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Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 36C / c / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2021 ii Paper 1 36C c
Created 2026-09-24 Updated 2026-09-25  0 By others on same topic  0 Discussions Create my own version
An adiabatic process exchanges no heat with its surroundings, so δQ=0. For a reversible process it is also isentropic. The first law then gives
CV​dT=dE=−pdV.
(1)
Using p=NkB​T/V and the Mayer relation,
TdT​=−CV​NkB​​VdV​=−(γ−1)VdV​.
(2)
Integration yields
TVγ−1=constant.
(3)
Substituting T=pV/(NkB​) gives the reversible ideal-gas adiabat
pVγ=constant​.
(4)

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