Stowe Thermodynamics and Statistical Mechanics Instructor Manual

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 terms, and N(N-1)=21 cross terms b) : N=3 square terms, and N(N-1)=32 cross terms 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)Cpf2 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 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/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 terms 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