PHYS 427_Fa08_HE01_ B_OpenBook_Solutions

# PHYS 427_Fa08_HE01_ B_OpenBook_Solutions - PHYS 427 1st...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PHYS 427 1st Midterm Exam‐Part B Solutions Fall 2008 PHYS 427 Thermal and Statistical Physics 1st Midterm Exam-Part B Solutions Zuanyi Li Problem 1 (a) The partition function is Z = ∑∑ e m n − Emn / τ =e − ω /τ ∑e m − m ω /τ ∑e n − n ω /τ =e − ω /τ 1 ⎛ ⎜ − ⎝ 1− e ω /τ e ω /τ ⎞ ⎟ = ω /τ − 1) 2 ⎠ (e 2 So the free energy is F = −τ ln Z = 2τ ln(e (b) The entropy is 1 ⎛ ∂F ⎞ ⎛ ω⎞ σ = −⎜ ⎟ = 2⎜ ⎟ − ⎝ ∂τ ⎠ N ,V ⎝ τ ⎠ 1− e ω /τ ω /τ − 1) − ω − 2 ln(e ω /τ − 1) (c) To sketch the entropy, we discuss two limits as below: (i) In the case of τ → 0 1 →1 1 − e− ω /τ ln(e So we get ω /τ − 1) → ln e ω /τ = ω τ σ → 2⎜ (ii) In the case of ω ⎛ ω⎞ =0 ⎟−2 τ ⎝τ ⎠ τ ω ⎛ ω⎞ 1 ⎛τ ⎞ σ ≈ 2⎜ − 2 ln( ω / τ ) = 2[1 + ln ⎜ ⎟ ⎟] ⎝ τ ⎠ ω /τ ⎝ ω⎠ So we finally get the plot of entropy vs temperature as below: 4 3 σ 2+2ln(τ /hω) σ 2 1 0 0.0 0.5 τ (hω) 1 / 4 1.0 1.5 2.0 PHYS 427 1st Midterm Exam‐Part B Solutions Fall 2008 (d) The energy is 2 ⎛ U = F + τσ = ω ⎜ − ⎝ 1− e So the heat capacity is e ω /τ ⎛ ∂U ⎞ ⎛ ω⎞ CV = ⎜ = 2⎜ ⎟ ⎟ ω /τ − 1) 2 ⎝ ∂τ ⎠V ⎝ τ ⎠ (e 2 ω /τ ⎞ − 1⎟ ⎠ (i) In the case of τ → 0 ω /τ ( ω / τ )2 ⎛ ω⎞ e =2 →0 CV ≈ 2 ⎜ ⎟ ω /τ 2 e ω /τ ⎝ τ ⎠ (e ) 2 (ii) In the case of τ ω ω /τ ⎛ ω⎞ e = 2e CV ≈ 2 ⎜ ⎟ 2 ⎝ τ ⎠ ( ω /τ ) 2 ω /τ → 2 So we can sketch the heat capacity vs temperature as below: 2.5 2.0 1.5 CV CV 1.0 0.5 0.0 0.0 0.5 τ (hω) 1.0 1.5 2.0 Problem 2 (a) For an ideal monatomic gas, we have PV = Nτ 3/ 2 ⎛ Mτ ⎞ nQ = ⎜ 2⎟ ⎝ 2π ⎠ 5 2 So we can express the entropy as a function of pressure and volume, 3 σ = N [ln(τ 3 / 2 ) + ln V + const.] = N [ln( P 3/ 2V 3/ 2 ) + ln V + const.] = N [ ln( PV 5/ 3 ) + const.] 2 σ = N [ln(nQ / n) + ] 2 / 4 PHYS 427 1st Midterm Exam‐Part B Solutions Fall 2008 The entropy will be constant if no heat is allowed to enter or leave the gas during the expansion. Thus, according to the above expression of the entropy, we obtain PV 5 / 3 is constant. (b) For an ideal monatomic gas, its thermal energy is just dependent on temperature τ 3 ( U = Nτ ). So it is constant if the temperature does not change, that is, ΔU = 0 . 2 (c) For an isothermal expansion, the change of the energy ( ΔU = Q + W ) is zero. Thus, we obtain the heat added to the system is Q = −W = ∫ So the change of its entropy is 3V0 V0 PdV = ∫ Q 3V0 V0 Nτ 0 dV = Nτ 0 ln 3 V Δσ = τ0 = N ln 3 Problem 3 Let J be the solar constant, then we have 2 J 4π rSE = σ BTS4 2 4π rS and hence ⎛ Jr 2 ⎞ TS = ⎜ SE2 ⎟ ⎝ σ B rS ⎠ 1/ 4 ⎛ 0.1× 104 × 1.52 ×1016 ⎞ =⎜ −8 2 10 ⎟ ⎝ 5.67 ×10 × 7 ×10 ⎠ 1/ 4 = 5335 K Problem 4 (a) The partition function of the molecule is Z = 1 + e −ε /τ + e −10ε /τ So the probability that level E1 and E2 are populated is P= 1 + e −ε /τ = 1 + e −ε /τ + e −10ε /τ 1 e −10ε /τ 1+ 1 + e −ε /τ Then, we can find that if τ ε P≈ 1 =1 1+ 0 (b) The average energy is 1 e −ε /τ + 10e −10ε /τ −ε /τ −10ε / τ ε + 10ε e )= E = (0i1 + ε e 1 + e −ε /τ + e −10ε /τ Z 3 / 4 PHYS 427 1st Midterm Exam‐Part B Solutions Fall 2008 (c) The heat capacity is −ε /τ + 100e −10ε /τ + 81e −11ε /τ ⎛ ∂E ⎞ ⎛ ε ⎞ e CV = ⎜ =⎜ ⎟ ⎟ (1 + e −ε /τ + e −10ε /τ ) 2 ⎝ ∂τ ⎠V ⎝ τ ⎠ 2 (d) We discuss the case of two limits: (i) τ → 0 , we have (ε / τ ) ⎛ε ⎞ CV ≈ ⎜ ⎟ (e −ε /τ + 100e −10ε /τ ) = ε /τ (1 + 100e −9ε /τ + 81e −10ε /τ ) → 0 ×1 = 0 e ⎝τ ⎠ 2 2 (ii) τ ε , we have ⎛ ε ⎞ 1 + 100 + 81 CV ≈ ⎜ ⎟ →0 2 ⎝ τ ⎠ (1 + 1 + 1) 2 So we can get the plot of the heat capacity vs temperature as below 0.5 0.4 0.3 CV CV 0.2 0.1 0.0 0 1 2 3 4 5 τ (ε) 6 7 8 9 10 11 12 4 / 4 ...
View Full Document

## This note was uploaded on 12/06/2009 for the course PHYS 427 taught by Professor Flynn during the Fall '08 term at University of Illinois, Urbana Champaign.

Ask a homework question - tutors are online