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.

The Ehrenfest equations assume a continuous transition with well-defined finite one-sided response functions. They are not generally valid at a critical point where response functions diverge, nor do they replace the Clapeyron equation when $\Delta s$ or $\Delta v$ is nonzero.

Solved Problems

1. Predict the sign of a solid-liquid coexistence slope

For melting, choose $\alpha=$ solid and $\beta=$ liquid. The entropy change is positive,

\[\Delta s=\frac{L_f}{T}>0.\]

For most substances the liquid has larger molar volume, $\Delta v=v_l-v_s>0$, and Clapeyron gives $dp/dT>0$. Ice is exceptional: liquid water is denser than ice near the melting point, so

\[\Delta v=v_l-v_s<0.\]

Therefore

\[\boxed{\left(\frac{dp}{dT}\right)_{\mathrm{ice-water}}<0}.\]

Increasing pressure consequently lowers the melting temperature of ice. The conclusion follows from signs alone and requires equilibrium coexistence.

2. Extract latent heat from two vapour-pressure measurements

Under the integrated Clausius-Clapeyron assumptions,

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

Solving for $L$ gives

\[\boxed{L= \frac{R\ln(p_2/p_1)}{1/T_1-1/T_2}}.\]

For $T_2>T_1$, equilibrium vapour pressure has $p_2>p_1$; numerator and denominator are both positive, so $L>0$. The logarithm contains a pressure ratio and is dimensionless; $R/(1/T)$ supplies $\mathrm{J\,mol^{-1}}$. The result is an interval-average latent heat if $L$ varies with temperature.

Descriptive Questions

Define first- and second-order transitions in terms of derivatives of molar Gibbs energy.
Explain why a first-order transition need not have latent heat if only molar volume is discontinuous.
State every approximation used to obtain the integrated Clausius-Clapeyron equation.
What discontinuities enter the two Ehrenfest equations, and why is molar volume treated as continuous?

Numerical Problems

Water has $p_1=101.325\ \mathrm{kPa}$ at $T_1=373.15\ \mathrm K$. Estimate its vapour pressure at $T_2=353.15\ \mathrm K$ using $L=40.7\ \mathrm{kJ\,mol^{-1}}$ as constant.

Final answer: $p_2=48.2\ \mathrm{kPa}$.
For ice melting at $273.15\ \mathrm K$, take $L_f=6.01\ \mathrm{kJ\,mol^{-1}}$ and $\Delta v=-1.63\times10^{-6}\ \mathrm{m^3\,mol^{-1}}$.

Final answer: $dp/dT=L/(T\Delta v)=-13.5\ \mathrm{MPa\,K^{-1}}$.
At a continuous transition, $\Delta C_p=10\ \mathrm{J\,mol^{-1}K^{-1}}$, $T=200\ \mathrm K$, $v=1.0\times10^{-4}\ \mathrm{m^3\,mol^{-1}}$, and $\Delta\alpha=2.0\times10^{-5}\ \mathrm{K^{-1}}$.

Final answer: the first Ehrenfest slope is $25.0\ \mathrm{MPa\,K^{-1}}$.
If instead $\Delta\alpha=2.0\times10^{-5}\ \mathrm{K^{-1}}$ and $\Delta\kappa_T=1.0\times10^{-12}\ \mathrm{Pa^{-1}}$, use the second equation.

Final answer: $dp/dT=20.0\ \mathrm{MPa\,K^{-1}}$.

The phase-transition Maxima worksheet verifies the Clapeyron signs, latent-heat substitution, Ehrenfest consistency, and every printed numerical answer.

References

Clausius-Clapeyron relation, Wikipedia.
H. B. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., Wiley, 1985, chapter 9.
P. Atkins, J. de Paula, and J. Keeler, Atkins’ Physical Chemistry, 11th ed., Oxford University Press, 2018, chapter “Physical transformations of pure substances.”