(Solution Manual) Physical Chemistry

Atkins,Trapp,Giunta,Cady, 8

Chemistry

Solution Manual

8th Edition

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Solution Manual for Physical Chemistry, The downloadable file consists of 543 pages, and the total size of the file is 18.35 MB. Purchase now only for highly discounted price.

Author(s): Atkins, Trapp, Giunta, Cady

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TH EDITION Student Solutions Manual to Accompany PHYSICAL CHEMISTRY PETER ATKINS • CHARLES TRAPP CARMEN GIUNTA • MARSHALL CADY STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY PHYSICAL CHEMISTRY EIGHTH EDITION STUDENT'S SOLUTIONS MANUAL TO ACCOMPANY PHYSIC~l CHEMISTRY Eighth Edition P. W. Atkins Professor of Chemistry , University of Oxford and Fellow of Uncoln College C. A. Trapp Professor of Chemistry, University of Louisville, Louisville, Kentucky, USA M. P. Cady Professor of Chemistry, Indiana University Southeas t, New Albany Indiana, USA C. Giunta Professor of Chemistry, Le Mayne College, Syracuse, NY, USA II w. H. Freeman and Company New York Student's solutions manual to accompany Physical Chemistry, Eighth Edition © Oxford University Press, 2006 All rights reserved ISBN-13: 978-0-7167-6206-5 ISBN-lO: 0-7167-6206--4 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. Library of Congress Cataloging in Publication Data Data available Second printing W. H. Freeman and Company 41 Madison Avenue New York, NY 10010 www.whfreeman.com Preface This manual provides detailed solutions to all the end-of-chapter (b) Exercises, and to the odd-numbered Discussion Questions and Problems. Solutions to Exercises and Problems carried over from previous editions have been reworked, modified, or corrected when needed. The solutions to the Problems in this edition rely more heavily on the mathematical and molecular modelling software that is now generally accessible to physical chemistry students, and this is partic­ ularly true for many of the new Problems that request the use of such software for their solutions. But almost all of the Exercises and many of the Problems can still be solved with a modem hand-held sci­ entific calculator . When a quantum chemical calculation or molecular modelling process has been called for, we have usually provided the solution with PC Spartan pro™ because of its common availability. In general, we have adhered rigorously to the rules for significant figures in displaying the final answers. However, when intermediate answers are shown, they are often given with one more figure than would be justified by the data. These excess digits are indicated with an overline. We have carefully cross-checked the solutions for errors and expect that most have been eliminated. We would be grateful to any readers who bring any remaining errors to our attention. We warmly thank our publishers for their patience in guiding this complex, detailed project to completion . P. W. A. e.A.T. M. P.e. e. G. Contents PART 1 Equilibrium 1 5 Simple mixtures 91 1 The properties of gases 3 Answers to discussion questions 91 Solutions to exercises 91 Solutions to problems 98 Answers to discussion questions 3 Solutions to numerical problems 98 Solutions to exercises 4 Solutions to theoretical problems 104 Solutions to problems 13 Solutions to applications 107 Solutions to numerical problems 13 Solutions to theoretical problems 18 Solutions to application s 20 6 Phase diagrams 112 Answers to discussion questions 112 2 The First Law 22 Solutions to exercises 113 Solutions to problems 119 Answers to discussion questions 22 Solutions to numerical problems 119 Solutions to exercises 23 Solutions to theoretical problems 124 Solutions to problems 33 Solutions to applications 124 Solutions to numerical problem s 33 Solutions to theoretical problems 41 7 Chemical equilibrium 127 Solutions to application s 47 Answers to discussion questions 127 3 The Second Law 50 Solutions to exercises 128 Solutions to problems 137 Answers to discussion questions 50 Solutions to numerical problems 137 Solutions to exercises 51 Solutions to theoretical problems 148 Solutions to problems 58 Solutions to applications 150 Solutions to numerical problems 58 Solutions to theoretical problems 68 Solutions to application s 74 PART 2 Structure 155 4 Physical transformations of pure 8 Quantum theory: introduction and substances 78 prinCiples 157 Answers to discussion questions 78 Answers to discussion questions 157 Solutions to exercises 80 Solutions to exercises 158 Solutions to problems 83 Solutions to problems 162 Solutions to numerical problems 83 Solutions to numerical problems 162 Solutions to theoretical problems 86 Solutions to theoretical problems 165 Solutions to applications 87 Solutions to applications 172 viii Contents 9 Quantum theory: techniques and 14 Molecular spectroscopy 2: applications 176 electronic transitions 280 Answers to discussion questions 176 Answers to discussion questions 280 Solutions to exercises 176 Solutions to exercises 281 Solutions to problems 183 Solutions to problems 284 Solutions to numerical problems 183 Solutions to numerical problems 284 Solutions to theoretical problems 186 Solutions to theoretical problems 289 Solutions to applications 195 Solutions to applicatio ns 292 10 Atomic structure and atomic 15 Molecular spectroscopy 3: spectra 199 magnetic resonance 297 Answers to discussion questions 199 Answers to discussion questions 297 Solutions to exercises 200 Solutions to exercises 299 Solutions to problems 207 Solutions to problems 305 Solutions to numerical problems 207 Solutions to numerical problems 305 Solutions to theoretical problems 211 Solutions to theoretical problems 309 Solutions to application s 218 Solutions to applications 311 11 Molecular structure 221 16 Statistical thermodynamics 1 : the concepts 315 Answers to discussion questions 221 Solutions to exercises 223 Answers to discussion questions 315 Solutions to problems 226 Solutions to exercises 315 Solutions to numerical problems 226 Solutions to problems 322 Solutions to theoretical problems 238 Solutions to numerical problems 322 Solutions to application s 241 Solutions to theoretical problems 326 Solutions to applications 329 12 Molecular symmetry 244 17 Statistical thermodynamics 2: Answers to discussion questions 244 applications 331 Solutions to exercises 245 Solutions to problems 249 Answers to discussion questions 331 Solutions to applications 255 Solutions to exercises 332 Solutions to problems 338 13 Molecular spectroscopy 1 : rotational Solutions to numerical problems 338 and vibrational spectra 259 Solutions to theoretical problems 345 Solutions to applications 353 Answers to discussion questions 259 Solutions to exercises 260 18 Molecular interactions 357 Solutions to problems 269 Solutions to numerical problems 269 Answers to discussion questions 357 Solutions to theoretical problems 275 Solutions to exercises 358 Solutions to applications 276 Solutions to problems 361 Contents ix Solutions to numerical problems 361 22 The rates of chemical reactions 440 Solutions to theoretical problems 366 Solutions to applications 368 Answers to discussion questions 440 Solutions to exercises 443 19 Materials 1: macromolecules and Solutions to problems 450 aggregates 370 Solutions to numerical problems 450 Solutions to theoretical problems 455 Answers to discussion questions 370 Solutions to applications 458 Solutions to exercises 372 Solutions to problems 375 23 The kinetics of complex reactions 464 Solutions to numerical problems 375 Solutions to theoretical problems 379 Answers to discussion questions 464 Solutions to applications 383 Solutions to exercises 465 Solutions to problems 468 20 Materials 2: the solid state 389 Solutions to numerical problems 468 Solutions to theoretical problems 471 Answers to discussion questions 389 Solutions to applications 478 Solutions to exercises 390 Solutions to problems 398 24 Molecular reaction dynamics 489 Solutions to numerical problems 398 Solutions to theoretical problems 405 Answers to discussion questions 489 Solutions to applications 408 Solutions to exercises 490 Solutions to problems 497 Solutions to numerical problems 497 Solutions to theoretical problems 502 PART 3 Change 411 Solutions to applications 506 21 Molecules in motion 413 25 Processes at solid surfaces 509 Answers to discussion questions 413 Answers to discussion questions 509 Solutions to exercises 414 Solutions to exercises 511 Solutions to problems 424 Solutions to problems 521 Solutions to numerical problems 424 Solutions to numerical problems 521 Solutions to theoretical problems 430 Solutions to theoretical problems 531 Solutions to applications 433 Solutions to applications 534 PART 1 Equilibrium The properties of gases Answers to discussion questions 01.1 An equation of state is an equation that relates the variables that define the state of a system to each other. Boyle, Charles, and Avogadro established these relations for gases at low pressures (perfect gases) by appropriate experiments. Boyle determined how volume varies with pressure (V ex lip), Charles how volume varies with temperature (V ex T), and Avogadro how volume varies with amount of gas (V ex n). Combining all of these proportionalities into one we find nT Vex -. p Inserting the constant of proportionality, R, yields the perfect gas equation RnT V = --or pV = nRT. p 01.3 Consider three temperature regions: (1) T < TB. At very low pressures , all gases show a compression factor, Z ~ I. At high pressures, all gases have Z > I, signifying that they have a molar volume greater than a perfect gas, which implies that repulsive forces are dominant. At intermediate pressures, most gases show Z < I, indicating that attractive forces reducing the molar volume below the perfect value are dominant. (2) T ~ TB. Z ~ I at low pressures, slightly greater than I at intermediate pressures, and significantly greater than I only at high pressures . There is a balance between the attractive and repulsive forces at low to intermediate pressures, but the repulsive forces predominate at high pressures where the molecules are very close to each other. (3) T > TB. Z > I at all pressures because the frequency of collisions between molecules increases with temperature. 01.5 The van der Waals equation 'corrects' the perfect gas equation for both attractive and repulsive interactions between the molecules in a real gas. See Justification 1.1 for a fuller explanation. The Bertholet equation accounts for the volume of the molecules in a manner similar to the van der Waals equation but the term representing molecular attractions is modified to account for the effect of temperature . Experimentally one finds that the van der Waals a decreases with increasing temperature. Theory (see Chapter 18) also suggests that intermolecular attractions can decrease with temperature . E1.1(b) E1.2(b) E1.3(b) 4 STUDENT'S SOLUTIONS MANUAL This variation of the attractive interaction with temperature can be accounted for in the equation of state by replacing the van der Waals a with a/To Solutions to exercises (a) The perfect gas law is pV = nRT implying that the pressure would be nRT P=­V All quantities on the right are given to us except n, which can be computed from the given mass of Ar. n = 25 g = 0.626 mol 39.95 g mol-1 (0.626 mol) x (8.31 x 10-2 dm3 bar K-1 mol-I) x (30 + 273 K) 1 - 1 so P = 3 = . 10.5 bar . 1.5dm not 2.0 bar. (b) The van der Waals equation is RT a P = V -b -V2 m m (8.31 X 10-2 dm3 bar K-1 mol-I) x (30 + 273) K sop = (1.53 dm3 /0.626 mol) -3.20 x 10-2 dm3 mol-1 (1.337dm6atmmol-2) x (1.013baratm -1) _I -1 - 3 -10.4 bar . (1.5 dm /0.626 mo1)2 (a) Boyle's law applies: P V = constant so pf Vf = Pi Vi and prVr (1.97 bar) x (2.14dm3) 1 1 Pi = --= = 1.07 bar Vi (2.14 + 1.80) dm3 (b) The original pressure in bar is ( 1 atm) (760 TOrr) I I Pi = (1.07 bar) x x = 803 Torr 1.013 bar I atm The relation between pressure and temperature at constant volume can be derived from the perfect gas law pV = nRT Pi so P ex: T and Ti Pr Tr E1.4(b) E1.S(b) E1.6(b) E1.7(b) THE PROPERTIES OF GASES 5 The final pressure, then, ought to be = pjTr = (125 kPa) x (II + 273) K = 1120 kPa I Pr Tj (23 + 273) K According to the perfect gas law, one can compute the amount of gas from pressure, temperature , and volume. Once this is done, the mass of the gas can be computed from the amount and the molar mass using pV = nRT pV (1.00atm) x (1.013 x 105 Paatm-I) x (4.00 x 103m3) 5 so n - - - = 1.66 x 10 mol . -RT - (8.3145J K-Imol-I) x (20+273) K and m = (1.66 x lOS mol) x (16.04 g mol-I) = 2.67 x 106g = 12.67 X 103 kg 1 Identifying Pex in the equation P = Pex + pgh [1.3] as the pressure at the top of the straw and P as the atmospheric pressure on the liquid, the pressure difference is P -Pex = pgh = (1.0 x 103 kg m-3) x (9.81 m s-2) x (0.15 m) = 11.5 x 103 Pa 1(= 1.5 X 10-2 atm) The pressure in the apparatus is given by P = Palm + pgh [1.3] Palm = 760 Torr = I atm = 1.013 x 105 Pa pgh = 13.55 g cm-3 x (/o~gg) x (1O:~m3) x 0.100 m x 9.806 m s-2 = 1.33 X 104 Pa P = 1.013 X 105 Pa + 1.33 x 104 Pa = 1.146 x 105 Pa = 1115 kPa I All gases are perfect in the limit of zero pressure. Therefore the extrapolated value of pVm/T will give the best value of R. m The molar mass is obtained from P V = nRT = -RT M . mRT RT which upon rearrangement gives M = - - = p- V P P The best value of M is obtained from an extrapolation of p / P versus P to P = 0; the intercept is M / RT. Draw up the following table 0.750000 0.082 00 14 1.428 59 0.500000 0.082 0227 1.428 22 0.250000 0.082 0414 1.427 90 From Figure l.l(a), (PVm) = I 0.082 0615 dm3 atm K-I mol-I I T p=O E1.8(b) 6 STUDENT'S SOLUTIONS MANUAL [1 .••. :I,.:.· •. · •. :,.!.: .••. I,.:· •. · •.• • .••. · •. ·rr=.·:,i.· .• • .••. i,.:· •. · •. · •. ·:!, .•• '··!.!j~!~.f ! ••• ! •••••.••.•••••• ::: ··i .. ·!···!···i .. ·!· .. !··· .. ······· .. . "'M ",,,.,,,.! ... ···'···I···I···,···j···j·· . 'i"'!"'!"'!'" .. ,." ..... . ... ~ ,,;8.202, ";"';";"';";"';"';" ... i ... i ... i ... ! . ... ............. ! ... ! ... ! ... ··-!-··!··-!-·-!-··!···,···,··!···.··! .. ·i .. ·! .. ~~ .. ~ ... ~ ... ~ ...... ~ ... ~ ... ! ... ! ... j "!"'! "!"-!-". 'i"+" "M ,,!"'!'''! ..... ,',,'!, .. !.,,!',,' ··i .. ·!· .. i .. ·!···!··· .. ". :2 ";8.200' ·; .. ·;··;··; .. ·; .. ·: .. Y·i ... : ... : ... : .. : ::!:··!:.·!: ::!:::!::i:::.::: ·:::~.~~:··::::·~.~~;.::::'.:~.~~: ... ; ... '1.0 .:t::::ttt :::::J:::':::;:::':::':: ::::.f.!~ti:!:':::::::':::'::t :· :::i::: Figure 1.I(a) From Figure 1.l(b), (~) -1.42755 g dm-3 atm-I p p=o- .. i .4288 .. ... : "': "': ·': .. ·1.4286' .. :. . ,,? .. ~ "f"'~ . .. , . ,: ... ~ .. ': .. , ... , .. ~ . ,! .. 'r"'~ ..~ ... ~ ... ~ ... ; .. ! .. ; .... : .. i .. i .... : .. j ••. ~ Figure 1.I(b) M = RT (~) (0.0820615 dm3 atm mol-I K-1) x (273.15 K) x (1.42755 g dm-3atm-l) p p=o = 131.9987 g mol-II The value obtained for R deviates from the accepted value by 0.005 percent. The error results from the fact that only three data points are available and that a linear extrapolation was employed . The molar mass, however, agrees exactly with the accepted value, probably because of compensating plotting errors. The mass density p is related to the molar volume Vrn by M p where M is the molar mass. Putting this relation into the perfect gas law yields pVrn = RT so pM =RT p E1.9(b) THE PROPERTIES OF GASES 7 Rearranging this result gives an expression for M; once we know the molar mass, we can divide by the molar mass of phosphorus atoms to determine the number of atoms per gas molecule RTp (8.314 Pa m3 mol-I) x [(100 + 273) K] x (0.6388kgm-3) M---------------------------~----------~----- P - 1.60xl()4Pa = 0.124 kg mol-I = 124 g mol-I The number of atoms per molecule is 124g mol-I ----'------,- = 4.00 31.0g mol-I suggesting a formula of ~ Use the perfect gas equation to compute the amount; then convert to mass. pV P V = nRT so n = RT We need the partial pressure of water, which is 53 percent of the equilibrium vapor pressure at the given temperature and standard pressure. p = (0.53) x (2.69 x 103 Pa) = 1.43 x 103 Pa (1.43 x 103 Pa) x (250 m3) -2 so n = = 1.45 x 10 mol (8.3145 J K-I mol-I) x (23 + 273) K or m = (l.45 x 102 mol) x (l8.0g mol-I) = 2.61 x 103 g = 12.61 kg 1 E1.10(b) (a) The volume occupied by each gas is the same, since each completely fills the container . Thus solving for V we have (assuming a perfect gas) nJRT 0.225 g V = --nNe = --------=-----,- PJ 20.18 g mol-I = 1.115x 10-2 mol, PNe= 8.87kPa, T=300K (!.l15 x 10-2 mol) x (8.314 dm3 kPa K-I mol-I) x 300 K) -3 V= =3.137dm 8.87 kPa =13.14dm31 (b) The total pressure is determined from the total amount of gas, n = nCH4 + nAr + nNe. 0.320 g --2 0.175 g nCH4 = I = 1.995 x 10 mol nAr = = 4.38 x 1O-3mol 16.04 g mol- 39.95 g mol-I n = (1.995 + 0.438 + 1.115) x 1O-2mol = 3.548 x 1O-2mol p = nRT [1.8] = (3.548 x 10-2 mol) x (8.314 d_m3 kPa K-I mol-I) x (300 K) V 3.137dm3 = 128.2 kPa 1 8 STUDENT'S SOLUTIONS MANUAL E1.11(b) This is similar to Exercise l.ll(a) with the exception that the density is first calculated. RT M = p-[Exercise 1.8(a)] p 33.5mg -p= ---3 =0.1340gdm-3, p= 152 Torr, T=298K 250cm (0.1340gdm-3) x (62.36dm3TorrK-1 mol-I) x (298K) I -II M= = 16.14gmol 152 Torr E1.12(b) This exercise is similar to Exercise 1.12(a) in that it uses the definition of absolute zero as that temperature at which the volume of a sample of gas would become zero if the substance remained a gas at low temperatures. The solution uses the experimental fact that the volume is a linear function of the Celsius temperature . Thus V = Vo + aVoO = Vo + bO, b = aVo At absolute zero, V = 0, or 0 = 20.oodm3 + 0.0741 dm3 °C-I x O(abs. zero) 20.00 dm3 t °C t O(abs. zero) = - 3 I = -270 0.0741 dm °C- which is close to the accepted value of -273°C. nRT E1.13(b) (a) P = V n = 1.0mol T = (i) 273.15K; (ii) 500K V = (i) 22.414dm3; (ii) 150cm3 (1.0 mol) x (8.206 x 10-2 dm3 atmK-I mol-I) x (273.15K) (i) P= 22.414dm 3 = t 1.0 atm t (1.0mol) x (8.206 x 10-2 dm3 atmK-I mol-I) x (5OOK) (ii) P = 0.150dm 3 = t 270 atm t (2 significant figures) (b) From Table (1.6) for H2S a = 4.484 dm6 atm mol-I b = 4.34 x 10-2 dm3 mol-I nRT an2 P=V-nb-V2' (1.0 mol) x (8.206 x 10-2 dm3 atm K-I mol-I) x (273.15 K) (i) P = 22.414 dm3 -(1.0 mol) x (4.34 x 10-2 dm3 mol I) (4.484 dm6 atm mol-I) x (1.0 mol)2 (22.414 dm3)2 = t 0.99 atm t THE PROPERTIES OF GASES 9 (ii) (1.0 mol) x (8.206 x 10-2 dm3 atm K-I mol-I) x (500K) p = 0.150dm3 -(1.0 mol) x (4.34 x 10-2 dm3 mol I) (4A84dm6atmmor1 ) x (1.0mol)2 (0.150 dm3)2 = 185.6atm ~ 1190 atm 1 (2 significant figures). E1.14(b) The conversions needed are as follows: Therefore , a = 1.32 atm dm6 mol-2 becomes, after substitution of the conversions a = 11.34 x 10-1 kg m5s-2mol-2 ~ and b = 0.0436 dm3 mol-I becomes b = 14.36 X 10-5 m3mol-1 I E1.1S(b) The compression factor is pVm Vm z=-=- RT V;:' (a) Because Vm = V;:' + 0.12 V;:' = (I. 12)V;:', we have Z = [Iill I Repulsive I forces dominate . (b) The molar volume is V = (1.12)V~ = (1.12) x (R;) (0.08206dm3 atmK-I mol-I) x (350K») 127 d 3 I-I 1 V=(1.12)x 12atm =.' m mo . E1.16(b) (a) o RT (8.314JK-I mol-I) x (298.15 K) V - - - --------------~----~- m -p -(200 bar) x (l05Pabar- l) = 1.24 x 10-4 m3 mol-I = I 0.124 dm3 mol-I I (b) The van der Waals equation is a cubic equation in Vm. The most direct way of obtaining the molar volume would be to solve the cubic analytically . However, this approach is cumbersome, so we proceed as in Example 104. The van der Waals equation is rearranged to the cubic form 3 ( RT) 2 (a) ab 3 ( RT) 2 (a) ab V m -b + P V m + p V m -p = 0 or x -b + P x + p x -p = 0 10 STUDENT'S SOLUTIONS MANUAL The coefficients in the equation are evaluated as b + RT = (3.183 x 1O-2dm3 mol-I) + (8.206 x 1O-2dm3 mol-I) x (298.15 K) P (200 bar) x (1.013 atm bar-I) a P = (3.183 X 10-2 + 0.1208) dm3 mol-I = 0.1526 dm3mol-1 1.360 dm6 atm mol-2 3 3 I 2 -----------::- = 6.71 x 10-(dm mol-) (200 bar) x (1.013 atm bar-I) ab (1.360 dm6 atmmol-2) x (3.183 x 1O-2dm3 mol-I) - 4 3 I 3 P -------~--.:......---__;_------.:. = 2.137 x 10-(dm mol-) (200 bar) x (1.013 atmbar-I) Thus, the equation to be solved is x3 -0.1526x2 + (6.71 x 1O-3)x -(2.137 x 10-4) = O. Calculators and computer software for the solution of polynomials are readily available. In this case we find x = 0.112 or Vrn = 1 0.112 dm3 mol-II The difference is about 15 percent. E1.17(b) The molar volume is obtained by solving Z = pVrn/RT [1.17], for Vrn, which yields ZRT (0.86) x (0.08206dm3atmK-lmol-l) x (300K) -3 -I Vrn = --= = 1.059dm mol p 20atm (a) Then, V = n Vrn = (8.2 x 10-3 mol) x (1.059 dm3 mol-I) = 8.7 x 10-3 dm3 = 18.7 cm31 (b) An approximate value of B can be obtained from eqn 1.19 by truncation of the series expansion after the second term, B/Vrn, in the series. Then, B = Vrn (p;; -I) = Vrn X (Z -I) = (1.059 dm3 mol-I) x (0.86 -I) = \-0.15 dm3mol-1 \ E1.18(b) (a) Mole fractions are nN 2.5 mol ~6 XN=-= =~ ntotal (2.5 + 1.5) mol Similarly, XH = 1 0.371 (c) According to the perfect gas law Ptotal V = ntotalRT ntotalRT so Ptotal = --V- _ (4.0 mol) x (0.08206 dm3 atm mol-I K-I) x (273.15 K) -140 1 - 22.4dm3 -.. atm. THE PROPERTIES OF GASES 11 (b) The partial pressures are PN = XNPtot = (0.63) x (4.0 atm) = 12.5 atm 1 and PH = (0.37) x (4.0 atm) = 11.5 atm 1 E1.19(b) The critical volume of a van der Waals gas is Vc = 3b so b = 1 Vc = 1(148cm3 mol-I) = 49.3cm3 mol-I = I 0.0493 dm3 mol-I I By interpreting b as the excluded volume of a mole of spherical molecules , we can obtain an estimate of molecular size. The centers of spherical particles are excluded from a sphere whose radius is the diameter of those spherical particles (i.e. twice their radius); that volume times the Avogadro constant is the molar excluded volume b (4n(2r)3 ) b=NA ---3 so r = ~ (~)1 /3 2 4nNA ( )IP I 3(49.3cm3mol-l) r=- I =1.94xlO-Scm=11.94xlO-IOml 2 4n(6.022 x 1023 mol-) The critical pressure is a Pc = 27b2 so a = 27Pcb2 = 27(48.20 atm) x (0.0493 dm3 mol-I)2 = 13.16 dm6 atm mol-21 But this problem is overdetermined. We have another piece of information 8a T. --­ c -27Rb According to the constants we have already determined , Tc should be However, the reported Tc is 305.4 K, suggesting our computed alb is about 25 percent lower than it should be. E1.20(b) (a) The Boyle temperatur e is the temperature at which Iimvm~oo dZ/(d(lIVm» vanishes. According to the van der Waals equation ( RT a ) Z = pVm = v:-=-h -~ Vm Vm a ------RT RT Vm -b VmRT 12 STUDENT'S SOLUTIONS MANUAL so dZ ( dZ) ( dVrn ) d(l/Vrn) = dVrn x d(l/Vrn) = -V~ Cd:rn) = -V~ CV:~rnb)2 + Vrnl-b + vtRT) V~b a (Vrn -b)2 RT In the limit of large molar volume, we have I. dZ a a 1m = b - -= 0 so -= b Vm ..... OO dO/Vrn) RT RT a (4.484 dm6 atm mol-2) and T = - = = 11259 K I Rb (O.08206dm3 atmK-I mol-I) x (O.0434dm3 mol-I) (b) By interpreting b as the excluded volume of a mole of spherical molecules, we can obtain an estimate of molecular size. The centres of spherical particles are excluded from a sphere whose radius is the diameter ofthose spherical particles (i.e. twice their radius); the Avogadro constant times the volume is the molar excluded volume b (47r(2r)3 ) b=NA ---3 so r = ~ (~)1 /3 2 47rNA ( )1/3 1 3(0.0434 dm3 mol-I) r = - I = 1.286 x 10-9 dm = 1.29 x 10-10 m = I 0.129 nm I 2 47r(6.022 x 1023 mol-) E1.21 (b) States that have the same reduced pressure, temperature, and volume are said to correspond. The reduced pressure and temperature for N2 at 1.0 atm and 25°C are P 1.0 atm Tr ___ T __ (25 + 273) K __ 2.36 Pr = - = = 0.030 and I, Pc 33.54 atm Tc 126.3 K The corresponding states are (a) For H2S P = PrPc = (0.030) x (88.3 atm) = 12.6 atm I T = TrTc = (2.36) x (373.2K) = 1881 K I (Critical constants of H2S obtained from Handbook of Chemistry and Physics.) (b) ForC02 P = PrPc = (0.030) x (12.85 atm) = 12.2 atm I T = TrTc = (2.36) x (304.2 K) = 1718 K I (c) For Ar P = PrPc = (0.030) x (48.00atm) = 11.4 attn I T = TrTc = (2.36) x (l50.12K) = 1356 K I THE PROPERTIES OF GASES 13 E1.22(b) The van der Waals equation is which can be solved for b RT -4 3 _, (8.3145JK-'mol-') x (288K) b = Vm ---a- = 4.00 x 10 m mol -(6 2) P+V2 4.0xI0 6Pa+ 0.76m Pamol- m (4.00 X 10-4 m3 mol-')2 = 11.3 x 10-4 m3 mol-' I The compression factor is z = pVm = (4.0 x 106 Pa) x (4.00 x 10-4 m3 mol-i) = 1 0.671 RT (8.3145 J K-' mol-') x (288 K) Solutions to problems Solutions to numerical problems P1.1 Since the Neptunians know about perfect gas behavior. we may assume that they will write p V = nRT at both temperatures. We may also assume that they will establish the size of their absolute unit to be the same as the oN. just as we write 1 K = 1°C. Thus pV(T,) = 28.0dm3 atm = nRT, = nR x (T, + OON). pV(T2) = 40.0dm3 atm = nRT2 = nR x (T, + 1000N). 28.0dm3 atm orT, =----­nR ° 40.0 dm3 atm T, + 100 N= ---­nR T, + 1000N 40.0dm3 atm - ° - . Dividing. = 3 = 1.429 or T[ + 100 N = 1.429T,. T, = 233 absolute UOltS. T, 28.0 dm atm As in the relationshi between our Kelvin scale and Celsius scale T = () -absolute zero(ON) so absolute zero (ON) = -233°N . COMMENT . To facilitate communication with Earth students we have converted the Neptunians' units of the pV product to units familiar to humans, which are dm3 atm. However, we see from the solution that only the ratio of pV products is required, and that will be the same in any civilization . Question. If the Neptunians' unit of volume is the lagoon (L). their unit of pressure is the poseidon (P). their unit of amount is the nereid (n). and their unit of absolute temperature is the titan (T). what is the value of the Neptunians' gas constant (R) in units of L. p. n. and T? P1.3 The value of absolute zero can be expressed in terms of a by using the requirement that the volume of a perfect gas becomes zero at the absolute zero of temperature . Hence 0= Vo[l + a(}(abs. zero)]. P1.S P1.7 14 STUDENT'S SOLUTIONS MANUAL 1 Then () (abs. zero) = --. ex All gases become perfect in the limit of zero pressure, so the best value of ex and, hence, () (abs. zero) is obtained by extrapolating ex to zero pressure. This is done in Fig. 1.2. Using the extrapolated value, ex = 3.6637 x 1O-3°C-I, or ()(abs.zero)=- I 3 I =1-272.95°cl, 3.6637 x 10-°C-· . which is close to the accepted value of -273.15°C. 3.672·· . i··~·· .; ..... ; .. i···;'··; . .. .... . . . . . . . . . ', .. !' .~ ... ~ ..... ~ .. : ... ~ .. : ... : .. ,! .. ~ .. ,! .. ~ :··"3.670' . .; ... ! .. -: ... ~ ..... ~ .. ! ...... ! .. -: .. ! .. ~ .. ~ •. ~ ... ~ .! .. : .... ,. '., .... ~ .. : .. ';' .. : .. ':' .. : .. ~ ... ~ ... " ... _." ;f'3.~$ · .;.: .. ~ .. ; .. ; ... ; . . . . 'U i •• ···;···i···; .. ·j···;···i.·,: .. i.·{·····I···:···I·· ~ G:~::rT: .~ .. : .. ':' .. ; .. ':' .. ; .. ; ... ~ ! ~.: .. ~ ... ; ..... ~ .. : ... : .. : .. : ... : .. ~ ... ; ~o 3.666· .~ .. ~ .. ~ .. : .. ~ ... " : -.: .. : ... ~ .. ···~··:··7·· : '''3.664' . . . . . . . . . . . -..•.• ~ •..••••. ~ "!"':- ":"~' ': .. ~ .. ': .. ~ ••. ~ .... _ ................ . · . . . . . . . . . . . . .. ~ .. ~ ... ~ .. : .. -:-.. : .. ~ .. ; .. , .. ,"! . .~ .. : .. ': .. : .. .:.. .. :. · ..... .. ~ ... , ................. , · . . . . . . .. ~ .. ; ... ~ .. : .. ~ ... ; .. ~ . . ,',. ,! .. ~ ...... .: .. ; ... ~ .. ~ .. : Figure 1.2 p nR. p P3 . - = constant, If n and V are constant. Hence, - = -, where P IS the measured pressure at T V T T3 temperature, T, and P3 and T3 are the triple point pressure and temperature, respectively. Rearranging , P=(~~)T. ~ ~~ I ... The ratio -is a constant = = 0.0245 kPa K-. Thus the change III P, t.p, IS proportIOnal to T3 273.16K the change in temperature, t.T: t.p = (0.0245 kPaK-I) x (t.T). (a) t.p = (0.0245 kPaK-I) x (LOOK) = I 0.0245 kPa I· (b) Rearranging,p= (!.-)P3 = (373.16K) x (6.69kPa) =19.14kPal. T3 273.16K (c) Since f is a constant at constant n and V, it always has the value 0.0245 kPa K -I ; hence T t.p = P374.15K -P373.15K = (0.0245 kPa K-1) x (1.00 K) = I 0.0245 kPa I· _ RT _ (8.206 x 10-2 dm3 atm K-I mol-I) x (350 K) -1125 d 3 I-I I (a) Vrn - - -. . m mo. P 2.30atm RT a RT (b) Fromp = ---2 [1.2Ib], we obtain Vrn = ( ) + b [rearrangeI.21b]. Vrn -b Vrn + a p V2 rn THE PROPERTIES OF GASES 15 Then, with a and b from Table 1.6, (8.206 x 1O-2dm3atmK-Imol-l) x (350K) 2 3 I Vm "'" 2 + (5.42 x 10-dm mol- ) (2.30 atm) + (6.260dm6 atm mol-2)/(12.5 dm3 mol-I) ) 28.72dm3 mol-I ( 2 3 ) I I "'" + 5.42 x 10-dm mol-I "'" 12.3dm3 mol-I. 2.34 Substitution of 12.3 dm3 mol-I into the denominator of the first expression again results in Vrn = 12.3 dm3 mol-I, so the cycle of approximation may be terminated . P1.9 As indicated by eqns 1.18 and 1.19 the compression factor of a gas may be expressed as either a virial expansion in p or in ( :m ). The virial form of the van der Waals equation is derived in Exercise 1.20(a) and is p = ~: { 1 + (b -RaT) x ( : m) + ... } Rearranging Z = pVm = I + (b -~) x (_1_) + ... 'RT RT Vm On the assumption that the perfect gas expression for V m is adequate for the second term in this expansion , we can readily obtain Z as a function of p. (a) Tc = 126.3 K. V m = (R;) x Z = R; + (b -:T) + ... (0.08206dm3 atm K-I mol-I) x (126.3 K) 1O.Oatm + {(0.0387 dm3 mol-I) _ ( 1.352 dm6 atm mol-2 ) } (0.08206 dm3 atm K-I mol-I) x (126.3 K) = (1.036 -0.092)dm3mol-1 =10.944dm3mol-ll. Z _ (l!...-) (V) _ (1O.0atm) x (0.944dm3 mol-I) - x m - =0.911. RT (0.08206dm3 atm K-I mol-I) x (126.3 K) (b) The Boyle temperature corresponds to the temperature at which the second virial coefficient is zero, hence correct to the first power in p, Z = 1, and the gas is close to perfect. However , if we assume that N2 is a van der Waals gas, when the second virial coefficient is zero, (b-~) =0 RTB ' a or TB =-. bR T 1.352 dm6 atm mol-2 B = = 426K. (0.0387dm3mol I) x (0.08206dm3atrnK-I mol-I) P1.11 16 STUDENT'S SOLUTIONS MANUAL The experimental value (Table 1.5) is 327.2 K. The discrepancy may be explained by two considerations. 1. Terms beyond the first power in p should not be dropped in the expansion for Z. 2. Nitrogen is only approximately a van der Waals gas. RT WhenZ = I, Vrn = -, and using TB = 327.2K p (0.08206 dm3 atm K-I mol-I) x 327.2 K 10.Oatm = 12.69dm3 mol-II and this is the ideal value of Vm. Using the experimental value of TB and inserting this value into the expansion for V rn above, we have 0.08206 dm3 atm K-1mol-1 x 327.2 K Vrn = ----------------------------10.Oatm { 3 I ( 1.352 dm6atm mol-2 ) } + 0.0387 dm mol-- 0.08206dm3atmK-lmol-1 x 327.2K = (2.685 -0.012) dm3mol-1 = 12.67 dm3 mol-II Vrn 2.67 dm3 mol-I and Z = -= = 0.992 ~ I. V;;' 2.69dm3 mol-I (c) TI = 621 K [Table 2.9]. 0.08206dm3atm K-Imol-I x 621 K Vrn = ----------~~------------1O.Oatm { 3 I ( 1.352 dm6atm mol-2 ) } + 0.0387 dm mol-- 0.08206 dm3 atm K -I mol-I x 621 K = (5.096 + 0.012) dm3 mol-I = 15.11 dm3 mol-II 5.11 dm3mol-1 and Z = = 1.002 ~ I. 5.lOdm3 mol-I Based on the values of TB and TJ given in Tables 1.4 and 2.9 and assuming that N2 is a van der Waals gas, the calculated value of Z is closest to I at I!i], but the difference from the value at T B is less than the accuracy of the method. _ molar mass _ M _ 18.02g mol-I -10 353 d 3 I-I 1 (a)Vm- . - - -.. 1 mmo . densIty p 1.332 x \02 gdm-3 _pVm _ (327.6atm) x (0.1353dm3mol-l) -1069571 (b) Z -[1.l7b] - I -. , RT (0.08206 dm3 atm K-I mol-) x (776.4 K) P1.13 THE PROPERTIES OF GASES 17 (e) Two expansions for Z based on the van der Waals equation are given in Problem 1.9. They are { 3 ( 5.464dm6 atmmol-2 )} = 1 + (0.0305dm mol-I)-(0.08206dm3 atm K-I mol-I) x (776.4 K) J x 0.J353dm 3mol-1 = 1-0.4084=0 .5916~0.59. Z = 1+ (RIT) x (b -:T) x (P) + ... I = I + -----,;--------;-----­ (0.08206dm3 atmK-1 mol-I) x (776.4K) x (0.0305 dm mol-I) - x 327.6 atm { 3 ( 5.464 dm6 atm mol-2 )} (0.08206dm3 atm K-I mol-I) x (776.4 K) = 1-0.2842 ~ lo.nl. I In this case the expansion in p gives a value close to the experimental value; the expansion in - Vrn is not as good. However, when terms beyond the second are included the results from the two expansions for Z converge. a Vc = 2b, Tc = 4bR [Table 1.7] Hence, with Vc and Tc from Table 1.5, b = ~ Vc = ~ x (118.8 cm3 mol-I) = 159.4 cm3 mol-I I. a = 4bRTc = 2RTc Vc Hence = (2) x (8.206 x 10-2 dm3atmK-1 mol-I) x (289.75K) x (118.8 x 10-3 dm3 mol-I) = 15.649 dm6 atm mol-21. (1.0 mol) x (8.206 x 10-2 dm3 atm K-I mol-I) x (298 K) (1.0dm3) -(1.0 mol) x (59.4 x 10-3 dm3 mol-I) x exp ( -(l.Omol) x (5.649dm6 atmmol-2) ) (8.206 x 10-2 dm3 atmK-1 mol-I) x (298K) x (1.0dm6atmmol-l) = 26.0atm x e-O.23T = 121 atm I. 18 STUDENT'S SOLUTIONS MANUAL Solutions to theoretical problems PU5 This expansion has already been given in the solutions to Exercise 1.20(a) and Problem 1.14; the result is C h· .. RT ( B C ) ompare t IS expansion wlthp = -1+ - + -+ ... [1.19] Vm Vm Vm2 and hence find I B = b -iT I and Ie = b21. Since C = 1200cm6 mol-2, b = CI/2 = 134.6 cm3 mol-II a = RT(b -B) = (8.206 x 10-2) x (273 dm3 atm mol-I) x (34.6 + 21.7) cm3 mol-I = (22.40 dm3 atm mol-I) x (56.3 x 10-3 dm3 mol-I) = 11.26 dm6 atm mol-21. P1.17 The critical point corresponds to a point of zero slope that is simultaneou sly a point of inflection in a plot of pressure versus molar volume. A critical point exists if there are values of p, V, and T that result in a point that satisfies these conditions. I " th, ,ri,kal poiOl. Th. -RTc V; + 2BVc -3C = 0 } at IS, 2 RTcVc -3BVc + 6C = 0 ~C~2 which solve to Vc = -, Tc = --. B 3RC Now use the equation of state to find Pc RTc B C (RB2) (B) ( B )2 ( B)3 iB3l Pc = V; -V2 + vg = 3RC x 3C -B 3C + C 3C = ~. PcVc (B3 ) (3C) (I) (3RC) III It follows that Zc = RTc = 27C2 x Ii x R x fi2 = llJ· P1.19 For a real gas we may use the virial expansion in terms of p [1.18] nRT RT I P = -(I + B'p + ... ) = p-(I + B P + ... ) V M