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Statistical Physics Exam 17th April 2015 Name Problem 1 Problem 2 Useful constants gas constant R Boltzmann constant kB Avogadro number NA speed of light c Student Number Problem 3 Problem 4 8.31J/ (K · mol) 1.38·10−23 J/K 6.02·1023 mol−1 300·106 m/s 1 Total Percentage Mark (25P) Problem 1 1. (1P) Write down the ideal gas equation of state. P V = N kB T 2. (2P) Find the entropy of one coin toss, and find the entropy of one dice rolling.The entropy S reads n X 1 S = kB pi ln , p i i=1 where pi is the probability that the system is in the i-th microstate, and n is the total number of the microstates. For the coin toss one finds pi = 1/2 and n = 2, yielding S = kB ln 2 . For the dice rolling one finds pi = 1/6 and n = 6, yielding S = kB ln 6 . 1. (4P) Which quantities are intensive: volume, temperature, particle number, pressure, chemical potential, entropy, density and mass? temperature, pressure, chemical potential, and density 2. (1P) Write down the value of the thermal energy at 300 K in units of pN·nm. kB × T = 0.0138 pN · nm/K × 300K = 4 3. (3P) Which quantities are constant in a (a) canonical ensemble ? N, V, T (b) grand-canonical ensemble ?µ, V, T (c) isobaric-isothermal ensemble?N, P, T 4. (1P) How are Gibbs free energy and chemical potential related?µ = energy per particle ∂G ∂N p,T ; µ: Gibbs free 5. (2P) How does the temperature of a solvent in a mixture change compared to the pure solvent (a) for boiling? higher in mixture (b) for melting ? lower in mixture 6. (2P) In a phase diagram, what is meant by a (a) triple point? Point (pressure and temperaure) where solid, liquid and gas phase coexist. (b) critical point? Point (pressure and temperaure) where liquid and gas phase cannot be distinduished (refelxion point in van der Waals isothermes). ∂S = ∂V 7. (3P) Derive the Maxwell relation − ∂P ∂T P . You can assume the number of particles T N to be fixed. We use ∂G ∂G dG = dPT + dT ∂P ∂T P = V dP − SdT (1) The mixed derivatives, which are equal for state functions are ∂S ∂V = ∂P T ∂T P 2 (2) 8. (1P) How does the heat capacity of a classical ideal monoatomic gas depend on temperature? Cv = 32 N kB no temperature dependence. 9. (3P) What corrections, compared to the ideal gas, does the van der Waals gas introduce ? Sketch the van der Waals Potential (Energy vs. distance). Gas particles have finite volume such that the total volume is reduced. This is a repelling interaction. In addition, there is an attractive interaction that can be understood as increased pressure N2 (3) p + a 2 (V − bN ) = N kB T V 10. (2P) Name an impotant assumption in the Einstein Theory for crystal phonons. All normal modes (global motions of all atoms together) have the same frequency. This Einstein frequency is different for different materials. 3 (25P) Problem 2 Consider a system of N spins subject to a magnetic field B. The spins are non-interacting, distinguishable, and have the non-degenerated energy eigenvalues m = −µBm + 0 per one spin, where µ is the magnetic moment per one spin, and m = −1 or 1. 1. (5P) Calculate the partition function ZN . The single-spin partition function Z1 : X Z1 = eβµBm−β0 m=−1,−1 = e−β0 eβµB + e−βµB | {z } 2 cosh(βµB) = 2e −β0 cosh(βµB). (4) The N -spin partition function ZN simply leads to ZN = = Z1 N −β0 N 2e cosh(βµB) . (5) N or = e−N β0 eβµB + e−βµB . (6) 2. (5P) Calculate the average energy E (2 points). For this, you may calculate E using ZN , or you may calculate E from the average of the total energy eigenvalues. The average energy is a monotonically increasing function of kB T : Find two asymptotic values of E in the limit of T → 0 and T → ∞ (2 points), and sketch E(kB T ) (1 point). The average energy E: ∂ ln(ZN ) E = − ∂β N,V = N [0 − Bµ tanh(βµB)] . (7) N (−Bµ − 0 ) e−βBµ−β0 + (Bµ − 0 ) eβBµ−β0 . e−βBµ−β0 + eβBµ−β0 or = − (8) or, using the average of the total energy eigenvalues, E = = = N h−µBm + 0 i P βµBm−β0 m=−1,1 (−µBm + 0 )e P N βµBm−β0 m=−1,1 e N (−Bµ − 0 ) e−βBµ−β0 + (Bµ − 0 ) eβBµ−β0 − . e−βBµ−β0 + eβBµ−β0 or = N [0 − Bµ tanh(βµB)] . For T → 0: E → −N µB + N 0 . 4 (9) For T → ∞: E → N 0 . (10) The average energy is plotted as below. Figure 1: The average energy E as a function of kB T 1. (6P) Sketch the heat capacity CB (kB T ). For this, you may calculate CB using E, or you may calculate CB using the energy fluctuations, or you may use the sketch of E(kB T ). The heat capacity CB : ∂E CB = ∂T B Bµ 0 0 0 Bµ Bµ Bµ 0 Bµ 0 0 kT − kT N (−Bµ − 0 ) e− kT − kT + (Bµ − 0 ) e kT − kT e− kT − kT kT + + e 2 kT 2 kT 2 − Bµ = 0 Bµ 0 e− kT − kT + e kT − kT 2 0 0 Bµ Bµ Bµ Bµ 0 0 N (−Bµ − 0 ) e− kT − kT kT + (Bµ − 0 ) e kT − kT kT 2 + kT 2 2 − kT 2 − 0 0 Bµ Bµ e− kT − kT + e kT − kT Bµ kT 2 2Bµ = = 4B 2 µ2 N e kT 2 2Bµ kT 2 e kT + 1 B 2 µ2 N sech2 Bµ kT kT 2 . (11) or, using the energy fluctuations, CB = 2 2 h∆E 2 i 2 h(−µBm + 0 ) i − h−µBm + 0 i = N kB T 2 kB T 2 2Bµ = = 4B 2 µ2 N e kT 2Bµ 2 kT 2 e kT + 1 B 2 µ2 N sech2 Bµ kT kT 2 5 . (12) 0.5 CB • kB 0.4 0.3 0.2 0.1 0.0 0 1 2 3 kBT•Bm 4 5 Figure 2: The heat capacity per the Boltzmann constant CB /kB as a function of kB T /Bµ 1. (4P) Write down the Helmholtz free energy A in terms of ZN and kB T (1 point). Calculate A (3 points). The Helmholtz free energy A: A = − = ln(ZN ) β −N kB T ln 2e−β0 cosh(βµB) . (13) 2. (5P) Sketch the magnetization M(kB T ). For this, you may calculate the average of the total magnetic moment h−N (m − 0 )/Bi, or you may calculate M using the Helmholtz free energy A. The magnetization M: ∂A M = − ∂B N,T = N µ tanh(βBµ). (14) or, using the average energy E, M = = −(E − N 0 )/B N µ tanh(βBµ). 6 (15) 1.0 M •N m 0.8 0.6 0.4 0.2 0.0 0 2 4 6 kBT•Bm 8 10 Figure 3: The magnetization per the total magnetic moment M/N µ as a function of kB T /Bµ (25P) Problem 3 Consider the grand-canonical partition function for non-relativistic, ideal Fermi gases and Bose gases, respectively, in logarithmic form ln Ξ = ± (2s + 1) Σp ln (1 ± exp [−β (εp − µ)]) (16) 2 p where εp = 2m is the energy corresponding to momentum state p, m is the mass of a particle, µ is the chemical potential, and β = kB1T with T temperature and kB Boltzmann’s constant 1. (4P) In the classical limit, the logarithm of the partition function can be expanded. (a) What does “classical limit” mean? Give a description in words and as a relation. The classical limit is low densities (large volume and/or low number of particles) and high temperatures, exp [−β (εp − µ)] 1 (b) Write down the first two terms of the expansion. Expand the logarithm as ln (x ± 1) ≈ 2 3 ±x − x2 ± x3 · · · for small x 1 ln Ξ = ± (2s + 1) X exp [−β (εp − µ)] ∓ p 1 exp [−2β (εp − µ)] + · · · 2 (17) 2. (6P) Show that the zeroth order term, assuming continuous space, for particles with s = 0, is ln Ξ = V exp (βµ) λ3 (18) 12 h2 with λ = 2πmk where h is Planck’s number and V is the volume. [You need the integral BT 2 √π ´∞ 2 dxx exp −x = 4 .] −∞ In continuous space ˆ X V ≈ 3 d3 p (19) h p Such that in zeroth order V ln Ξ ≈ 3 h ˆ d3 p exp [−β (εp − µ)] 7 (20) Plugging in βεp = p2 2mkB T and integrating over the sphere V ln Ξ ≈ 3 4π h We now substitute x2 = p2 2mkB T ln Ξ = ˆ dp · p exp − 2 p2 exp [βµ] 2mkB T (21) and obtain a standard integral 3 V 4π (2mkB T ) 2 3 h ˆ ∞ dx · x2 exp −x2 exp [βµ] −∞ | {z } (22) 1√ π 4 such that ln Ξ = = 3 V (2πmkB T ) 2 exp [βµ] h3 V exp [βµ] λ3 (23) (24) 3. (2P) Use eq. 18 to express the chemical potential as a function of number of particles. For the grand-canonical partition function we have P V = kB T ln Ξ and for an ideal gas PV kB T = N , hence ln Ξ = N in the classical limit. Therefore V exp [βµ] λ3 V exp [βµ] = N λ3 V βµ = ln N λ3 V µ = kB T ln N λ3 N = 4. Use the first-order corrected logarithmic partition function V 1 V exp (2βµ) ln Ξ = ± (2s + 1) 3 exp (βµ) ∓ 5 λ λ3 22 (25) (a) (4P) Calculate a first-order corrected number of particles . We need N= 1 ∂ ln Ξ β ∂µ (26) and get N 1 ∂ V 1V 1 = ± (2s + 1) 3 exp [βµ] ∓ exp [2βµ] β ∂µ λ 2 λ3 2 32 V 1 V = ± (2s + 1) 3 exp [βµ] ∓ 3 3 exp [2βµ] λ 22 λ 8 (27) V (b) (4P) Obtain a first-order corrected expression for the pressure P in the form NPkB T for 1 PV Fermi particles with s = 2 and for Bose gas particles with s = 0. We again use kB T = ln Ξ and the N obtained above. to obtain the correction we determine the difference in the first-order corrections ln Ξ1 and N1 (in zeroth order ln Ξ = N ) 1 V 1 V ln Ξ1 − N1 = ± (2s + 1) ∓ 5 3 exp [2βµ] ± 3 3 exp [2βµ] 22 λ 22 λ 1 1 V = ± (2s + 1) ∓ 5 ± 3 exp [2βµ] (28) 22 2 2 λ3 Hence, ln Ξ = N ± and therefore Fermi: Bose: 2s + 1 V exp [2βµ] 5 2 2 λ3 (29) pV 2s + 1 V exp [2βµ] =1± 5 N kB T 2 2 N λ3 (30) pV 1 V =1+ 3 exp [2βµ] N kB T 2 2 N λ3 (31) pV 1 V =1− 5 exp [2βµ] N kB T 2 2 N λ3 (32) (c) (3P) Write down a first-order h energy for an ideal Bose gas. i For ideal gases we have E = 3 3 V 1 V N k T . Hence, E = k T exp [βµ] + exp [2βµ] . Also E = 32 P V such that 3 λ3 B 2 2 B λ3 22 E = = = = = 1 V 3 N kB T 1 − 5 exp [2βµ] 2 2 2 N λ3 3 1 V kB T N − 5 3 exp [2βµ] 2 22 λ V 1 V 3 kB T exp [βµ] + 3 3 exp [2βµ] − 2 λ3 22 λ 3 V 2 V kB T exp [βµ] + 5 3 exp [2βµ] − 2 λ3 22 λ V 3 1 V kB T exp [βµ] + exp [2βµ] 5 2 λ3 2 2 λ3 1 V exp [2βµ] 5 2 2 λ3 1 V exp [2βµ] 5 2 2 λ3 (33) 5. (2P) How does the first-order corrected pressure of a Fermi gas/Bose gas compare to a classical gas (lower, higher, equal)? The corrected pressure, compared to the classical ideal gas, is larger for a Fermi gas (plus sign in partition function) and smaller for a Bose gas (minus sign in partition function). 9 (25P) Problem 4 1. (4P) Consider a system, divided into two sub-systems 1 and 2, each characterised by their entropy, volume, and number of particles (S1 , V1 , N1 ) and (S2 , V2 , N2 ), respectively. The total system is isolated and is at equilibrium. the two subsystems are also in equilibrium with each other and can exchange energy and particles, nut the total energy and total number of particles remains constant. The subsystems can also change their volume prvided the total volume is fixed. Show that at equilibrium the two subsystems have the same pressure P1 = P2 , the same temperature T1 = T2 , and the same chemical potential µ1 = µ2 . For the total system we always have S = S1 + S2 V = V1 + V2 N = N1 + N2 E = E1 + E2 (34) since the total system is isolated. This means that also dS = dS1 + dS2 = 0 dV = dV1 + dV2 = 0 dN = dN1 + dN2 = 0 dE = dE1 + dE2 = 0 (35) Starting from the first law of thermodynamics we write dE1 = T1 dS1 − P1 dV1 + µ1 dN1 (36) = −T2 dS2 + P2 dV2 − µ2 dN2 (37) and −dE2 such that with dE1 = −dE2 T1 dS1 − P1 dV1 + µ1 dN1 = −T2 dS2 + P2 dV2 − µ2 dN2 (38) = −T2 (−dS1 ) − P2 (−dV1 ) + µ2 (−dN1 ) = T2 dS1 + P2 dV1 − µ2 dN1 which holds only for T1 = T2 P1 = P2 µ1 = µ2 (39) 2. (10P) A cylinder contains an ideal gas in thermodnamic equilibrium at pressure P , Volume V , temperature T , internal energy E, and entropy S. The cylinder is surrounded by a very large heat reservoir at the same temperature T. By moving a piston a small change in volume ±∆V can be made. The cylinder walls and the piston can be switched to 10 be either perfect thermal conductors or perfect thermal insulators. For each of the five processes below, state whether the changes in the quantities P, V, T, E, S (after re-establishment of the equilibrium) have been positive, negative or zero. “Slow” (“fast”) means during the volume change, the speed of the piston is much less (greater) than the speed of the gas particles at temperature T . (a) slow volume increase, thermal conductor isothermal expansion∆T = 0; ∆E = 0; ∆S = ∆V V > 0;∆P < 0 (b) slow volume increase, thermal insulator reversible, adiabatic expansion∆Q = 0, hence ∆S = 0 ; ∆E < 0 because of external work; ∆P < 0; ∆T < 0 (c) fast volume increase, thermal insulator adiabatic free expansion, irreversible∆S > 0;∆P < 0;∆T = 0; ∆E = 0 (d) fast volume increase thermal conductor isothermal free expansion, irreversible∆S > 0;∆P < 0;∆T = 0; ∆E = 0; (e) fast volume decrease, thermal conductor isothermal (free) compression, ∆P > 0;∆T = 0; ∆E = 0; ∆S < 0 3. (7P) Draw (and label) a thermodynamic cycle consisting of three processes involving an ideal gas in a pressure-volume diagram. (a) Leg 1-2: quasi-static adiabatic compression Leg 2-3: isobaric cooling Leg 3-1: quasi-static isothermal expansion (b) Determine the sign of the heat and work transfers and change in internal energy for each leg and for the cycle as a whole. Leg 1-2: quasi-static adiabatic compression 3 Q12 = 0, W12 = (P2 V2 − P1 V1 ), because P2 < P1 and V2 < V1 then W12 < 0. 2 On the other hand, we know: E12 = Q12 − W12 =⇒ E12 > 0 (c) Leg 2-3: isobaric cooling 3 W23 = P (V3 − V2 ) and E23 = (P3 V3 − P2 V2 ), because P3 = P2 and V3 < V2 then W32 < 0 2 and E32 < 0. Then regarding to E32 = Q32 − W32 =⇒ Q32 < 0 (d) Leg 3-1: quasi-static isothermal expansion E31 = 0, Q31 = W31 . For work done by system we have: W 31 = W 31 = V ´1 V3 V ´3 P dV and for ideal gas we could rewrite: V1 N kB T V1 dV = N kB T ln( ), because V3 < V1 then W31 > 0 and then Q31 > 0. V V3 11 4. (4P) Consider an ideal monoatomic gas with a mass of m=28g/mol and a heat capacity of J J cV = 4 mol·K . For simplicity, in this task, use a gas constant of R = 3 mol·K . (a) How much heat is required to heat 1kg of that gas from 210K to 420K at constant pressure? For constant pressure we need cp = cV + R Q = n (cv + R) ∆T with n = 1000g 28g/mol , hence 1000g J (4 + 3) · 210K 28g/mol mol · K 1000 J = mol · 210K 4 mol · K = 52500J Q = (b) How much has the internal energy of that gas increased? ∆E = ncv · ∆T ∆E 4J 1000g · · 210K 28g/mol mol · K 1000 J = mol · 210K 7 mol · K = 30000J = (c) How much external work was done? ∆E W = Q−W = Q − ∆E = 52500J − 30000J = 22500J (d) The gas is heated by a heat pump, what is the efficiency of that heat pump? Effficiency of a heat pump ηH ηH = = = 12 Thigh Thigh − Tlow 420K 420K − 210K 2