Problem Solutions

2 Problem SolutionsProblem Solutions 1

Problem SolutionsChapter 1 1. - 2. a) 4/52 b) 7.7x104 3. The result of 20 flips is closer to 50-50 on average. The probability of exactly 10 heads is small (0.18) 4. They stick together and lose mobility. 5. - 6. 225 7. Foolish. The probability each time is 1/2, regardless of past history. 8. Molecule: 3.1x10-10 m. Atom: about 2x10-10 m. Nucleus: about 4x10-15 m 9. 12 10. 7.3x10-11 m , 4.0x10-14 m, 9.5x10-37 m 11. a) 0.2 nm b) 3.3x10-24 kgm/s c) electron: 3.6x106 m/s, proton: 2.0x103 m/s d) electron: 38 eV, proton: 0.021 eV e) 6200 eV 12. a) 0.42 nm b) 1.5x1010 /m c) 6.3 (=2) d) 130 m, 0.0483 /m, 6.3 (=2) 13. Dpx=6.6x10-24 kgm/s, Dvx=7.3x106 m/s 14. a ) 3.0x105 b) 27 15. 13 16. 7.8x1080 17. a) 1.6x10-14 m b) 3.2 MeV 18. a) 38 eV b) 7.3x106 m/s (It could be going in either direction.) 19. a) x2 b) x2 c) x8 20. - 21. a) 7.6x1054 states/J = 1.2x1036 states/eV b) 8.4 states/eV 22. a) 35, 66, 90 b) 45, 90 c) 55 23. They are in a state with orbital angular momentum of l=1, and oriented opposite to the orientation of the two parallel spins. 24. || = 1.61x10-23 J/T, z = 9.27x10-24 J/T, U=9.27x10-24 J 25. || = 2.44x10-26 J/T, z = 1.41x10-26 J/T, U=1.41x10-26 J 26. about 10364 27. 7.4x10-24 J, 3.7x10-24 J, 0 J 28. a) 9.2x10-8 N b) 1.8x103 N/m c) 4.5x1016 /s d) 30 eV e) about 3 times greater 29. If the kinetic energy were zero, the momentum would be zero, the wavelength would be infinite, and so the particle would not be contained within the potential well 30. 4, 8, 21024 = 103.0x1023 31. 77 (2771023)Chapter 2 1. a) b) 2. 16 3. a) 33 1/6, b) 5 1/6 c) -4/3 d) 63 1/2 4. a) 4 1/2 b) 23 1/2 5. 7 7/8 6. -(1/2)B 7. a) 3/8 b) 6 c) hhtt, htht, htth, thht, thth, tthh, yes 8. a) 0.59 b) 0.33 c) 0.073 9. 1/36 10. a) 0.48 b) 0.39 11. a) 0.0042, 56 b) 0.0046 12. 0.156, 5 13. a) 0.313 b) 10 14. 0.165 15. 1820 16. 10375 17. 0.0252 18. 0.0224 19. a) 6.7x10-6 b) 4.5x10-2 c) 2.1x10-35 20. a) 4.0% low b) 1.7% low c) 0.8% low d) 0.4% low 21. 1.0x1029 22. a) 0.130 b) 0.072 c) 0.026 23. a) 0.062 b) 0.062 24. 25. - 26. a) no b)yes 27. 3/40 (There are 40 unseen cards and 3 of them are queens) 28. a) 1/13 b) 1/4 c) yes d) 1/52 29. 5.8 cents 30. a) 0.0467 b) 0.155 31. 12.25 32. a) 7 b) 7Chapter 31. a) 16.7 b) 3.7 c) 0.22 2. a) 1.67x107 b) 3.7x103 c) 2.2x10-4 3. (Justify each step in the Equation 3.3.) 4. a) 1018 b) 109 c) 10-9 5. a) 10 b) 2.6 c) 0.26 6. a) 1027 b) 2.6x1013 c) 2.6x10-14 7. a) 50 b) 5 c) 0.080 d) 0.067 8. a) 1014 b) 107 9. a) (1) Definition of mean values with (2) (3) binomial expansion, b) c) Just do the derivatives and evaluate them at p+q=1 (q=1-p, etc.). 10. a) 100 b) 9.1 c) A=0.0437, B=0.00600 d) 0.0437 e) 0.0326 11. 1011612. a) b) - c) - d) - e) - 13. We know with certainty (i.e., probability=1) that a system must be in one of its possible cofigurations. Therefore, the sum over all configurations must give a total probability of one. 14. a) A=0.0798, B=0.0200 b) 0.0798 c) 0.0737 d) 0.0484 e) 0.0108 f) 0.0016 15. 0.607 16. a) 200 b) 10 c) 0.0399 d) 3.3x10-4 17. a) 60 b) 7.07 c) 0.0564 d) 7.9x10-3 18. a) 50 b) 1.08x10-3 19. a) 0.0133 b) 5x10-198 20. a) 0.0798 b) 10-199 21. a) Gaussian=0.44, binomial=0.40 b) Gaussian= 0.040, binomial=0.054 c) Gaussian=0.24, binomial=0.34 22. a) Gaussian=0.138, binomial=0.137 b) Gaussian = 0.109, binomial = 0.116 c) Gaussian = 0.031, binomial=0.031 d) Gaussian=3.4x10-4, binomial1.8x10-5 23. a) Gaussian=0.178, binomial=0.176 b) Gaussian = 0.120, binomial = 0.120 c) Gaussian = 0.015, binomial = 0.015 d) Gaussian = 0.00030, binomial =0.00018 24. 25. 26. a) B=0.0139 b) A=0.0665 c) 0.0273 27. a) 0.20 m and 0.872 m b) 80 m and 17.4 m c) 4.4 and 0.22 28. a) 0.4 m and 1.2 m b) 160 m and 24 m c) 3 and 0.15 29. a) 0 b) 3.33x10-11 m2 c) 0 d) 0.026 m e) 1.08x105 sec = 30 hours 30. a) 0 and 5.77x10-12 m b) 0 and 2.8x10-4 m c) 0 and 0.21 m 31. a) 0 and 1.73x10-10 m b) 0 and 5.5x10-4 m 32. a) 0.01 m, 0.01 m b) 3 m, 0.17 m c) A=4.0x108 m-1, B=5.0x1017 m-2, x0=10-16 m 33. a) 2.5x10-14 m b) 5x10-10 m c) 12.6 A 34. a) 0 b) 5.8x10-3 m c) 1.5x1022 d) 4.6x106 years 35. 4.64 m, 36.2 m, 176 m (Using the standard deviation as the characteristic spread.) 36. a) : N=2 square terms, and N(N-1)=21 cross terms b) : N=3 square terms, and N(N-1)=32 cross terms 37. Chapter 41. - 2. Potential energy in the ice crystal is lower, because energy is released from the water even though its temperature remains unchanged as it freezes. (In section F we will learn that a second possibility is that the number of degrees of freedom per molecule decreases. However, this isn't the case for water that freezes.) 3. a) f'(0)=0, f"(0)=+2 b) f(x)=-2+0x+x2+x3+0x4 c) (-.57, +.54)4. a) f'(0)=0, f"(0)=+2 b) f(x)=-1+0x+x2+0x3-(1/2)x4+... c) (-.67, +.67)5. 6. a) 1.25x1029 b) 4x109 J c) W=(1/2)Cpf2 d) 2.8x1013 Pa = 2.8x108 atm e) 5x10-23 eV/Pa7. B, C, A 8. a) 0.04 nm, 0.02 nm b) 1.66x10-23 kgm/s, 3.32x10-23 kgm/s c) 0.11 eV d) about 1300 K 9. a) 1.42x10-46 kgm2 b) 4.9x10-4 eV c) 5.7 K 10. a) 1.95x10-46 kgm2 b) 3.6x10-4 eV c) 4.1 K 11. a) 4.58x10-48 kgm2 b) 1.5x10-2 eV c) 176 K12. 6 13. H2O is not a linear molecule as is N2. Therefore, H2O has appreciable moments of inertia about all three rotational axes, and so rotations about all three axes can be excited. For N2, however, the rotational inertia about one axis (the one that passes through both atomic nuclei) is extremely small, and so the energy of the even the first excited rotational state about this axis is too high to be reached. 14. 3 15. a) 6 b) 3 c) no d) decreased e) The iron atoms are released from the potential wells in which they were bound when in the solid state, so their potential energy in the liquid state is higher (although still negative). 16. At very low temperatures, only the 3 translational degrees of freedom. At intermediate temperatures, 2 rotational degrees of freedom become accessible, making 5 altogether. At very high temperatures, two vibrational degrees of freedom (kinetic and potential energy due to their vibrational motion along the molecular axis) become accessible, making 7 degrees of freedom altogether. 17. 20.8 J/moleK = 5.0 cal/moleK 18. a) -0.421 eV b) +0.097 eV c) -0.32 eV 19. a) 0.0707 eV b) 0.0707 eV c ) 0.0625 eV d) Du0=+0.0625 eV e) remains same 20. a) 510 m/s b) 1910 m/s c) 8.9x10-34 kgm2/s d) 8.4 21. a) 35.3 J (1.74x1022 degrees of freedom) b) 420 J (2.076x1023 degrees of freedom) 22. 0.70 kg food and 0.42 kg water.

Chapter 51. a) yes b) yes c) Although the air and walls emit more radiation altogether, only a small fraction of that hits the rock. 2. a) No, because if the pressure is zero, then no work is done (pdV=0).b) If it is not under pressure, then the molecules are not colliding with the walls. So their energy is not affected by the walls' motions. 3. a) It decreases, because the system does work, rather than having work done on it. b) As the system contracts, the potential energy of each particle decreases. (That is why it contracts. The particles are seeking the configuration of lower potential energy.) Like a ball rolling downhill, each particle's average kinetic (hence thermal) energy increases as the potential well deepens. But the gain in thermal energy is less than the loss in potential energy. 4. 172J or 41 cal, which is about 8% of the latent heat. 5. a) 120 J b) 796 K 6. a) -1.4x10-2 J b) +6.7x10-3 K7. u0 is negative, because thermal energy is released when the H2SO4 molecule enters the solution. The change in chemical potential is negative, because the molecules go into solution rather than out of solution. They go in the direction that lowers their chemical potential. 8. The intermolecular attractive forces are much stronger for the water molecules, making their potential well u0 much deeper. Correspondingly more energy is required to liberate them. 9. a) The temperature would fall because the same thermal energy would be distributed over more degrees of freedom, meaning less thermal energy per degree of freedom. b) The temperature would fall because the gain in potential energy leaves less energy to be distributed among the thermal degrees of freedom. 10. 194 K 11. 2760 cal/mol 12. 4150 cal/mol 13. 2560 cal/mol 14. a) +0.012 eV b) +0.006 eV c) -0.006 eV 15. 1.6x109 K 16. -0.51 eV 17. n=18.1 18. 0.029 eV 19. a) In liquid water the molecules are closer together. Their mutually attractive forces are stronger, and therefore the potential well is deeper at these closer distances. b) In ice, their reduced thermal motion allows them to orient and space themselves in a way that lowers their potential energy. c) Compared to molecules of oil, the water molecule is much more highly polarized, making stronger electrostatic attraction between the charged parts of the water molecule and the charged salt ions. 20. a) -0.174 eV b) -0.608 eV 21. a) 20.3 J b) 4.9 cal c) u0 rises. (Less thermal energy but the same total internal energy.) 22. a) 1550 J b) Internal energy includes only those things that are entirely internal to the object, and not interactions with things external to the object, such as an external magnetic or gravitational field. The torque that the external field exerts on the atomic magnets does tend to flip them into alignment, and this work done on them does increase their kinetic energy (work energy theorem), which gets distributed through all thermal degrees of freedom. So the thermal energy of the system does increase. 23. a) 1.52x105 J b) DT=-22 K 24. a) 0.98 liter b) 506 J c) -0.12 0C 25. a) exact b) exact c) inexact d) exact e) exact 26. a) exact b) inexact c) exact d) exact 27. (1) 47 (2) 47 yes 28. (1) 69 1/2 (2) 54 1/2 no 29. a) pi(Vf-Vi) b) pf(Vf-Vi) c) yes 30. a) 34 b) 106 c) 64 (In parts a and b, break each integral into two pieces. For each piece, one of the variables is constant. For part c, write y in terms of x and integrate over x only.) 31. a) 142 b) 142 c) 142 32. a) w2/y3 b) x3/w c) d) x3/z e) 33. a) 2w/y3 b) c) d) 34. a) x/v+2v2 b) u2v(u2+2v2) c) d) 1/v e) 2uv f) 35. y(z+1) 36. y2ey 37. 38. (3/2)VChapter 61. a) b) c) 2. a) b) c) 3. a) 10 b) 1012 c) 10900 d) 10900 e) 4. In equilibrium, the probabilities for all 16 possible arrangements are the same. That doesn't mean that the distribution is even. For example, in equilibrium there is 1 chance in 16 that all 4 molecules might be in the front half of the room, 4 chances in 16 that there are 3 in front and 1 in back, etc. We should expect to see these very uneven distributions the appropriate share of the time if the system is in equilibrium. 5. a) +++, ++-, +-+, -++, +--, -+-, --+, --- b) 3/8 c) 2/8 6. a) 6/16 (same as binomial prediction) b) 1/16 c) 2/6 7. a) 1.3x10-24 b) 0.30 c) 0.48 8. 9. a) 2.9x10-25 eV b) 11.4x10-25 eV c) 8.6x10-25 eV 10. 1.7x10-13 eV 11. a) hh, ht, th, tt b) hhh, hht, hth, thh, htt, tht, tth, ttt c) hhhh, hhht, hhth, hthh, thhh, hhtt, htht, htth, thht, thth, tthh, httt, thtt, ttht, ttth, tttt d) yes 12. a) 36 b) 21 13. a) 64 b) 7 14. a) 7776 b) 288 c) 15. a) 1.1x109 b) 3.5x109 c) 3.7x1018 16. a) 6 b) 3 17. a) 20, 10 b) 9900, 4950 c) 9.7x105 , 1.6x105 d) 1.0x109 1.7x108 18. a) microscopic b) 2 19. a) macroscopic b) 4.6x1026 20. a) b) 21. a) 8.4x1024 b) 1.26 c) 22. a) 3.8x104 J b) 1.05x10-66 (kgm/s)3 c) 5.4x1032 d) 6.7x108 e) 23. a) 3.0x107 J b) 4.64x10-67 (kgm/s)3 c) 2.0x1035 d) 1.82x108 e) f) wc(N2)=2.29x108 , wc(O2)=8.63x108, W= 24. a) 1.22x103 J b) 1.16x10-66 (kgm/s)3 c) 3990 d) 3990 e) 25. When it melts, the volume in coordinate space accessible to each molecule increases immensely, because the molecules become mobile and can move throughout the fluid. This volume is much larger and affords access to many more states than did the much more restricted volume that was represented in the three potential energy degrees of freedom of the solid state. 26. a) b) c) The factor (N/e)N is much larger than (by a factor like NN-1/2).Chapter 71. a) Set df/dE=0 for each, which gives , where n is the first exponent, and m the second. For all three cases the ratio n/(n+m) is 0.4, giving E=2. b) f(2)/f(1) = 1.7, 187, , f(2)/f(3)= 1.5, 58, 2. 3. a) (E1,E2,W1,W2,W0)= (0,4,0,16,0), (1,3,1,9,9), (2,2,2.8,4,11.3), (3,1,5.2,1,5.2), (4,0,8,0,0) total=25.5 b) 0.20 c) (2,2), 0.44 4. a) (E1,E2,W1,W2,W0)= (0,4,0,64,0), (1,3,1,27,27), (2,2,5.7,8,45.3), (3,1,15.6,1,15.6), (4,0,32,0,0), total=87.9 b) 0.31 c) (2,2), 0.52 5. a) (E1,E2,W1,W2,W0)= (0,6,0,216,0), (1,5,1,125,125), (2,4,4,64,256), (3,3,9,27,243), (4,2,16,8,128), (5,1,25,1,25), (6,0,36,0,0), total=777 b) (2,4), 0.336. a) (E1,E2,W1,W2,W0)= (0,6,0,2.21x1023,0), (1,5,1,9.31x1020, 9.31x1020), (2,4,1.05x106,1.15x1018,1,21x1024), (3,3,3.49x109,2.06x1014,7.19x1023), (4,2,1.10x1012,1.07x109,1.18x1021), (5,1,9.54x1013,1,9.54x1013), (6,0,3.66x1015,0,0), total=1.93x1024 b) (2,4), 0.63 7. a) (E1,E2,E3,W1,W2,W3,W0)= (4,0,0,16,0,0,0), (3,1,0,9,1,0,0), (3,0,1,9,0,1,0), (2,2,0,4,5.7,0,0), (2,1,1,4,1,1,4), (2,0,2,4,0,8,0), (1,3,0,1,15.6,0,0), (1,2,1,1,5.7,1,5.7), (1,1,2,1,1,8,8), (1,0,3,1,0,27,0), (0,4,0,0,32,0,0), (0,3,1,0,15.6,1,0), (0,2,2,0,5.7,8,0), (0,1,3,0,1,27,0), (0,0,4,0,0,64,0), total=17.7 b) (1,1,2), 0.458. a) (E1,E2,W1,W2,W0)= (0,4,0,1024,0), (1,3,1,243,243), (2,2,4,32,128), (3,1,9,1,9), (4,0,16,0,0), total=380 b) 0.64 9. All states have to be equally probable in order for the respective probabilities to be proportional to the number of accessible states. 10. 11. a) 10 b) c) 12. a) 10 b) 1010 c) 100 d) 10900 e) f) 13. 14. Same as Table 7.3 except that for every power of 10 that appears there, you double the power. For example, , etc. 15. a) (E1,E2,W1,W2,W0)= (0,5,0,exp6.99x1024,0), (1,4,1,exp6.02x1024,exp6.02x1024), (2,3,exp3.61x1024,exp4.77x1024,exp8.38x1024), (3,2,exp5.73x1024,exp3.01x1024,exp8.74x1024), (4,1,exp7.22x1024,1,exp7.22x1024), (5,0,exp8.39x1024,0,0) 16. The change is a factor of 17. 18. m1/(m1+m2) is the fraction of the total number of degrees of freedom that is in system #1. So the energy in system #1 is in proportion to the number of degrees of freedom there. 19. a) (E1,E2,W1,W2,W0)= (0,4,0,exp3.61x1024,0), (1,3,1,exp2.86x1024,exp2.86x1024), (2,2,exp0.60x1024,exp1.81x1024,exp2.41x1024), (3,1,exp0.95x1024,1,exp0.95x1024), (4,0,exp1.20x1024,0,0), total=exp2.86x1024 b) , or one chance in 20. a) ax=(10log10a)x=10xlog10a b) Use the previous result with a=e c) The Taylor series expansion of ln(1+e) around the point e=0 gives ln(1+e)=0+e-e2/2+...=e(1-e/2+...)e, for e0. 8. This must be a non equilibrium process where the system gains volume in coordinate or momentum space without losing volume in the other. (Ex. removing a partition, or a chemical explosion.) 9. The condition 9.3a applies to systems interacting thermally only, which precludes particles entering or leaving (i.e. diffusive interactions). 10. If the system expands, the temperature may drop even if heat is added. Etc. 11. crystal2(H2O) b) The valence electrons are in lower energy orbitals in the H2O than in the H2 and O2 molecules, on average, so the potential energy reference level, u0, is lower. 21. a) 2.04x10-19 J b) 41.7 K, 0.288x10-19 J c) 0.141 for both 22. a) 0.202 J b) 4.17x10-8 K, 2.88x10-11 J c) 1.41x10-10 for both 23. a) 3780 J b) 3.78x10-9 J c) 1.0x10-12 24. a) 3740 J b) 3.94x10-9 J c) 1.05x10-12 15. 2.25 mm 26. a) 3.01x1024 b) 154 K c) 8.2x10-11% 27. a) 3.35x1010 b) 3.2x10-6 28. - 29. a) 6.1x10-3 b) 6.1x10-12 30. a) 1.2x1010 b) 1.1x105 c) 9x10-6 31. - 32. - 33. a) You could seal the oven and use the air pressure as a measure of its temperature. b) Yes. (You wouldn't be able to get food in or out without breaking the pressure seal.) 34. Constant volume. If you used constant pressure, the volume would decrease at lower temperatures, meaning that interatomic spacings would get smaller and interactions would get stronger. This would mean greater deviation from the ideal gas law (pressure varies linearly with temperature) and would also mean that the gas liquefies sooner. 35. a) 46 C b) -281 C 36. a) water b) water c) The greater pressure favors the phase of lower volume, which is liquid for water, but not for glass. d) At lower pressure, it is easier to expand, so the water doesn't have so much trouble freezing. e) Higher pressure favors the denser liquid phase, so the boiling point rises under increased pressure, and falls under reduced pressure. 37. - 38. a) A2 acquires particles because 2 is lower. b) N11+N22 c) , , DG= 39. Whatever the values of TS and pV for the system, these values could be have been obtained in an infinite number of different ways, with different heat transfers and different amounts of work. For example, imagine starting with a system in a canister at fixed volume near absolute zero, where the pressure is nearly zero (pV0). Then simply add heat until the pressure rises to some arbitrary value. This value of pV was obtained without any work at all having been done. Similarly, you could get any value of TS through volume changes without any heat at all having been added. 40. a) 189.3 J/K b) Dp=2.1x107 Pa 41. Take the differential and subtract dE=TdS--pdV+dN. 42. - 43. In each case, write out the full derivative of the appropriate function in Equation 9.14' (e.g., dF=-pdV-Vdp+dN+Nd) and then substitute in -SdT+Vdp for Nd. 44. In each case use the integrated energy expression (9.12) in for E. 45. Just carry out the steps for these proofs that are outlined in Sections F.2, F.3, and F.4. 46. dF=-SdT-pdV+dN. dF=-pdV=work done only if dT=dN=0 47. dH=TdS+Vdp+dN. dH=TdS=heat added only if dp=dN=0 48. Two interacting systems. DF0=DF1+DF2=-(p1-p2)DV+(1-2)DN, since DT=0. Make A2 a reservoir so that p2 and 2 are constant. Write =p1+Dp/2 and =1+D/2. To first order DF0=0, because p1=p2 and 1=2 in equilibrium. To second order. DF0=-DpDV+DDN, and both these terms must be positive according to the results 9.3. If first order terms are zero and second order terms are positive, then the function is a minimum. 49. Two interacting systems. DG0=DG1+DG2=(1-2)DN, since DT=Dp=0. Make A2 a reservoir so that 2 is constant. Write =1+D/2. To first order DG0=0, because 1=2 in equilibrium. To second order. DG0=DDN/2, and this term must be positive according to the results 9.3c. If first order terms are zero and second order terms are positive, then the function is a minimum. 50. f/z)y=g/y)z 51. Each has the form dw=fdx+gdy+hdz, so you will get the relations corresponding to f/y)x,z=g/x)y,z, f/z)x,y=h/x)y,z , g/z)x,y=h/y)x,z 52. a) Change the volume by DV adiabatically, and measure DT . b) Add heat (DS=DQ/T) at constant volume, and measure Dp. c) Add heat (DS=DQ/T), and measure how much the volume must change (DV) in order to keep the temperature the same. d) Add heat to a system held at constant volume and measure the changes in pressure and temperature (Dp,DT). 53. a) Add DN particles. To ensure that T and p retain their original values, two things must be adjusted, such as the volume and adding or removing heat until the two are at their original values. Then record the net change in volume, DV b) Add heat (DS=DQ/T) to a system of fixed number of particles. adjust the volume (DV) so that the pressure returns to its original value. c) Add heat to a system of a fixed number of particles kept at constant volume, Record both Dp and DT. 54. a) (M12) b) From Nd=-SdT+Vdp, we get (/T)p,N=-(S/N)=0, because S=0 at absolute zero. 55. Solve for p, with wc=1 and the given V/N. KE= p2/2m. a) 6.4 eV b) 0.30 MeVChapter 101. 4 2. a) p=NkT/V, p/V)T,N=-NkT/V2=-p/V b) E=(3/2)NkT, E/p)T,N=03. a) 1/T=CV2/E or E=CV2T b) p/T=2CVlnE c) E=CV2T, E/V)T=2CVT=2E/Vd) p=(2x/V)ex, where x=S/CV2 p/V)S=(-2x/V2)(3+2x)ex 4. example: m=-CT/Np, where C is a positive constant 5. E=(n/2)pV, E/p)V,N=(n/2)V 6. - 7. - 8. -9. a) 2 b) 2N c) 41.4 J d) W=(CAE/N2)N e) S=Nkln(CAE/N2) 10. a) N b) c) d) dE=TdS-FdL, e) FL=NkT11. a) S=klnC+bkV4/5+2NklnE b) E=2NkT c) 4N d) pV1/5=(4/5)bkT 12. pressure=-force/area, with force=Dp/Dt, Dp=2mvx, Dt=2X/vx, area =YZ. Finally, use the average value of (1/2)mvx2=(1/2)kT. 13. a) N/2 b) flux=(density)x(velocity)=number per area per second. So (density)x(velocity)x(area)=number per second=(N/2V)(vx)(A) c) 2mvx d) force=Dp/Dt = (number per second)x(impulse per particle) =(N/2V)(vx)(A)( 2mvx) e) divide force by the area; p=(N/V)mvx2 f) use the average value of (1/2)mvx2=(1/2)kT 14. We can separate the units by using ln(ab)=ln(a)+ln(b). For example, if E is 4J, then ln(E)=ln(4)+ln(J). At the end, combine all the unit terms into one, and they all cancel, giving us ln(1)=0 for the units term. 15. m=kT[(n+2)/2-lnwc] = (E+pV-TS)/N 16. Changes in volume cause changes in T, which is reflected in the equation E=(3/2)NkT. 17. S=klnC+kaV1/2+kbVlnE and use Equations 8.10 a) T=E/(kbV) b) p=kT[(a/2V1/2)+blnE] 18. S=klnC+kaNV2+kbNVlnE and use Equations 8.10 a) T=E/(kbNV) b) p=NkT[2aV+blnE] c) m=-kT[aV2+bVlnE] 19. H=E+pV=(6/2)RT+RT =25,700 J 20. 6.0x1027 molecules, 3.4x10-9 m 21. 28.97 g, 28.64 g 22. 3260 m3 23.a) wc=3.8x106, S=126 J/Kmole b) wc=1.3x1011, S=213 J/Kmole c) wc=250, S=45.9 J/Kmole 24. E=6130 J, F=8580 J, H=-56,700 J, G=-54,300 J 25. a) DH=7540 J/mole, DS=23.5 J/Kmole, DE=7540 J/mole b) DH=4.07x104 J/mole, DS=109 J/Kmole, DE=3.76x104 J/mole c) DH=3570 J/mole, DS=8.5 J/Kmole, DE=2740 J/mole 26. p(V-b)=NkT 27. a) 3.1x10-10 m b) 33x10-10 m c ) 11 28. a) acetone b) water 29. Add coulomb repulsion, 1/r21/u2/3, to the pressure term. (p+a/v2-c/v2/3)(v-b)=RT 30. (p-c/v2/3)v=RT 31. water: v=0.018 l = 0.6b ethyl alcohol: v=0.058 l = 0.7b 32. [p+a(v-2b)/v3](v-b)=RT or [p+(a/v2)(1-e-c(v-2b))][v-b]=RT , etc.33. a) From p/T=S/V)E,N get S=(AV/T)[1+B(1-V/2V0)]+f(E,N) (or other equivalent forms) where f is any const. or function of E and N. b) W=eS/k, where S is given in part (a). 34. u0= -0.441eV+6k(T-273K) 35. a) xi'= b) mi= 36. a) 193 m/s b) 116,000 m/s, which is about 600 times faster than the motion of the atoms 37. -4.7 eV 38. a) 1.06x1022 b) 5.9 mm/s c) vrms=116,000 m/s is 2x1010 times larger. 39. a) 2.04x10-21 J b) k=82 N/m c) f=6.8x1012 Hz 40. k=1/p=0.99x10-5 m2/N 41. cp-cv = pv/T)p+NA[u0/T)p-u0/T)V] 42. cp-cV = 16.3 J/Kmole, 530 K 43. a) polyatomic b) 8 44. Write pV/T)p=pVb = 3.02x10-5 J/K, whereas cp=3R=24.9 J/K. So cp is larger than cV by only 1.2x10-4 %. 45. a) dV=(1/a)[-(2/p)dp+(1/3T)dT] b) k=2/(apV) c) b=1/(3aVT) 46. b=R/pvB, where B=[1--a/pv2+2ab/pv3] 47. a) 0 b) -p/V c) -gp/V 48. a) dv=(R/pB)dT-[vA/pB]dp, where A=(1-b/v) and B=[1--a/pv2+2ab/pv3] b) - c) R/Av 49. You should get the differential form; V2dp+(2pV-aT)dV-(aV+b)dT=0, from which you can read off the answers to parts a and b. a) b=(a+b/V)/(2pV-aT) b) k=V/(2pV-aT) 50. a) cp-cV=2.5x10-5R, compared to the real value of 4.5x10-2R b) b=1.13x10-4/K, compared to the real value of 2.1x10-4/K 51. a) p=83.5 atm b) b=2.1x10-3 /K 52. differential form: Adp+Bdv=RdT, with , a) b=R/vB b) k=A/vB c) cp-cV=pR/B 53. differential form: Cdp + Ddv = RdT , with C=eBvv , D=eBv(p+pvB+AB) a) b=R/vD b) k=C/vD c) cp-cV=pR/D 54. a) 3.03 b) 18.1

Chapter 111. definition of Cp 2. definition of CV 3. S/p)T=-V/T)p=-Vb 4. =(S/T)V(T/p)V =(CV/T)[-(V/p)T/(V/T)p]=(CV/T)Vk/Vb 5. =(S/p)T(p/V)T=(S/p)T/(V/p)T=(-Vb)/(-Vk)=b/k 6. =(S/T)p/(V/T)p=(Cp/T)/Vb 7. (T/p)S=(V/S)p=(V/T)p/(S/T)p=Vb/(Cp/T)8. =-(V/p)T/(V/T)p=Vk/Vb 9. =-(p/S)V=-1/(answer to problem 4) 10. =1/(V/T)p=1/Vb11. =-(S/V)p/(S/p)V=-(Cp/TVb)/(kCV/bT) 12. =1/(V/p)T=-1/Vk13. dV=(V/p)T dp+(V/T)pdT=-Vkdp+VbdT 14. dS=(S/T)pdT+(S/p)Tdp=(Cp/T)dT-Vbdp , dV=(V/T)pdT+(V/p)Tdp=VbdT-Vkdp together: dE=TdS-pdV=(Cp-pVb)dT+V(pk-Tb)dp15. dE=T(S/T)VdT+[T(S/V)T-p]dV=CVdT+[Tb/k-p]dV 16. dV=(V/S)TdS+(V/T)SdT= (k/b)dS-(kCV/Tb)dT 17. dE=T(S/p)Vdp+[T(S/V)p-p]dV=(CVk/b)dp+[(Cp/Vb)-p]dV18. dE=[T-p(V/S)T]dS-p(V/T)SdT=[T-pk/b]dS+(CVpk/Tb)dT 19. dT=(T/p)Vdp+(T/V)pdV= (k/b)dp+(1/Vb)dV 20. dS=(S/T)pdT+(S/p)Tdp=(Cp/T)dT-Vbdp 21. a) dE=[T-p(V/S)p]dS-p(V/p)Sdp, solve for dS b) dS=(1/A)dE-(pVCVk/CpA)dp, with A=T(1-pVb/Cp) 22. a) dN=dV=0 b) 1 c) dp=(p/T)VdT d) dp=(b/k)dT 23. a) dN=dV=0 b) dS=(S/T)VdT c) dS=(CV/T)dT 24. a) dV=(V/E)SdE+(V/S)EdS=-(1/p)dE+(T/p)dS b) dV=(V/p)Sdp+ (V/S)pdS=-(VCVk/Cp)dp+(TVb/Cp)dS c) dV=(V/p)Tdp+(V/T)pdT=Vkdp+VbdT d) dS=(S/p)Tdp+(S/T)pdT=-Vbdp+(Cp/T)dT 25. a) 1 b) dS=(S/N)T,pdN c) dS=-(/T)p,NdN 6. a) dE=[T(S/N)T,p-p(V/N)T,p+]dN b) dE=[-T(/T)p,N-p(/p)T,N+]dN 27. b=4.0x10-5 /K (Ignore the smaller terms of order DT2 and DT3.) 28. a) and b) Add heat to the system in the cylinder and let it expand under constant pressure. You must record the values of (V,DV,DT) to calculate b=(1/V)(DV/DT)p. c) This will be done the same as parts (a) and (b), but with a complication for measuring the volume and change in volume for the solid (Vs and DVs). Put water in the cylinder and measure its volume, Vw. Then add the solid, and the increase in volume is its volume, Vs. Now add the heat and record the change in temperature and volume (DT and DVtotal). Since you already know bw from the previous measurements, you can calculate DVw=bDT and subtract this from DVtotal to get DVs. (DVs=DVtotal-DVw) 29. a) Hold the gas in the cylinder at constant volume. Add heat DQ and measure the temperature change, DT. CV=DQ/DT)V b) Put the liquid in the cylinder and let the volume change to keep the pressure constant as you add DQ and measure DT. Cp=DQ/DT)p. c) Put volume V of the liquid in the cylinder and measure the change in volume DV as you change the pressure by Dp. You will have to add or remove heat to keep the temperature constant as you do this. k=(1/V)(DV/Dp)T d) Immerse the solid in the liquid in the cylinder, and measure Vtotal and DV total as above. To get Vs and DVs for the solid alone, you have to subtract that of the liquid, determined as above. (Vs=Vtotal-Vw , DVs=DVtotal-DVw) k=(1/V)(DV/Dp)T 30. dz=0=(z/x)ydx+(z/y)xdy Now use this to solve for the ratio dx/dy when dz=0. 31. dE=TdS-pdV=T(S/T)VdT+[T(S/V)T-p]dV=CVdT+[Tb/k-p]dV 32. Show that (Tb/pk)-1=(RT/pvA)-1, and you sthen juggle the van der Waals equation around to get RT/pvA=(1+a/pv2). 33. a) 0.019 b) 13 34. a) dE=TdS-pdV, with dS=(S/T)pdT+(S/p)Tdp and dV= (V/T)pdT+(V/p)Tdp. Collect terms. b) Use Table 11.1 to convert the partial derivatives into (Cp,CV,k,b,T,p,V). 35. Write DS=(S/p)VDp+(S/V)pDV and then use Table 11.1 to convert the two partials into (Cp,CV,k,b,T,p,V) 36. Use 2S/VT=2S/TV and Maxwell's relation (M4) for S/V)T. 37. Use 2S/pT=2S/Tp and Maxwell's relation (M10) for S/p)T. 38. a) p=NkT/V, so p/T)V=Nk/V, and 2p/T2)V=0 b) p=RT/(v-b)-a/v2, so p/T)V=R/(v-b), and 2p/T2)V=0 39. a) - b) 4.9x106 Pa, 4.2x105 Pa 40. a) - b) 19.7 J/Kmole 41. a) V=NkT/p, so V/T)p=Nk/p, and 2V/T2)p=0 b) From the van der Waals equation, Avdp+Bpdv=RdT, where A=(1-b/v) and B=(1-a/pv2+2ab/pv3), we get (v/T)p=R/pB. The second derivative is (2v/T2)p=-(R/pB2)(B/T)p, where (B/T)p=(2a/pv3)(1-3b/v)(v/T)p, and this last derivative is given above (R/pB). 42. (replace p by -B, and V by M everywhere) T/M)S=B/S)M (M1), S/M)T=-B/T)M (M4), -T/B)S=M/S)B (M7), S/B)T=M/T)B (M10) 43. DE=T(S/T)MDT+[T(S/M)T+B]DMChapter 121. a) dS=(Cp/TVb)dV b) dS=(Cp/T)dT 2. a) dE=[(Cp/Vb)-p]dV b) dE=[Cp-TVb]dT 3. dV=VbdT4. a) dH=CpdT b) dF=-[S+pVb]dT c) dG=-SdT 5. a) dH=(Cp/Vb)dV b) dF=-[(S/Vb)+p]dV c) dG=-(S/Vb)dV 6. dp=dT=0 7. a)dE=[(Cp/Vb)-p]dV+[T(S/N)p,V+]dNb) dE=[T(S/)p,N-p(V/)p,N]d+[T(S/N)p,-p(V/N)p,+]dN 8. a) dS=(b/k)dV b) dS=-Vbdp 9. a) dE=[(Tb/k)-p]dV b) dE=[-TVb+pVk]dp 10. dV=-Vkdp 11. a) dH=[-TVb+V]dp b) dF=pVkdp c) dG=Vdp 12. a) dH=[(Tb/k)-(1/k)]dV b) dF=-pdV c) dG=-(1/k)dV 13. (for an ideal gas, b=1/T, k=1/p) 14. a/v2 15. a) 530 K b) 800 K c) 61 atm 16. a) nRTln(Vf/Vi) b) nRTln(Vf/Vi) 17. a) Use pV=NkT to eliminate the volume V, and integrate. b) 5.5 km 18. - 19. 9 20. DW=[piVi/(g-1)][1-(Vi/Vf)g-1] 21. a) Start with the rearranged first law, dQ=dE+pdV, with E=(Nn/2)kT, and use the definitions CV=Q/T)V, and Cp=Q/T)p. b) D(E+pV)=D[(n/2)NkT+NkT] =[(n+2)/2]NkDT=CpDT 22. a) T=410 K, p=1 atm b) T=273 K, p=0.667 atm c) T=232 K, p=0.567 atm 23. a) 507 J, 410 J, 379 J b) 1770 J, 410 J, 0 J c) 1260 J, 0J, -379 J d) 1770 J, 0J, -531 J 24. Tf/Ti=exp[(Vb/C)(pf-pi)] Tf/Ti=exp[-(b/Ck)(Vf-Vi)] Vf/Vi=exp[-k(pf-pi)] 25. a) dE=(pVCVk/Cp)dp b) dE=-pdV c) dE=(pCVk/Tb)dT 26. a) 283 K b) inversion 27. a) 1.01x104 Pa/m b) Use DT/T=(vb/C)Dp and get DTadiabatic=0.8x10-4 K/m, which is greater than the lapse rate, so it is stable. 28. Use b=(1/V)(V/T)p and k=-(1/V)(V/p)T with pV=NkT to get b=1/T, k=1/p 29. As moist air rises and cools adiabatically, some of the moisture condenses, releasing latent heat. Consequently, rising moist air cools less rapidly with altitude than rising dry air. If the lapse rate falls between these two, then the moist air will continue to rise, whereas the dry air won't. 30. Use dT=(T/p)Sdp dT=(T/V)SdV dV=(V/p)Sdp and Table 11.1 31. V=.226 l, T=531 K, Work=205 J. 32. a) B=gp b) 341 m/s c) 33. yes 34. no 35. While moving, a part of the motions of all molecules is coherent - all moving the same direction together. Friction turns this coherent motion into random thermal motion. To go back, this random thermal motion would have to turn back into synchronized coherent motion - all molecules going the same direction together. This is very unlikely - a state of very low entropy. Like flipping 1024 coins and having them all land heads. 36. Although the initial kinetic energy of the book is not part of the internal energy of either system individually, it is part of the internal energy of the combined system. This kinetic energy in the combined system gets turned into thermal (internal) energy in the individual systems. 37. -0.32 C 38. a) 141 K b) 273 K 39. a) 324 K b) 3.92x10-3 /K c) 0.23 C/atm 40. a) 2v(2av-bpv2-3ab)/D, where D=(pv3-av+2ab)(n+2)R b) -2a/(v2nR) 41. Tf-Ti=(2a/nR)(1/vf - 1/vi) 42. a) 796 K b) 721 K 43. a) DE=DQ=DW=0 b) DE=-DW=-[(NkTi)/(g-1)][1-(Vi/Vf)g-1], with g=(n+2)/n, DQ=0 c) DE=0, DQ=DW=NkTiln(Vf/Vi) 44. Yes, no heat enters or leaves the gas. No. Volume is gained under unequal pressures, so entropy increases. (Or, the accessible volume in coordinate space increases with no change in the volume in momentum space, so the number of states per particle increases.) 45. a) 256 b) 25 c) 9 46. a) wc2N b) (2wc)2N (The accessible volume per particle doubles for each gas.) c) wc2N d) The volume doubles, but so does the number of particles, and wc depends on the ratio of the two. 47. series: parallel: 48. a) 6.6x10-2 W/m2 b) about 2700 times greater 49. a) Rw=3.5x10-2 K/W b) Ri=9.5x10-2 K/W c) Rs=21.6x10-2 K/W d) Rtotal=2.30x10-2 K/W e) 6.9x109 J f) electricity: $286, gas: $95 50. Take the second derivative with respect to x, and compare it to 1/K times the first derivative with respect to t. 51. Take the second derivative with respect to x, and compare it to 1/K times the first derivative with respect to t. Then use the fact that the Gaussian factor at t=0 is an infinitely narrow spike beneath which the area is 1 to show that at T(x,t=0)=f(x).Chapter 131. a) DQ= [(n+2)/2](nRTi/Vi)(Vf-Vi) , DW=(nRTi/Vi)(Vf-Vi) b) DQ=DW=nRTiln(Vf/Vi)c) DQ=0, DW=(nnRTi/2)[1-(Vi/Vf)2/n] [(g-1)=2/n] 2. One way is to use DW=-DE=-CV(Tf-Ti)=-(n/2)nR(Tf-Ti)=-(n/2)(pfVf-piVi) with n/2=1/(g-1). Or you might integrate pdV with p=const/Vg, and const =piVig=pfVfg. 3. 1730 J 4. The diagram will have "-F" on the vertical axis and "L" on the horizontal axis. 1) Isothermal heat addition - down and to left. 2) Adiabatic contraction - down steeper to left. 3) Isothermal heat removal - up to right. 4) Adiabatic extension - up steeper to right. 5. The diagram will have "-F" on the vertical axis and "L" on the horizontal axis. 1) Constant length heat addition - straight up. 2) Adiabatic contraction - down and to left. 3) Constant length heat removal - straight down. 4) Adiabatic extension - up to right. 6. a) 0.071 moles b) 571 K, 8.47 atm c) 385 J d) 1371 K e) 1180 J f) 924 J g) 0.46 7. a) p2,V1 ; p2(V1/V3)g,V3 ; p1(V1/V3)g,V3 ; p1,V1 b) DQ=(p2-p1)V1/(g-1), DW=0 ; DQ=0, DW=[p2V1/(g-1)][1-(V1/V3)g-1] ; DQ=(p1-p2)(V1/V3)gV3/(g-1), DW=0 ; DQ=0, DW=[p1V1/(g-1)][(V1/V3)g-1-1] 8. a) p1,V2 ; p1(V2/V3)g,V3 ; p1(V2/V3)g,(V1V3/V2) ; p1,V1 b) DQ=[g/(g-1)]p1(V2-V1) , DW=p1(V2-V1) ; DQ=0 , DW=[p1V2/(g-1)][1-(V2/V3)g-1]; DQ=[g/(g-1)]p1(V2/V3)g-1(V1-V2), DW=p1(V2/V3)g-1(V1-V2); DQ=0, DW=[p1V1/(g-1)][(V2/V3)g-1-1] 9. They will look like the diagrams of Figure 13.5. 10. (DQ,DW,DE)=(5,0,5), (0,4,-4), (-2,0,-2), (0,-1,1) ; e=0.6 11. a) straight across, slope down, straight down, slope up to left (Adiabatic is steeper than isothermal, and both are concave upward.) b) Slope up, straight across, slope down to left, straight up. (Isochoric is steeper than isobaric, and both are concave upward.) c) (DQ,DW,DE)=: (1) +++, (2) ++0, (3) -0-, (4) 0-+ 12. a) slope down, straight down, straight to left, slope up to left. (Adiabatic is steeper than isothermal, and both are concave upward.) b) straight across, down to left, down to left, straight up. (Isochoric is steeper than isobaric, and both are concave upward.) c) (DQ,DW,DE)=: (1) ++0, (2) -0-, (3) ---, (4) 0-+ 13. a) slope down, slope down, straight to left, straight up. (Adiabatic is steeper than isothermal, and both are concave upward.) b) straight across, straight down, slope down to left, slope up to right. (Isochoric is steeper than isobaric, and both are concave upward.) c) (DQ,DW,DE)=: (1) ++0, (2) 0+-, (3) ---, (4) +0+ 14. a) A rectangle whose corners are at (p0V0), (p0,2V0), (p0/2,2V0), p0/2,V0) b) T0=p0V0/nR c) In terms of (nRT0), (DQ, DW, DE)= ((n+2)/2, 1, n/2), (-n/2,0,- n/2), (-(n+2)/4,-1/2, - n/4), (n/4,0,n/4) d) 2/(3n+4) e) 3/4 15. a) 6 C b) 600 kg/s 16. negative 17. The gas turbine's hot reservoir is slightly hotter than the coil, because heat is flowing from the reservoir to the coil. Vice versa for the refrigerator. 18. a) 0,1 b) 0, 19. a) 2T/(nR) b) 2T/[(n+2)R] c) 0 d) The isochoric and isobaric cases curve upward because they slope upward and the slope increases with increasing T. 20. a) 0 b) -p/V c) -gp/V d) The isothermal and adiabatic cases are concave upward because they slope downward and the slope decreases with decreasing p and increasing V. 21. a) DQ=DW=pdV, with p=(nRT)/V b) DS=DQ/T (constant T) c) replace (V2/V1)(p1/p2), because V=const/p. 22. a) DS=dQ/T , with dQ=CpdT b) DW=pDV with pV=nRT (constant p) c) DS=Cpln(V2/V1) , DW=p(V2-V1) 23. a) DW=nRTln(Vf/Vi)=piViln(Vf/Vi) b) pf=pi(Vi/Vf) 24. a) DW=(5piVi/2)[1-(Vi/Vf)2/5] b) pf=pi(Vi/Vf)7/5 25. a) DW=p1V1{ln(V2/V1)+(5/2)[1-(V2/V3)2/5]} b) p3=(p1V1/V2)(V2/V3)7/5 26. a) p=p1V1/V b) p=p1(V1/V)g c) From p=(nRT)/V for isothermal and p=p1(V1/V)g for adiabatic, you should get the slopes as follows: isothermal dp/dV=-p/V , adiabatic dp/dV=-gp/V. 27. a) 2.5 J b) 21 J 28. DW=p(V2-V1) , DQ=Cp(p6/b)(eaV2-eaV1) , (DE=DQ-DW) , DS=aCp(V2-V1) 29. a) 1020 J b) 582 J c) 873 J d) 436J e) 0.43 30. a) 22,370 J b) 12,400 J -28,280 J d) -32,050 J e) -35,060 J 31. a) DQ=DW=161 J b) DQ=1400 J, DW=400 J c) DQ=0, DW=119 J32. a) DQ=0, DW=(5/2)p1v1[1-(v1/v2)2/5] ; DQ=(7/2)p1(v1/v2)7/5(v1-v2) , DW=p1(v1/v2)7/5(v1-v2) ;DQ=(5/2)p1v1[1-(v1/v2)2/5] , DW=0 b) DQ=DW=p1v1ln(v2/v1) ; DQ=(7/2)p1(v1/v2)(v1-v2), DW=p1(v1/v2)(v1-v2) ; DQ=(5/2)p1v1[1-(v1/v2)] , DW=0 c) DQ=0, DW=(5/2)p1v1[1-(v1/v2)2/5]; DQ=DW=p1(v1/v2)7/5v2ln(v1/v2) ; DQ=(5/2)p1v1[1-(v1/v2)2/5], DW=0 d) DQ=DW=p1v1ln(v2/v1) ; DQ=(5/2)p1v1[(v1/v2)2/5-1] , DW=0 ; DQ=0, DW=(5/2)p1v1[(v1/v2)2/5-1] e) (a) slope down, straight left, straight up (b) slope down, straight left, straight up (c) slope down, slope up to left, straight up (d) slope down, straight down, slope up to left. (Adiabatic slopes are steeper than isothermal, and all slopes are concave upward.) f) (a) straight down, slope down to left, slope up (b) straight across, slope down to left, slope up (c) straight down, straight left, slope up (d) straight across, slope down to left, straight up. (Isochoric slopes are steeper than isobaric, and all slopes are concave upward.) 33. a) DQ=0, DW=(5/2)R(T1-T2) ; DQ=DW=(5/2)RT2ln((T2/T1) ; DQ=(5/2)R(T1-T2) , DW=0 b) DQ=0, DW=(5/2)R(T1-T2) ; DQ=(7/2)RT2[(T2/T1)5/2-1], DW=RT2[(T2/T1)5/2-1] ; DQ=(5/2)RT1[1-(T2/T1)7/2], DW=0 34. a) DQ=(7/2)R(T2-T1), DW=(7/2)R(T2-T1) ; DQ=0, DW=(5/2)R(T2-T3) ; DQ=(5/2)R(T1-T3), DW=0; DQ=DW=RT1[(5/2)ln(T3/T2)+ln(T1/T2)] b) straight across, sloping down, straight down, sloping up and to left. (Adiabatic slopes steeper than isothermal, and both are concave upwards.) c) slope up, straight down, slope down to left, straight to left (Isochoric slopes steeper than isobaric, and both are concave upward.) 35. a) In units of p0V0, (DQ, DW, DE)=(7/2,1,5/2), (ln2,ln2,0), (0,5/2,-5/2) b) In units of p0V0/nR, T= 2, 1, 0.76 36. a) 18,000 J/mole b) (15,000, 11,220, -29,890, -33,560, -37,240, -43,350, -45,870) J/mole 37. Start with H=E+pV, and then take the differential form with dQ=0, so dH=Vdp. Then integrate using pVg=constant. 38. (DQ,DW,DE)=: 4,4,0; 0,1,-1; -1,-1,0; 0,-1,1; e=0.2539. a) [p(105Pa),V(10-3m3),T(K)]=: (2,2,600), (1.13,3,510), (2.26,1.5,510), (4,1,600) b) (DQ,DW,DE) (in Joules)=: (277,277,0), (0,150,-150), (-236,-236,0), (0,-150,150) c) 0.1540. a) 0.12 J b) 0.41 J c) heat pump is more efficient 41. If DQ goes from cold to hot, then DS=DQ/Th-DQ/Tc=[DQ/(ThTc)](Tc-Th)