 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/ii/paper-4/35a/c/solution

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

δ Q = 0

. With work done by the gas taken as positive, the first law of thermodynamics gives

d E = δ Q - p d V = - p d V .

Using

d E = C_{V} d T

and

p = N k_{B} T / V

C_{V} \frac{d T}{T} = - N k_{B} \frac{d V}{V} .

Integration, followed by the Mayer relation

N k_{B} / C_{V} = γ - 1

, yields

T V^{γ - 1} = constant .

Eliminating

T

p V^{γ} = constant, γ = \frac{C _{p}}{C _{V}} .

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