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

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- Solution c

Solution

An adiabatic process exchanges no heat with its surroundings, so

δ Q = 0

. For a reversible process it is also isentropic. The first law then gives

C_{V} d T = d E = - p d V .

Using

p = N k_{B} T / V

and the Mayer relation,

\frac{d T}{T} = - \frac{N k _{B}}{C _{V}} \frac{d V}{V} = - (γ - 1) \frac{d V}{V} .

Integration yields

T V^{γ - 1} = constant .

Substituting

T = p V / (N k_{B})

gives the reversible ideal-gas adiabat

p V^{γ} = constant .

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