 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/36b/c/i/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

With

r = ∣ x ∣

r = V^{1/3} (k_{B} T)^{1/ (2 n)} u .

Then

\int e^{- β U (x)} d^{3} x = 4 πV (k_{B} T)^{3/ (2 n)} I_{n} .

Part (b) gives

Z_{1} = \frac{4}{π ( ℏ c ) ^{3}} V (k_{B} T)^{3 + 3/ (2 n)} I_{n} .

For

N

identical classical non-interacting particles,

Z_{N} = \frac{Z _{1}^{N}}{N !} .

Using

lo g N! = N lo g N - N + o (N)

F = - N k_{B} T [lo g V + A lo g (k_{B} T) + lo g I_{n} + B] + o (N),

where

A = 3 + \frac{3}{2 n}, B = 1 + lo g (\frac{4}{π N ( ℏ c ) ^{3}}) .

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