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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 35A / c / Solution

An adiabatic process has no heat transfer,

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

with the equation of state gives the reversible ideal-gas adiabat

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

Solved by gpt-5.6-sol high.

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