 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-2/37a/c/solution

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

Substitution of the canonical probabilities into the Gibbs entropy gives

S = - k_{B} \sum_{n} p_{n} (- β E_{n} - lo g Z) = k_{B} lo g Z + k_{B} βE .

Hence

S = k_{B} lo g Z + \frac{E}{T} .

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