Stowe Thermodynamics And Statistical Mechanics Instructor Manual 31616a

Problem Solutions 1

Problem Solutions Chapter 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 kg m/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) 10 1.5x10 /m c) 6.3 (=2π) d) 130 m, 0.0483 /m, 6.3 (=2π) 13. p x=6.6x10-24 kgm/s, vx=7.3x106 m/s 14. a ) 3.0x105 b) 27 15. 13 16. 7.8x1080 17. -14 6 a) 1.6x10 m b) 3.2 MeV 18. a) 38 eV b) 7.3x10 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) ±35º, ±66º, 90º b) ±45º, 90º c) ±55º 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. |µ| = 1.61x10-23 J/T, µz = ±9.27x10-24 J/T, U=±9.27x10-24 J 25. |µ| = 2.44x10-26 J/T, µz = ±1.41x1026 J/T, U=±1.41x10-26 J 26. about 10364 27. ±7.4x10-24 J, ±3.7x10-24 J,

0J 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 (27710 23) Chapter 2 f + γ = ∑ Πσ(φσ + γ σ) = ∑ Πσφσσ + ∑ Πσγ σ = φ+ γ cf = ∑ Πσχφσ = χ∑ Πσφσ = χφ 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)µB 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. 375 -6 1820 16. 10 17. 0.0252 18. 0.0224 19. a) 6.7x10 b) 4.5x10-2 -35 c) 2.1x10 20. a) 4.0% low b) 1.7% low c) 0.8% low d) 0.4% low 29 21. 1.0x10 22. a) 0.130 b) 0.072 c) 0.026 23. a) 0.062 b) 0.062 24. Ν! ( ρ−2)Ν − ν− µ P N (n,m ) = ρΝ ν!µ !(Ν − ν − µ )! 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) 7 Chapter 3 1. 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

2 Problem Solutions 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. 10116 12. 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) -4 -3 3.3x10 17. a) 60 b) 7.07 c) 0.0564 d) 7.9x10 18. a) 50 b) 1.08x10-3 19. a) 0.0133 b) 5x10-198 -199 20. a) 0.0798 b) 10 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 -12 and 5.77x10 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 -4 5.5x10 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 , and N(N-1)=21 cross b) : N=3 square , and N(N-1)=32 cross 37. Chapter 4 1. 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) 2 3 4 4. a) f'(0)=0, f"(0)=+2 b) f(x)=-1+0x+x +0x -(1/2)x +... c) (-.67, +.67) 5. 6. a) 1.25x1029 b) 4x109 J c) W=(1/2)f2 d) 2.8x1013 Pa = 2.8x108 atm e) 5x10-23 eV/Pa 7. B, C, A 8. a) 0.04 nm, 0.02 nm b) 1.66x10-23 kgm/s, 3.32x10-23 kgm/s c) 0.11 eV d) about 1300 K 9. a) 1.42x10-46 kgm2 b) 4.9x10-4 eV c) 5.7 K 10. a)

Problem Solutions 3 1.95x10-46 kgm2 b) 3.6x10-4 eV c) 4.1 K 11. a) 4.58x10-48 kgm2 b) -2 1.5x10 eV c) 176 K 12. 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 es 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/moleK = 5.0 cal/moleK 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) u0=+0.0625 eV e) remains same 20. a) 510 m/s b) 1910 m/s c) 8.9x10-34 kgm2/s d) 8.4 21. a) 35.3 J (1.74x1022 degrees of 23 freedom) b) 420 J (2.076x10 degrees of freedom) 22. 0.70 kg food and 0.42 kg water.

4 Problem Solutions Chapter 5 1. 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 K 7. 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. =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) T=-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 of x and integrate over x only.) 31. a) 142 2 3 3 3 b) 142 c) 142 32. a) w /y b) x /w c) d) x /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. 2 y y(z+1) 36. y e 37. 38. (3/2)V Chapter 6 1. a) b) c)

2. a) b)

c)

Problem Solutions 5 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) -25 -13 8.6x10 eV 10. 1.7x10 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 9 13. a) 64 b) 7 14. a) 7776 b) 288 c) 15. a) 1.1x109 b) 3.5x10 c) 18 16. a) 6 b) 3 17. a) 20, 10 b) 9900, 4950 c) 9.7x105 , 3.7x10 1.6x105 d) 1.0x109 1.7x108 18. a) microscopic b) 2 19. a) 26 macroscopic b) 4.6x10 20. a) b) 21. a) 8.4x1024 b) 1.26 c) 22. 4 b) 1.05x10-66 (kgm/s)3 c) 5.4x1032 d) 6.7x108 a) 3.8x10 J 3.0x107 J b) 4.64x10-67 (kgm/s)3 c) 2.0x1035 d) 1.82x108 8 8 c(N2)=2.29x10 , c(O2)=8.63x10 , = 24. a) 1.22x103 J

e) e)

23. a) f)

b) 1.16x10-66 (kgm/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 7 1. 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,1,2,0)= (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,1,2,0)= (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,1,2,0)= (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), 0.33 6. a) (E1,E2,1,2,0)= (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,1,2,3,0)= (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), 0.45 8. a) (E1,E2,1,2,0)= (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)

6 Problem Solutions c)

12. a) 10

b) 1010

c) 100

d) 10900

e)

f)

2.00000000000000000009 x10 20

13. 10 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) 24 (E1,E2,1,2,0)= (0,5,0,exp6.99x10 ,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,1,2,0)= (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+ ) around the point =0 gives ln(1+ )=0+ 2/2+...=(1-/2+...) for<<1 d) for (1+)x take the logarithms, use the preceding result, and then exponentiate 21. 10219,976 22. a) 1018 b) 1.7x1013 m3 c) 55 m , about the same d) 2.0x10-3 J, 2.0x10-12 J 23. a) At the peak, the distribution has a maximum, so its derivative is zero. b) -, Because each degree of freedom carries the same energy, on average, the total energy E0 is apportioned in proportion to the number of degrees of freedom. c) The second gives -n1/E12-n2/E22, which you manipulate using n1/E1=n2/E2=n0/E0. d) Substitute the results for the first and second derivatives into Equation 7.5 and exponentiate it. 24. You want to do an integral of the form, where P(x) is of the form . To make this difficult integral easier, first square it, . Then combine the integrands and turn the area integral, dx dy, into an area integral in polar coordinates. This integral is easy to do and should give you the desired result that . 25. 26. Thermal energy would have to decrease (temperature falls) if potential energy increases, because total energy is conserved. 27. Heat flowing from cold to hot, fluids flowing from lower pressure to higher pressure, particles going from regions of low concentration to high concentration, friction speeding things up and cooling them off, etc. In short, have energy conserved, but flowing the wrong way. 28. It takes about 40 seconds to write 100 zeros, or about 0.4 seconds per zero. At that rate it would take about 1.3x1016 years to write out 1024 zeros, or about 106 times the age of the Universe. 29. a) 8 b) 0.96x10-23 J/K , 1.91x10-23 J/K c) 2.87x10-23 J/K 30. a) 80,000 b) 7.32x10-23 J/K , 8.27x10-23 J/K c) 15.59x10-23 J/K 31. a) b) 63.6 J/K , 95.4 J/K c) 159.0 J/K 32. a) 120 b) 0, 0.693k , 1.099k 1.386k, 1.609k c) 4.787k, 4.787k 33. a) b) S=kln(f/i)=0.41 J/K 34. a) b) 35. a) 16J b) 1151 J/K, 3499 J/K, 4650 J/K 36. S=kln(f)-kln(i)=kln(f/i)=kln(f/i)n/2= (n/2)kln(f/i), where n is the number of degrees of freedom. Solving for n gives n=2S/[kln(Ef/Ei)]

a) 1.3x1022

b) 2.1x1025 37. 6.21x105 J b) c) 6.9 J/K 38. 153k 39. greater volume in momentum space (For molecules with the same kinetic energy, the ones with the larger mass have the greater momentum.)

Problem Solutions 7 Chapter 8 1. a) 3N/2 b) Write S=kln with =CE3N/2, and use this in Equation 8.2 for 1/T. c) Use S=kln and Equation 8.2 to get E= N kT. 2. a) 5x1028 b) 1.02x108 J 3. a) 20, 16 b) E1=5.56x10-19 J, E2=4.44x10-19 J c) 2.7x10-21 J/K d) 4020 K 4. a) 515 ºR b) 322 ºR c) 5.15x10-21 ºR d) 0.0322 ºR 5. 3.77x10-23 J/K, 137 K 6. 1.380x10-23 J/K 7. a) 0.079 J/K b) 8. 2.65x10-21 J 9. +3.67x10-3 J/K b) -3.53x103 J/K c) +0.14x10-3 J/K 10. a) +13.7 J/K b) -13.2 J/K c) 11. a) Use p/T= S/ V) E,N where S=kln to get pV=NkT b) R=NAk 12. Use p/T=S/V)E,N where S=kln to get: a) pV=[1+2bV2]NkT b) pV=[(1/2)+bV]NkT c) pV11=10bkT 13. Use 1/T=S/E)V,N where S=kln to get: a) E=NkT b) E=2NkT c) E=3NkT 14. Use /T=-S/N)E,V where S=kln to get: a) =-[bV2+ln(EV)]kT b) c) =  - 3kT lnE 15. a) 0=CV1N1(V0-V1)N2 b) dS0=0=dS1+dS2=(S1/V1-S2/V2)dV1 c) p1/T1=p2/T2 16. 110 J/K 17. 250 J/K 18. a) inexact, y b) exact c) inexact, 1/y d) inexact, 1/x e) exact (would also be exact if multiplied by xy) f) exact g) inexact 1/p 19. a) 7 8 about 10 to 10 J, depending on the room's size b) 660 J c) 60 J d) 2.8x1026 J 20. a) increase b) 21. a) 3.2x10-3 J b) 314 K 22. dE, dV, dp, dT, d, dN, dS; (1/T)dQ, (1/p)dW 23. 1/T , 1/p 24. a) +0.72x10-2 J -5 -3 b) 7.5x10 J/K 25. a) 3.3x10 J/K b) 26. row 1: 1.16, ; row 2: 2.9x1025, ; row 3: 1.56x1025, 3.18 J/K; row 4: 0.0137 J/K, 27. a) 3.57x10-4 J/K b) 3.57 x10-4 J/K c) for each 28. a) 1000 -7 -8 b) 10 29. a) 3.38x10 J/K b) c) 3.48x10 J/K , 30. a) 3.35x10-9 J/K b) 31. 9.6x10-24 J/K 32. 4.9x10-26 m3 33. a) 0 b) Although there is greater room in momentum space, there is correspondingly less room in coordinate space, due to the smaller volume. 34. 3140 K 35. -0.25 eV 36. T=0, Q≠0, so the derivative is infinite (undefined). 37. a) 3.4 J/K b) 0.10 J/K c) 0.71 J/K d) 1.4 J/K 38. Water. Water's high specific heat means that it can absorb solar heat (daytime, summer) or release it (nighttime, winter) with relatively small changes in temperature. So the oceans greatly moderate the climate. 39. would be larger. If held at constant pressure and allowed to expand, the expansion would cause some cooling, resulting in a smaller net rise in temperature as heat is added. Smaller rise in temperature means larger heat capacity. 40. a) 0.74 cal/gK b) Table 8.1 has specific heats at constant pressure, not constant volume. (See above answer). Later, we will learn that should be 5/3 times higher than cV for helium, and it is. 41. For a mole, Nk=R. Write the first law dE=dQ-pdV=RdT. Then look at dQ/dT for dV=0 (cV) and for pdV=RdT (). 42. a) 5.77 J/K b) 43. a) 7.26 J/K b) 44. 8.56 J/K 45. a) 13,700 J b) The heat released goes to some other system whose entropy increases as a result of the heat transfer. The total entropy of the combined system rises. 46. 5670 J 47. a) +0.376 eV b) 0.136 J/K 48. a) +0.269 eV b) 2.94 J/K (work done is irrelevant) 49. 8730 J/K Chapter 9 1. a) 1 b) 1

2. a) 3

b) 2

c) f=w2+10w(1+z)+110z

d) f=x(x-

8 Problem Solutions 10)+10z+10xz e) z=(f-w2-10w)/[10(w+11)] 3. Perhaps some manifestation of: heat flowing from cold to hot, something moving backward to the net applied force, particles diffusing from low chemical potential to high chemical potential. 4. a) x=2 b) 0 c) negative 5. a) (0,0) b) 0 c) 0 d) negative e) negative 6. Write dE1=-dE2, and then use the first law for each dE. 7. Carry out the derivation from Equation 9.4 to Equation 9.6, with the constraint that dS0>0. 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. µcrystal<µamorphous 12. µdissolved<µcrystal<µair 13. Mix in some water and shake it. The sugar moves from the oil to the water, where its chemical potential is lower. Then let it sit so that the water separates, and remove the water. If you want to recover the sugar, evaporate the water. 14. a) 30 ºC b) 0.209R=1.73 J/K c) 0.0168R=0.14 J/K d) They are not engaging in mechanical interactions, so their volumes are constant, not their pressures. So is irrelevant. 15. 16. No because the condition is for purely mechanical interactions. As heat is removed from the water, it is engaging in thermal interactions, too. 17. No, because the condition is for purely thermal interactions. The volume of melting ice is changing, so it is engaging in mechanical interactions as well. If the volume were held constant, the pressure would fall and the freezing point would rise (T>0) as heat was added (Q>0) to melt the ice. The condition would be met. 18. They will both move closer together. 19. You must explain how to measure T and Q for thermal interactions, p and V for mechanical interactions, or µ and N for diffusive interactions. (µ will probably be the most difficult, unless you are using voltmeters for the diffusion of charged particles.) 20. a) [2µ(H2)+µ(O2)]>2µ(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 -8 -11 -10 b) 4.17x10 K, 2.88x10 J c) 1.41x10 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) -12 24 1.05x10 15. 2.25 m 26. a) 3.01x10 b) 154 K c) 8.2x10-11% 10 -6 -3 27. a) 3.35x10 b) 3.2x10 28. 29. a) 6.1x10 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 ºC b) -281 ºC 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 µ2 is lower. b) N1µ1+N2µ2 c) , , G= 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

Problem Solutions 9 with a system in a canister at fixed volume near absolute zero, where the pressure is nearly zero (pV0). 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) p=2.1x107 Pa 41. Take the differential and subtract dE=TdS--pdV+µdN. 42. 43. In each case, write out the full derivative of the appropriate function in Equation 9.14' (e.g., dF=-pdV-Vdp+µdN+Ndµ) and then substitute in -SdT+Vdp for Ndµ. 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=-SdTpdV+µdN. dF=-pdV=work done only if dT=dN=0 47. dH=TdS+Vdp+µdN. dH=TdS=heat added only if dp=dN=0 48. Two interacting systems. F0=F1+F2=-(p1-p2)V+(µ1-µ2)N, since T=0. Make A2 a reservoir so that p2 and µ2 are constant. Write =p1+p/2 and =µ1+µ/2. To first order F0=0, because p1=p2 and µ1=µ2 in equilibrium. To second order. F0=-pV µN, and both these must be positive according to the results 9.3. If first order are zero and second order are positive, then the function is a minimum. 49. Two interacting systems. G0=G1+G2=(µ1-µ2)N, since T=p=0. Make A2 a reservoir so that µ is constant. Write µ =µ +µ/2. To first order G =0, because 2

1

0

µ1=µ2 in equilibrium. To second order. G0=µN/2, and this term must be positive according to the results 9.3c. If first order are zero and second order 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 V adiabatically, and measure T . b) Add heat (S=Q/T) at constant volume, and measure p. c) Add heat (S=Q/T), and measure how much the volume must change (V) 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 (p,T). 53. a) Add N 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, V b) Add heat (S=Q/T) to a system of fixed number of particles. adjust the volume (V) 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 p and T. 54. a) (M12) b) From Ndµ=-SdT+Vdp, we get (∂µ/∂T)p,N=-(S/N)=0, because S=0 at absolute zero. 55. Solve for p, with c=1 and the given V/N. KE= p2/2m.

a) 6.4 eV

b) 0.30 MeV

Chapter 10 1. 4 2. a) p=NkT/V, ∂p/∂V)T,N=-NkT/V2=-p/V 3. a) 1/T=CV2/E or E=CV2T ∂E/∂V)T=2CVT=2E/V

b) p/T=2CVlnE

b) E=(3/2)NkT, ∂E/∂p)T,N=0 c) E=CV2T,

d) p=(2x/V)ex, where x=S/CV2 ∂p/∂V)S=(-2x/V2)(3+2x)ex 4. example: =CT/Np, where C is a positive constant 5. E=(/2)pV, ∂E/∂p)V,N=(/2)V 6. 7. 8. 9. a) 2 b) 2N c) 41.4 J d) =(CAE/N2)N e) S=Nkln(CAE/N2) 10. a) N b) c) d) dE=TdS-FdL, e) FL=NkT 11. a) S=klnC+bkV4/5+2NklnE b) E=2NkT c) 4N d) pV1/5=(4/5)bkT 12. pressure=-force/area, with force=p/t, p=2mvx, t=2X/vx, area =YZ. Finally,

10 Problem Solutions 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=p/t = (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 into one, and they all cancel, giving us ln(1)=0 for the units term. 15. =kT[(+2)/2-lnc] = (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) 1/2 b) p=kT[(a/2V )+blnE] 18. S=klnC+kaNV2+kbNVlnE and use Equations 8.10 a) T=E/(kbNV) b) p=NkT[2aV+blnE] c) =-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) c=3.8x106, S=126 J/K·mole b) c=1.3x1011, S=213 J/K·mole c) c=250, S=45.9 J/K·mole 24. E=6130 J, F=8580 J, H=-56,700 J, G=-54,300 J 25. a) H=7540 J/mole, S=23.5 J/K·mole, E=7540 4 J/mole b) H=4.07x10 J/mole, S=109 J/K·mole, E=3.76x104 J/mole c) H=3570 J/mole, S=8.5 J/K·mole, E=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/r21/  2/3, to the pressure term. (p+a/v2-c/v2/3)(v-b)=RT 30. (p2/3 c/v )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) =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 m/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. =1/p=0.99x10-5 m2/N 41. -cv = p∂v/∂T)p+NA[∂u0/∂T)p-∂u0/∂T)V] 42. -cV = 16.3 J/Kmole, 530 K 43. a) polyatomic b) 8 44. Write p∂V/∂T)p=pV  = 3.02x10-5 J/K, whereas =3R=24.9 J/K. So is larger than cV by only 1.2x10-4 %. 45. a) dV=(1/a)[-(2/p)dp+(1/3T)dT] b) =2/(apV) c)  3 =1/(3aVT) 46. =R/pvB, where B=[1--a/pv +2ab/pv ] 47. a) 0 b) -p/V c) -p/V 48. a) dv=(R/pB)dT-[vA/pB]dp, where A=(1-b/v) andB=[1- a/pv+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) =(a+b/V)/(2pV-aT) b) =V/(2pV-aT) 50. a) -cV=2.5x10-5R, compared to the real value of 4.5x102R

b) =1.13x10-4/K, compared to the real value of 2.1x10-4/K 51. a) -3 p=83.5 atm b) =2.1x10 /K 52. differential form: Adp+Bdv=RdT, with , a) =R/vB b) =A/vB c) -cV=pR/B 53. differential form: Cdp + Ddv = RdT , with C=eBvv , D=eBv(p+pvB+AB) cV=pR/D 54. a) 3.03 b) 18.1

a) =R/vD

b) =C/vD

c) -

Problem Solutions 11 Chapter 11 1. definition of 2. definition of CV 3. ∂S/∂p)T=-∂V/∂T)p=-V  ∂S/∂T)V∂T/∂p)V =(CV/T)[-∂V/∂p)T/∂V/∂T)p]=(CV/T)V/V   ∂S/∂p)T(∂p/∂V)T=∂S/∂p)T/(∂V/∂p)T=(-V)/(V)=/ 6. =∂S/∂T)p/(∂V/∂T)p=(/T)/V ∂T/∂p)S=∂V/∂S)p=∂V/∂T)p/∂S/∂T)p=V/ (/T) 8. =-∂V/∂p)T/∂V/∂T)p=V/V   ∂p/∂S)V=-1/(answer to problem 4) 10. =1/∂V/∂T)p=1/V       ∂ S/∂V)p/∂S/∂p)V=-(/TV)/(CV/T) 12. =1/∂V/∂p)T=-1/V 13. dV=∂V/∂p)T dp+(V/∂T)pdT=-Vdp+VdT 14. dS=(∂S/∂T)pdT+ (∂S/∂p)Tdp=(/T)dT-Vdp , dV=(∂V/∂T)pdT+(∂V/∂p)Tdp=VdT-Vdp together: dE=TdS-pdV=(-pVb)dT+V(p-T)dp 15. dE=T(∂S/∂T)VdT+[T(∂S/∂V)T-p]dV=CVdT+[T/-p]dV 16. dV=(∂V/∂S)TdS+ (∂V/∂T)SdT= (/)dS-(CV/T)dT 17. dE=T(∂S/∂p)Vdp+[T(∂S/∂V)pp]dV=(CV/)dp+[(/V)-p]dV 18. dE=[T-p(∂V/∂S)T]dS-p(∂V/∂T)SdT=[T-p/]dS+(CVp/T)dT 19. dT=(∂T/∂p)Vdp+(∂T/∂V)pdV= ( )dp+(1/V)dV 20. dS=(∂S/∂T)pdT+ (∂S/∂p)Tdp=(/T)dT-Vdp 21. a) dE=[T-p(∂V/∂S)p]dS-p(∂V/∂p)Sdp, solve for dS b) dS=(1/A)dE-(pVCV/A)dp, with A=T(1-pV/) 22. a) dN=dV=0 b) 1 c) dp=(∂p/∂T)VdT d) dp=( )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=-(VCV/)dp+(TV/)dS c) dV=(∂V/∂p)Tdp+ (∂V/∂T)pdT=Vdp+VdT d) dS=(∂S/∂p)Tdp+(∂S/∂T)pdT=-Vdp+(/T)dT 25. a) 1 b) dS=(∂S/∂N)T,pdN c) dS=-(∂µ/∂T)p,NdN 6. a) dE=[T(∂S/∂N)T,p-p(∂V/∂N)T,p+µ]dN b) dE=[T(∂µ/∂T)p,N-p(∂µ/∂p)T,N+µ]dN

27. =4.0x10-5 /K (Ignore the smaller of

order T2 and T3.) 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,V,T) to calculate =(1/V)(V/T)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 Vs). 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 (T and Vtotal). Since you already know w from the previous measurements, you can calculate Vw=T and subtract this from Vtotal to get Vs. (Vs=Vtotal-Vw) 29. a) Hold the gas in the cylinder at constant volume. Add heat Q and measure the temperature change, T. CV=Q/T)V b) Put the liquid in the cylinder and let the volume change to keep the pressure constant as you add Q and measure T. =Q/T)p. c) Put volume V of the liquid in the cylinder and measure the change in volume V as you change the pressure by p. You will have to add or remove heat to keep the temperature constant as you do this. =(1/V)(V/p)T d) Immerse the solid in the liquid in the cylinder, and measure Vtotal and V total as above. To get Vs and Vs for the solid alone, you have to subtract that of the liquid, determined as above. (Vs=Vtotal-Vw , Vs=VtotalVw) =(1/V)(V/p)T 30. dz=0=(∂z/∂x)ydx+(∂z/∂y)xdy Now use this to solve

12 Problem Solutions for the ratio dx/dy when dz=0. 31. dE=TdS-pdV=T(∂S/∂T)VdT+[T(∂S/∂V)Tp]dV=CVdT+[T  -p]dV 32. Show that (T /p -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 . b) Use Table 11.1 to convert the partial derivatives into (,CV,,,T,p,V). 35. Write S=(∂S/∂p)Vp+(∂S/∂V)pV and then use Table 11.1 to convert the two partials into (,CV,,,T,p,V) 36. Use ∂2S/∂V∂T=∂2S/∂T∂V and Maxwell's relation (M4) for ∂S/∂V)T.

37. Use

∂2S/∂p∂T=∂2S/∂T∂p and Maxwell's relation (M10) for ∂S/∂p)T. 38. a) p=NkT/V, so 2 2 2 ∂p/∂T)V=Nk/V, and ∂ p/∂T )V=0 b) p=RT/(v-b)-a/v , so ∂p/∂T)V=R/(v-b), and 2 2 ∂ p/∂T )V=0 39. a) - b) 4.9x106 Pa, 4.2x105 Pa 40. a) - b) 19.7 J/K·mole 2 2 41. a) V=NkT/p, so ∂V/∂T)p=Nk/p, and ∂ V/∂T )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. E=T(∂S/∂T)MT+[T(∂S/∂M)T+B]M Chapter 12 1. a) dS=(/TV)dV b) dS=(/T)dT 2. a) dE=[(/V)-p]dV b) dE=[TV]dT 3. dV=VdT 4. a) dH=dT b) dF=-[S+pV]dT c) dG=-SdT 5. a) dH=(/V)dV b) dF=-[(S/V)+p]dV c) dG=-(S/V)dV 6. dp=dT=0 7. a)dE=[(/V)-p]dV+[T(∂S/∂N)p,V+µ]dN b) dE=[T(∂S/∂µ)p,N-p(∂V/∂µ)p,N]dµ+[T(∂S/∂N)p,µ-p(∂V/∂N)p,µ+µ]dN 8. a) dS=( )dV b) dS=-Vdp 9. a) dE=[(T )-p]dV b) dE=[-TV+pV]dp 10. dV=-Vdp 11. a) dH=[-TV+V]dp b) dF=pVdp c) dG=Vdp 12. a) dH=[(T )-(1/)]dV b) dF=-pdV c) dG=(1/)dV 13. (for an ideal gas, =1/T, =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. W=[piVi/(-1)][1-

(Vi/Vf)- 1] 21. a) Start with the rearranged first law, dQ=dE+pdV, with E=(N/2)kT, and use the definitions CV=∂Q/∂T)V, and =∂Q/∂T)p. b) (E+pV)=[(/2)NkT+NkT] =[( /2]Nk  T=T 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[(V/C)(pf-pi)] Tf/Ti=exp[-(/C)(Vf-Vi)] Vf/Vi=exp[-(pf-pi)] 25. a) dE=(pVCV/)dp b) dE=-pdV c) dE=(pCV/T)dT 26. a) 283 K b) inversion 27. a) 1.01x104 Pa/m b) Use T/T=(v/C)p and get Tadiabatic=0.8x10-4 K/m, which is greater than the lapse rate, so it is stable. 28. Use =(1/V)(∂V/∂T)p and =-(1/V)(∂V/∂p)T with pV=NkT to get =1/T, =1/p 29. As moist air rises and cools adiabatically, some of the moisture condenses, releasing latent heat. Consequently, rising moist air

Problem Solutions 13 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=p 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 ºC 38. a) 141 K b) 273 K 39. a) 324 K b) 3.92x10-3 /K c) 0.23 ºC/atm 2 3 40. a) 2v(2av-bpv -3ab)/D, where D=(pv -av+2ab)(+2)R b) -2a/(v2R) 41. TfTi=(2a/R)(1/vf - 1/vi) 42. a) 796 K b) 721 K 43. a) E=Q=W=0

b) E=-W=-[(NkTi)/(-1)][1-(Vi/Vf)-1], with =(+2)/, Q=0 c) E=0, Q=W=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) 2N 2N c b) (2c) (The accessible volume per particle doubles for each gas.) c) c2N d) The volume doubles, but so does the number of particles, and c 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 Ri=9.5x10-2 K/W

c) Rs=21.6x10-2 K/W

d) Rtotal=2.30x10-2 K/W

b)

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 13 1. a) Q= [(+2)/2](nRTi/Vi)(Vf-Vi) , W=(nRTi/Vi)(Vf-Vi) Q=W=nRTiln(Vf/Vi)

b)

c) Q=0, W=(nRTi/2)[1-(Vi/Vf)2/] [(-1)=2/] 2. One way is to use W=-E=-CV(Tf-Ti)=-(/2)nR(Tf-Ti)=-(/2)(pfVf-piVi) with /2=1/(-1). Or you

might integrate ∫pdV with p=const/V, and const =piVi=pfVf. 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),V3 ; p1(V1/V3),V3 ; p1,V1 1)][1-(V1/V3)-1] ;

b) Q=(p2-p1)V1/(-1), W=0 ; Q=0, W=[p2V1/(-

Q=(p1-p2)(V1/V3)V3/(-1), W=0 ;

Q=0, W=[p1V1/(-

14 Problem Solutions 1)][(V1/V3)-1-1]

8. a) p1,V2 ; p1(V2/V3),V3 ; p1(V2/V3),(V1V3/V2) ; p1,V1 b) Q=[/(-1)]p1(V2-V1) , W=p1(V2-V1) ; Q=0 , W=[p1V2/(-1)][1-(V2/V3)1];

Q=[/(-1)]p1(V2/V3)-1(V1-V2), W=p1(V2/V3)-1(V1-V2); Q=0,

W=[p1V1/(-1)][(V2/V3)-1-1] 9. They will look like the diagrams of Figure 13.5. 10. (Q,W,E)=(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) (Q,W,E)=: (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) (Q,W,E)=: (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) (Q,W,E)=: (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 of (nRT0), (Q, W, E)= ((+2)/2, 1, /2), (-/2,0,- /2), (-(+2)/4,-1/2, - /4), (/4,0,/4) d) 2/(3+4) e) 3/4 15. a) 6 ºC 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/(R) b) 2T/[(+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) -p/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) Q=W=pdV, with p=(nRT)/V b) S=Q/T (constant T) c) replace (V2/V1) (p 1/p2), because V=const/p. 22. a) S=dQ/T , with dQ=dT b) W=pV with pV=nRT (constant p) c) S=ln(V2/V1) , W=p(V2-V1) 23. a) W=nRTln(Vf/Vi)=piViln(Vf/Vi) b) pf=pi(Vi/Vf) 24. a) W=(5piVi/2)[1(Vi/Vf)]

b) pf=pi(Vi/Vf)

25. a) W=p1V1{ln(V2/V1)+(5/2)[1-(V2/V3)2/5]} b) p3=(p1V1/V2)(V2/V3)7/5 a) p=p1V1/V

26.

b) p=p1(V1/V) c) From p=(nRT)/V for isothermal and p=p1(V1/V) for adiabatic, you should get the slopes as follows: isothermal dp/dV=-p/V , adiabatic dp/dV=p/V. 27. a) 2.5 J b) 21 J 28. W=p(V2-V1) , Q=(p6/b)(eaV2-eaV1) , (E=Q-W) , S=a(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) Q=W=161 J b) Q=1400 J, W=400 J c) Q=0, W=119 J 32. a) Q=0, W=(5/2)p1v1[1-(v1/v2)2/5] ; Q=(7/2)p1(v1/v2)7/5(v1-v2) , W=p1(v1/v2)7/5(v1-v2) ; Q=(5/2)p1v1[1-(v1/v2)2/5] , W=0 b) Q=W=p1v1ln(v2/v1) ; Q=(7/2)p1(v1/v2)(v1-v2), Wp1(v1/v2)(v1-v2) ; Q=(5/2)p1v1[1-(v1/v2)] , W=0 c) Q=0, W=(5/2)p1v1[1-(v1/v2)2/5]; Q=W=p1(v1/v2)7/5v2ln(v1/v2) ; Q=(5/2)p1v1[1-(v1/v2)2/5], W=0 d) Q=W=p1v1ln(v2/v1) ;

Problem Solutions 15 Q=(5/2)p1v1[(v1/v2)2/5-1] , W=0 ; Q=0, W=(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) Q=0, W=(5/2)R(T1-T2) ; Q=W=(5/2)RT2ln((T2/T1) ; Q=(5/2)R(T1-T2) , W=0

b) Q=0, W=(5/2)R(T1-T2) ; Q=(7/2)RT2[(T2/T1)5/2-1],

W=RT2[(T2/T1)5/2-1] ; Q=(5/2)RT1[1-(T2/T1)7/2], W=0 34. a) Q=(7/2)R(T2-T1), W=(7/2)R(T2-T1) ; Q=0, W=(5/2)R(T2-T3) ; Q=(5/2)R(T1T3), W=0; Q=W=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, (Q, W, E)=(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 pV=constant. 38. (Q,W,E)=: 4,4,0; 0,1,-1; -1,-1,0; 0,-1,1; e=0.25 39. a) [p(105Pa),V(10-3m3),T(K)]=: (2,2,600), (1.13,3,510), (2.26,1.5,510), (4,1,600) b) (Q,W,E) (in Joules)=: (277,277,0), (0,150,-150), (-236,-236,0), (0,-150,150) c) 0.15 40. a) 0.12 J b) 0.41 J c) heat pump is more efficient 41. If Q goes from cold to hot, then S=Q/Th-Q/Tc=[Q/(ThTc)](Tc-Th) <0, violating second law. 42. You should be able to juggle it into the form Weng-Wfrig=Weng' +(a-b)(Qh/Th)(TaTc). This last term is negative [(a-b)≤0], so Weng-Wfrig≤Weng'. That is, you get less work output with the refrigeration than without it. 43. Suppose that on your pV diagram you have isothermal expansion from V1 to V2, adiabatic expansion from V2 to V3, isothermal compression from V3 to V4, and adiabatic compression from V4 to V1. Then from the two isothermal parts you should be able to show that Qh/Th=nRln(V2/V1) and Qc/Tc=nRln(V3/V4). And from the two adiabatic lines you should bet ThV2=TcV3 and ThV1=TcV4, from which (V2/V1)=(V3/V4). Plug this

into your isothermal results. 44. a) 3.70x1024 b) 3.37x107 J c) 7.2x106 J d) 0.21 45. a) 0.48 b) 0.44 46. High, because that would give higher efficiency, and get more work per unit of energy input. 47. Gasoline burns fast, so that the piston doesn't move much (and therefore, the volume doesn't change much) during the combustion. In turbine engines, the heat is added while the fluid is in an open tube in the heat exchanger, so the pressure is the same from one end of the tube to the other. 48. gasoline
16 Problem Solutions phase, horizontal across the mixed phase, and then more gently downward in the gas phase. 54. The line slopes up until it reaches the mixed phase, then horizontal across the mixed phase, then up again in the gas phase. 55. a) 2.9x1024 J b) 20 2.8x10 J c) 0.01% d) 0.1% e) 7% f) 0.7%56. about 104 times larger (Reception = 9x1016 W, consumption =9x1012 W) Chapter 14 1. a) 0.096 eV b) -0.294 eV c) -0.372 eV d) -0.390 eV e) -0.443 eV f) 0.33 cal into thermal energy, 0.67 cal into u0. 2. Chemical potential decreases. =- kTln2, from =2-1, and =u0+kTln+f(T). 3. Liquid water’s chemical potential falls faster, because above 0 ºC the molecules prefer the liquid phase, so it must have the lower chemical potential. 4. -dN=TdS-dE-pdV 5. (1,2,2/1)=: a) (2,3,1.5) b) (3,6,2) c) (3,4,1.33) d) (6,10,1.67) e) (10,20,2) f) yes g) yes 6. a) Use Stirling’s formula and ignore 1 in comparison to N b&c) Show that the derivatives with respect to N and n are positive. It is easier to take the derivative of the logarithms. If the logarithms are increasing, so are the functions. 7. S=kln=Nklnc. So if you multiply your answers by k, you should find that ∂S/∂E=Nk/2E=1/T, ∂S/∂V=Nk/V=p/T, and ∂S/∂N=k[lnc-(+2)/2]. If you also multiply your answers by T, you should get TS=E+pV-N, where =kT[(+2)/2-lnc]=-kTlnc, as in Equation 14.2. 8. a) - b) - 9. Vp=2.9x1068 (kgm/s)3. eV/N=8.1x10-29m3.  =(e/N)(V V /h3)=8000. 10. a) c r p -68 3 -25 3 3 Vp=8.7x10 (kgm/s) . eV/N=1.0x10 m . c=(e/N)(VrVp/h )=3.0x107. b) s=klnc=2.4x10-22J/K = 1.5x10-3eV/K

c) ==-0.35 eV 11. a) E=+60.2 eV, W=-0.013 eV, N=+400 eV, Q=E+W-N=-339.8 eV b) E=-60.2 eV, W=+0.013 eV, N=-720 eV, Q=E+W-N=+659.8 eV c) 219.8 eV d) Use =-kTlnc+ and solve for c. c,A=1470 and c,B=40. 12. 12.7, The second (Calculate the average thermal energy of a particle in each system, and use the fact that cEtherm)3. 13. 2.7 g/m3 14. 5.6 km 15. a) 1.66x10-29 m3, 2.44x10-29 m3, 9.67x10-29 m3, 2.99x10-29 m3 b) 1.05x10-5 eV, 1.54x10-5 eV, 6.12x10-5 eV, 1.89x10-5 eV c) 2420, 1650, 416, 1350 16. a) u0,A is deeper. b) 0.185 eV (Use Eq.

14.5’.) 17. a) -0.505 eV b) 16 (From Eq. 14.5’.) 18. 3x10-3 kg/m3 19. 0.338 eV 20. 0.0005 eV 21. 2.02x103 Pa, -40 Pa 22. 0.57 ºC, -2.1 ºC 23. 4.9x106 Pa 6 5 (48 atm) 24. 5.4x10 Pa, 530 m 25. a) 10 Pa, or about 1 atm b) Toward the hot side. c) somewhere around 200,000 to 1,000,000 K 26. =0, since T and p are fixed (Equation 14.1). 27. No. A+B<C. So to minimize G, the system stays as A+B, rather than going to C. 28. - (H2SO4) - 2(NaOH) + (Na2SO4) + 2(H2O) = 0 29. To the right, because 2D+3E<3A+B+4C. 30. a) 3A+B rel="nofollow">2C. b) As A and B become C, 2C rises and (3A+B) falls. So the two approach each other in value. When they become equal, the reaction stops. 31. a) 45.23 cal/mole·K b) 1.959x10-3 eV/K, 1.965x10-3 eV/K c) =-3 3.92x10 eV (=-T) 32. 1.05x107 Pa 33. a) (H+)=10-7 mole/l b) 7 c) 6.51 34. a) E=0.45 mole/l b) If A decreases by 0.03, then B decreases by 0.01, C decreases by 0.02, and D increases by 0.01. A=0.17, B=0.49, C=0.08, D=0.21,

Problem Solutions 17 giving E=0.27. So E dropped by 0.18 mole/l. 35. a) Use Eq. 14.4 and b) c) - d) – 36. 0.39 eV 37. The line separating the solid and liquid phases should be vertical. Pressure favors neither phase, so increased pressure has no effect on the temperature of the phase transition. 38. Water molecules are highly polarized, resulting in strong intermolecular attraction. Consequently, water molecules in the liquid phase are in a deeper potential well, which requires larger thermal energies (higher temperatures) for escape. 39. Tc higher for water. The self attraction is stronger for the water molecules, so larger thermal energies are required to oppose the tendency for molecules to stick together and condense into a liquid. 40. a) 43.5 cal/K b) 49.2 cal/K c) 75.2 cal/K d) 41.4 cal/K 41. a) 3.6 atm b) 16 atm c) 97 atm d) 0.082 atm e) 0.0082 atm f) 36% error. We assumed L is a constant, independent of the temperature. 42. We ignore vliquid compared to vgas. Near the critical point, the two molar volumes do not differ by much, so we cannot ignore vliquid in this region. 43. a) vsolid = 7.154x10-6 m3, vliquid=7.904x10-6 m3

b) v=+0.750x10-6 m3

d) 7.6x108 Pa (Integrate Equation 14.19, dp=(L/v)dT/T.) 44. a) -6 3 -6 3 vsolid=19.63x10 m , vliquid=18.00x10 m b) v=-1.63x10-6 m3 d) 4.3x108 Pa (Integrate Equation 14.19, dp=(L/v)dT/T.) 45. -5 ºC

c) increase c) increase

46. a) 1.36x107 Pa b) 52 mm2 c) no d) yes e) Skates have smaller area of , and therefore exert greater pressure on the ice. So the ice melts faster and provides more water for lubrication of the slide. 47. pT-B/ReA/RT=constant 48. a) Tc=8a/(27bR), pc=a/(27b2) b) water: 647 K, 218 atm. methane: 190 K, 46 atm 49. More volume in position space means more accessible states – hence increased entropy. During adiabatic expansion, the system cools, so as volume in position space increases, that in momentum space decreases, with the net change in accessible states being zero. 50. Because g/p)T=v. 51. a) - b) Show that the second derivative of -TSm is positive for all f. 52. The molten state is a homogenous mixture of the two ingredients, whereas the solid state consists of a heterogeneous mixture of microscopic crystals of tin and crystals of lead. The entropy of mixing favors the homogeneously mixed molten state. 53. Thermal energy is released when the particles fall into deeper potential wells. So the attraction between unlike particles is stronger than that between like particles. 54. –35.4 J/K 55. 56. a) - b) (2X-A-B)
2. a) 2.9 b) 6.4x108 c) 4.3x1011 3. 3.2x1010 Pa = 3.2x105 atm 4. 17 2.2x10 Pa 5. a) E=+10.2 eV, pV=2.5x10-5 eV b) 4.2x1010 Pa=4.1x105 atm 6. a) 0.572 b) 0.572 c) 0.245 7. a) 0.371 b) 0.371 c) 0.233 8. a) 0.550 b) 0.247 c) 0.111 9. a) 0.798 b) 0.798 c) 0.161 10. a) 0.832 b) 0.832 c) 0.168 11. 4.635 b) 0.990 c) 0.990 d) 9.3x10-5 12. 1160 K 13. a) 0.765 b) 0.235 14. 3.5 K 15. 10.2 times 16. a) 0.622 b) 422 K 17. a) 1010 K b) 2320 K 18. a) 0.901 b) 0.088 c) 8.7x10-3

18 Problem Solutions 19. 11,600 K 20. a) 0.154 b) 131 K 21. 348 K 22. 5.4 km 23 a) 4.56x10-3 b) 0.332, 0.333, 0.335 c) 2.84x10-26 J/T d) 0.017 J/T 24. The new constant is C=[ e -(  s-0)]-1 25. a) 0.368 b) 0.288 26. a) 0.182 b) 9.5x10-3 27. a) 0.08006 eV b) 0.043 28. 1670 K 29. dV=0, so dQ/dT=dE/dT with E=(NA/2)kT 30. a) 1.41x10-46 kg·m2 b) 4.94x10-4 eV c) 5.7 K

31. a) 1.16x10-48 kg·m2 b) 0.0599 eV

c) 694 K 32. For N2, 7 9 5.7x10 K. For H2, 6.9x10 K 33. a) 2320 K b) 8.1 K c) 3R 34. a) 2.3x1024 eV, 4.6x10-24 eV b) 2.7x10-20 K 35. 0.24 K 36. 298 K 37. E/N=[ei/kT ]/[e-i/kT] 38. Tighter binding means narrower confinement, shorter i wavelengths, higher energies, and therefore higher excitation temperatures for the bound atoms. 39. a) 0.772 b) 0.076 c) 0.228 40. 836 K 41. 3x10-7 -174 42. a) 4x10 b) 56,800 K 43. a) gl()=nl/(b-a), gu()=nu/(d-c) b) [(b-a)/(d-c)](nu/nl)(ec-e-d)/(e-a-e-b)

c) (nu/nl)e- c- a)

 d) [(b-a)/(d-c)](nu/nl)e-c-a)

44. a) 0.450 b) 0.310

c)

0.252 45. a) 3 b) 3.8x10-2 eV=6.1x10-21 J c) 508 m/s d) 1920 m/s 5 7 46. a) 4.94x10 m/s b) 2.13x10 m/s 47. (See next.) 48. . Write the numerator as with 49. 2.8x10-5 m/s 50. a) 0.311 mm/s b) 27 m 51. a) -21 -11 -11 4.07x10 J b) 2.0x10 radians or 4.1x10 m Chapter 16 1. 0.006 2. (1.386mkT)1/2 3. a) 0.01 b) 0.81 4. 5. Use Eq. E.1 with =m/2. 6. a)  =(1/ m)1/2 b) yes c) (1/2)kT 7. With d3v=v2dvsin d d , integrate over the angles to get 4 , and use Eq. E.2 for the integral over v (from 0 to ) with n=1, m/2. 8. (2kT/m)1/2 9. a) 0.39 b) 0.11 c) 7.4x10-3 10. Use Eqs. E.4 with n=1 and E.2 with n=2. 11. a) 0 b) 470 m/s c) 510 m/s d) 416 m/s 12. a) 4(2kT/m)3/2 b) 1460 m/s 13. 1.45 14. a) (2kT/m)1/2 b) 1/2 15. Use Eq. E.4 with n=0. 16. a) 27 2 19 21 3.14x10 particles/m /s b) 9.86x10 particles/s 17. a) 3.11x10 particles/s b) 2.88x10-6 /s c) N=N0e-Ct, N0=1.08x1027, C=2.88x10-6 /s d) 67 hours 18. a) 2kT (mean value of KE in x-direction is kT, twice as much as the mean values in the y and z directions) b) dT/dt=-(T3/2/3V)(k/2πm)1/2 (Write E=(3/2)NkT, so that dE/dt=(3/2) [(dN/dt)kT+Nk(dT/dt)]. From part (a) you know that dE/dt=(2kT)dN/dt) 19. 122 s 20. Use Eq. 16.8 for dJx with =N/V, and P(vx)dvx given by Eq. 16.3'. Integrate over vx>0 using Eq.E.2 with n=1. 21. a) 20 m/s b) 14 m/s c) 14 m/s d) 0 e) those moving south 22. a) b) 1 (doesn't change) 23. Water vapor. 1.25 times 24. (N2, He): a) 1.93x10-10m, 1.79x10-10m b) 5.6x10-8m, 6.5x10-8m c) 8.1x109/s, 1.8x1010/s 25. Put v1=V+u/2, v2=V-u/2 in Eq. 16.15. 26. a) 1/2 b) /4 c) 1/3 d) 0 27. a) + b) -x 28. Increased by 29. It would proceed more slowly, because fatter molecules don't go as far between collisions, and therefore transfer Q over smaller distances. 30. a) slower b) slower c) slower d) faster 31. a) D=4.8x10-6 m2/s b) K=4.2x10-3 W/m·K c)

Problem Solutions 19 =5.8x10 -6 kg/m·s2 32. We assumed that all the heat gets transferred in one collision (i.e., one mean free path). It must take about 5 collisions on average. 33. 2 43 Watts/m 34. As they move in the x direction, they carry their y velocities with them. This is the way that the different y velocities in neighboring layers get mixed and evened out. 35. a) #/m3 b) #/m6 c) m2/s d) 3Da 36. a) =(2Dt)1/2 1/2 b) 0/(4πDt) Chapter 17 1. a) out of page b) radially inward, adds to c) increase, increases, opposes d) out of page; radially outward, detracts from; decrease, decrease, opposes 2. a) The magnetic moment of a domain () is much larger than that of an individual atom, so that zB is much larger than kT. Consequently, the magnetic moments line up with B and thermal agitation is ineffective in randomizing their orientations. b) At very low temperatures, or in extremely strong magnetic fields, where kT is small compared to zB, even for the individual atoms. 3. Overall neutral, but the negative charge is distributed farther out, and the positive charge more centrally. (So the negative charge has a larger orbit as the particle spins, making the negative charge dominate the magnetic moment.) 4. Overall it has more positive charge than negative, but the negative charge is distributed farther out, and the positive charge more centrally. (See above.) 5. a) 45º b) 35º c) 30º 6. There must be an even number of electrons. (Any even combination of ±1/2 gives an integer.) 7. 1.88x10-28 kg 8. a) l=0,1,2; s=0,1 b) all 15 possible combinations of [(2,1,0,-1,-2),(1,0,-1)] 9. -27 -28 6.16x10 J/T 10. 8.0x10 J/T 11. Put z=-(lx+2sz)B and x=BB/kT into Equation 17.10.

12. Use e-1-  for <<1 with =(l x+2sz)x. Eliminate

linear in , because lz and sz both come in ± pairs. 13. Show that enx>>e(n-1)x for x>>1. That should justify keeping only the largest term in each sum in Equation 17.10. 14. 3.4 K 15. a) kT = 2.5x10-2 eV, BB=5.8x10-5 eV , thermal energy dominates b) 3.7x10-4 K 16. a) 3.75 J·K/T2·mole b) 1.27x10-2 J/T 17. a) 2 -3 -5 1.88 J·K/T ·mole b) 3.75x10 J/T, 0.375 J/T c) 6.7x10 K 18. a) 2.50 J·K/T2·mole b) 1.83x10-2 J/T c) 0.959 d) 3.11 J/T 19. a) 1.007 b) 9.53 J/T 20. a) 11.25 J·K/T2·mole b) 3.81x10-6 J/T c) 6.33x10-30 J/T 21. a) 1.11x10-6 J·K/T2·mole b) 2.31x10-5 J·K/T2·mole c) -8 7.83x10 J/T·mole 22. a) 3.75 J·K/T2·mole b) 21.3 J·K/T2·mole c) 7.50 J·K/T2·mole 23. a) 2 5.58 J/T b) 22.3 J/T c) 11.2 J/T 24. 4.13 J/T 25. a) 9.38 J·K/T ·mole b) 1.07 J/T c) 91.7 J/T Chapter 18 1. a) 1 +1.6x10-17 b) 1.68 2. a) 3.03 b) 0.330 c) 0.330 d) 0.0203 eV S/k e) 1.46 3. Use P =e , where the entropy of the small system in a given state is 1 and that of the reservoir is S0-S, where S is the loss in entropy due to supplying energy s to the small system. 4. a) 0.0205, 0.126 b) 0.0021 eV, 0.0126 eV

c) yes

5. With Z=se-s, show that the

indicated operation gives sPsEs2, with Ps=(1/Z)e-s. 6. Show that the 2 2 - indicated operation gives sPsEs -(sPsEs) with Ps=(1/Z)e s. 7. -kT2∂/∂T 8. (Use ∂/∂=-kT2∂/∂T)9. a) (Use ∂/∂=-kT2∂/∂T) b) - /kT - /kT -∞ -∞ c) Z=1+e 1 +e 2 +...1+e +e +...=1d) The indicated operation should

20 Problem Solutions give (kT/Z)[0-(1/kT2)e-1/kT-(2/kT2)e-2/kT-...], and so you have to show that of the form e-axx20 as x∞. e) 10. a) eCVT/k b) p=CT2 11. a) (V/V0)N b) p=NkT/V c) =-kTln(V/V0) 12. a) (V/V0)NF(T), where F(T)=se-fs/kT b) p=NkT/V c) =-kTln(V/V0) 13. a) -3N/2 N -C Z=C( 0) (V/V0) , where C=se s b) (3N/2)kT c) NkT/V d) -kT[(3/2)ln(0)+ln(V/V0)] e) f) (3/2)Nk 14. a) 3NkT b) c) NkT[3ln( 0)-(V/V0)] d) Nk[-3ln( 0)+(V/V0)+3] e) p=NkT/V0f) =kT[3ln( 0)-(V/V0)] 15. a) 2 b) 4 c) 8 16. a) 6 b) 1 17. a) 24 b) 418. a) 4 b) 3 c) no 19. a) 9, 6, no b) 10,000, 5050, no c) case b 20. a) 0.0521 b) 5.21 eV21. a) [e(1+e-580/T)/N]N b) -NkTln[e(1+e-580/T)/N] c) =kT{ln[e(1+e-580/T)/N]-1} 22. 23. a) 3480 K b) 2.3x109 K 24. 25. a) 137 N/m b) (Multiply both sides by (1-a).) −βηω −3 βηω ) − 1) c) Z = (1 − ε d) 3hω /(ε 26. a) (3N/2)kT b) k[lnZ+3N/2], where 2 3 /2 Ν

Z = [(ες / Ν)(2πµκΤ / η )

]

c) p=NkT/V

d) =- kTln( /N) with

2 3/ 2

ς = (V / N)(2pmkT / h )

e) 2=(N/2)(kT)2 27. a) 2.08x10-46 kg·m2 b) 3.35x10-4 eV c) 3.9 K - - 28. ave=[l(2l+1)e l l]/[l(2l+1)e l ] , where el=(/2I)l(l+1) 29. a) -x 0.196 eV b) 2280 K 30. 31. Use e ı-x for x<<1. 32. (3/2)kT, kT, kT, They are the same. 33. Carry out the steps indicated between Equations 18.18 and 18.19. 34. –0.37 eV 35. a) - b) - c) 153 J/K·mole Chapter 19 1. a) 0.597 b) 0.403 c) 0.981 d) 0.019 2. a) 2.08 b) 0.020 c) 8.7x10 183. No 4. a) 0.460 b) 5.77 5. a) 25.9 b) 2.72 c) 1.001 d) 1.006. a) 6360 K b) 1060 K c) 381 K 7. a) 1,0 b) 0,1,2 c) boson d) fermions, fermions8. See Figures 19.1 9. (e x+2)/(e2x+ex+1), where x=(-) 10. -2x -5x -2x -5x (2e +5e )/(1+e +e ), where x=(-) 11. -0.173 eV 12. a) ∞ b) They carry no energy. c) kT 13. - 14. a) 1/2 b) 1 c) ∞ 15. a) 0.693 b) 1.5 16. a) 0.239 b) 0.314 c) 0.457 d) 0.0206, 0.0210, 0.0215 17. a) 8 b) 4/3 c) 1/4 d) 4 e) 1/2 18. a) 36 b) 18 c) 1/6 d) 21 e) 2/7 19. a) s=12, d=4 b) s=6, d=2 c) s=6, d=4 d) s=6, d=0 20. a) (12,0,0), (0,12,0), (0,0,12), (1,2,0), (2,1,0), (1,0,2), (2,0,1), (0,1,2), (0,2,1) b) (11,0,0), (0,11,0), (0,0,11), (1,1,0), (1,0,1), (0,1,1) c) (1,1,0), (1,0,1), (0,1,1) 21. a) 1/3 b) 1/2 c) 022. a) (123,0), (0,123), (12,3), (3,12), (23,1), (1, 23), (13,2), (2,13) b) (111,0), (0,111), (11,1), (1,11) c) none 23. a) 1/4 b) 1/4 c) 1/2 d) 0 24. 2.5x1034 electrons/m3, 4 about 2.5x10 times denser 25. a) yes b) 0.02 K 26. a) 1.19x10-29 m3 b) 2.29x10-10 m c) no (rp/h=0.03) d) no (rp/h=0.12) e) yes (rp/h=12) 27. yes (rp/h=9) 28. a) 7.0x1010 K b) 9.4x1011 K 29. a) 4.64x1012 /s b) 35 K c) large 30. a) 1.1x10-14 eV b) 1.3x10-10 K c) small 31. a) [8πV/h3]p2dp 3/2 3 1/2 b) g()=[4πV(2m) /h ](-0) c) dN/d= [4πV(2m)3/2/h3](-0)1/2/ [e (  -  ) +1]

32. a) 3.99

b) 0.051 eV c) 9.09, 0.28 eV

33. a) 4.05 b)

Problem Solutions 21 d/[e (  -  ) ±1] , E=C   d/[e (  -  ) ±1] 0.130 eV c) 17.0, 0.93 eV 34. N=C 35. Write N = dN = g()dand integrate. Solve for . 36.  37. a) fermion 5 5 5 9 28 eV, 2.2x10 K b) 23 eV, 1.8x10 K 38. a) 3.3x10 eV , 2.5x10 K b) 180 eV, 1.4x106 K c) 1.2x108 eV, 9.5x1011 K 39. a) 2.4x1010 K b) 3.9x1014 kg/m3

Chapter 20 3 3 2 2 4 1. g(ε) = λ(4πς / η χ ) ε − µ χ ε

2. 1, 2, 3 eV

3. about 1 [V0  -29 m3, p3=10-72

(kg·m/s)3, h3=10-100 (J·s)3] 4. a) 1.23x1054 states/joule = 1.97x1035 states/eV b) 6.71x1040 states/joule = 1.07x1022 states/eV 5. a) 6.49x1034 particles/eV 22 b) 0.96x10 particles/eV6. 7. a) C C kT b) C C (kT)2 8. a) b) 1

3

µ = −kT ln[((2s +1)V / Nh )(2pmkT)

2

1

2

3/ 2

] 9. Use Equation 19.14 and ex1+x for x<<1. 10. 11. a) fx=Nvx/2V b) 2mvx c) press.=(N/V)mvx2 d) press.=(N/V)m ave/3=(2/3)(N/V)ave e) fx=Nc·cos/2V, px=2p·cos, press.=(N/V)pc·cos2, then average over cos2 and p. 12. There are an infinite number of photons in this lowest state. 13. – 14. The density of states increases with energy. So there are more states (hence more fermions) at higher energies than lower ones. So the average is weighted towards higher energies. 15. - 16. Use the substitutions x=/kT, y=µ/kT 17. –0.39eV 18. - 19. a) 6.8 eV, 4.1 eV, 53,000 K b) 5.6 eV, 3.4 eV, 44,000 K c) -7.6x10-5 eV, -9.2x10-5 eV 20. a) 79 keV, 6.1x108 K b) 43 eV, 330,000 K c) 30 MeV, 2.3x1011 K 21. a) 7.9 eV b) 4.7 eV c) 61,000 K Chapter 21 1. a) Use ex1+x for x<<1. b) Use L’Hospital’s rule on Cx3/(ex-1), where x=. 2. a) Use ex1+x for x<<1. b) Use L’Hospital’s rule on Cx5/(ex-1), where x= hc/ . 3. Write the distribution as Cx3/(ex-1), where x= Set the derivative equal to zero to get x=3(1-e-x). Show that it is satisfied by x=2.82. 4. Write the distribution as 5 x Cx /(e -1), where x=hc/ Set the derivative equal to zero to get x=5(1-e-x). Show that it is satisfied by x=4.965=1/(0.201). 2.9 mm·K 5. No. Using the chain rule d=(-hc/2)d, so the distribution in  has an extra factor of -hc/ compared to the distribution in . This extra factor shifts the position of the maximum. 6. a) 5.11 mm·K b) 1.87 mm c) 0.881 m d) 1.06 mm, 0.500 m 7. (4π/h3c3)3/(e+1) 8. No 9. Energy absorbed: green and nothing else. Energy emitted: a blackbody spectrum at temperature T. 10. -13 3 -7 3 5 4.23x10 J/m , 1.5x10 J/m , which is about 3.5x10 times larger. 11. 4.0x1066 J, 20 1.5x10 K 12. a) 3.95x1026 W b) 1.09x1041 J c) 8.7x106 years 13. a) 16 b) 2 c) 16 14. a) See Figure 19.1 with =0. b) You should get an integral of the form CT3x  2dx/(ex-1), where C is a constant and the integral is a standard integral and independent of the temperature. c) You should get an integral of the form CT4x  3x/(ex-1), where C is a constant and the integral is a standard integral and independent of the temperature. d) The average energy E/N, increases linearly in T. 15. Ir/I0 has a peak in the red. Ia/I0 = Ie/I0 = 1-Ir/I0 have a dip in the red. 16. 1.39x103 W/m2 17. Wait 2 minutes before adding the water, because you want to radiate away as much heat as possible, and the coffee radiates heat faster when hotter. 18. a) 188 W/m2 b) 9.68x1016 W c) 54% 19. a) about 300 K and 2 m2 b) 920 W

22 Problem Solutions c) 1.9x104 kcal d) 47 e) Slightly cooler skin temperature greatly reduces the rate of heat loss through radiation. 20. a) 117 W b) 2400 food Calories 21. a) 3 net power radiated = 4AT T b) d(T)/dt = - C1T , with C1=4AT3/C c) Write this equation as d(T)/T = C1dt and integrate, getting T=C2 e- C1t, where C is the temperature difference at the beginning. 22. Shiny surfaces are poor absorbers, and therefore poor emitters, so that the tea will stay hot longer. Black surfaces are good absorbers and therefore good emitters. So the heat will be released faster. 23. The absorptivity would wiggle between 0 and 1 (like the ratio of the Sun’s curve to that of the 5800 K black body) for wavelengths shorter than green, but would be about equal to 1 for the green and longer. reflectivity = 1-absorptivity. 24. - The z-component of a photon’s velocity is c·cos, so to find the average velocity of those photons moving in the +zdirection we would have to average c·cos over the +z-hemisphere. 25. Double glazing reduces the rate of heat loss 26. a) 289 W b) 144 W c) 48 27. 214K, 253K, 279K 28. 29. Volume in coordinate space is gained without corresponding loss in momentum space due to the reduction in momentum due to collisions with receding walls. (Free expansion is a non-equilibrium process, and so entropy increases.) 30. 1.1x106K 1.1x1020K 31. W=-E=Ei-Ef=a(ViTi4-VfTf4), where VT3 is constant (constant entropy) 32. 2.8x10-3 volts 33. the mean square voltage noise would be half as large. Chapter 22 1. 81 N/m 2. a) 0 b) V(x)=-V0(1-x2) c) 0.005% 4% 3. a) 6R 4. a) Nk b) (Set kT ≈ ηω ) c) No. The number of degrees of freedom decreases as more and more oscillators become confined to their ground states. 5. a) b) 8.6x10-3 eV, 13 x 1.3x10 /s c) x/(e -1) d) It starts at one and slopes downward (positive curvature), approaching 0 as x∞. High temperature limit. 6. a) Use ex-1ex for x>>1, and ex1-x for x<<1. b) (3Na02/kT2)e-0/kT, and 3Nak. 7. D(T)=(3/x3)t3dt/(et-1) 8. 58 N/m, 1800 N/m 9. a) cs=/k b) =pcs 10. a) Every other atom is up and every other one down. Atoms are 2.5 wavelengths apart. b) No. Same as above but 1.5 wavelengths apart. c) 2 11. 0.013 eV 12. a) 3.42x109 /m b) 1020 m/s 13. 0.025 eV, 3.8x1013 /s 14. a) B, 2 b) A, 8 15. 16. a) g()=(4πA/h2cs2) , D=hcs(Na/πA)1/2-,  6x10-4K  D=59 K, CV=(20.8 J/kg·K4)T3 19. a) 90 MeV b) 1012 K 20. 6x10-9 m [Use p=h/r, p2/2m  kT.] 21. a) =(h2/2m)(Ne/2V)2/3 b) off by (3/4π)2/30.38 c) =p2/2m where p=h/, so that =(8πV/3Ne)1/3=2.03(V/Ne)1/3=2.03x(spacing)

22. a) 4.14x10-6 J

b) 2x1015, 16.6x10-6 J 23. a) 0.4 K b) 0.04 K 24. 6.25x10-10 /K2 25. a) 1.02x1044 /m3, 6.40x1043 /m3 b) 5.0x1011 K, 3.7x1011 K 26. Density of protons or electrons is 89.8x1030 /m3, so the Fermi temperature is 4600 K for protons, and 0.85x107 K for electrons. So the protons are not degenerate, but the electrons are marginal. (You could also use condition 19.9, in which case you find rp/h=0.7 for the electrons and 31 fore the protons.) 27. 0.07 K, No 28. 0.0036 29. -, 0.0027 28 3 30. 5.9 eV 31. a) 5.90x10 /m b) f=5.5 eV 32. 7.0 eV 33. It

Problem Solutions 23 is larger by 4.2x10-3% c) electron, 100 times

34. a) Kl=1.56x10 J/K4, Ke=3.54x10-4 J/K2 b) 1.5 K 35. A slope of 0.05x10-3 J/K4 gives D=0.029 eV, and an

intercept of 0.68x10-3 J/K2 gives f=5.2 eV .

36. a) 8.37x1028 /m3 for both b) D=0.029 eV D=4.4x1013 /s D=340 K c) 3R d) Kl=5.0x10-5 J/K4 Ke=5.1x10-4 J/K2 e) Kl is the same, Ke is about 25% low f) 3.2 K 37. a) 3.49x10-3 J/K b) 1.29x10-3 J/K 38. 39. It should begin at zero and increase linearly (electrons). Then after a few degrees it increases as T3 (lattice) and then approaches 3 (lattice saturated), whereupon it nearly levels off, increasing linearly (slowly, electrons) after that until it reaches 4.5 (electrons saturated), whereupon it levels off for good. Chapter 23 1. a) Outer orbits overlap with the electron orbits of neighboring atoms, whereas inner orbits may not. b) Outer orbits have greater overlap and therefore greater splitting. 2. platinum: C=2.2x109 /·m, =-0.96; copper: C=4.6x1010 /·m, =-1.2 3. Same, but the band is full, rather than half full, and  moves up to between the two bands. 4. In the divalent metals, the Fermi surface lies at a minimum in the density of states, meaning both fewer electrons and fewer empty states in the Fermi tail, so the electrons are fewer and have less mobility. 5. a) Pt=6x10-4 m2/V·s, Cu=4x10-3 m2/V·s b) Most conduction electrons in conductors are stuck well below the Fermi level, and cannot respond at all to imposed electric fields. The average mobility is heavily weighted by these non-mobile electrons. 6. a) Use ex>>1 for x>>1. b) - c) 7. 2.1x1029 /m3 and 1.8x1029 /m3. 3 3 2.1x10 and 1.8x10 times larger, respectively. 8. 0, 3.5x1012 /m3, 4.5x1021 /m3 9. a) Set ne=nh and solve for . v<<c, and |-|>>kT. b) 0.013 eV 10. a) 4.9x1011 b) -3.22 eV c) 2.8x1017, -3.15 eV 11. a) ne=nh=4.9x1016 /m3 b) ne=nh=1.0x108 /m3 c) ne=nh=1.1x1021 /m3 12. a) 0.014 eV above b) 0.033 eV above 13. a) 1.0x1016 /m3 b) 1.0x1015 /m3 14. a) 3.3 A/V·m -5 15 19 b) 8.7x10 A/V·m c) 127 A/V·m 15. a) 6.7x10 , 5.9x10 , 2.5x1013, 7.7x1023 /m3 b) 2.0x10-4, 5.5, 3.5x10-6, 1.0x106 A/V·m 16. 200 K 17. 1.05 eV 18. a) 249 K b) 294 K c) 18% 19. 9 K 20. x6.9 21. 6.4 eV 22. a) at (The occupation number for the donor states must be 1/2, because all Nd/2 electrons are in the Nd donor states.)

b) 153 K

c) 1021 /m3 2.3x1016 /m3 (Use law of mass

action with ne=1021, to find number of holes in valence band.) b) 1.25x1021 /m3

d) 430 K

23. a) 0.256 eV d) 0.460 eV, 1.25x1021 /m3, 1.9x1019 /m3 e) 590 K 24. a) 0.375 eV b) 1.26x1019 /m3 c) 1.7x1013 /m3 d) 0.55 eV [Actually, it is 0.548 eV, because the material is not quite purely intrinsic. We must have (no. of holes in valence band)+(no. of donor electrons)=(no. of electrons in conduction band). Use law of mass action to find the product nh,v·ne,c=1040.38, and then set ne,c=nh,v+1.25x1019.],

c) 1.7x1011 /m3

1.62x1020 /m3, 1.49x1020 /m3

e) 420 K

24 Problem Solutions 25. a) 0.256 eV b) 1.25x1021 /m3 c) 1.8x1016 /m3 d) 0.4 eV [Actually, it is 0.394 eV. See the explanation in the solution to 27d.], 5.7x1021 /m3, 4.4x1021 /m3 e) 440 K 26. a) 1.00 eV b) 2x1021 /m3 c) 0.262 eV 27. a) 0.405 eV b) 0.202 eV c) 162 K 28. 3.5x1020 /m3 29. 20 K (Only 10-5 of the donors are ionized, so the Fermi level lies almost on the donor levels.) 30. a) Use the fact that the probability to cross the barrier is 1 for those approaching from one side, and e  for those approaching from the other. b) 1.9x1015 /m3 (Use equation from part a) c) 0.202 eV, 0.602 eV 31. a) 0.7 eV b) 8.0x1021 /m3, 5.5x109 /m3 c) 5.5x109 /m3 , 8.0x1021 /m3 d) 2.0x1022 /m3, 1.9x1011 /m3 32. a) Those flowing downhill are coming from the side of higher potential energy and lower concentration. b) Of the n charge carriers, only half (n/2) are going in the positive zdirection. Averaging v·cos over the positive z hemisphere gives . c) (8kT/m*π)1/2 33. a) 1.3x1011 /m3 b) 1.3x1011 /m3 c) 1.1x105 m/s d) 1.15x10-3 A/m2 34. a) -1.95x10-4 A/m2, -6.00x10-4 A/m2 b) 2.9x10-4 A/m2, 1.56 A/m2 Chapter 24 1. a) P1/P0=e-(1-0)=e-50=10-21.7 b) 1160 K 1.16 K for excitation = 10-17)

3. 2.4x10-3

2. a) Yes (probability

4. a) 10-477

b) No b) 10-24 5. (M,S)=: ±6, 0; ±4, kln(6); ±2, kln(15); 0, kln(20) 6. a) 0 b) kln(6) 7. 5.4 cm/s 8. Set Nj/Ni=Pj/Pi, and solve for T. 9. Decreases. Decreases. Increases. Decreases. Decreases. 10. 290, 152 K 11. a) 4.6x10-3 b) 3.5x107 12. Cooling via expansion. Cooling via evaporation. 13. 1.2x10-7 b) 8.6x106 times more 14. Because any heat transfer would be in the wrong direction. 15. 8x10-4 K 16. a) 6.25x10-6 eV b) 6.0x10-5 eV/K c) 0.10 K 17. a) 0.555x1022 b) 22 0.550x10 c) 1.46x10-4 J/K d) 0.012 K 18. a) 0.79 K b) 0.0117 J/K c) 0.00927 J d) yes e) 1.6x10-6 K 19. 5.3x10-5 K 20. a) 7.7x10-3 K b) 5.6x10-4 21. 0.9999999999 22. a) 1.87x10-24 kg·m/s b) 4.7x10-26 kg·m/s c) 8x10-5 eV 23. a) 11.0x10-10 m b) spacing = 3.5x10-10, vibrations are 3x larger 24. Infinite thermal conductivity makes the temperature is the same throughout, but the boiling point is lowest at the top, where the pressure is smallest. 25. The normal fluid component slows down and stops, but the superfluid component keeps on going without friction (i.e., flowing through the normal component). 26. 3.2 K 27. 0.65 -19 28. a)  b) - c) =10.8x10 J or 6.8 eV, n=3.60x1015 d) =6.0x10-34 J or 3.8x10-15 eV, Te=4.4x10-11 K e) no 29. 30. Use = ∫p2g(p)dp/∫g(p)dp, where g(p)=(4πVr/h3)p2

31. -

32. -

33. 2050 km, 5.8 km

35. a) x=0.885(M/Ms)1/3 and x=0.356(M/Ms)1/3

34. 1.9 Ms, 7.3 Ms 4.1 km c) 1.2 Ms , 4.7 Ms

b) 600 km and