperature, at least over reasonably limited ranges, as illustrated in the following example. Although the individual heat capacities may vary, their difference varies less significantly. In some cases the temperature dependence of heat capacities is taken into account by using eqn 2.25. 2.10 EXACT AND INEXACT DIFFERENTIALS 57 Example 2.6 Using Kirchhoff’s law The standard enthalpy of formation of gaseous H2O at 298 K is −241.82 kJ mol−1. Estimate its value at 100°C given the following values of the molar heat capacities at constant pressure: H2O(g): 33.58 J K−1 mol−1; H2(g): 28.84 J K−1 mol−1; O2(g): 29.37 J K−1 mol−1. Assume that the heat capacities are independent of temperature. Method When ∆C 7p is independent of temperature in the range T1 to T2, the integral in eqn 2.36 evaluates to (T2 − T1)∆rC 7p. Therefore, ∆r H 7(T2) = ∆r H 7(T1) + (T2 − T1)∆rC 7p To proceed, write the chemical equation, identify the stoichiometric coefficients, and calculate ∆rC 7p from the data. Answer The reaction is H2(g) + –12 O2(g) → H2O(g), so 7 7 7 ∆rC 7p = C p,m (H2O, g) − {C p,m (H2, g) + –12 C p,m (O2, g)} = −9.94 J K−1 mol−1 It then follows that ∆ f H 7(373 K) = −241.82 kJ mol−1 + (75 K) × (−9.94 J K−1 mol−1) = −242.6 kJ mol−1 Self-test 2.7 Estimate the standard enthalpy of formation of cyclohexene at 400 K from the data in Table 2.5. [−163 kJ mol−1] State functions and exact differentials We saw in Section 2.2 that a ‘state function’ is a property that is independent of how a sample is prepared. In general, such properties are functions of variables that define the current state of the system, such as pressure and temperature. The internal energy and enthalpy are examples of state functions, for they depend on the current state of the system and are independent of its previous history. Processes that describe the preparation of the state are called path functions. Examples of path functions are the work and heating that are done when preparing a state. We do not speak of a system in a particular state as possessing work or heat. In each case, the energy transferred as work or heat relates to the path being taken between states, not the current state itself. We can use the mathematical properties of state functions to draw far-reaching conclusions about the relations between physical properties and establish connections that may be completely unexpected. The practical importance of these results is that we can combine measurements of different properties to obtain the value of a property we require. 2.10 Exact and inexact differentials Consider a system undergoing the changes depicted in Fig. 2.20. The initial state of the system is i and in this state the internal energy is Ui. Work is done by the system as it expands adiabatically to a state f. In this state the system has an internal energy Uf and the work done on the system as it changes along Path 1 from i to f is w. Notice our use of language: U is a property of the state; w is a property of the path. Now consider another process, Path 2, in which the initial and final states are the same as those in Path 1 but in which the expansion is not adiabatic. The internal energy of both the Fig. 2.20 As the volume and temperature of a system are changed, the internal energy changes. An adiabatic and a non-adiabatic path are shown as Path 1 and Path 2, respectively: they correspond to different values of q and w but to the same value of ∆U. 58 2 THE FIRST LAW initial and the final states are the same as before (because U is a state function). However, in the second path an energy q′ enters the system as heat and the work w′ is not the same as w. The work and the heat are path functions. In terms of the mountaineering analogy in Section 2.2, the change in altitude (a state function) is independent of the path, but the distance travelled (a path function) does depend on the path taken between the fixed endpoints. If a system is taken along a path (for example, by heating it), U changes from Ui to Uf, and the overall change is the sum (integral) of all the infinitesimal changes along the path: # dU f ∆U = (2.38) i The value of ∆U depends on the initial and final states of the system but is independent of the path between them. This path-independence of the integral is expressed by saying that dU is an ‘exact differential’. In general, an exact differential is an infinitesimal quantity that, when integrated, gives a result that is independent of the path between the initial and final states. When a system is heated, the total energy transferred as heat is the sum of all individual contributions at each point of the path: # f q= dq (2.39) i, path Notice the difference between this equation and eqn 2.38. First, we do not write ∆q, because q is not a state function and the energy supplied as heat cannot be expressed as qf − qi. Secondly, we must specify the path of integration because q depends on the path selected (for example, an adiabatic path has q = 0, whereas a nonadiabatic path between the same two states would have q ≠ 0). This path-dependence is expressed by saying that dq is an ‘inexact differential’. In general, an inexact differential is an infinitesimal quantity that, when integrated, gives a result that depends on the path between the initial and final states. Often dq is written pq to emphasize that it is inexact and requires the specification of a path. The work done on a system to change it from one state to another depends on the path taken between the two specified states; for example, in general the work is different if the change takes place adiabatically and non-adiabatically. It follows that dw is an inexact differential. It is often written pw. Example 2.7 Calculating work, heat, and internal energy Consider a perfect gas inside a cylinder fitted with a piston. Let the initial state be T, Vi and the final state be T, Vf. The change of state can be brought about in many ways, of which the two simplest are the following: Path 1, in which there is free expansion against zero external pressure; Path 2, in which there is reversible, isothermal expansion. Calculate w, q, and ∆U for each process. Method To find a starting point for a calculation in thermodynamics, it is often a good idea to go back to first principles, and to look for a way of expressing the quantity we are asked to calculate in terms of other quantities that are easier to calculate. We saw in Molecular interpretation 2.2 that the internal energy of a perfect gas depends only on the temperature and is independent of the volume those molecules occupy, so for any isothermal change, ∆U = 0. We also know that in general ∆U = q + w. The question depends on being able to combine the two 2.11 CHANGES IN INTERNAL ENERGY expressions. In this chapter, we derived a number of expressions for the work done in a variety of processes, and here we need to select the appropriate ones. Answer Because ∆U = 0 for both paths and ∆U = q + w, in each case q = −w. The work of free expansion is zero (Section 2.3b); so in Path 1, w = 0 and q = 0. For Path 2, the work is given by eqn 2.11, so w = −nRT ln(Vf /Vi) and consequently q = nRT ln(Vf /Vi). These results are consequences of the path independence of U, a state function, and the path dependence of q and w, which are path functions. Self-test 2.8 Calculate the values of q, w, and ∆U for an irreversible isothermal expansion of a perfect gas against a constant nonzero external pressure. [q = pex ∆V, w = −pex ∆V, ∆U = 0] 2.11 Changes in internal energy We begin to unfold the consequences of dU being an exact differential by exploring a closed system of constant composition (the only type of system considered in the rest of this chapter). The internal energy U can be regarded as a function of V, T, and p, but, because there is an equation of state, stating the values of two of the variables fixes the value of the third. Therefore, it is possible to write U in terms of just two independent variables: V and T, p and T, or p and V. Expressing U as a function of volume and temperature fits the purpose of our discussion. (a) General considerations When V changes to V + dV at constant temperature, U changes to U′ = U + A ∂U D dV C ∂V F T The coefficient (∂U/∂V)T , the slope of a plot of U against V at constant temperature, is the partial derivative of U with respect to V (Fig. 2.21). If, instead, T changes to T + dT at constant volume (Fig. 2.22), then the internal energy changes to Fig. 2.21 The partial derivative (∂U/∂V)T is the slope of U with respect to V with the temperature T held constant. Fig. 2.22 The partial derivative (∂U/∂T)V is the slope of U with respect to T with the volume V held constant. 59 60 2 THE FIRST LAW U′ = U + A ∂U D dT C ∂T F V Now suppose that V and T both change infinitesimally (Fig. 2.23). The new internal energy, neglecting second-order infinitesimals (those proportional to dVdT), is the sum of the changes arising from each increment: U′ = U + A ∂U D A ∂U D dV + dT C ∂V F T C ∂T F V As a result of the infinitesimal changes in conditions, the internal energy U′ differs from U by the infinitesimal amount dU, so we an write U′ = U + dU. Therefore, from the last equation we obtain the very important result that An overall change in U, which is denoted dU, arises when both V and T are allowed to change. If second-order infinitesimals are ignored, the overall change is the sum of changes for each variable separately. Fig. 2.23 dU = The internal pressure, πT , is the slope of U with respect to V with the temperature T held constant. (2.40) The interpretation of this equation is that, in a closed system of constant composition, any infinitesimal change in the internal energy is proportional to the infinitesimal changes of volume and temperature, the coefficients of proportionality being the two partial derivatives. In many cases partial derivatives have a straightforward physical interpretation, and thermodynamics gets shapeless and difficult only when that interpretation is not kept in sight. In the present case, we have already met (∂U/∂T)V in eqn 2.15, where we saw that it is the constant-volume heat capacity, CV. The other coefficient, (∂U/∂V)T, plays a major role in thermodynamics because it is a measure of the variation of the internal energy of a substance as its volume is changed at constant temperature (Fig. 2.24). We shall denote it πT and, because it has the same dimensions as pressure, call it the internal pressure: πT = Fig. 2.24 A ∂U D A ∂U D dV + dT C ∂V F T C ∂T F V A ∂U D C ∂V F T [2.41] In terms of the notation CV and πT, eqn 2.40 can now be written dU = π T dV + CV dT (2.42) (b) The Joule experiment When there are no interactions between the molecules, the internal energy is independent of their separation and hence independent of the volume of the sample (see Molecular interpretation 2.2). Therefore, for a perfect gas we can write πT = 0. The statement πT = 0 (that is, the internal energy is independent of the volume occupied by the sample) can be taken to be the definition of a perfect gas, for later we shall see that it implies the equation of state pV = nRT. If the internal energy increases (dU > 0) as the volume of the sample expands isothermally (dV > 0), which is the case when there are attractive forces between the particles, then a plot of internal energy against volume slopes upwards and π T > 0 (Fig. 2.25). James Joule thought that he could measure π T by observing the change in temperature of a gas when it is allowed to expand into a vacuum. He used two metal vessels immersed in a water bath (Fig. 2.26). One was filled with air at about 22 atm and the other was evacuated. He then tried to measure the change in temperature of the water of the bath when a stopcock was opened and the air expanded into a vacuum. He observed no change in temperature. 2.11 CHANGES IN INTERNAL ENERGY Fig. 2.25 For a perfect gas, the internal energy is independent of the volume (at constant temperature). If attractions are dominant in a real gas, the internal energy increases with volume because the molecules become farther apart on average. If repulsions are dominant, the internal energy decreases as the gas expands. Fig. 2.26 A schematic diagram of the apparatus used by Joule in an attempt to measure the change in internal energy when a gas expands isothermally. The heat absorbed by the gas is proportional to the change in temperature of the bath. The thermodynamic implications of the experiment are as follows. No work was done in the expansion into a vacuum, so w = 0. No energy entered or left the system (the gas) as heat because the temperature of the bath did not change, so q = 0. Consequently, within the accuracy of the experiment, ∆U = 0. It follows that U does not change much when a gas expands isothermally and therefore that π T = 0. Joule’s experiment was crude. In particular, the heat capacity of the apparatus was so large that the temperature change that gases do in fact cause was too small to measure. From his experiment Joule extracted an essential limiting property of a gas, a property of a perfect gas, without detecting the small deviations characteristic of real gases. (c) Changes in internal energy at constant pressure Partial derivatives have many useful properties and some that we shall draw on frequently are reviewed in Appendix 2. Skilful use of them can often turn some unfamiliar quantity into a quantity that can be recognized, interpreted, or measured. As an example, suppose we want to find out how the internal energy varies with temperature when the pressure of the system is kept constant. If we divide both sides of eqn 2.42 by dT and impose the condition of constant pressure on the resulting differentials, so that dU/dT on the left becomes (∂U/∂T)p, we obtain A ∂U D A ∂V D = πT + CV C ∂T F p C ∂T F p It is usually sensible in thermodynamics to inspect the output of a manipulation like this to see if it contains any recognizable physical quantity. The partial derivative on the right in this expression is the slope of the plot of volume against temperature (at 61 62 2 THE FIRST LAW Synoptic Table 2.8* Expansion coefficients (α) and isothermal compressibilities (κT) at 298 K Benzene α /(10−4 K−1) κ T /(10−6 bar−1) 12.4 90.9 Diamond 0.030 0.185 Lead 0.861 2.18 Water 2.1 49.0 constant pressure). This property is normally tabulated as the expansion coefficient, α, of a substance,7 which is defined as α= 1 A ∂V D [2.43] V C ∂T F p and physically is the fractional change in volume that accompanies a rise in temperature. A large value of α means that the volume of the sample responds strongly to changes in temperature. Table 2.8 lists some experimental values of α and of the isothermal compressibility, κT (kappa), which is defined as κT = − * More values are given in the Data section. 1 A ∂V D [2.44] V C ∂p F T The isothermal compressibility is a measure of the fractional change in volume when the pressure is increased by a small amount; the negative sign in the definition ensures that the compressibility is a positive quantity, because an increase of pressure, implying a positive dp, brings about a reduction of volume, a negative dV. Example 2.8 Calculating the expansion coefficient of a gas Derive an expression for the expansion coefficient of a perfect gas. Method The expansion coefficient is defined in eqn 2.43. To use this expression, substitute the expression for V in terms of T obtained from the equation of state for the gas. As implied by the subscript in eqn 2.43, the pressure, p, is treated as a constant. Answer Because pV = nRT, we can write α= 1 A ∂(nRT/p) D VC ∂T Fp = 1 V × nR dT p dT = nR pV = 1 T The higher the temperature, the less responsive is the volume of a perfect gas to a change in temperature. Self-test 2.9 Derive an expression for the isothermal compressibility of a perfect gas. [κT. = 1/p] When we introduce the definition of α into the equation for (∂U/∂T)p, we obtain A ∂U D = απTV + CV C ∂T F p (2.45) This equation is entirely general (provided the system is closed and its composition is constant). It expresses the dependence of the internal energy on the temperature at constant pressure in terms of CV, which can be measured in one experiment, in terms of α, which can be measured in another, and in terms of the quantity π T. For a perfect gas, π T = 0, so then A ∂U D = CV C ∂T F p 7 (2.46)° As for heat capacities, the expansion coefficients of a mixture depends on whether or not the composition is allowed to change. Throughout this chapter, we deal only with pure substances, so this complication can be disregarded. 2.12 THE JOULE–THOMSON EFFECT That is, although the constant-volume heat capacity of a perfect gas is defined as the slope of a plot of internal energy against temperature at constant volume, for a perfect gas CV is also the slope at constant pressure. Equation 2.46 provides an easy way to derive the relation between Cp and CV for a perfect gas expressed in eqn 2.26. Thus, we can use it to express both heat capacities in terms of derivatives at constant pressure: Cp − CV = A ∂H D A ∂U D − C ∂T F p C ∂T F p (2.47)° Then we introduce H = U + pV = U + nRT into the first term, which results in Cp − CV = A ∂U D A ∂U D + nR − = nR C ∂T F p C ∂T F p (2.48)° which is eqn 2.26. We show in Further information 2.2 that in general Cp − CV = α 2TV κT (2.49) Equation 2.49 applies to any substance (that is, it is ‘universally true’). It reduces to eqn 2.48 for a perfect gas when we set α = 1/T and κT = 1/p. Because expansion coefficients α of liquids and solids are small, it is tempting to deduce from eqn 2.49 that for them Cp ≈ CV. But this is not always so, because the compressibility κT might also be small, so α 2/κT might be large. That is, although only a little work need be done to push back the atmosphere, a great deal of work may have to be done to pull atoms apart from one another as the solid expands. As an illustration, for water at 25°C, eqn 2.49 gives Cp,m = 75.3 J K−1 mol−1 compared with CV,m = 74.8 J K−1 mol−1. In some cases, the two heat capacities differ by as much as 30 per cent. 2.12 The Joule–Thomson effect We can carry out a similar set of operations on the enthalpy, H = U + pV. The quantities U, p, and V are all state functions; therefore H is also a state function and dH is an exact differential. It turns out that H is a useful thermodynamic function when the pressure is under our control: we saw a sign of that in the relation ∆H = qp (eqn 2.19). We shall therefore regard H as a function of p and T, and adapt the argument in Section 2.10 to find an expression for the variation of H with temperature at constant volume. As set out in Justification 2.2, we find that for a closed system of constant composition, dH = −µCpdp + CpdT (2.50) where the Joule–Thomson coefficient, µ (mu), is defined as µ= A ∂T D C ∂p F H [2.51] This relation will prove useful for relating the heat capacities at constant pressure and volume and for a discussion of the liquefaction of gases. Justification 2.2 The variation of enthalpy with pressure and temperature By the same argument that led to eqn 2.40 but with H regarded as a function of p and T we can write A ∂H D A ∂H D E dp + B dH = B E dT C ∂p F T C ∂T F p (2.52) 63 64 2 THE FIRST LAW The second partial derivative is Cp; our task here is to express (∂H/∂p)T in terms of recognizable quantities. The chain relation (see Further information 2.2) lets us write A ∂H D 1 B E =− (∂p/∂T)H (∂T/∂H)p C ∂p F T and both partial derivatives can be brought into the numerator by using the reciprocal identity (see Further information 2.2) twice: A ∂H D (∂T/∂p)H A ∂T D A ∂H D B E =− =B E B E = −µCp ∂p (∂T/∂H) C FT p C ∂p F H C ∂T F p (2.53) We have used the definitions of the constant-pressure heat capacity, Cp, and the Joule–Thomson coefficient, µ (eqn 2.51). Equation 2.50 now follows directly. Fig. 2.27 The apparatus used for measuring the Joule–Thomson effect. The gas expands through the porous barrier, which acts as a throttle, and the whole apparatus is thermally insulated. As explained in the text, this arrangement corresponds to an isenthalpic expansion (expansion at constant enthalpy). Whether the expansion results in a heating or a cooling of the gas depends on the conditions. The analysis of the Joule–Thomson coefficient is central to the technological problems associated with the liquefaction of gases. We need to be able to interpret it physically and to measure it. As shown in the Justification below, the cunning required to impose the constraint of constant enthalpy, so that the process is isenthalpic, was supplied by Joule and William Thomson (later Lord Kelvin). They let a gas expand through a porous barrier from one constant pressure to another, and monitored the difference of temperature that arose from the expansion (Fig. 2.27). The whole apparatus was insulated so that the process was adiabatic. They observed a lower temperature on the low pressure side, the difference in temperature being proportional to the pressure difference they maintained. This cooling by isenthalpic expansion is now called the Joule–Thomson effect. Justification 2.3 The Joule–Thomson effect Here we show that the experimental arrangement results in expansion at constant enthalpy. Because all changes to the gas occur adiabatically, q = 0, which implies ∆U = w Consider the work done as the gas passes through the barrier. We focus on the passage of a fixed amount of gas from the high pressure side, where the pressure is pi, the temperature Ti, and the gas occupies a volume Vi (Fig. 2.28). The gas emerges on the low pressure side, where the same amount of gas has a pressure pf, a temperature Tf, and occupies a volume Vf. The gas on the left is compressed isothermally by the upstream gas acting as a piston. The relevant pressure is pi and the volume changes from Vi to 0; therefore, the work done on the gas is w1 = −pi(0 − Vi) = piVi The gas expands isothermally on the right of the barrier (but possibly at a different constant temperature) against the pressure pf provided by the downstream gas acting as a piston to be driven out. The volume changes from 0 to Vf , so the work done on the gas in this stage is w2 = −pf (Vf − 0) = −pfVf The total work done on the gas is the sum of these two quantities, or w = w1 + w2 = piVi − pfVf 2.12 THE JOULE–THOMSON EFFECT 65 It follows that the change of internal energy of the gas as it moves adiabatically from one side of the barrier to the other is Uf − Ui = w = piVi − pfVf Reorganization of this expression gives Uf + pfVf = Ui + piVi, or Hf = Hi Therefore, the expansion occurs without change of enthalpy. The property measured in the experiment is the ratio of the temperature change to the change of pressure, ∆T/∆p. Adding the constraint of constant enthalpy and taking the limit of small ∆p implies that the thermodynamic quantity measured is (∂T/∂p)H, which is the Joule–Thomson coefficient, µ. In other words, the physical interpretation of µ is that it is the ratio of the change in temperature to the change in pressure when a gas expands under conditions that ensure there is no change in enthalpy. The modern method of measuring µ is indirect, and involves measuring the isothermal Joule–Thomson coefficient, the quantity µT = A ∂H D C ∂p F T [2.54] which is the slope of a plot of enthalpy against pressure at constant temperature (Fig. 2.29). Comparing eqns 2.53 and 2.54, we see that the two coefficients are related by: µT = −Cp µ (2.55) To measure µ T , the gas is pumped continuously at a steady pressure through a heat exchanger (which brings it to the required temperature), and then through a porous plug inside a thermally insulated container. The steep pressure drop is measured, and the cooling effect is exactly offset by an electric heater placed immediately after the plug (Fig. 2.30). The energy provided by the heater is monitored. Because the energy transferred as heat can be identified with the value of ∆H for the gas (because Fig. 2.29 The isothermal Joule–Thomson coefficient is the slope of the enthalpy with respect to changing pressure, the temperature being held constant. Fig. 2.30 A schematic diagram of the apparatus used for measuring the isothermal Joule–Thomson coefficient. The electrical heating required to offset the cooling arising from expansion is interpreted as ∆H and used to calculate (∂H/∂p)T, which is then converted to µ as explained in the text. Fig. 2.28 The thermodynamic basis of Joule–Thomson expansion. The pistons represent the upstream and downstream gases, which maintain constant pressures either side of the throttle. The transition from the top diagram to the bottom diagram, which represents the passage of a given amount of gas through the throttle, occurs without change of enthalpy. 66 2 THE FIRST LAW Synoptic Table 2.9* Inversion temperatures (TI), normal freezing (Tf) and boiling (Tb) points, and Joule–Thomson coefficient (µ) at 1 atm and 298 K Ar CO2 TI/K Tf /K 723 83.8 Tb/K µ /(K bar−1) 87.3 +1.10 1500 194.7 He 40 4.2 −0.060 N2 621 77.4 +0.25 63.3 * More values are given in the Data section. Fig. 2.31 The sign of the Joule–Thomson coefficient, µ, depends on the conditions. Inside the boundary, the shaded area, it is positive and outside it is negative. The temperature corresponding to the boundary at a given pressure is the ‘inversion temperature’ of the gas at that pressure. For a given pressure, the temperature must be below a certain value if cooling is required but, if it becomes too low, the boundary is crossed again and heating occurs. Reduction of pressure under adiabatic conditions moves the system along one of the isenthalps, or curves of constant enthalpy. The inversion temperature curve runs through the points of the isenthalps where their slope changes from negative to positive. ∆H = qp), and the pressure change ∆p is known, we can find µT from the limiting value of ∆H/∆p as ∆p → 0, and then convert it to µ. Table 2.9 lists some values obtained in this way. Real gases have nonzero Joule–Thomson coefficients. Depending on the identity of the gas, the pressure, the relative magnitudes of the attractive and repulsive intermolecular forces (see Molecular interpretation 2.1), and the temperature, the sign of the coefficient may be either positive or negative (Fig. 2.31). A positive sign implies that dT is negative when dp is negative, in which case the gas cools on expansion. Gases that show a heating effect (µ < 0) at one temperature show a cooling effect (µ > 0) when the temperature is below their upper inversion temperature, TI (Table 2.9, Fig. 2.32). As indicated in Fig. 2.32, a gas typically has two inversion temperatures, one at high temperature and the other at low. The ‘Linde refrigerator’ makes use of Joule–Thompson expansion to liquefy gases (Fig. 2.33). The gas at high pressure is allowed to expand through a throttle; it cools and is circulated past the incoming gas. That gas is cooled, and its subsequent expansion cools it still further. There comes a stage when the circulating gas becomes so cold that it condenses to a liquid. For a perfect gas, µ = 0; hence, the temperature of a perfect gas is unchanged by Joule–Thomson expansion.8 This characteristic points clearly to the involvement of intermolecular forces in determining the size of the effect. However, the Joule– Thomson coefficient of a real gas does not necessarily approach zero as the pressure is reduced even though the equation of state of the gas approaches that of a perfect gas. The coefficient behaves like the properties discussed in Section 1.3b in the sense that it depends on derivatives and not on p, V, and T themselves. Fig. 2.32 The inversion temperatures for three real gases, nitrogen, hydrogen, and helium. 8 Fig. 2.33 The principle of the Linde refrigerator is shown in this diagram. The gas is recirculated, and so long as it is beneath its inversion temperature it cools on expansion through the throttle. The cooled gas cools the high-pressure gas, which cools still further as it expands. Eventually liquefied gas drips from the throttle. Simple adiabatic expansion does cool a perfect gas, because the gas does work; recall Section 2.6. CHECKLIST OF KEY IDEAS 67 Molecular interpretation 2.3 Molecular interactions and the Joule–Thomson effect The kinetic model of gases (Molecular interpretation 1.1) and the equipartition theorem (Molecular interpretation 2.2) imply that the mean kinetic energy of molecules in a gas is proportional to the temperature. It follows that reducing the average speed of the molecules is equivalent to cooling the gas. If the speed of the molecules can be reduced to the point that neighbours can capture each other by their intermolecular attractions, then the cooled gas will condense to a liquid. To slow the gas molecules, we make use of an effect similar to that seen when a ball is thrown into the air: as it rises it slows in response to the gravitational attraction of the Earth and its kinetic energy is converted into potential energy. We saw in Section 1.3 that molecules in a real gas attract each other (the attraction is not gravitational, but the effect is the same). It follows that, if we can cause the molecules to move apart from each other, like a ball rising from a planet, then they should slow. It is very easy to move molecules apart from each other: we simply allow the gas to expand, which increases the average separation of the molecules. To cool a gas, therefore, we allow it to expand without allowing any energy to enter from outside as heat. As the gas expands, the molecules move apart to fill the available volume, struggling as they do so against the attraction of their neighbours. Because some kinetic energy must be converted into potential energy to reach greater separations, the molecules travel more slowly as their separation increases. This sequence of molecular events explains the Joule–Thomson effect: the cooling of a real gas by adiabatic expansion. The cooling effect, which corresponds to µ > 0, is observed under conditions when attractive interactions are dominant (Z < 1, eqn 1.17), because the molecules have to climb apart against the attractive force in order for them to travel more slowly. For molecules under conditions when repulsions are dominant (Z > 1), the Joule–Thomson effect results in the gas becoming warmer, or µ < 0. Checklist of key ideas 1. Thermodynamics is the study of the transformations of energy. 2. The system is the part of the world in which we have a special interest. The surroundings is the region outside the system where we make our measurements. 3. An open system has a boundary through which matter can be transferred. A closed system has a boundary through which matter cannot be transferred. An isolated system has a boundary through which neither matter nor energy can be transferred. 4. Energy is the capacity to do work. The internal energy is the total energy of a system. 5. Work is the transfer of energy by motion against an opposing force, dw = −Fdz . Heat is the transfer of energy as a result of a temperature difference between the system and the surroundings. 6. An exothermic process releases energy as heat to the surroundings. An endothermic process absorbs energy as heat from the surroundings. 7. A state function is a property that depends only on the current state of the system and is independent of how that state has been prepared. 8. The First Law of thermodynamics states that the internal energy of an isolated system is constant, ∆U = q + w. 9. Expansion work is the work of expansion (or compression) of a system, dw = −pexdV. The work of free expansion is w = 0. The work of expansion against a constant external pressure is w = −pex ∆V. The work of isothermal reversible expansion of a perfect gas is w = −nRT ln(Vf /Vi). 10. A reversible change is a change that can be reversed by an infinitesimal modification of a variable. 11. Maximum work is achieved in a reversible change. 68 2 THE FIRST LAW 12. Calorimetry is the study of heat transfers during physical and chemical processes. 13. The heat capacity at constant volume is defined as CV = (∂U/∂T)V. The heat capacity at constant pressure is Cp = (∂H/∂T)p. For a perfect gas, the heat capacities are related by Cp − CV = nR. 14. The enthalpy is defined as H = U + pV. The enthalpy change is the energy transferred as heat at constant pressure, ∆H = qp. 15. During a reversible adiabatic change, the temperature of a perfect gas varies according to Tf = Ti(Vi/Vf)1/c, c = CV,m/R. The pressure and volume are related by pV γ = constant, with γ = Cp,m/CV,m. 16. The standard enthalpy change is the change in enthalpy for a process in which the initial and final substances are in their standard states. The standard state is the pure substance at 1 bar. 17. Enthalpy changes are additive, as in ∆subH 7 = ∆fus H 7 + ∆ vap H 7. 18. The enthalpy change for a process and its reverse are related by ∆forwardH 7 = −∆reverseH 7. 19. The standard enthalpy of combustion is the standard reaction enthalpy for the complete oxidation of an organic compound to CO2 gas and liquid H2O if the compound contains C, H, and O, and to N2 gas if N is also present. 20. Hess’s law states that the standard enthalpy of an overall reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided. 21. The standard enthalpy of formation (∆ f H 7) is the standard reaction enthalpy for the formation of the compound from its elements in their reference states. The reference state is the most stable state of an element at the specified temperature and 1 bar. 22. The standard reaction enthalpy may be estimated by combining enthalpies of formation, ∆ r H 7 = ∑Productsν∆ f H 7 − ∑Reactantsν∆f H 7. 23. The temperature dependence of the reaction enthalpy is given by Kirchhoff’s law, ∆r H 7(T2) = ∆r H 7(T1) + # ∆ C dT. T2 r 7 p T1 24. An exact differential is an infinitesimal quantity that, when integrated, gives a result that is independent of the path between the initial and final states. An inexact differential is an infinitesimal quantity that, when integrated, gives a result that depends on the path between the initial and final states. 25. The internal pressure is defined as πT = (∂U/∂V)T . For a perfect gas, πT = 0. 26. The Joule–Thomson effect is the cooling of a gas by isenthalpic expansion. 27. The Joule–Thomson coefficient is defined as µ = (∂T/∂p)H. The isothermal Joule–Thomson coefficient is defined as µT = (∂H/∂p)T = −Cp µ. 28. The inversion temperature is the temperature at which the Joule–Thomson coefficient changes sign. Further reading Articles and texts Sources of data and information P.W. Atkins and J.C. de Paula, Physical chemistry for the life sciences. W.H. Freeman, New York (2005). M.W. Chase, Jr. (ed.), NIST–JANAF thermochemical tables. Published as J. Phys. Chem. Ref. Data, Monograph no. 9. American Institute of Physics, New York (1998). G.A. Estèvez, K. Yang, and B.B. Dasgupta, Thermodynamic partial derivatives and experimentally measurable quantities. J. Chem. Educ. 66, 890 (1989). I.M. Klotz and R.M. Rosenberg, Chemical thermodynamics: basic theory and methods. Wiley–Interscience, New York (2000). G.N. Lewis and M. Randall, Thermodynamics. Revised by K.S. Pitzer and L. Brewer. McGraw–Hill, New York (1961). J. Wisniak, The Joule–Thomson coefficient for pure gases and their mixtures. J. Chem. Educ. 4, 51 (1999). J.D. Cox, D.D. Wagman, and V.A. Medvedev, CODATA key values for thermodynamics. Hemisphere Publishing Corp., New York (1989). D.B. Wagman, W.H. Evans, V.B. Parker, R.H. Schumm, I. Halow, S.M. Bailey, K.L. Churney, and R.L. Nuttall, The NBS tables of chemical thermodynamic properties. Published as J. Phys. Chem. Ref. Data 11, Supplement 2 (1982). R.C. Weast (ed.), Handbook of chemistry and physics, Vol. 81. CRC Press, Boca Raton (2000). M. Zabransky, V. Ruzicka Jr., V. Majer, and E. S. Domalski. Heat capacity of liquids. Published as J. Phys. Chem. Ref. Data, Monograph no. 6. American Institute of Physics, New York (1996). 69 FURTHER INFORMATION Further information Further information 2.1 Adiabatic processes Consider a stage in a reversible adiabatic expansion when the pressure inside and out is p. The work done when the gas expands by dV is dw = −pdV; however, for a perfect gas, dU = CV dT. Therefore, because for an adiabatic change (dq = 0) dU = dw + dq = dw, we can equate these two expressions for dU and write CV dT = −pdV We are dealing with a perfect gas, so we can replace p by nRT/V and obtain CV dT =− T nRdV V To integrate this expression we note that T is equal to Ti when V is equal to Vi, and is equal to Tf when V is equal to Vf at the end of the expansion. Therefore, # Tf # dT CV = −nR T Ti Vf dV Vi V (We are taking CV to be independent of temperature.) Then, because ∫dx/x = ln x + constant, we obtain CV ln Tf Ti = −nR ln Vf Vi Because ln(x /y) = −ln(y/x), this expression rearranges to CV nR ln Tf Ti = ln Vi Vf With c = CV /nR we obtain (because ln x a = a ln x) A Tf D A Vi D ln B E = ln B E C Ti F C Vf F c which implies that (Tf /Ti)c = (Vi /Vf) and, upon rearrangement, eqn 2.28. The initial and final states of a perfect gas satisfy the perfect gas law regardless of how the change of state takes place, so we can use pV = nRT to write piVi pfVf = Ti Tf However, we have just shown that A Vf D =B E Tf C Vi F Ti 1/c A Vf D =B E C Vi F γ −1 where we use the definition of the heat capacity ratio where γ = Cp,m/CV,m and the fact that, for a perfect gas, Cp,m – CV,m = R (the molar version of eqn 2.26). Then we combine the two expressions, to obtain A Vf D = ×B E pf Vi C Vi F pi Vf γ −1 A Vf D E =B C Vi F γ which rearranges to piV γi = pfV γf , which is eqn 2.29. Further information 2.2 The relation between heat capacities A useful rule when doing a problem in thermodynamics is to go back to first principles. In the present problem we do this twice, first by expressing Cp and CV in terms of their definitions and then by inserting the definition H = U + pV: A ∂H D A ∂U D E −B E Cp − CV = B C ∂T F p C ∂T F V A ∂U D A ∂(pV) D A ∂U D E E +B E −B =B C ∂T F p C ∂T F p C ∂T F V We have already calculated the difference of the first and third terms on the right, and eqn 2.45 lets us write this difference as απTV. The factor αV gives the change in volume when the temperature is raised, and πT = (∂U/∂V)T converts this change in volume into a change in internal energy. We can simplify the remaining term by noting that, because p is constant, A ∂(pV) D A ∂V D B E = p B E = αpV C ∂T F p C ∂T F p The middle term of this expression identifies it as the contribution to the work of pushing back the atmosphere: (∂V/∂T)p is the change of volume caused by a change of temperature, and multiplication by p converts this expansion into work. Collecting the two contributions gives Cp − CV = α(p + πT)V (2.56) As just remarked, the first term on the right, α pV, is a measure of the work needed to push back the atmosphere; the second term on the right, απTV, is the work required to separate the molecules composing the system. At this point we can go further by using the result we prove in Section 3.8 that A ∂p D πT = T B E − p C ∂T F V When this expression is inserted in the last equation we obtain A ∂p D Cp − CV = αTV B E C ∂T F V (2.57) We now transform the remaining partial derivative. It follows from Euler’s chain relation that A ∂p D A ∂T D A ∂V D B E B E B E = −1 C ∂T F V C ∂V F p C ∂p F T Comment 2.8 The Euler chain relation states that, for a differentiable function z = z(x,y), A ∂y D A ∂x D A ∂z D B E B E B E = −1 C ∂x F z C ∂z F y C ∂y F x For instance, if z(x,y) = x2y, 70 2 THE FIRST LAW A ∂y D A ∂(z/x 2)D d(1/x 2) 2z =− 3 B E =B E =z dx x C ∂x F z C ∂x F z Comment 2.9 A ∂z D A ∂(x 2y) D dy B E =B E = x2 = x2 dy C ∂y F x C ∂y F x For example, for the function z(x,y) = x 2y, A ∂x D A ∂(z/y)1/2 D 1 dz1/2 1 = E = 1/2 B E =B dz 2(yz)1/2 C ∂z F y C ∂z F y y Multiplication of the three terms together gives the result −1. and therefore that A ∂p D 1 B E =− C ∂T F V (∂T/∂V)p(∂V/∂p)T Unfortunately, (∂T/∂V)p occurs instead of (∂V/∂T)p. However, the ‘reciprocal identity’ allows us to invert partial derivatives and to write A ∂p D (∂V/∂T)p α B E =− = C ∂T F V (∂V/∂p)T κT The reciprocal identity states that A ∂y D 1 B E = C ∂x F z (∂x/∂y)z A ∂y D A ∂(z/x 2) D d(1/x 2) 2z B E =B E =z =− 3 dx x C ∂x F z C ∂x F z We can also write x = (z/y)1/2, in which case A ∂x D A ∂(z/y)1/2 D d(1/y1/2) B E =B E = z1/2 dy C ∂y F z C ∂y F z =− z1/2 2y 3/2 =− z1/2 2 3/2 2(z/x ) =− x3 2z which is the reciprocal of the coefficient derived above. Insertion of this relation into eqn 2.57 produces eqn 2.49. Discussion questions 2.1 Provide mechanical and molecular definitions of work and heat. 2.2 Consider the reversible expansion of a perfect gas. Provide a physical γ 2.5 Explain the significance of the Joule and Joule–Thomson experiments. What would Joule observe in a more sensitive apparatus? interpretation for the fact that pV = constant for an adiabatic change, whereas pV = constant for an isothermal change. 2.6 Suggest (with explanation) how the internal energy of a van der Waals gas 2.3 Explain the difference between the change in internal energy and the 2.7 In many experimental thermograms, such as that shown in Fig. 2.16, the change in enthalpy accompanying a chemical or physical process. baseline below T1 is at a different level from that above T2. Explain this observation. 2.4 Explain the significance of a physical observable being a state function and should vary with volume at constant temperature. compile a list of as many state functions as you can identify. Exercises Assume all gases are perfect unless stated otherwise. Unless otherwise stated, thermochemical data are for 298.15 K. 2.1(a) Calculate the work needed for a 65 kg person to climb through 4.0 m on the surface of (a) the Earth and (b) the Moon (g = 1.60 m s−2). 2.1(b) Calculate the work needed for a bird of mass 120 g to fly to a height of 50 m from the surface of the Earth. 2.2(a) A chemical reaction takes place in a container of cross-sectional area 100 cm2. As a result of the reaction, a piston is pushed out through 10 cm against an external pressure of 1.0 atm. Calculate the work done by the system. 2.2(b) A chemical reaction takes place in a container of cross-sectional area 50.0 cm2. As a result of the reaction, a piston is pushed out through 15 cm against an external pressure of 121 kPa. Calculate the work done by the system. 2.3(a) A sample consisting of 1.00 mol Ar is expanded isothermally at 0°C from 22.4 dm3 to 44.8 dm3 (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely (against zero external pressure). For the three processes calculate q, w, ∆U, and ∆H. 2.3(b) A sample consisting of 2.00 mol He is expanded isothermally at 22°C from 22.8 dm3 to 31.7 dm3 (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely (against zero external pressure). For the three processes calculate q, w, ∆U, and ∆H. 2.4(a) A sample consisting of 1.00 mol of perfect gas atoms, for which CV,m = –32 R, initially at p1 = 1.00 atm and T1 = 300 K, is heated reversibly to 400 K at constant volume. Calculate the final pressure, ∆U, q, and w. 2.4(b) A sample consisting of 2.00 mol of perfect gas molecules, for which CV,m = –52 R, initially at p1 = 111 kPa and T1 = 277 K, is heated reversibly to 356 K at constant volume. Calculate the final pressure, ∆U, q, and w. 2.5(a) A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against a constant external pressure of 200 Torr until its volume has increased by 3.3 dm3. (b) Calculate the work that would be done if the same expansion occurred reversibly. 2.5(b) A sample of argon of mass 6.56 g occupies 18.5 dm3 at 305 K. (a) Calculate the work done when the gas expands isothermally against a EXERCISES constant external pressure of 7.7 kPa until its volume has increased by 2.5 dm3. (b) Calculate the work that would be done if the same expansion occurred reversibly. 2.6(a) A sample of 1.00 mol H2O(g) is condensed isothermally and reversibly to liquid water at 100°C. The standard enthalpy of vaporization of water at 100°C is 40.656 kJ mol−1. Find w, q, ∆U, and ∆H for this process. 2.6(b) A sample of 2.00 mol CH3OH(g) is condensed isothermally and reversibly to liquid at 64°C. The standard enthalpy of vaporization of methanol at 64°C is 35.3 kJ mol−1. Find w, q, ∆U, and ∆H for this process. 2.7(a) A strip of magnesium of mass 15 g is dropped into a beaker of dilute hydrochloric acid. Calculate the work done by the system as a result of the reaction. The atmospheric pressure is 1.0 atm and the temperature 25°C. 71 2.14(b) A sample of 5.0 mol CO2 is originally confined in 15 dm3 at 280 K and then undergoes adiabatic expansion against a constant pressure of 78.5 kPa until the volume has increased by a factor of 4.0. Calculate q, w, ∆T, ∆U, and ∆H. (The final pressure of the gas is not necessarily 78.5 kPa.) 2.15(a) A sample consisting of 1.0 mol of perfect gas molecules with CV = 20.8 J K−1 is initially at 3.25 atm and 310 K. It undergoes reversible adiabatic expansion until its pressure reaches 2.50 atm. Calculate the final volume and temperature and the work done. 2.15(b) A sample consisting of 1.5 mol of perfect gas molecules with Cp,m = 20.8 J K−1 mol−1 is initially at 230 kPa and 315 K. It undergoes reversible adiabatic expansion until its pressure reaches 170 kPa. Calculate the final volume and temperature and the work done. 2.7(b) A piece of zinc of mass 5.0 g is dropped into a beaker of dilute 2.16(a) A certain liquid has ∆ vapH 7 = 26.0 kJ mol−1. Calculate q, w, ∆H, and hydrochloric acid. Calculate the work done by the system as a result of the reaction. The atmospheric pressure is 1.1 atm and the temperature 23°C. 2.16(b) A certain liquid has ∆ vapH 7 = 32.0 kJ mol−1. Calculate q, w, ∆H, and 2.8(a) The constant-pressure heat capacity of a sample of a perfect gas was −1 ∆U when 0.50 mol is vaporized at 250 K and 750 Torr. ∆U when 0.75 mol is vaporized at 260 K and 765 Torr. found to vary with temperature according to the expression Cp /(J K ) = 20.17 + 0.3665(T/K). Calculate q, w, ∆U, and ∆H when the temperature is raised from 25°C to 200°C (a) at constant pressure, (b) at constant volume. Calculate its standard enthalpy of combustion. 2.8(b) The constant-pressure heat capacity of a sample of a perfect gas was Calculate its standard enthalpy of combustion. found to vary with temperature according to the expression Cp /(J K−1) = 20.17 + 0.4001(T/K). Calculate q, w, ∆U, and ∆H when the temperature is raised from 0°C to 100°C (a) at constant pressure, (b) at constant volume. 2.9(a) Calculate the final temperature of a sample of argon of mass 12.0 g that is expanded reversibly and adiabatically from 1.0 dm3 at 273.15 K to 3.0 dm3. 2.9(b) Calculate the final temperature of a sample of carbon dioxide of mass 16.0 g that is expanded reversibly and adiabatically from 500 cm3 at 298.15 K to 2.00 dm3. 2.10(a) A sample of carbon dioxide of mass 2.45 g at 27.0°C is allowed to expand reversibly and adiabatically from 500 cm3 to 3.00 dm3. What is the work done by the gas? 2.10(b) A sample of nitrogen of mass 3.12 g at 23.0°C is allowed to expand reversibly and adiabatically from 400 cm3 to 2.00 dm3. What is the work done by the gas? 2.11(a) Calculate the final pressure of a sample of carbon dioxide that expands reversibly and adiabatically from 57.4 kPa and 1.0 dm3 to a final volume of 2.0 dm3. Take γ = 1.4. 2.11(b) Calculate the final pressure of a sample of water vapour that expands reversibly and adiabatically from 87.3 Torr and 500 cm3 to a final volume of 3.0 dm3. Take γ = 1.3. 2.12(a) When 229 J of energy is supplied as heat to 3.0 mol Ar(g), the temperature of the sample increases by 2.55 K. Calculate the molar heat capacities at constant volume and constant pressure of the gas. 2.12(b) When 178 J of energy is supplied as heat to 1.9 mol of gas molecules, the temperature of the sample increases by 1.78 K. Calculate the molar heat capacities at constant volume and constant pressure of the gas. 2.13(a) When 3.0 mol O2 is heated at a constant pressure of 3.25 atm, its temperature increases from 260 K to 285 K. Given that the molar heat capacity of O2 at constant pressure is 29.4 J K−1 mol−1, calculate q, ∆H, and ∆U. 2.13(b) When 2.0 mol CO2 is heated at a constant pressure of 1.25 atm, its temperature increases from 250 K to 277 K. Given that the molar heat capacity of CO2 at constant pressure is 37.11 J K−1 mol−1, calculate q, ∆H, and ∆U. 2.14(a) A sample of 4.0 mol O2 is originally confined in 20 dm3 at 270 K and then undergoes adiabatic expansion against a constant pressure of 600 Torr until the volume has increased by a factor of 3.0. Calculate q, w, ∆T, ∆U, and ∆H. (The final pressure of the gas is not necessarily 600 Torr.) 2.17(a) The standard enthalpy of formation of ethylbenzene is −12.5 kJ mol−1. 2.17(b) The standard enthalpy of formation of phenol is −165.0 kJ mol−1. 2.18(a) The standard enthalpy of combustion of cyclopropane is −2091 kJ mol−1 at 25°C. From this information and enthalpy of formation data for CO2(g) and H2O(g), calculate the enthalpy of formation of cyclopropane. The enthalpy of formation of propene is +20.42 kJ mol−1. Calculate the enthalpy of isomerization of cyclopropane to propene. 2.18(b) From the following data, determine ∆f H 7 for diborane, B2H6(g), at 298 K: (1) B2H6(g) + 3 O2(g) → B2O3(s) + 3 H2O(g) (2) 2 B(s) + –32 O2(g) → B2O3(s) (3) H2(g) + –12 O2(g) → H2O(g) ∆r H 7 = −1941 kJ mol−1 ∆r H 7 = −2368 kJ mol−1 ∆r H 7 = −241.8 kJ mol−1 2.19(a) When 120 mg of naphthalene, C10H8(s), was burned in a bomb calorimeter the temperature rose by 3.05 K. Calculate the calorimeter constant. By how much will the temperature rise when 10 mg of phenol, C6H5OH(s), is burned in the calorimeter under the same conditions? 2.19(b) When 2.25 mg of anthracene, C14H10(s), was burned in a bomb calorimeter the temperature rose by 1.35 K. Calculate the calorimeter constant. By how much will the temperature rise when 135 mg of phenol, C6H5OH(s), is burned in the calorimeter under the same conditions? (∆cH 7(C14H10, s) = −7061 kJ mol−1.) 2.20(a) Calculate the standard enthalpy of solution of AgCl(s) in water from the enthalpies of formation of the solid and the aqueous ions. 2.20(b) Calculate the standard enthalpy of solution of AgBr(s) in water from the enthalpies of formation of the solid and the aqueous ions. 2.21(a) The standard enthalpy of decomposition of the yellow complex H3NSO2 into NH3 and SO2 is +40 kJ mol−1. Calculate the standard enthalpy of formation of H3NSO2. 2.21(b) Given that the standard enthalpy of combustion of graphite is −393.51 kJ mol−1 and that of diamond is −395.41 kJ mol−1, calculate the enthalpy of the graphite-to-diamond transition. 2.22(a) Given the reactions (1) and (2) below, determine (a) ∆r H 7 and ∆rU 7 for reaction (3), (b) ∆ f H 7 for both HCl(g) and H2O(g) all at 298 K. (1) H2(g) + Cl2(g) → 2 HCl(g) (2) 2 H2(g) + O2(g) → 2 H2O(g) (3) 4 HCl(g) + O2(g) → Cl2(g) + 2 H2O(g) ∆rH 7 = −184.62 kJ mol−1 ∆rH 7 = −483.64 kJ mol−1 72 2 THE FIRST LAW 2.22(b) Given the reactions (1) and (2) below, determine (a) ∆r H 7 and ∆rU 7 for reaction (3), (b) ∆f H 7 for both HCl(g) and H2O(g) all at 298 K. (1) H2(g) + I2(s) → 2 HI(g) (2) 2 H2(g) + O2(g) → 2 H2O(g) (3) 4 HI(g) + O2(g) → 2 I2(s) + 2 H2O(g) ∆r H 7 = +52.96 kJ mol−1 ∆r H 7 = −483.64 kJ mol−1 2.23(a) For the reaction C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g), ∆rU 7 = −1373 kJ mol−1 at 298 K. Calculate ∆r H 7. 2.23(b) For the reaction 2 C6H5COOH(s) + 13 O2(g) → 12 CO2(g) + 7 −1 7 6 H2O(g), ∆rU = −772.7 kJ mol at 298 K. Calculate ∆r H . 2.24(a) Calculate the standard enthalpies of formation of (a) KClO3(s) from the enthalpy of formation of KCl, (b) NaHCO3(s) from the enthalpies of formation of CO2 and NaOH together with the following information: 2 KClO3(s) → 2 KCl(s) + 3 O2(g) NaOH(s) + CO2(g) → NaHCO3(s) ∆r H 7 = −89.4 kJ mol−1 ∆r H 7 = −127.5 kJ mol−1 2.24(b) Calculate the standard enthalpy of formation of NOCl(g) from the enthalpy of formation of NO given in Table 2.5, together with the following information: 2 NOCl(g) → 2 NO(g) + Cl2(g) ∆r H 7 = +75.5 kJ mol−1 2.25(a) Use the information in Table 2.5 to predict the standard reaction enthalpy of 2 NO2(g) → N2O4(g) at 100°C from its value at 25°C. 2.25(b) Use the information in Table 2.5 to predict the standard reaction enthalpy of 2 H2(g) + O2(g) → 2 H2O(l) at 100°C from its value at 25°C. 2.26(a) From the data in Table 2.5, calculate ∆rH 7 and ∆rU 7 at (a) 298 K, (b) 378 K for the reaction C(graphite) + H2O(g) → CO(g) + H2(g). Assume all heat capacities to be constant over the temperature range of interest. 2.26(b) Calculate ∆r H 7 and ∆rU 7 at 298 K and ∆r H 7 at 348 K for the hydrogenation of ethyne (acetylene) to ethene (ethylene) from the enthalpy of combustion and heat capacity data in Tables 2.5 and 2.7. Assume the heat capacities to be constant over the temperature range involved. 2.27(a) Calculate ∆rH 7 for the reaction Zn(s) + CuSO4(aq) → ZnSO4(aq) + 2.28(b) Set up a thermodynamic cycle for determining the enthalpy of hydration of Ca2+ ions using the following data: enthalpy of sublimation of Ca(s), +178.2 kJ mol−1; first and second ionization enthalpies of Ca(g), 589.7 kJ mol−1 and 1145 kJ mol−1; enthalpy of vaporization of bromine, +30.91 kJ mol−1; dissociation enthalpy of Br2(g), +192.9 kJ mol−1; electron gain enthalpy of Br(g), −331.0 kJ mol−1; enthalpy of solution of CaBr2(s), −103.1 kJ mol−1; enthalpy of hydration of Br−(g), −337 kJ mol−1. 2.29(a) When a certain freon used in refrigeration was expanded adiabatically from an initial pressure of 32 atm and 0°C to a final pressure of 1.00 atm, the temperature fell by 22 K. Calculate the Joule–Thomson coefficient, µ, at 0°C, assuming it remains constant over this temperature range. 2.29(b) A vapour at 22 atm and 5°C was allowed to expand adiabatically to a final pressure of 1.00 atm; the temperature fell by 10 K. Calculate the Joule–Thomson coefficient, µ, at 5°C, assuming it remains constant over this temperature range. 2.30(a) For a van der Waals gas, πT = a/V 2m. Calculate ∆Um for the isothermal expansion of nitrogen gas from an initial volume of 1.00 dm3 to 24.8 dm3 at 298 K. What are the values of q and w? 2.30(b) Repeat Exercise 2.30(a) for argon, from an initial volume of 1.00 dm3 to 22.1 dm3 at 298 K. 2.31(a) The volume of a certain liquid varies with temperature as V = V′{0.75 + 3.9 × 10−4(T/K) + 1.48 × 10−6(T/K)2} where V′ is its volume at 300 K. Calculate its expansion coefficient, α, at 320 K. 2.31(b) The volume of a certain liquid varies with temperature as V = V′{0.77 + 3.7 × 10−4(T/K) + 1.52 × 10−6(T/K)2} where V′ is its volume at 298 K. Calculate its expansion coefficient, α, at 310 K. 2.32(a) The isothermal compressibility of copper at 293 K is 7.35 × 10−7 atm−1. Calculate the pressure that must be applied in order to increase its density by 0.08 per cent. 2.32(b) The isothermal compressibility of lead at 293 K is 2.21 × 10−6 atm−1. Cu(s) from the information in Table 2.7 in the Data section. Calculate the pressure that must be applied in order to increase its density by 0.08 per cent. 2.27(b) Calculate ∆r H 7 for the reaction NaCl(aq) + AgNO3(aq) → AgCl(s) + 2.33(a) Given that µ = 0.25 K atm−1 for nitrogen, calculate the value of its NaNO3(aq) from the information in Table 2.7 in the Data section. 2.28(a) Set up a thermodynamic cycle for determining the enthalpy of hydration of Mg2+ ions using the following data: enthalpy of sublimation of Mg(s), +167.2 kJ mol−1; first and second ionization enthalpies of Mg(g), 7.646 eV and 15.035 eV; dissociation enthalpy of Cl2(g), +241.6 kJ mol−1; electron gain enthalpy of Cl(g), −3.78 eV; enthalpy of solution of MgCl2(s), −150.5 kJ mol−1; enthalpy of hydration of Cl−(g), −383.7 kJ mol−1. isothermal Joule–Thomson coefficient. Calculate the energy that must be supplied as heat to maintain constant temperature when 15.0 mol N2 flows through a throttle in an isothermal Joule–Thomson experiment and the pressure drop is 75 atm. 2.33(b) Given that µ = 1.11 K atm−1 for carbon dioxide, calculate the value of its isothermal Joule–Thomson coefficient. Calculate the energy that must be supplied as heat to maintain constant temperature when 12.0 mol CO2 flows through a throttle in an isothermal Joule–Thomson experiment and the pressure drop is 55 atm. PROBLEMS 73 Problems* Assume all gases are perfect unless stated otherwise. Note that 1 atm = 1.013 25 bar. Unless otherwise stated, thermochemical data are for 298.15 K. Numerical problems 2.1 A sample consisting of 1 mol of perfect gas atoms (for which CV,m = –32 R) is taken through the cycle shown in Fig. 2.34. (a) Determine the temperature at the points 1, 2, and 3. (b) Calculate q, w, ∆U, and ∆H for each step and for the overall cycle. If a numerical answer cannot be obtained from the information given, then write in +, −, 0, or ? as appropriate. Table 2.2. Calculate the standard enthalpy of formation of ethane at 350 K from its value at 298 K. 2.8 A sample of the sugar d-ribose (C5H10O5) of mass 0.727 g was placed in a calorimeter and then ignited in the presence of excess oxygen. The temperature rose by 0.910 K. In a separate experiment in the same calorimeter, the combustion of 0.825 g of benzoic acid, for which the internal energy of combustion is −3251 kJ mol−1, gave a temperature rise of 1.940 K. Calculate the internal energy of combustion of d-ribose and its enthalpy of formation. 2.9 The standard enthalpy of formation of the metallocene bis(benzene)chromium was measured in a calorimeter. It was found for the reaction Cr(C6H6)2(s) → Cr(s) + 2 C6H6(g) that ∆rU 7(583 K) = +8.0 kJ mol−1. Find the corresponding reaction enthalpy and estimate the standard enthalpy of formation of the compound at 583 K. The constant-pressure molar heat capacity of benzene is 136.1 J K−1 mol−1 in its liquid range and 81.67 J K−1 mol−1 as a gas. 2.10‡ From the enthalpy of combustion data in Table 2.5 for the alkanes methane through octane, test the extent to which the relation ∆cH 7 = k{(M/(g mol−1)}n holds and find the numerical values for k and n. Predict ∆cH 7 for decane and compare to the known value. Fig. 2.34 2.2 A sample consisting of 1.0 mol CaCO3(s) was heated to 800°C, when it decomposed. The heating was carried out in a container fitted with a piston that was initially resting on the solid. Calculate the work done during complete decomposition at 1.0 atm. What work would be done if instead of having a piston the container was open to the atmosphere? 2.3 A sample consisting of 2.0 mol CO2 occupies a fixed volume of 15.0 dm3 at 300 K. When it is supplied with 2.35 kJ of energy as heat its temperature increases to 341 K. Assume that CO2 is described by the van der Waals equation of state, and calculate w, ∆U, and ∆H. 2.4 A sample of 70 mmol Kr(g) expands reversibly and isothermally at 373 K from 5.25 cm3 to 6.29 cm3, and the internal energy of the sample is known to increase by 83.5 J. Use the virial equation of state up to the second coefficient B = −28.7 cm3 mol−1 to calculate w, q, and ∆H for this change of state. 2.5 A sample of 1.00 mol perfect gas molecules with Cp,m = –27R is put through the following cycle: (a) constant-volume heating to twice its initial volume, (b) reversible, adiabatic expansion back to its initial temperature, (c) reversible isothermal compression back to 1.00 atm. Calculate q, w, ∆U, and ∆H for each step and overall. 2.6 Calculate the work done during the isothermal reversible expansion of a van der Waals gas. Account physically for the way in which the coefficients a and b appear in the final expression. Plot on the same graph the indicator diagrams for the isothermal reversible expansion of (a) a perfect gas, (b) a van der Waals gas in which a = 0 and b = 5.11 × 10−2 dm3 mol−1, and (c) a = 4.2 dm6 atm mol−2 and b = 0. The values selected exaggerate the imperfections but give rise to significant effects on the indicator diagrams. Take Vi = 1.0 dm3, n = 1.0 mol, and T = 298 K. 2.7 The molar heat capacity of ethane is represented in the temperature range 298 K to 400 K by the empirical expression Cp,m /(J K−1 mol−1) = 14.73 + 0.1272(T/K). The corresponding expressions for C(s) and H2(g) are given in 2.11 It is possible to investigate the thermochemical properties of hydrocarbons with molecular modelling methods. (a) Use electronic structure software to predict ∆cH 7 values for the alkanes methane through pentane. To calculate ∆cH 7 values, estimate the standard enthalpy of formation of CnH2(n+1)(g) by performing semi-empirical calculations (for example, AM1 or PM3 methods) and use experimental standard enthalpy of formation values for CO2(g) and H2O(l). (b) Compare your estimated values with the experimental values of ∆cH 7 (Table 2.5) and comment on the reliability of the molecular modelling method. (c) Test the extent to which the relation ∆cH 7 = k{(M/(g mol−1)}n holds and find the numerical values for k and n. 2.12‡ When 1.3584 g of sodium acetate trihydrate was mixed into 100.0 cm3 of 0.2000 m HCl(aq) at 25°C in a solution calorimeter, its temperature fell by 0.397°C on account of the reaction: H3O+(aq) + NaCH3CO2 · 3 H2O(s) → Na+(aq) + CH3COOH(aq) + 4 H2O(l). The heat capacity of the calorimeter is 91.0 J K−1 and the heat capacity density of the acid solution is 4.144 J K−1 cm−3. Determine the standard enthalpy of formation of the aqueous sodium cation. The standard enthalpy of formation of sodium acetate trihydrate is −1064 kJ mol−1. 2.13‡ Since their discovery in 1985, fullerenes have received the attention of many chemical researchers. Kolesov et al. reported the standard enthalpy of combustion and of formation of crystalline C60 based on calorimetric measurements (V.P. Kolesov, S.M. Pimenova, V.K. Pavlovich, N.B. Tamm, and A.A. Kurskaya, J. Chem. Thermodynamics 28, 1121 (1996)). In one of their runs, they found the standard specific internal energy of combustion to be −36.0334 kJ g−1 at 298.15 K Compute ∆c H 7 and ∆ f H 7 of C60. 2.14‡ A thermodynamic study of DyCl3 (E.H.P. Cordfunke, A.S. Booji, and M. Yu. Furkaliouk, J. Chem. Thermodynamics 28, 1387 (1996)) determined its standard enthalpy of formation from the following information (1) DyCl3(s) → DyCl3(aq, in 4.0 m HCl) (2) Dy(s) + 3 HCl(aq, 4.0 m) → DyCl3(aq, in 4.0 m HCl(aq)) + –32 H2(g) (3) –12 H2(g) + –12 Cl2(g) → HCl(aq, 4.0 m) Determine ∆ f H 7(DyCl3, s) from these data. * Problems denoted with the symbol ‡ were supplied by Charles Trapp, Carmen Giunta, and Marshall Cady. ∆r H 7 = −180.06 kJ mol−1 ∆r H 7 = −699.43 kJ mol−1 ∆r H 7 = −158.31 kJ mol−1 74 2 THE FIRST LAW 2.15‡ Silylene (SiH2) is a key intermediate in the thermal decomposition of silicon hydrides such as silane (SiH4) and disilane (Si2H6). Moffat et al. (H.K. Moffat, K.F. Jensen, and R.W. Carr, J. Phys. Chem. 95, 145 (1991)) report ∆f H 7(SiH2) = +274 kJ mol−1. If ∆ f H 7(SiH4) = +34.3 kJ mol−1 and ∆ f H 7(Si2H6) = +80.3 kJ mol−1 (CRC Handbook (2004)), compute the standard enthalpies of the following reactions: (a) SiH4(g) → SiH2(g) + H2(g) (b) Si2H6(g) → SiH2(g) + SiH4(g) 2.16‡ Silanone (SiH2O) and silanol (SiH3OH) are species believed to be important in the oxidation of silane (SiH4). These species are much more elusive than their carbon counterparts. C.L. Darling and H.B. Schlegel (J. Phys. Chem. 97, 8207 (1993)) report the following values (converted from calories) from a computational study: ∆f H 7(SiH2O) = −98.3 kJ mol−1 and ∆f H 7(SiH3OH) = −282 kJ mol−1 . Compute the standard enthalpies of the following reactions: (a) SiH4(g) + –12 O2(g) → SiH3OH(g) (b) SiH4(g) + O2(g) → SiH2O(g) + H2O(l) (c) SiH3OH(g) → SiH2O(g) + H2(g) Note that ∆ f H 7(SiH4, g) = +34.3 kJ mol−1 (CRC Handbook (2004)). 2.17 The constant-volume heat capacity of a gas can be measured by observing the decrease in temperature when it expands adiabatically and reversibly. If the decrease in pressure is also measured, we can use it to infer the value of γ = Cp /CV and hence, by combining the two values, deduce the constant-pressure heat capacity. A fluorocarbon gas was allowed to expand reversibly and adiabatically to twice its volume; as a result, the temperature fell from 298.15 K to 248.44 K and its pressure fell from 202.94 kPa to 81.840 kPa. Evaluate Cp. 2.18 A sample consisting of 1.00 mol of a van der Waals gas is compressed 3 3 from 20.0 dm to 10.0 dm at 300 K. In the process, 20.2 kJ of work is done on the gas. Given that µ = {(2a/RT) − b}/Cp,m, with Cp,m = 38.4 J K−1 mol−1, a = 3.60 dm6 atm mol−2, and b = 0.44 dm3 mol−1, calculate ∆H for the process. 2.19 Take nitrogen to be a van der Waals gas with a = 1.352 dm6 atm mol−2 and b = 0.0387 dm3 mol−1, and calculate ∆Hm when the pressure on the gas is decreased from 500 atm to 1.00 atm at 300 K. For a van der Waals gas, µ = {(2a/RT) − b}/Cp,m. Assume Cp,m = –27R. Theoretical problems 2.20 Show that the following functions have exact differentials: (a) x 2y + 3y 2, (b) x cos xy, (c) x 3y 2, (d) t(t + es) + s. 2.21 (a) What is the total differential of z = x 2 + 2y 2 − 2xy + 2x − 4y − 8? (b) Show that ∂2z/∂y∂x = ∂2z/∂x∂y for this function. (c) Let z = xy − y + ln x + 2. Find dz and show that it is exact. expressing (∂H/∂U)p as the ratio of two derivatives with respect to volume and then using the definition of enthalpy. 2.26 (a) Write expressions for dV and dp given that V is a function of p and T and p is a function of V and T. (b) Deduce expressions for d ln V and d ln p in terms of the expansion coefficient and the isothermal compressibility. 2.27 Calculate the work done during the isothermal reversible expansion of a gas that satisfies the virial equation of state, eqn 1.19. Evaluate (a) the work for 1.0 mol Ar at 273 K (for data, see Table 1.3) and (b) the same amount of a perfect gas. Let the expansion be from 500 cm3 to 1000 cm3 in each case. 2.28 Express the work of isothermal reversible expansion of a van der Waals gas in reduced variables and find a definition of reduced work that makes the overall expression independent of the identity of the gas. Calculate the work of isothermal reversible expansion along the critical isotherm from Vc to xVc. 2.29‡ A gas obeying the equation of state p(V − nb) = nRT is subjected to a Joule–Thomson expansion. Will the temperature increase, decrease, or remain the same? 2 2.30 Use the fact that (∂U/∂V)T = a/V m for a van der Waals gas to show that µCp,m ≈ (2a/RT) − b by using the definition of µ and appropriate relations between partial derivatives. (Hint. Use the approximation pVm ≈ RT when it is justifiable to do so.) 2.31 Rearrange the van der Waals equation of state to give an expression for T as a function of p and V (with n constant). Calculate (∂T/∂p)V and confirm that (∂T/∂p)V = 1/(∂p/∂T)V. Go on to confirm Euler’s chain relation. 2.32 Calculate the isothermal compressibility and the expansion coefficient of a van der Waals gas. Show, using Euler’s chain relation, that κT R = α(Vm − b). 2.33 Given that µCp = T(∂V/∂T)p − V, derive an expression for µ in terms of the van der Waals parameters a and b, and express it in terms of reduced variables. Evaluate µ at 25°C and 1.0 atm, when the molar volume of the gas is 24.6 dm3 mol−1. Use the expression obtained to derive a formula for the inversion temperature of a van der Waals gas in terms of reduced variables, and evaluate it for the xenon sample. 2.34 The thermodynamic equation of state (∂U/∂V)T = T(∂p/∂T)V − p was quoted in the chapter. Derive its partner A ∂H D A ∂V D B E = −T B E + V ∂p C FT C ∂T F p from it and the general relations between partial differentials. 2.35 Show that for a van der Waals gas, Cp,m − CV,m = λR 1 λ =1− (3Vr − 1)2 4V 3rTr and evaluate the difference for xenon at 25°C and 10.0 atm. 2.22 (a) Express (∂CV /∂V)T as a second derivative of U and find its relation to (∂U/∂V)T and (∂Cp /∂p)T as a second derivative of H and find its relation to (∂H/∂p)T. (b) From these relations show that (∂CV /∂V)T = 0 and (∂Cp /∂p)T = 0 for a perfect gas. heat capacities γ by cs = (γ RT/M)1/2. Show that cs = (γ p/ρ)1/2, where ρ is the mass density of the gas. Calculate the speed of sound in argon at 25°C. 2.23 (a) Derive the relation CV = −(∂U/∂V)T (∂V/∂T)U from the expression 2.37‡ A gas obeys the equation of state Vm = RT/p + aT 2 and its constant- for the total differential of U(T,V) and (b) starting from the expression for the total differential of H(T,p), express (∂H/∂p)T in terms of Cp and the Joule–Thomson coefficient, µ. 2.24 Starting from the expression Cp − CV = T(∂p/∂T)V (∂V/∂T)p, use the appropriate relations between partial derivatives to show that Cp − CV = T(∂V/∂T)p2 (∂V/∂T)T Evaluate Cp − CV for a perfect gas. 2.25 (a) By direct differentiation of H = U + pV, obtain a relation between (∂H/∂U)p and (∂U/∂V)p. (b) Confirm that (∂H/∂U)p = 1 + p(∂V/∂U)p by 2.36 The speed of sound, cs, in a gas of molar mass M is related to the ratio of pressure heat capacity is given by Cp,m = A + BT + Cp, where a, A, B, and C are constants independent of T and p. Obtain expressions for (a) the Joule–Thomson coefficient and (b) its constant-volume heat capacity. Applications: to biology, materials science, and the environment 2.38 It is possible to see with the aid of a powerful microscope that a long piece of double-stranded DNA is flexible, with the distance between the ends of the chain adopting a wide range of values. This flexibility is important because it allows DNA to adopt very compact conformations as it is packaged in a chromosome (see Chapter 18). It is convenient to visualize a long piece PROBLEMS of DNA as a freely jointed chain, a chain of N small, rigid units of length l that are free to make any angle with respect to each other. The length l, the persistence length, is approximately 45 nm, corresponding to approximately 130 base pairs. You will now explore the work associated with extending a DNA molecule. (a) Suppose that a DNA molecule resists being extended from an equilibrium, more compact conformation with a restoring force F = −kF x, where x is the difference in the end-to-end distance of the chain from an equilibrium value and kF is the force constant. Systems showing this behaviour are said to obey Hooke’s law. (i) What are the limitations of this model of the DNA molecule? (ii) Using this model, write an expression for the work that must be done to extend a DNA molecule by x. Draw a graph of your conclusion. (b) A better model of a DNA molecule is the onedimensional freely jointed chain, in which a rigid unit of length l can only make an angle of 0° or 180° with an adjacent unit. In this case, the restoring force of a chain extended by x = nl is given by F= A1+νD ln B E 2l C1−νF kT ν = n/N where k = 1.381 × 10−23 J K−1 is Boltzmann’s constant (not a force constant). (i) What are the limitations of this model? (ii) What is the magnitude of the force that must be applied to extend a DNA molecule with N = 200 by 90 nm? (iii) Plot the restoring force against ν, noting that ν can be either positive or negative. How is the variation of the restoring force with end-to-end distance different from that predicted by Hooke’s law? (iv) Keeping in mind that the difference in end-to-end distance from an equilibrium value is x = nl and, consequently, dx = ldn = Nldν, write an expression for the work of extending a DNA molecule. (v) Calculate the work of extending a DNA molecule from ν = 0 to ν = 1.0. Hint. You must integrate the expression for w. The task can be accomplished easily with mathematical software. (c) Show that for small extensions of the chain, when ν << 1, the restoring force is given by F≈ νkT l = nkT Nl Hint. See Appendix 2 for a review of series expansions of functions. (d) Is the variation of the restoring force with extension of the chain given in part (c) different from that predicted by Hooke’s law? Explain your answer. 2.39 There are no dietary recommendations for consumption of carbohydrates. Some nutritionists recommend diets that are largely devoid of carbohydrates, with most of the energy needs being met by fats. However, the most common recommendation is that at least 65 per cent of our food calories should come from carbohydrates. A –34 -cup serving of pasta contains 40 g of carbohydrates. What percentage of the daily calorie requirement for a person on a 2200 Calorie diet (1 Cal = 1 kcal) does this serving represent? 2.40 An average human produces about 10 MJ of heat each day through metabolic activity. If a human body were an isolated system of mass 65 kg with the heat capacity of water, what temperature rise would the body experience? Human bodies are actually open systems, and the main mechanism of heat loss is through the evaporation of water. What mass of water should be evaporated each day to maintain constant temperature? 2.41 Glucose and fructose are simple sugars with the molecular formula C6H12O6. Sucrose, or table sugar, is a complex sugar with molecular formula C12H22O11 that consists of a glucose unit covalently bound to a fructose unit (a water molecule is given off as a result of the reaction between glucose and fructose to form sucrose). (a) Calculate the energy released as heat when a typical table sugar cube of mass 1.5 g is burned in air. (b) To what height could you climb on the energy a table sugar cube provides assuming 25 per cent of the energy is available for work? (c) The mass of a typical glucose tablet is 2.5 g. Calculate the energy released as heat when a glucose tablet is burned in air. (d) To what height could you climb on the energy a cube provides assuming 25 per cent of the energy is available for work? 2.42 In biological cells that have a plentiful supply of O2, glucose is oxidized completely to CO2 and H2O by a process called aerobic oxidation. Muscle cells may be deprived of O2 during vigorous exercise and, in that case, one 75 molecule of glucose is converted to two molecules of lactic acid (CH3CH (OH)COOH) by a process called anaerobic glycolysis (see Impact I7.2). (a) When 0.3212 g of glucose was burned in a bomb calorimeter of calorimeter constant 641 J K−1 the temperature rose by 7.793 K. Calculate (i) the standard molar enthalpy of combustion, (ii) the standard internal energy of combustion, and (iii) the standard enthalpy of formation of glucose. (b) What is the biological advantage (in kilojoules per mole of energy released as heat) of complete aerobic oxidation compared with anaerobic glycolysis to lactic acid? 2.43 You have at your disposal a sample of pure polymer P and a sample of P that has just been synthesized in a large chemical reactor and that may contain impurities. Describe how you would use differential scanning calorimetry to determine the mole percentage composition of P in the allegedly impure sample. 2.44‡ Alkyl radicals are important intermediates in the combustion and atmospheric chemistry of hydrocarbons. Seakins et al. (P.W. Seakins, M.J. Pilling, J.T. Niiranen, D. Gutman, and L.N. Krasnoperov, J. Phys. Chem. 96, 9847 (1992)) report ∆ f H 7 for a variety of alkyl radicals in the gas phase, information that is applicable to studies of pyrolysis and oxidation reactions of hydrocarbons. This information can be combined with thermodynamic data on alkenes to determine the reaction enthalpy for possible fragmentation of a large alkyl radical into smaller radicals and alkenes. Use the following set of data to compute the standard reaction enthalpies for three possible fates of the tert-butyl radical, namely, (a) tert-C4H9 → sec-C4H9, (b) tert-C4H9 → C3H6 + CH3, (c) tert-C4H9 → C2H4 + C2H5. Species: C2H5 sec-C4H9 tert-C4H9 ∆ f H 7/(kJ mol−1) +121.0 +67.5 +51.3 2.45‡ In 1995, the Intergovernmental Panel on Climate Change (IPCC) considered a global average temperature rise of 1.0–3.5°C likely by the year 2100, with 2.0°C its best estimate. Predict the average rise in sea level due to thermal expansion of sea water based on temperature rises of 1.0°C, 2.0°C, and 3.5°C given that the volume of the Earth’s oceans is 1.37 × 109 km3 and their surface area is 361 × 106 km2, and state the approximations that go into the estimates. 2.46‡ Concerns over the harmful effects of chlorofluorocarbons on stratospheric ozone have motivated a search for new refrigerants. One such alternative is 2,2-dichloro-1,1,1-trifluoroethane (refrigerant 123). Younglove and McLinden published a compendium of thermophysical properties of this substance (B.A. Younglove and M. McLinden, J. Phys. Chem. Ref. Data 23, 7 (1994)), from which properties such as the Joule–Thomson coefficient µ can be computed. (a) Compute µ at 1.00 bar and 50°C given that (∂H/∂p)T = −3.29 × 103 J MPa−1 mol−1 and Cp,m = 110.0 J K−1 mol−1 . (b) Compute the temperature change that would accompany adiabatic expansion of 2.0 mol of this refrigerant from 1.5 bar to 0.5 bar at 50°C. 2.47‡ Another alternative refrigerant (see preceding problem) is 1,1,1,2tetrafluoroethane (refrigerant HFC-134a). Tillner-Roth and Baehr published a compendium of thermophysical properties of this substance (R. TillnerRoth and H.D. Baehr, J. Phys. Chem. Ref. Data 23, 657 (1994)), from which properties such as the Joule–Thomson coefficient µ can be computed. (a) Compute µ at 0.100 MPa and 300 K from the following data (all referring to 300 K): p/MPa 0.080 0.100 0.12 Specific enthalpy/(kJ kg−1) 426.48 426.12 425.76 (The specific constant-pressure heat capacity is 0.7649 kJ K−1 kg−1.) (b) Compute µ at 1.00 MPa and 350 K from the following data (all referring to 350 K): p/MPa 0.80 1.00 1.2 Specific enthalpy/(kJ kg−1) 461.93 459.12 456.15 (The specific constant-pressure heat capacity is 1.0392 kJ K−1 kg−1.) 3 The direction of spontaneous change 3.1 The dispersal of energy 3.2 Entropy I3.1 Impact on engineering: Refrigeration 3.3 Entropy changes accompanying specific processes 3.4 The Third Law of thermodynamics The Second Law The purpose of this chapter is to explain the origin of the spontaneity of physical and chemical change. We examine two simple processes and show how to define, measure, and use a property, the entropy, to discuss spontaneous changes quantitatively. The chapter also introduces a major subsidiary thermodynamic property, the Gibbs energy, which lets us express the spontaneity of a process in terms of the properties of a system. The Gibbs energy also enables us to predict the maximum non-expansion work that a process can do. As we began to see in Chapter 2, one application of thermodynamics is to find relations between properties that might not be thought to be related. Several relations of this kind can be established by making use of the fact that the Gibbs energy is a state function. We also see how to derive expressions for the variation of the Gibbs energy with temperature and pressure and how to formulate expressions that are valid for real gases. These expressions will prove useful later when we discuss the effect of temperature and pressure on equilibrium constants. Concentrating on the system 3.5 The Helmholtz and Gibbs energies 3.6 Standard reaction Gibbs energies Combining the First and Second Laws 3.7 The fundamental equation 3.8 Properties of the internal energy 3.9 Properties of the Gibbs energy Checklist of key ideas Further reading Further information 3.1: The Born equation Further information 3.2: Real gases: the fugacity Discussion questions Exercises Problems Some things happen naturally; some things don’t. A gas expands to fill the available volume, a hot body cools to the temperature of its surroundings, and a chemical reaction runs in one direction rather than another. Some aspect of the world determines the spontaneous direction of change, the direction of change that does not require work to be done to bring it about. A gas can be confined to a smaller volume, an object can be cooled by using a refrigerator, and some reactions can be driven in reverse (as in the electrolysis of water). However, none of these processes is spontaneous; each one must be brought about by doing work. An important point, though, is that throughout this text ‘spontaneous’ must be interpreted as a natural tendency that may or may not be realized in practice. Thermodynamics is silent on the rate at which a spontaneous change in fact occurs, and some spontaneous processes (such as the conversion of diamond to graphite) may be so slow that the tendency is never realized in practice whereas others (such as the expansion of a gas into a vacuum) are almost instantaneous. The recognition of two classes of process, spontaneous and non-spontaneous, is summarized by the Second Law of thermodynamics. This law may be expressed in a variety of equivalent ways. One statement was formulated by Kelvin: No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work. For example, it has proved impossible to construct an engine like that shown in Fig. 3.1, in which heat is drawn from a hot reservoir and completely converted into work. All real heat engines have both a hot source and a cold sink; some energy is always discarded into the cold sink as heat and not converted into work. The Kelvin 3.1 THE DISPERSAL OF ENERGY statement is a generalization of another everyday observation, that a ball at rest on a surface has never been observed to leap spontaneously upwards. An upward leap of the ball would be equivalent to the conversion of heat from the surface into work. Hot source Flow of energy The direction of spontaneous change What determines the direction of spontaneous change? It is not the total energy of the isolated system. The First Law of thermodynamics states that energy is conserved in any process, and we cannot disregard that law now and say that everything tends towards a state of lower energy: the total energy of an isolated system is constant. Is it perhaps the energy of the system that tends towards a minimum? Two arguments show that this cannot be so. First, a perfect gas expands spontaneously into a vacuum, yet its internal energy remains constant as it does so. Secondly, if the energy of a system does happen to decrease during a spontaneous change, the energy of its surroundings must increase by the same amount (by the First Law). The increase in energy of the surroundings is just as spontaneous a process as the decrease in energy of the system. When a change occurs, the total energy of an isolated system remains constant but it is parcelled out in different ways. Can it be, therefore, that the direction of change is related to the distribution of energy? We shall see that this idea is the key, and that spontaneous changes are always accompanied by a dispersal of energy. 77 Heat Engine Work The Kelvin statement of the Second Law denies the possibility of the process illustrated here, in which heat is changed completely into work, there being no other change. The process is not in conflict with the First Law because energy is conserved. Fig. 3.1 3.1 The dispersal of energy We can begin to understand the role of the distribution of energy by thinking about a ball (the system) bouncing on a floor (the surroundings). The ball does not rise as high after each bounce because there are inelastic losses in the materials of the ball and floor. The kinetic energy of the ball’s overall motion is spread out into the energy of thermal motion of its particles and those of the floor that it hits. The direction of spontaneous change is towards a state in which the ball is at rest with all its energy dispersed into random th