 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ii/paper-3/35c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-10-14

Z_{1} = \int_{0}^{\infty} K V ε^{1/2} e^{- βε} d ε = \frac{K V π}{2} β^{- 3/2} .

The classical indistinguishable noninteracting gas has

Z_{N} = Z_{1}^{N} / N!

; for labeled particles omit

N!

, which changes none of the requested derivatives. With

β = (k_{B} T)^{- 1}

E = - \partial_{β} lo g Z_{N} = \frac{3}{2} N k_{B} T, C_{V} = \frac{3}{2} N k_{B} .

The allowed equation of state gives

P V = N k_{B} T

, so the enthalpy is

E + P V = 5 N k_{B} T /2

and

C_{P} = 5 N k_{B} /2

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