《物理化学》课程参考书籍：《ATKINS' Physical Chemistry》PDF电子书英文原版，Eighth Edition，2006（Peter Atkins、Julio de Paula）.pdf_P6-P10

ABOUT THE BOOK ix Notes on good practice 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 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. 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. q 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. 5.8 The activities of regular solutions 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 γA = βxB 2 ln γB = βxA 2 (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 ∆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 + βxAxB 2 + βxBxA 2} = 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

x ABOUT THE BOOK Further information 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. Appendices 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. Synoptic tables and the Data section 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. 966 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 = y dx + x dy where the symbol ∂ is used (instead of d) to denote a parti df is also called the differential of f. For example, if f = ax3 y y = 3ax2 y x = ax3 + 2by D F ∂f ∂y A C D F ∂f ∂x A C D F ∂f ∂y A C D F ∂f ∂x A C 1000 DATA SECTION Table 2.8 Expansion coefficients, α, and isothermal compressibilities, κT a/(10 − 4 K−1) kT /(10 −6 atm−1) Liquids Benzene 12.4 92.1 Carbon tetrachloride 12.4 90.5 Ethanol 11.2 76.8 Mercury 1.82 38.7 Water 2.1 49.6 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). Table 2.9 Inversion temperatures, no points, and Joule–Thomson coefficient TI/K Tf/K Air 603 Argon 723 83.8 Carbon dioxide 1500 194.7s Helium 40 Hydrogen 202 14.0 Krypton 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). 0.2 0.4 0.6 0.8 1.0 0 Potential, /( / ) Z rD 0 0.5 Distan 0.3 1 3 Fig. 5.36 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. Exploration Write an expression f unshielded and shielded Coulom Then plot this expression against rD and interpretation for the shape of the plot. 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 MpXq, we write Gm Gm ideal we = (pµ+ + qµ −) − (pµ+ ideal + qµ − ideal) = 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± = s = p + q (5.73) 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 zi e in a medium of permittivity ε is φi = Zi = (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 φi = e−r/rD (5.75) Zi r zt e 4πε Zi r we sRT 5 4 6 4 7 5 4 4 6 4 4 7 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

ABOUT THE BOOK xi 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. Appendices 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. Mathematics and Physics support Comment 1.2 A hyperbola is a curve obtained by plotting y against x with xy = constant. e n s r e , 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 2 y, then y = y = y = 2yx Partial derivatives are reviewed in Appendix 2. dx 2 dx D E F ∂[x 2 y] ∂x A B C D E F ∂z ∂x A B C e e 978 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 A3.3 The trajectory in terms of the energy The velocity, V, of a particle is the rate of change of its po V = 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 p = mV Like the velocity vector, the linear momentum vector poi of the particle (Fig. A3.1). In terms of the linear momentu ticle is 2 dr dt p pz px py A3.1 The linear momentum of a particle is a vector property and points in the direction of motion. 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 == = 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 calculation 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 ] 21 kPa 7.9 × 104 kPa kg mol−1 pO2 KO2 Problem solving Illustrations 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 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 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. Self-test 3.12 Calculate the change in Gm for ice at −10°C, with density 917 kg m−3 , when the pressure is increased from 1.0 bar to 2.0 bar. [+2.0 J mol−1 ] 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 == = Substitution of the data gives N = = 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] (5.60 × 10−7 m) × (100 J s−1 ) × (1.0 s) (6.626 × 10−34 J s) × (2.998 × 108 m s−1 ) λP∆t hc P∆t h(c/λ) E hν Worked examples 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. Self-tests 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. Discussion questions 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 xiii Exercises and Problems 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. max Molar absorption coefficient, Wavenumb ( ) {1 max max ~ Exercises 14.1a The term symbol for the ground state of N2 + is 2Σg. 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 g 2 1σu 2 1πu 3 1π g 1 . 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 . 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 Fig. 14.49 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 = 3 –2R) 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. 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? Fig. 2.34 Isotherm 1.00 Pressure, /atm 0.50 p 22.44 44.88 Volume, /dm V 3 1 2 3 Table 2.2. Calculate the standard enthalpy of from its value at 298 K. 2.8 A sample of the sugar d-ribose (C5H10O 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 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. 2.10‡ From the enthalpy of combustion dat alkanes methane through octane, test the ext ∆cH7 = k{(M/(g mol−1 )}n holds and find the Predict ∆cH7 for decane and compare to the 2.11 It is possible to investigate the thermoc hydrocarbons with molecular modelling me software to predict ∆cH7 values for the alkan calculate ∆cH7 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 ∆cH7 (Table 2.5) and the molecular modelling method. (c) Test th ∆cH7 = k{(M/(g mol−1 )}n holds and find the 2 12‡ When 1 3584 g of sodium acetate trih