The Clapeyron Equation

Ice skaters do not glide on a slippery solid. They glide on a thin film of liquid water, produced partly by pressure and mostly by friction, sitting atop the ice. The pressure argument traces back to one deceptively simple equation: the Clapeyron equation, which tells you exactly how the melting temperature of ice shifts as pressure changes. The Clapeyron equation is the master relation for any phase boundary on a one-component phase diagram. It connects the slope of a coexistence curve to the entropy and volume changes of the transition. Every other phase-equilibrium formula students memorize, including the Clausius–Clapeyron equation for vaporization, is a special case of this one relation. This article derives $dp/dT = \Delta S/\Delta V$ from the equality of chemical potentials along a phase boundary. It then applies the result to solid–liquid and solid–solid lines, where the approximations used for vapor equilibria break down. You will see the full derivation. Then two worked numerical examples. Finally, limiting cases, common traps, and a compact exam sheet.

Key takeaways

The Clapeyron equation, $dp/dT = \Delta S/\Delta V$ , gives the exact slope of any coexistence curve in a one-component phase diagram.
It follows directly from equating chemical potentials of two phases in equilibrium and demanding that equality persists along the boundary.
For solid–liquid equilibria, $\Delta V$ is small and can be positive or negative, so melting curves are nearly vertical and can even tilt backward.
Water is the classic anomaly: ice is less dense than liquid water, so its melting curve has a negative slope.
Solid–solid transitions obey the same equation, with $\Delta S$ and $\Delta V$ taken between two crystalline forms.
Key idea: equal chemical potentials along a phase boundary force $dp/dT$ to equal the ratio of the transition entropy to the transition volume, exactly, with no vapor-phase approximations required.

The core insight

Along any two-phase coexistence line, the chemical potentials of the two phases must remain equal as $p$ and $T$ both change together, and demanding that this equality survive an infinitesimal step generates the exact differential equation $dp/dT = \Delta S/\Delta V$ , which is the Clapeyron equation.

The physical picture

Consider a one-component system with two phases, $\alpha$ and $\beta$ , in equilibrium. From the equilibrium criterion built on chemical potential, coexistence requires

\mu_\alpha(p,T) = \mu_\beta(p,T)

at every point on the phase boundary. This is one equation in two variables, $p$ and $T$ , so it defines a curve — the coexistence line you see on a phase diagram, whether it is the fusion line, the vaporization line, or the sublimation line. The chemical potential of a pure substance is just its molar Gibbs energy, $\mu = G_m$ . Its differential is

d\mu = -S_m\,dT + V_m\,dp

where $S_m$ and $V_m$ are the molar entropy and molar volume of that phase. As you move along the coexistence curve, both phases experience the same $dp$ and $dT$ (they are, after all, in contact at the same pressure and temperature), but each phase's chemical potential changes by an amount governed by its own $S_m$ and $V_m$ . Because $\mu_\alpha$ must track $\mu_\beta$ exactly along the curve, the changes in chemical potential must match too. That constraint, worked through below, delivers the Clapeyron equation. The key qualitative point for solid–liquid and solid–solid boundaries is that $\Delta V$ is small — typically a few percent of the molar volume — and can be positive (most substances expand on melting) or negative (water, bismuth, gallium, and a few others contract on melting, i.e., the solid floats). Since $\Delta S_{fus}$ is always positive (melting increases disorder), the sign of $\Delta V_{fus}$ alone determines whether the fusion curve leans forward or backward.

Figure

Chemical potential surfaces crossing along the coexistence curve

The two curves in the figure are $\mu$ vs $T$ at fixed pressure for each phase; their crossing point defines the melting temperature at that pressure. Raise the pressure and both curves shift (via the $V_m\,dp$ term), moving the crossing point — that motion of the crossing point, tracked continuously, traces out the coexistence curve on the $p$ – $T$ diagram.

Figure

Fusion line slope for a normal substance versus water

The core relations

\frac{dp}{dT} = \frac{\Delta S}{\Delta V} = \frac{\Delta_{trs} H}{T\,\Delta_{trs} V}

where: $\Delta S$ and $\Delta V$ are the molar entropy and volume changes of the transition ( $\beta$ minus $\alpha$ ), and $\Delta_{trs}H = T\,\Delta_{trs}S$ at equilibrium.

\Delta p \approx \frac{\Delta_{fus} H}{\Delta_{fus} V}\,\frac{\Delta T}{T_{fus}}

where: this is the linearized (integrated, small-range) form used for solid–liquid and solid–solid lines, valid when $\Delta_{fus}H$ and $\Delta_{fus}V$ are treated as pressure- and temperature-independent constants.

\mu_\alpha(p,T) = \mu_\beta(p,T) \quad \text{along the entire coexistence curve}

where: this is the equilibrium condition that generates the Clapeyron equation upon differentiation.

Derivation AFrom equal chemical potentials to dp/dT = ΔS/ΔV

Goal. Show that the slope of any two-phase coexistence curve in a one-component system is exactly $\Delta S/\Delta V$ , with no assumptions about which phases are involved.

Step 1 — State the equilibrium condition. Two phases $\alpha$ and $\beta$ of a pure substance coexist at pressure $p$ and temperature $T$ only if their molar Gibbs energies (chemical potentials) are equal:

\mu_\alpha(p,T) = \mu_\beta(p,T)

This holds not just at one point but along an entire curve in the $p$ – $T$ plane, since a pure substance has a continuous locus of coexistence conditions (this is exactly the content of the phase rule, $F = C - P + 2 = 1$ for $C=1, P=2$ : one degree of freedom remains, so fixing $T$ fixes the equilibrium $p$ ).

Step 2 — Perturb along the curve. Move to a neighboring point on the same coexistence curve, $(p+dp,\, T+dT)$ . Equilibrium still requires equal chemical potentials there:

\mu_\alpha + d\mu_\alpha = \mu_\beta + d\mu_\beta

Subtracting the Step 1 equation from this one gives

d\mu_\alpha = d\mu_\beta

The changes in chemical potential of the two phases must be identical as you move along the boundary, even though the phases themselves have different molar entropies and volumes.

Step 3 — Insert the fundamental equation for $d\mu$ . For a pure substance, $\mu = G_m$ , and the Gibbs energy differential $dG = -S\,dT + V\,dp$ becomes, per mole,

d\mu = -S_m\,dT + V_m\,dp

Apply this to both phases:

-S_{m,\alpha}\,dT + V_{m,\alpha}\,dp = -S_{m,\beta}\,dT + V_{m,\beta}\,dp

Step 4 — Collect terms. Rearrange to group the $dT$ and $dp$ terms:

(S_{m,\beta} - S_{m,\alpha})\,dT = (V_{m,\beta} - V_{m,\alpha})\,dp

Define $\Delta S = S_{m,\beta} - S_{m,\alpha}$ and $\Delta V = V_{m,\beta} - V_{m,\alpha}$ for the transition $\alpha \to \beta$ . Then

\Delta S\,dT = \Delta V\,dp

Step 5 — Solve for the slope. Divide both sides by $dT\,\Delta V$ :

\frac{dp}{dT} = \frac{\Delta S}{\Delta V}

Step 6 — Rewrite using enthalpy. Because the transition occurs reversibly at constant $T$ and $p$ (equilibrium), $\Delta S = \Delta H/T$ . Substituting gives the equivalent, often more convenient, form:

\boxed{\dfrac{dp}{dT} = \dfrac{\Delta_{trs} H}{T\,\Delta_{trs} V}}

This result is exact — no ideal-gas assumption, no approximation for either molar volume, valid for any pair of coexisting phases: solid–liquid, liquid–vapor, solid–vapor, or solid–solid polymorphs. The Clausius–Clapeyron equation you meet next is what results when you additionally assume the vapor phase is ideal and $V_{m,gas} \gg V_{m,liquid}$ — an approximation that is not available for condensed-phase transitions, which is exactly why the full Clapeyron equation is indispensable for melting and solid–solid lines.

Worked examples

Example 1 — Foundational: the slope of the ice–water fusion line. At $T = 273.15\ \text{K}$ and $p = 1\ \text{bar}$ , the molar enthalpy of fusion of ice is $\Delta_{fus}H = 6.01\ \text{kJ mol}^{-1}$ . The molar volumes are $V_m(\text{ice}) = 19.65\ \text{cm}^3\,\text{mol}^{-1}$ and $V_m(\text{liquid}) = 18.02\ \text{cm}^3\,\text{mol}^{-1}$ (densities $0.9167\ \text{g cm}^{-3}$ and $0.9998\ \text{g cm}^{-3}$ for $M = 18.02\ \text{g mol}^{-1}$ ). Step 1: compute $\Delta_{fus}V$ (liquid minus solid, since melting is solid → liquid):

\Delta_{fus}V = 18.02 - 19.65 = -1.63\ \text{cm}^3\,\text{mol}^{-1} = -1.63\times10^{-6}\ \text{m}^3\,\text{mol}^{-1}

Step 2: apply the Clapeyron equation.

\frac{dp}{dT} = \frac{\Delta_{fus}H}{T\,\Delta_{fus}V} = \frac{6.01\times10^{3}\ \text{J mol}^{-1}}{(273.15\ \text{K})(-1.63\times10^{-6}\ \text{m}^3\,\text{mol}^{-1})}

\frac{dp}{dT} = \frac{6010}{-4.453\times10^{-4}}\ \text{Pa K}^{-1} = -1.35\times10^{7}\ \text{Pa K}^{-1} = -134\ \text{bar K}^{-1}

Step 3: estimate the pressure needed to lower the melting point by $1\ \text{K}$ (linearizing, valid for small $\Delta T$ ):

\Delta p \approx \left(\frac{dp}{dT}\right)\Delta T = (-134\ \text{bar K}^{-1})(-1\ \text{K}) = +134\ \text{bar}

So increasing the pressure by about $134$ bar lowers the melting point of ice by roughly $1\ \text{K}$ . Sanity check: this matches the textbook rule of thumb that it takes on the order of $100$ – $140$ atm per degree to depress ice's melting point — consistent with the very small, negative $\Delta_{fus}V$ . The negative sign is the whole story behind ice skating and glaciers: squeezing ice can melt it, a behavior essentially unique among common substances.

Example 2 — Applied: locating a solid–solid transition line for tin. Tin exists in two solid allotropes: gray tin (α, diamond-cubic, the low-temperature form) and white tin (β, tetragonal, the ordinary metallic form), with the classic α ⇌ β transition near room temperature at $1$ bar (the culprit behind "tin pest" in old organ pipes). Take the following approximate literature-consistent values for the α → β transition at $p = 1\ \text{bar}$ , $T_0 = 286.2\ \text{K}$ ( $13.05\,^\circ\text{C}$ ):

\Delta_{trs}H = +2.1\ \text{kJ mol}^{-1}, \qquad \Delta_{trs}V = V_m(\beta) - V_m(\alpha) = -2.6\ \text{cm}^3\,\text{mol}^{-1}

(White tin is denser than gray tin, so $\Delta V$ is negative for α → β, consistent with $\Delta H > 0$ meaning heat is absorbed on warming through the transition, i.e. β is the high-temperature, high-entropy form.) Step 1: compute the slope at $T_0$ .

\frac{dp}{dT} = \frac{\Delta_{trs}H}{T_0\,\Delta_{trs}V} = \frac{2.1\times10^{3}\ \text{J mol}^{-1}}{(286.2\ \text{K})(-2.6\times10^{-6}\ \text{m}^3\,\text{mol}^{-1})}

\frac{dp}{dT} = \frac{2100}{-7.44\times10^{-4}}\ \text{Pa K}^{-1} = -2.82\times10^{6}\ \text{Pa K}^{-1} = -28.2\ \text{bar K}^{-1}

Step 2: estimate the pressure needed to shift the transition temperature by $+10\ \text{K}$ (from $286.2\ \text{K}$ to $296.2\ \text{K}$ ), using the linearized integrated form:

\Delta p \approx \frac{\Delta_{trs}H}{\Delta_{trs}V}\cdot\frac{\Delta T}{T_0} = (-28.2\ \text{bar K}^{-1})(10\ \text{K}) = -282\ \text{bar}

So raising the α→β transition temperature by $10\ \text{K}$ requires lowering the pressure by about $282$ bar (or, equivalently, applying about $282$ bar of extra pressure lowers the transition temperature by $10\ \text{K}$ , stabilizing the denser β-phase less and — since here increasing $p$ favors the smaller-volume phase, β — actually raising pressure stabilizes β further). Sanity check on direction: Le Chatelier's principle says increasing pressure favors the phase of smaller molar volume (β, since $V_m(\beta) < V_m(\alpha)$ ); the negative $dp/dT$ found here is consistent with a phase boundary that tilts backward in $p$ – $T$ space, exactly analogous to the ice line in Example 1, and both trace back to the same sign rule: $\Delta H>0$ and $\Delta V<0$ together force $dp/dT<0$ .

Limiting cases & common traps

Situation	What happens to the Clapeyron equation
Solid–liquid (fusion) line	$\Delta V$ small (a few %), can be $+$ or $-$ ; slope is very steep (often tens to hundreds of bar per kelvin)
Solid–solid (polymorphic) line	Same equation applies directly; both phases condensed, so again a steep line, sign set by which phase is denser
Liquid–vapor / solid–vapor	$\Delta V \approx V_m(gas)$ , huge and always positive; slope is shallow — this is where the Clausius–Clapeyron approximation becomes useful, not the full equation itself
Near a critical point	$\Delta V \to 0$ and $\Delta S \to 0$ simultaneously; the ratio stays finite, but the linearized $\Delta p$ formula breaks down and must be integrated properly
Water's fusion line specifically	$\Delta_{fus}V < 0$ , so $dp/dT < 0$ : melting point decreases with pressure, unlike almost every other substance

The chemist behind it

Émile Clapeyron was a French engineer who, working in the tradition of Sadi Carnot, first expressed the connection between the latent heat of a phase change and the slope of its pressure–temperature boundary in 1834 — decades before Clausius and Kelvin put the second law on rigorous footing. Clapeyron's original derivation used Carnot-cycle reasoning applied to a substance undergoing vaporization, extracting work from the volume change across an infinitesimal phase transition. It was Rudolf Clausius who, in the 1850s, rebuilt the argument on the newly established entropy concept and extended it into the exponential form used for vapor pressures — the equation now bearing both names when applied to condensed-vapor equilibria. The general form derived above, however, requires no gas-phase approximation at all and is properly Clapeyron's alone.

Origins · 1 minBenoît Paul Émile Clapeyron and how this was discovered

Benoît Paul Émile Clapeyron

1799–1864 · French engineer who derived the pressure–temperature slope of phase boundaries from Carnot's cycle, decades before the second law was fully formalized

Ask the tutor: How does the Clapeyron equation simplify into the Clausius–Clapeyron equation once one phase is a gas?

Is the Clapeyron equation exact, or is it an approximation like the Clausius–Clapeyron equation?

It is exact. The derivation uses only the equality of chemical potentials and the exact differential $d\mu = -S_m\,dT + V_m\,dp$ ; no assumption about ideal gas behavior or negligible molar volume enters anywhere. The Clausius–Clapeyron equation is a further approximation, valid only for condensed-phase–vapor equilibria, obtained by assuming the vapor is ideal and that its molar volume vastly exceeds the condensed phase's. For solid–liquid and solid–solid lines that approximation is inapplicable, which is exactly why the full Clapeyron form is used there instead.

Why is the fusion line on a phase diagram almost vertical compared to the vaporization line?

The slope is $\Delta H/(T\Delta V)$ . For fusion, $\Delta V$ is only a few cm³ per mole — both solid and liquid are condensed phases with similar densities — so the denominator is tiny and the slope is very steep, often on the order of $100$ bar per kelvin, as in the ice example above. For vaporization, $\Delta V$ is close to the molar volume of the gas, thousands of times larger, so the slope is shallow, giving the familiar gently rising vaporization curve.

Can Δ_fus V be zero, and what would that mean?

In principle, yes, at a special point where the solid and liquid molar volumes coincide, although this is rare for real substances across normal conditions. If $\Delta V \to 0$ while $\Delta S$ stays finite, the Clapeyron equation predicts $dp/dT \to \infty$ : the fusion line becomes momentarily vertical in the $p$ – $T$ plane, then can even reverse slope on the other side. Something close to this happens for water's fusion curve at very high pressure, where multiple ice polymorphs (ice III, V, VI) meet and the simple monotonic picture breaks down.

How does the Clapeyron equation relate to the phase rule and the number of independent variables?

The phase rule gives $F = C - P + 2$ . For a one-component system with two phases coexisting, $C=1$ and $P=2$ , so $F=1$ : only one variable, say $T$ , can be chosen independently, and $p$ is then fixed by the equilibrium condition. The Clapeyron equation is precisely the differential statement of that one remaining degree of freedom — it tells you how the dependent variable $p$ must move as you sweep the independent variable $T$ along the coexistence curve, consistent with $F=1$ .

Why does applying pressure to ice only lower its melting point by a modest amount, contrary to some popular explanations of ice skating?

Because $\Delta_{fus}V$ for water is so small, it takes roughly $134$ bar to lower the melting point by just $1\ \text{K}$ , as computed above. A skater's blade, even concentrating body weight onto a small contact area, typically generates only a few tens of bar — enough for a fraction of a degree of melting-point depression, not several degrees. Frictional heating, not pressure melting, is now understood to be the dominant mechanism producing the thin lubricating water film beneath skate blades.