 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2020/ii/paper-2/36a/solution

Codex  0 Created 2026-09-24 Updated 2026-09-29

For a simple system of fixed composition, the first law of thermodynamics and

G = E - TS + P V

give

d G = - S d T + V d P .

Thus

S = - (\frac{\partial G}{\partial T})_{P}, V = (\frac{\partial G}{\partial P})_{T} .

(\frac{\partial S}{\partial P})_{T} = - (\frac{\partial V}{\partial T})_{P} .

C_{V} = (\frac{\partial E}{\partial T})_{V} = T (\frac{\partial S}{\partial T})_{V},

C_{P} = (\frac{\partial H}{\partial T})_{P} = T (\frac{\partial S}{\partial T})_{P}, H = E + P V .

Regarding

S

T, V

and differentiating along a constant-pressure curve gives

(\frac{\partial S}{\partial T})_{P} = (\frac{\partial S}{\partial T})_{V} + (\frac{\partial S}{\partial V})_{T} (\frac{\partial V}{\partial T})_{P} .

The Maxwell relation from the Helmholtz free energy is

(\frac{\partial S}{\partial V})_{T} = (\frac{\partial P}{\partial T})_{V} .

C_{P} - C_{V} = T (\frac{\partial V}{\partial T})_{P} (\frac{\partial P}{\partial T})_{V} .

G_{l} = G_{g}

along the curve gives

- S_{l} d T + V_{l} d P = - S_{g} d T + V_{g} d P .

\frac{d P}{d T} = \frac{S _{g} - S _{l}}{V _{g} - V _{l}} = \frac{L}{T ( V _{g} - V _{l} )},

where

L = T (S_{g} - S_{l})

is the molar latent heat.

V_{g} ≫ V_{l}

and the gas is ideal, then

V_{g} = RT / P

, so

\frac{d P}{d T} ≃ \frac{L P}{R T ^{2}} .

For constant

L

, integration yields

lo g P = C - \frac{L}{RT}, P (T) = P_{0} e^{- L / (RT)},

where

P_{0}

is constant over the range in which the approximations apply.

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