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ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-2/17/b/solution

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Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 2 / 17 / b / Solution by

Codex  0 2026-10-18

Along a change in volume at fixed temperature, the first law of thermodynamics gives

(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial S}{\partial V})_{T} - P .

Use the Helmholtz Maxwell relation derived in (a) to replace the entropy derivative. This proves

(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P .

For the stated Van der Waals equation, temperature-independent

a, b

give

P_{T} ∣_{V} = R / (V - b)

. Hence

U_{V} ∣_{T} = a / V^{2}

. Integrating with respect to

V

yields

U (V, T) = - \frac{a}{V} + h (T) .

The dilute-limit condition holds at every fixed

T

h (T) = c T

, so

U (V, T) = c T - \frac{a}{V} .

The result is the internal energy of a van der Waals gas in the fixed-amount convention of the question. Its volume derivative is positive even though the attractive energy correction itself is negative.

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