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hw9solv4

# hw9solv4 - Homework 9 Solution Physics 404 Spring 2011...

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Homework 9 Solution Physics 404 - Spring 2011 Homework 9 Solutions Question 1 K+K Chapter 6, Problem 9 a) We can calculate the chemical potential using eq(6.48): μ = τ (log( n/n Q ) - log Z int ) (1) Z int = X int. states e - ε int = 1 + e - Δ (2) μ = τ log n/n Q 1 + e - Δ (3) b) The partition function in the presence of internal degrees of freedom is Z = Z IG × Z int . As a result the free energy, which is proportional to the logarithm of Z is additive: F = F IG + F int . F IG is the known ideal gas result. Let us calculate the contribution from the internal degrees of freedom. F int = - log Z int (4) F int = - log 1 + e - Δ (5) F = F IG + F int = log n/n Q 1 + e - Δ - (6) c) The entropy can be easily calculated using the free energy obtained in part b using the formula: σ int = ∂F int ∂τ V (7) = - ∂τ - log 1 + e - Δ (8) = N log 1 + e - Δ + Δ e Δ + 1 (9) Just like the free energy, entropy is additive as well: σ = σ IG + σ int (10) = N log ( n Q /n ) + 5 / 2 + log 1 + e - Δ + Δ e Δ + 1 (11) = N log n Q (1 + e - Δ ) n + 5 2 + Δ e Δ + 1 (12) 1

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Homework 9 Solution Physics 404 - Spring 2011 d) Pressure is not effected by the existence of the internal degrees of freedom: p = - ∂F ∂V τ,N (13) = - ∂V N log n Q (1 + e - Δ ) n + 5 2 + Δ e Δ + 1 (14) = V (15) e) Heat capacity at constant pressure C p = τ ∂σ ∂τ p = ∂U ∂τ p + p ∂V ∂τ p (16) U = F + τσ = · · · = 3 2 + N Δ e Δ + 1 (17) ∂U ∂τ p = 3 2 N + N 2 2 ) e Δ ( e Δ + 1 ) 2 (18) C p = 3 2 N + N 2 2 ) e Δ ( e Δ + 1 ) 2 + p N p (19) where in the last term of the last equation we used the ideal gas law to obtain ∂V ∂τ . Hence the final answer reads C p = 5 2 N + N 2 2 ) e Δ ( e Δ + 1 ) 2 (20) Question 2 K+K Chapter 6, Problem 12 a) In K+K Chapter 7, Problem 1, part b we calculate the density of states for an electron in 2 dimensions. We can borrow that result here but we need
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hw9solv4 - Homework 9 Solution Physics 404 Spring 2011...

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