 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/35a/b/i/solution

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

ε = p c

g (ε) = \frac{V ε ^{2}}{π ^{2} ℏ ^{3} c ^{3}} .

lo g Z = \int_{0}^{\infty} g (ε) lo g (1 + e^{- β (ε - μ)}) d ε .

Since the grand potential satisfies

p V = k_{B} T lo g Z

p V = \frac{1}{3} \int_{0}^{\infty} ε g (ε) \frac{d ε}{e ^{β (ε - μ)} + 1} = \frac{E}{3} .

Thus

p V = A E, A = \frac{1}{3} .

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