Solutions to Problem Set 4

Cornell University, Physics Department Fall 2014 PHYS-3341 Statistical Physics Prof. Itai Cohen Solutions to Problem Set 4 David C. sang, Woosong Choi 4.1 Equilibrium Fluctuations (a From lecture we had Ω(E, V, N = 1 3N N! 2E ( 4πEV 2/3 3N/2 e de 3h 2 N E 3N/2 1 V N de At equilibrium we have that the number of states of the total system is maximum thus we have his gives For large N we have d ln Ω total de = d ln Ω de + d ln Ω r de = 3N/2 1 β r = 0 E 3N/2 1 Ē = β r = 1 k Ē = 3 2 Nk aylor exanding ln Ω (3N/2 ln E + N ln V and ln Ω r about E = Ē, ln Ω tot (E ln Ω(Ē + β(e Ē + 1 ( β (E 2 E Ē2 + O((E Ē3 + ln Ω r (Ē P (E Ω(E Ω(Ēe( β β(e Ē e 2( 1 3N (E Ē2 2Ē2 Where there is only a first order term for ln Ω r = β(e tot E, as the temperature of the reservoir (and therefore β is a constant. hus we have (normalizing where P (E = 1 2πσE e (E Ē2 /2σ 2 E σe 2 = 2 Ē2 3N = 3 2 Nk2 2 is the variance of E. With standard deviation 3N σ E = 2 k 1

(c Similarly we have P (E, V Ω tot (E, V = exp[ln Ω tot ] P (E, V e 1 2 At equilibrium we must have exp [ ln Ω(Ē, V + ln Ω r (Ē, V ( p β p β(v V + 1 ( β (E 2 E Ē2 + 1 ( ] pβ (E 2 Ē2 ( β E (E Ē2 + 1 2 ( pβ (E Ē2 ln Ω tot = ln Ω + ln Ω r = N V p r β r = 0 Giving us pβ = N V which gives an equilibrium volume of V = Nk p Noting that β = 0 = pβ E we must have E and V statistically independent. hus P (E, V = P (E P (V with P (E = P (E = 1 2πσE e (E Ē2 /2σ 2 E the same as before, with the same variance, and P (V = P (E = 1 2πσV e (V V 2 /2σ 2 V with the variance σv 2 = V 2 N = N ( k p 2 with standard deviation (and fluctuations of the size σ V = Nk/p. 2

4.2 Calorimetry Reif 4.2: A 750-g copper calorimeter can containing 200 g of water is in equilibrium at a temperature of 20 C. An experimenter now places 30 g of ice at 0 C in the calorimeter and encloses the latter with a heat-insulating shield. (a When all the ice has melted and equilibrium has been reached, what will be the temperature of the water? (he specific heat of copper is 0.418 joules g 1 deg 1. Ice has a specific gravity of 0.917 and its heat of fusion is 333 joules g 1 ; ie it requires 333 joules of heat ot convert 1 g of ice to water at 0 C. (b Compute the total entropy change resulting from the process of part (a. (c After all the ice has melted and equilibrium has been reached, how much work, in joules, must be supplied to the system (e.g., by means of a stirring rod to restore all the water to 20 C? (a Since the calorimeter is enclosed in a heat insulating shield we have 0 = Q = m cu c cu ( f 20 C + m w c w ( f 20 C + m i c w ( f 0 C + m i L f solving for f we find where c w = 4.186J/g C. his gives (b he entropy for the ice melting is f = m cuc cu 20 C + m w c w 20 C m i L f m cu c cu + (m w + m i c w f = 10.205 C = 283.355K S melt = Q melt = m il f 273.15K = 36.57J/K. he net entropy change for heating a substance from temperature to temperature f (with no phase changes is hus we have S = f mcd = mc ln f / (1 S total = S melt + m i c w ln f /273.15K + (m w c w + m cu + c cu ln f /293.15K = 2.0727J/K (c he extra work required to restore the water to 20 C is W = Q = (m cu c cu + (m i + m w c w (20 C 10.205 C = 12501.6J = 12.5kJ. 3

4.3 Pulling on a Polymer (a For N 1, we make a Gaussian approximation for the number of miccrostates. his problem is analogous to the random walk, and to Reif 2.4 which we covered in section. We have Ω = N! n 1!n 2! dn 1 except we wish to write Ω in terms of E, N, and al only. Hence we have Ω(E = Applying sterling s approximation we see ( N 2 E N! (! N + E 2! de ln n! n ln n n + 1 2 ln(2πn ln Ω(E N ln N n 1 ln n 1 n 2 ln n 2 + 1 2 [ln(2πn ln(2πn 1 ln(2πn 2 ] + ln n 1 ln n 1 N n 2 ln n 2 N + 1 2 ln N + ln de 2πn 1 n 2 ( N 2 E ( 1 ln 2 E ( N 2Nal 2 + E + 1 ( N 2 ln 1 [ ( 1 2 ln N + ln 2π 2 2 E 2Nal ( 1 ln ln 2 + E 2Nal ( 1 2 + E ] 2Nal ( de + ln de expanding in terms of E/Nal, and only keeping factors up to size E/N we eventually get ln Ω(E N ln 2 1 ( 2 [ ] 1 E 1 ln 2Nπ + ln de 2 N al 2 which gives Ω(E 2 N al 2πN e E 2 2Nl 2 a 2 (b We have [ 2 N de S = k b ln Ω = k b ln al 2πN k ln 2 N al 2πN k 2N ( E al ] k 2N 2 ( E 2 al 2 4

(c o find the equilibrium value Ē we see that at thermal equilibrium ln Ω E = β = 1 k where is the temperature of the solution. this yields Ē = Nl2 a 2 k with fluctuations characterized by the variance σ 2 = Nl 2 a 2 Note that this expression for the variance is only valid in the limit of β, Ē 0, because of our Gaussian expansion. A more careful analysis without using the Gaussian expansion yields σ 2 = Na 2 l 2 cosh 2 (alβ. (d We have the force F x = E = E = +a dl x x N In terms of the average length and energies we have Which gives x N = Ē a = βnl2 a 2 F x = a = x N βnl 2 5

4.4 Ideal Work Reif 5.2: he molar specific heat at constant volume of a monatomic ideal gas is known to be 3 R. Suppose that one mole of such a gas is subjected to a cyclic 2 quasistatic process which appears as a circle on the diagram of pressure p versus volume V shown in the figure. Find the following quantities: (a he net work (in joules done by the gas in one cycle. (b he internal energy difference (in joules of the gas between state C and state A. (c he heat absorbed (in joules by the gas in going from A to C via the path ABC of the cycle. (a he work done by the gas is the area under the lines in the p-v diagram W = ABCD dw = ABCD pdv = π(10 3 cm 3 (10 6 dynes cm 2 in joules, W = π 10 9 erg = π 10 2 J (b he internal energy difference per mole is given by the difference in energy E = 3 2 Rn in joules E CA = 6 10 2 J. E CA = E C E A = 9 10 9 3 10 9 = 6 10 9 erg (c he heat absorbed in going from A to C can be calculated from dq = de + pdv. de integrated from A to C is already calculated in (b, and pdv is ABC ( π pdv = 2 + 4 10 9 erg ABC Add this with the result of (b we get ( π Q ABC = E CA + W ABC = 2 + 10 10 2 J. 6

4.5 Elastic Rod Reif 5.14: In a temperature range near absolute temperature, the tension force F of a stretched plastic rod is realted to its length L by the expresssion F = a 2 (L L 0 (2 where a and L 0 are positive constants, L 0 being the unstretched length of the rod. When L = L 0, the heat capacity C L of the rod (measured at constant length is given by the relation C L = b, where b is a constant. (a Write down the fundamental thermodynamic relation for this system, expressing ds in terms of de and dl. (b he entropy S(, L of the rod is a function of and L. Compute ( S/ L. (c Knowing S( 0, L 0, find S(, L at any other temperature and length L. (It is most convenient to calculate first the change of entropy with temperature at the length L 0 where the heat capacity is known. (d If one starts at = i and L = L i and stretches the thermally insulated rod quasi-statically until it attains the length L f what is the final temperature f? Is f larger or smaller than i? (e Calculate the heat capacity C L (L, of the rod when its length is L instead of L 0. (f Calculate S(, L by writing S(, L S( 0, L 0 = [S(, L S( 0, L] + [S( 0, L S( 0, L 0 ] and using the result of part (e to compute the first term in square brackets. Show that the final answer agrees with that found in (c. (a Using F as an analog for p and L as an analog for V we see de = ds + F dl where we ve taken the positive sign in the last term since F is the force on the rod, not the force by the rod. (b aking the energy to be E(S(, L E(, L we have de = E d + E L L dl he entropy then must be ds = de F dl = 1 E d + 1 L ( E L F dl 7

which gives the partial derivatives S = 1 E S L L L = 1 ( E L F Remembering that F = E we have E = F which gives us L S L = 2F = 2a (L L 0 (c We also know E L = C L and C Lo = b hence ds = 2a (L L 0 dl + 1 C Ld Starting at S( 0, L 0 and first increasing in temperature at fixed L = L 0 and then increasing in L with fixed we have S(, L = S( 0, L 0 + = S( 0, L 0 +,L0 0,L 0 ds + 0 bd,l ds,l 0 L L 0 2a (L L 0 dl Giving us S(, L = S( 0, L 0 + b( 0 a (L L 0 2 (3 (d By quasi-statically stretching a thermally insulated rod we see that the process must be adiabatic, thus S = 0 and we must have S( 0, L 0 + b( i 0 a i (L i L 0 2 = S( 0, L 0 + b( f 0 a f (L f L 0 2 Canceling terms we see [b a(l i L 0 2 ] i = [b a(l f L 0 2 ] f f = b a(l i L 0 2 b a(l f L 0 2 i hus if we have L f > L i > L 0, then we must have f > i. (e We have from part (c ds d = b a(l L 0 2 2a (L L 0 dl d 8

We also have from above which gives ds = C L d 2a (L L 0dL C L = ds d + 2a 2 (L L 0 dl d C L = b a (L L 0 2 (4 (f S(, L = S( 0, L 0 + = S( 0, L 0 0,L 0,L 0 ds + L,L ds 0,L 2a (L L 0 dl + (b a(l L 0 2 d L 0 0 S(, L = S( 0, L 0 + b( 0 0 a (L L 0 2 (5 the same as above. 9

4.6 Van der Waals Gas (a At constant E we have 0 = de = ds pdv he change in entropy S(, V can be expressed as ds = S d + S V dv = C V d + p dv V his gives ( 0 = C V d + p p dv V d = p V p dv C V = p V p E C V (b Again fixing Energy we have (c From part (a we know 0 = de = ds pdv ds = P dv S = P E d = p V p dv C V From the Van der Waals equation of state we have p = Nk V Nb an2 V 2 p = V Nk V Nb giving us d = 1 C V = an 2 C V ( an 2 V dv 2 ( 1 V 2 1 V 1 (6 10