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 Incoming links: Mayer relation

Past exam of the mathematics course of the University of Cambridge / 2021 / ii / Paper 1 / 36C / b / Solution Created 2026-09-24 Updated 2026-09-25

At constant volume, the First law of thermodynamics gives

δ Q = d E = C_{V} d T .

At constant pressure,

δ Q = C_{p} d T = d E + p d V .

Differentiating

p V = N k_{B} T

at fixed

p

gives

p d V = N k_{B} d T .

Consequently,

C_{p} d T = (C_{V} + N k_{B}) d T,

and the Mayer relation is

C_{p} - C_{V} = N k_{B} .

The specific-heat ratio is therefore

γ = \frac{C _{p}}{C _{V}} = 1 + \frac{N k _{B}}{C _{V}} .

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Past exam of the mathematics course of the University of Cambridge / 2023 / ii / Paper 4 / 35A / c / Solution Created 2026-09-24 Updated 2026-09-25

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}} .

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