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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-2/35d/iii/solution

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Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 35D / iii / Solution by

Codex  0 Created 2026-09-24 Updated 2026-10-14

Free expansion into vacuum performs no work and receives no heat, so the initial and final internal energies agree. If

C_{V}

is temperature independent, the preceding energy expression gives

C_{V} (T_{2} - T_{1}) = a N^{2} (\frac{1}{V _{2}} - \frac{1}{V _{1}}), T_{2} = T_{1} - \frac{a N ^{2}}{C _{V}} (\frac{1}{V _{1}} - \frac{1}{V _{2}}) .

(1)

The gas cools for positive

a

and

V_{2} > V_{1}

. Genuine free expansion is irreversible and not quasistatic; the endpoint energy calculation remains valid without the prompt's quasistatic qualification. Use of equilibrium thermodynamic identities only requires the two endpoint states to be equilibrated.

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