4.4
Physics Foundations
§4 · Free energy and phase transitions

4.4 The Clausius–Clapeyron relation

The coexistence curves of the phase diagram are not free to take any shape. Their slope at every point is fixed by two measurable quantities — the latent heat of the transition and the volume change across it — through the Clausius–Clapeyron relation.

The slope of a coexistence curve

▶ dp/dT = L/(T Δv) Derivation

Along a coexistence curve the two phases keep equal Gibbs free energy per unit mass, g1=g2g_1 = g_2, so any move along the curve changes them equally, dg1=dg2dg_1 = dg_2. With dg=−s dT+v dpdg = -s\,dT + v\,dp for each phase,

−s1 dT+v1 dp  =  −s2 dT+v2 dp,-s_1\,dT + v_1\,dp \;=\; -s_2\,dT + v_2\,dp,

and rearranging gives the slope of the curve in terms of the jumps across it:

dpdT  =  s2−s1v2−v1  =  ΔsΔv.\frac{dp}{dT} \;=\; \frac{s_2 - s_1}{v_2 - v_1} \;=\; \frac{\Delta s}{\Delta v}.

The entropy jump is the latent heat over the temperature, Δs=L/T\Delta s = L/T, so

  dpdT  =  LT Δv.  \boxed{\;\frac{dp}{dT} \;=\; \frac{L}{T\,\Delta v}.\;}

✓

The vapour-pressure curve

For boiling, the volume change is dominated by the vapour, Δv≈vvapour\Delta v \approx v_\text{vapour}, and if the vapour is treated as an ideal gas, vvapour=RT/(Mp)v_\text{vapour} = RT/(Mp). The relation becomes a separable equation for the vapour pressure,

dpdT  ≈  LMpRT2⟹ln⁡p  =  −LMRT+const.\frac{dp}{dT} \;\approx\; \frac{L M p}{R T^2} \quad\Longrightarrow\quad \ln p \;=\; -\frac{L M}{R T} + \text{const}.

The vapour pressure rises exponentially with temperature, controlled by the Boltzmann-like factor e−LM/RTe^{-LM/RT} — the same activation form seen throughout thermal physics, here with the latent heat per particle playing the role of the energy barrier to escape the liquid.

280300320340360380102103104105106T (K)pv (Pa)20 °C37 °C (body)100 °C (boiling)
L (latent heat)2.45 MJ/kg
L M / R5304 K
pv(20°C)2094 Pa
pv(100°C)101.3 kPa

The curve is exponential in 1/T; on log-y vs linear-T it looks like a steep S. The Clausius-Clapeyron form ln p = -LM/(RT) + const captures the bare exponential; the measured water data sit on it to high accuracy. Lowering L flattens the curve; the latent heat is what makes vapour pressure so sensitive to temperature.

The dependence on 1/T1/T is steep: water’s vapour pressure climbs from 0.6 kPa0.6\,\text{kPa} at 0∘0^\circC to 101 kPa101\,\text{kPa} at 100∘100^\circC, a factor of more than a hundred over a hundred kelvin — which is what boiling at 100∘100^\circC under one atmosphere means, the temperature at which the vapour pressure reaches the ambient pressure. Slide the latent heat to see the curvature steepen: a larger LL binds the liquid more tightly and makes the vapour pressure rise more sharply.