23 Jul 2025

Phase Transitions, Clausius-Clapeyron, and Ehrenfest Equations

First- and second-order transitions, coexistence slopes, latent heat, and the two Ehrenfest equations.

bsc semester-iv mj-6 heat-and-thermodynamics phase-transitions clapeyron-equation

At fixed temperature and pressure, two phases $\alpha$ and $\beta$ coexist when their molar Gibbs free energies, or chemical potentials, are equal:

\[g_\alpha(T,p)=g_\beta(T,p).\]

For one mole,

\[dg=-s\,dT+v\,dp,\]

where $s$ and $v$ are molar entropy and molar volume.

First-order transition and Clapeyron equation

Along the coexistence curve, equality of the two Gibbs energies must persist:

\[dg_\alpha=dg_\beta.\]

Therefore

\[-s_\alpha\,dT+v_\alpha\,dp =-s_\beta\,dT+v_\beta\,dp.\]

With $\Delta s=s_\beta-s_\alpha$ and $\Delta v=v_\beta-v_\alpha$,

\[\boxed{\frac{dp}{dT}=\frac{\Delta s}{\Delta v} =\frac{L}{T\Delta v}},\]

where $L=T\Delta s$ is the molar latent heat for $\alpha\to\beta$. This is the Clapeyron equation. Its units are

\[\frac{\mathrm{J\,mol^{-1}}} {\mathrm{K\,m^3\,mol^{-1}}} =\mathrm{Pa\,K^{-1}}.\]

A first-order transition has continuous $g$ but a discontinuity in at least one first derivative: $s=-(\partial g/\partial T)_p$ and/or $v=(\partial g/\partial p)_T$. An entropy jump gives latent heat $L=T\Delta s$; a volume jump may also occur. Thus latent heat is present when $\Delta s\ne0$, not merely from the classification label alone.

For liquid-vapour coexistence far below the critical point, $v_g\gg v_l$ and the vapour may be treated as ideal. Then $\Delta v\simeq v_g=RT/p$, and

\[\boxed{\frac{d\ln p}{dT}=\frac{L}{RT^2}}.\]

If $L$ is approximately constant between $T_1$ and $T_2$,

\[\boxed{\ln\!\frac{p_2}{p_1} =-\frac{L}{R}\left(\frac1{T_2}-\frac1{T_1}\right)}.\]

This integrated Clausius-Clapeyron form requires an ideal vapour, negligible condensed-phase molar volume, and nearly constant latent heat.

Second-order transition and Ehrenfest equations

In the classical Ehrenfest classification, a second-order transition has continuous $g$, $s$, and $v$, but discontinuities in second derivatives such as $C_p$, the expansion coefficient $\alpha$, or the isothermal compressibility $\kappa_T$. There is no latent heat because $\Delta s=0$.

In the following equations $C_p$ is the molar heat capacity, consistent with the molar quantities $g$, $s$, and $v$.

On the transition curve, $\Delta s=0$. Differentiate this condition along the curve and use

\[ds=\frac{C_p}{T}\,dT -\left(\frac{\partial v}{\partial T}\right)_pdp =\frac{C_p}{T}\,dT-v\alpha\,dp.\]

Because $v$ is continuous at the transition,

\[0=d(\Delta s)=\frac{\Delta C_p}{T}\,dT-v\Delta\alpha\,dp.\]

The first Ehrenfest equation is therefore

\[\boxed{\left(\frac{dp}{dT}\right)_{\mathrm{tr}} =\frac{\Delta C_p}{Tv\Delta\alpha}}.\]

Likewise $\Delta v=0$ on the curve. Since

\[dv=v\alpha\,dT-v\kappa_T\,dp,\]

we obtain

\[0=d(\Delta v)=v\Delta\alpha\,dT-v\Delta\kappa_T\,dp,\]

and hence the second Ehrenfest equation,

\[\boxed{\left(\frac{dp}{dT}\right)_{\mathrm{tr}} =\frac{\Delta\alpha}{\Delta\kappa_T}}.\]

Here every $\Delta$ means the value in phase $\beta$ minus that in phase $\alpha$, evaluated on the two sides of the same transition.