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 Incoming links: Gibbs free-energy Maxwell relation

Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 3 / 33C / b / ii / Solution Created 2026-09-24 Updated 2026-10-06

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Combining

d H = T d S + V d p

with the preceding Gibbs free-energy Maxwell relation gives

d H = C_{p} d T + [V - T (\frac{\partial V}{\partial T})_{p}] d p .

(1)

Setting

d H = 0

and solving for

d T / d p

yields the Joule-Thomson coefficient

μ = (\frac{\partial T}{\partial p})_{H} = \frac{V}{C _{p}} [\frac{T}{V} (\frac{\partial V}{\partial T})_{p} - 1] .

(2)

When

μ > 0

, a negative pressure change produces cooling.

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Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 2 / 36A / Solution Created 2026-09-24 Updated 2026-09-29

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

G = E - TS + P V

(1)

give

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

(2)

Thus

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

(3)

Equality of mixed derivatives gives the Gibbs free-energy Maxwell relation

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

(4)

The heat capacity at constant volume and heat capacity at constant pressure are

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

(5)

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

(6)

Regarding

S

as a function of

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} .

(7)

The Maxwell relation from the Helmholtz free energy is

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

(8)

Therefore the difference between constant-pressure and constant-volume heat capacities is

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

(9)

Along the liquid-gas phase coexistence curve, the molar Gibbs free energies agree. Differentiating

G_{l} = G_{g}

along the curve gives

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

(10)

Hence the Clausius-Clapeyron relation is

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

(11)

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}} .

(12)

For constant

L

, integration yields

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

(13)

where

P_{0}

is constant over the range in which the approximations apply.

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