140a Final Exam, Fall 2007., κ T 1 V P. (? = P or V ), γ C P C V H = U + PV, F = U TS G = U + PV TS. T v. v 2 v 1. exp( 2πkT.

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1 4a Final Exam, Fall 27 Data: P 5 Pa, R = J/kmol K = N A k, N A = particles/kilomole, T C = T K du = TdS PdV + i µ i dn i, U = TS PV + i µ i N i Defs: 2 β ( ) V V T ( ) /dq C? dt P? /dq/t ext S 2 S =, κ T V ( V P 2 ) T /dq R /T, κ S V (? = P or V ), γ C P C V = (f + 2)/f. ( ) V, P S c = C/n, v = V/n, s = S/n, P β κ K T κ T V av V. Ideal gas: PV = nrt = NkT, C P = C V + nr. PV γ = const.. η W / Q 2 T /T 2, ω Q / W. H = U + PV, F = U TS G = U + PV TS. W mech (U T S + P V ), (U T S + P V ). F( v) = dp dt = s 2 s v 2 v = Free particles: Φ = 4 l T v. ( m ) 3/2 exp( 2πkT 2 m v2 /kt), N V v, P = N 3 V mv2. p(x) = σ 2 2π e (x x) /2σ 2, (x x) n p(x) = π 2 n/2 σ n Γ( ( + n)) for n even 2 for n odd

2 dt t z e at = Γ(z)a z, x4 e x dx (e x ) = 2 4π4 /5. where Γ(z + ) = zγ(z) and Γ() = (so Γ(n) = (n )! for integer n) and Γ(/2) = π. (p + q) N = i= N N = ( N N ) p N q N N. N = Np, N 2 = N2 p 2 + Npq. n! ( n ) n 2πn for n. e n p N i n i p(n i }) = N! N i!, where p i =. g(ɛ)dɛ 4πV 2 (2π h) 3 m3/2 ɛ /2 dɛ ω(n i }) B.E. = i S = k ln Ω k lnω max. ω(n i }) M.B. = i= (for free particle in box) n i= g N i i N i!. (N i + g i )! N i!(g i )! bosons ω(n i }) F.D. = i g i! N i!(g i N i )! fermions. N i g i = where α µ/kt and β /kt. MB e α βɛ where a = i + a bosons fermions k k lnω(ni i g ie α+βɛ i = kn MB ) + kαn + kβu k i g i ln( e α+βɛ i ) BE k. i g i ln( + e α+βɛ i ) FD Z(T, V ) i g i e βɛ i, ( ) µ MB = kt ln(n/z), U MB = NkT 2 T lnz V F MB = NkT ( + ln(z/n)). Z V ( ) 3/2 2πmkT ideal monatomic gas h 2 2

3 Z d SHO = n= e (n+ 2 )hν/kt = e hν/2kt e hν/kt. U SHO = kt 2 T lnz = N(ν)dν = with x m hν m /kt θ D /T. g(ν)dν e hν/kt = C V = 9Nkx 3 m [ 2 hν + hν 9Nν 3 m e hν/kt ν 2 dν e hν/kt ]. ν ν m ν > ν m. xm x 4 e x (e x ) 2dx, 3

4 . A closed system undergoes three reversible steps. In the first step, the system expands, while in thermal contact with a heat bath, absorbing heat Q and doing 2J of work. In the second step, the system s volume is held fixed and heat Q 2 is removed. In the third step, the system is thermally insulated and its volume is compressed; this requires 3J of work. After the three steps, the system has the same pressure and volume as it had initially. Assuming the system is an ideal gas, solve for the numerical values of Q and Q 2. [5 points] 2. The latent heat of melting ice is L per unit mass. A bucket contains a mixture of water and ice, at the ice point (absolute temperature T ). It is desired to use a cyclic refrigerator to freeze an additional mass m of water in the bucket. The refrigerator also rejects heat Q R, which all goes into warming up a body of constant heat capacity C and, initially, also at temperature T. (a) What is the change in the entropies in the bucket, the refrigerator, and the body in this process? Write your answers as inequalities if appropriate. [5 points] (b) What is the change in the Gibbs function in the bucket in this process? [5 points] (c) Write an inequality for Q R (it s more interesting than just saying it s positive or negative). [5 points] (d) What is the minimum work required to run the refrigerator for this process? [5 points] 3. Vessel and vessel 2 each contain N molecules of identical, ideal monatomic gas, each at initial pressure P. Initially the vessels are separated, and have temperatures T and T 2. Then they are directly connected, and eventually the system reaches equilibrium. (a) What is the final pressure? [5 points] (b) What is the final temperature? [5 points] (c) What is the total entropy change of the universe in this process? Please write your answer in terms of N and T and T 2 only. [5 points] 4. Photons in a box at temperature T have internal energy U = V σt 4 and pressure P = 3 σt 4. (a) Compute the entropy of the photons. (Take S(T = ) = ). [5 points] (b) Compute C V and C P. [5 points] 5. The specific Gibbs function of a gas is given by g = RT ln(p/p ) AT 2 P 3, where A is a constant. 4

5 (a) What is the equation of state of the gas? [5 points] (b) What is the specific Helmholtz function? (It is OK to write it as a function of the wrong variables, as long as you write what the right ones are, and write out which variables should, in principle, be solved for in terms of which other variables.) [5 points] 6. The distribution of particle speeds of a certain gas inside of a box is given by f(v)dv = Ave v/v dv where A and v are constants. (Lots of partial credit for setting up the correct mathematical expressions, so don t spend too much time doing mathematics if you re pressed for time.) (a) Write a general expression for v n. [5 points] (b) Write a general expression for ve n, where v e is the speed of particles escaping through a small hole in the box. [5 points] 8. A certain thermodynamic system has nondegenerate energy levels, with energies, ɛ, 3ɛ, 5ɛ, 9ɛ, 4ɛ,.... Suppose that there are four particles, with total energy U = 9ɛ. Identify the possible distribution of particles, and evaluate their ω(n i }) and Ω. Also, compute the average occupation number N of the ground state. (a) When the particles are distinguishable. [5 points] (b) When the particles are gaseous identical bosons. [5 points] (c) When the particles are gaseous identical fermions. [5 points] 9. Consider a system of N distinguishable particles, at temperature T, with two available energy levels: the groundstate is nondegenerate, with energy ɛ =, and the excited energy level is doubly degenerate, with energy ɛ 2 ɛ. (a) Determine the equilibrium values of the occupation numbers N and N 2 (such that N + N 2 = N) as a function of temperature. [5 points] (b) Determine the energy U of the system, as a function of temperature. [5 points] (c) Determine the specific heat (at constant volume) as a function of temperature. [5 points] 5

140a Final Exam, Fall 2006., κ T 1 V P. (? = P or V ), γ C P C V H = U + PV, F = U TS G = U + PV TS. T v. v 2 v 1. exp( 2πkT.

140a Final Exam, Fall 2006., κ T 1 V P. (? = P or V ), γ C P C V H = U + PV, F = U TS G = U + PV TS. T v. v 2 v 1. exp( 2πkT. 40a Final Exam, Fall 2006 Data: P 0 0 5 Pa, R = 8.34 0 3 J/kmol K = N A k, N A = 6.02 0 26 particles/kilomole, T C = T K 273.5. du = TdS PdV + i µ i dn i, U = TS PV + i µ i N i Defs: 2 β ( ) V V T ( )

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