Statistical Physics & Thermodynamics: Entropy, Susceptibility, & Bose-Einstein Gas, Exams of Mechanics

Download Statistical Physics & Thermodynamics: Entropy, Susceptibility, & Bose-Einstein Gas and more Exams Mechanics in PDF only on Docsity! Suggested Solution for Statistical Physics and Thermodynamics AY2007-08 Semester 2 NUS Physics Society 1. (a) (Note that the subscript 1, 2 indicating the subsystem 1 and subsystem 2 respectively. Whereas the variables without subscript refer to variables of whole system) S1 + S2 = S V1 + V2 = V E1 + E2 = E N1 + N2 = N (1) Hence, ∂V2 ∂V1 = −1 as ∂V ∂V1 = 0( ∂S1 ∂V1 ) E1N1 = ( ∂S2 ∂V2 ) E2N2 (2) The pressure of each subsystem is defined as Pi = Ti ∂Si ∂Vi where i = 1 or 2. Hence the equilibrium condition is achieved when T1 = T2 and P1 = P2 from equation ??. (b) dS dt = ∂S1 ∂V1 dV1 dt + ∂S2 ∂V2 dV2 dt = ( P1 T1 − P2 T2 ) dV1 dt > 0 from the second law of thermodynamics. Hence, if P1 > P2, dV1 dt > 0 according to the second law. 2. (a) dS = dQ T dQ = C · dT ∆S = C ∫ T0 T1 1 T dT = C ln( T0 T1 ) (b) Change in entropy of water = mC · ln(T0 T1 ) = 1000× 4.2× ln(373/273) = 1311JK−1 Change in entropy of heat reservoir = −1311JK−1 while change in entropy of the entire system = 0 3. (a) S = − ( ∂F ∂T ) V = NK ln [ eV N ( 2πmkT h2 )3/2 ] + 3NKT 3/2 2 1 (b) 4. (a) Vapour is much less dense than water, ∆V = Vvapour − Vliquid ≈ Vvapour PV = RT per one mole Hence, L = T∆V dP dT = RT 2 P dP dT (b) dP dT = (788− 733.7)mmHg (373− 272)K = 54.3mmHg 2K = 7239PaK−1 L = RT 2 P dP dT = 8.31× 3732 760× 133.3 × 7239 = 82614J/mole 5. (a) Fermi energy, F achived at the condition T = 0K. The step function of the equation is ignored. Thus, N(F ) = 4πV h3 (2m)3/2 ∫ F 0 1/2d = 4πV h3 2 3  3/2 F Hence, after rearranging, F = h2 2m ( 3N 8πV )2/3 (b) F = k · TF F = 12mV 2 F where k is Botlzmann constant. (c) Given N V = 4.7× 1022cm−3 = 4.7× 1028m−3. F = h2 2m ( 3 8π N V )2/3 = (6.626× 10−34)2 2× 9.11× 10−31 ( 3 8π × 4.7× 1028)2/3 = 7.608× 10−19J TF = 7.608× 10−19 1.381× 10−23 = 5.509× 104K VF = √ 2F m = √ 2× 7.608× 10−19 9.11× 10−31 = 1.292× 106m/s 2