 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-314/2/d/solution

Codex  0 Created 2026-10-03 Updated 2026-10-05

Put

S (P) = [s] / c_{v}

. Differentiating the expression in part (c) gives

S^{'} (P) = \frac{( γ ^{2} - 1 ) ( P - 1 ) ^{2}}{P [( γ + 1 ) P + γ - 1 ] [( γ - 1 ) P + γ + 1 ]} .

For

P = 1 + δ

4 γ^{2} + O (δ)

, so

S^{'} (1 + δ) = \frac{γ ^{2} - 1}{4 γ ^{2}} δ^{2} + O (δ^{3}) .

Integrating from the unshocked state, where

S (1) = 0

\frac{[ s ]}{c _{v}} = \frac{γ ^{2} - 1}{12 γ ^{2}} δ^{3} + O (δ^{4}) .

The first-order and second-order terms vanish: a weak shock agrees with reversible compression through second order.

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