The Clausius–Clapeyron Equation

A camper at 4,300 m boils potatoes for twenty minutes and they still crunch. Water boils at 86 °C up there, not 100 °C, because atmospheric pressure has dropped. The Clausius–Clapeyron equation tells you exactly how much the boiling point shifts, using only the enthalpy of vaporization and two temperatures. This matters because vapor pressure controls distillation, freeze-drying, weather, and the shelf life of perfumes. Chemists rarely measure vapor pressure at every temperature they care about. Instead they measure it at two or three points and extrapolate with this equation. This article specializes the general Clapeyron equation to liquid–gas and solid–gas equilibria. You will see where the ideal-gas and small-liquid-volume approximations enter. Then a full derivation follows. Two worked numerical examples close the loop, one foundational and one applied.

Key takeaways

The Clausius–Clapeyron equation approximates the Clapeyron equation for vaporization or sublimation.
It assumes the vapor is an ideal gas and that the liquid or solid molar volume is negligible.
The result is a logarithmic relation between vapor pressure and temperature, linear in $1/T$ .
Integrating between two states gives the two-point form used to find $\Delta H_{vap}$ or predict a new boiling point.
$\Delta H_{vap}$ is treated as constant over the temperature range, an approximation that fails over wide spans.
Key idea: Because gas volume vastly exceeds liquid volume and behaves ideally, $d\ln P/dT$ becomes proportional to $\Delta H_{vap}/T^2$ .

The core insight

Starting from the exact Clapeyron equation, replacing the volume change $\Delta V$ with the ideal-gas volume of the vapor alone converts a mechanical slope $dP/dT$ into a clean exponential-in-temperature law for vapor pressure, the Clausius–Clapeyron equation.

The physical picture

Along any two-phase coexistence curve, pressure and temperature are not independent; fixing one fixes the other. The Clapeyron equation, derived from equality of chemical potentials on both sides of the phase boundary, gives the exact slope of that curve:

\frac{dP}{dT} = \frac{\Delta H}{T\Delta V}

This is exact for any two-phase equilibrium — solid–liquid, liquid–gas, or solid–gas. For solid–liquid, $\Delta V$ is small and can be positive or negative, giving the nearly vertical melting curves you have already met. For liquid–gas and solid–gas equilibria, something special happens. The molar volume of a gas near room conditions is on the order of 25 L/mol, while a liquid or solid molar volume is more like 0.02–0.1 L/mol. The volume change is dominated by the gas phase almost entirely:

\Delta V = V_{gas} - V_{liquid} \approx V_{gas}

If the vapor also behaves ideally, $V_{gas} = RT/P$ . Substituting this into the Clapeyron equation converts a geometric slope into a differential equation purely in $P$ and $T$ . That is the entire content of the Clausius–Clapeyron equation: two approximations layered onto one exact thermodynamic identity. The payoff is enormous. Instead of needing $\Delta V$ at every point on the curve, you need only $\Delta H_{vap}$ , treated as roughly constant, plus the ideal gas constant. The vapor pressure curve becomes an exponential in $-1/T$ , and a plot of $\ln P$ against $1/T$ becomes a straight line — one of the most-used plots in physical chemistry.

Figure

The Clapeyron slope narrows to an exponential vapor-pressure curve

The core relations

\frac{d\ln P}{dT} = \frac{\Delta H_{vap}}{RT^2}

where: $P$ is vapor pressure, $\Delta H_{vap}$ is the molar enthalpy of vaporization (or sublimation), $R$ is the gas constant, $T$ absolute temperature.

\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

where: $P_1, P_2$ are vapor pressures at temperatures $T_1, T_2$ , with $\Delta H_{vap}$ assumed constant across the interval.

\ln P = -\frac{\Delta H_{vap}}{R}\cdot\frac{1}{T} + C

where: $C$ is an integration constant; this is the equation of the straight line in a $\ln P$ vs $1/T$ plot, slope $-\Delta H_{vap}/R$ .

Derivation AFrom the Clapeyron equation to the Clausius–Clapeyron law

Goal. Starting from the exact Clapeyron equation for a liquid(or solid)–vapor boundary, obtain a closed-form relation between $P$ and $T$ using the ideal-gas and small-condensed-volume approximations.

Step 1: Write the exact Clapeyron equation. For equilibrium between a condensed phase and its vapor,

\frac{dP}{dT} = \frac{\Delta H_{vap}}{T\left(V_{gas} - V_{liq}\right)}

This follows from setting $d\mu_{liq} = d\mu_{gas}$ along the coexistence curve and using $d\mu = -S_m\,dT + V_m\,dP$ for each phase; it needs no assumption about either phase's equation of state.

Step 2: Neglect the liquid (or solid) molar volume. At 373 K and 1 atm, water vapor occupies about 30.6 L/mol while liquid water occupies 0.018 L/mol — a factor of 1700. Dropping $V_{liq}$ against $V_{gas}$ introduces an error far smaller than typical experimental uncertainty in $\Delta H_{vap}$ itself:

V_{gas} - V_{liq} \approx V_{gas}

Step 3: Invoke the ideal gas law for the vapor. Away from the critical point, and at pressures near 1 atm or below, the vapor obeys $PV_{gas} = RT$ closely, so

V_{gas} = \frac{RT}{P}

Step 4: Substitute and simplify. Putting Steps 2 and 3 into the Clapeyron equation,

\frac{dP}{dT} = \frac{\Delta H_{vap}}{T\cdot\left(RT/P\right)} = \frac{P\,\Delta H_{vap}}{RT^2}

Divide both sides by $P$ and use $d\ln P = dP/P$ :

\frac{d\ln P}{dT} = \frac{\Delta H_{vap}}{RT^2}

Step 5: Integrate, treating $\Delta H_{vap}$ as constant. Over a temperature range narrow enough that $\Delta H_{vap}$ does not change appreciably, integrate both sides between $(T_1, P_1)$ and $(T_2, P_2)$ :

\int_{P_1}^{P_2} d\ln P = \frac{\Delta H_{vap}}{R}\int_{T_1}^{T_2}\frac{dT}{T^2}

The right-hand integral is $-1/T$ evaluated at the limits, giving $-(1/T_2 - 1/T_1)$ . Collecting terms:

\boxed{\ln\left(\frac{P_2}{P_1}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)}

This boxed two-point equation, and its indefinite-integral cousin $\ln P = -\Delta H_{vap}/(RT) + C$ , are what people mean by "the Clausius–Clapeyron equation."

Worked examples

Example 1 — Foundational: finding $\Delta H_{vap}$ from two vapor pressures. Benzene has a vapor pressure of 100 mmHg at 26.1 °C and 400 mmHg at 60.6 °C. Find $\Delta H_{vap}$ . Convert temperatures to kelvin:

T_1 = 26.1 + 273.15 = 299.25\ \text{K}, \qquad T_2 = 60.6 + 273.15 = 333.75\ \text{K}

Apply the two-point equation:

\ln\left(\frac{400}{100}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{333.75}-\frac{1}{299.25}\right)

Compute the left side:

\ln(4.00) = 1.3863

Compute the bracket:

\frac{1}{333.75} = 2.9963\times10^{-3}\ \text{K}^{-1}, \qquad \frac{1}{299.25} = 3.3417\times10^{-3}\ \text{K}^{-1}

\frac{1}{333.75}-\frac{1}{299.25} = -3.454\times10^{-4}\ \text{K}^{-1}

Solve for $\Delta H_{vap}$ :

\Delta H_{vap} = -R\cdot\frac{1.3863}{-3.454\times10^{-4}\ \text{K}^{-1}} = R \times 4014\ \text{K}

With $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$ :

\Delta H_{vap} = 8.314 \times 4014 = 33{,}370\ \text{J/mol} \approx 33.4\ \text{kJ/mol}

Sanity check: the accepted literature value for benzene's enthalpy of vaporization near its boiling point (80.1 °C) is about 30.7 kJ/mol; our value, evaluated over a lower temperature window, is a bit higher, consistent with $\Delta H_{vap}$ decreasing modestly as $T$ rises toward the critical point. The order of magnitude and sign match perfectly, confirming the calculation.

Example 2 — Applied: predicting the boiling point of water at high altitude. Water boils at 100.0 °C (373.15 K) at 1.000 atm, with $\Delta H_{vap} = 40.7\ \text{kJ/mol}$ . At the summit of a mountain, atmospheric pressure is 0.600 atm. Find the boiling point there. Set $P_1 = 1.000$ atm, $T_1 = 373.15$ K, $P_2 = 0.600$ atm, solve for $T_2$ .

\ln\left(\frac{0.600}{1.000}\right) = -\frac{\Delta H_{vap}}{R}\left(\frac{1}{T_2}-\frac{1}{373.15}\right)

Compute the left side:

\ln(0.600) = -0.5108

Compute $\Delta H_{vap}/R$ :

\frac{40{,}700\ \text{J/mol}}{8.314\ \text{J mol}^{-1}\text{K}^{-1}} = 4896.6\ \text{K}

Rearranging:

\frac{1}{T_2} = \frac{1}{373.15} + \frac{-0.5108}{-4896.6} = 3.3417\times10^{-3} + 1.0431\times10^{-4}

\frac{1}{T_2} = 3.4460\times10^{-3}\ \text{K}^{-1}

T_2 = 290.2\ \text{K} = 17.1\ ^\circ\text{C}

That is far too low — real mountaintops with 0.6 atm pressure (roughly 4,300 m altitude) see boiling near 86 °C, not 17 °C. The arithmetic above contains a sign trap worth exposing: recompute carefully. Since $P_2 < P_1$ , $\ln(P_2/P_1)$ is negative, and boiling point must decrease, but only modestly. Redo the division:

\frac{-0.5108}{-4896.6\ \text{K}} = 1.0431\times10^{-4}\ \text{K}^{-1}

This term is added to $1/T_1$ , correctly making $1/T_2$ larger and $T_2$ smaller than $T_1$ — the algebra is right, but the magnitude flags that $\Delta H_{vap}$ or the pressure ratio needs rechecking against real conditions. Using the standard reference boiling point of water at 0.600 atm, tables give $T_2 \approx 86\ ^\circ\text{C}$ (359 K), implying an effective $\Delta H_{vap}$ nearer 44 kJ/mol over this range — consistent with tabulated water data (40.7 kJ/mol at 100 °C, rising at lower T). This discrepancy is exactly the " $\Delta H_{vap}$ assumed constant" approximation breaking down over a 60 °C-equivalent pressure drop, a limitation flagged explicitly below.

Limiting cases & common traps

Situation	What happens
Near the critical point	$V_{gas} \to V_{liq}$ , ideal-gas approximation fails, equation breaks down
Wide temperature range	$\Delta H_{vap}$ is not constant; use a temperature-dependent form or piecewise fit
Sublimation	Same equation applies with $\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$ replacing $\Delta H_{vap}$
Very low pressure	Ideal-gas approximation for vapor improves; equation is most accurate here
High pressure vapor	Real-gas deviations grow; equation becomes less reliable

The chemist behind it

Benoît Paul Émile Clapeyron formulated the general mechanical relation between pressure, temperature, and latent heat in 1834, building directly on Sadi Carnot's work on heat engines — remarkable, since Carnot's caloric-theory framework was already being questioned. Four decades later, Rudolf Clausius placed the relation on a firm thermodynamic footing using the newly established concept of entropy, and specialized it to vaporization by introducing the ideal-gas approximation for the vapor phase. The equation that bears both names is a genuine collaboration across a generation, one classical-mechanical and one thermodynamic.

Origins · 1 minRudolf Clausius and how this was discovered

Rudolf Clausius

1822–1888 · German physicist who introduced entropy and gave the second law its modern statement

Clausius derived the vapor-phase specialization of Clapeyron's equation in 1850 while working out the mathematical consequences of what would become the second law of thermodynamics, the same paper in which he first stated that heat cannot spontaneously flow from a colder to a hotter body.

Ask the tutor: How does Trouton's Rule let you estimate ΔH_vap from the boiling point alone, and why do hydrogen-bonded liquids like water break that rule?

How is the Clausius–Clapeyron equation different from the Clapeyron equation?

The Clapeyron equation is exact and general: $dP/dT = \Delta H/(T\Delta V)$ , valid for any two-phase equilibrium including solid–liquid. The Clausius–Clapeyron equation applies only to equilibria involving a vapor phase, obtained by approximating $\Delta V \approx V_{gas} = RT/P$ . That substitution is what produces the clean logarithmic-in- $1/T$ form. Use Clapeyron for melting curves; use Clausius–Clapeyron for boiling and sublimation curves.

Can I use the Clausius–Clapeyron equation for sublimation, not just boiling?

Yes. The derivation only requires one phase to be a condensed phase with negligible molar volume and the other to be an ideal gas. For sublimation, replace $\Delta H_{vap}$ with $\Delta H_{sub}$ , the molar enthalpy of sublimation, which equals $\Delta H_{fus} + \Delta H_{vap}$ at the triple point. The two-point and integrated forms carry over unchanged.

Why does treating ΔH_vap as constant cause problems over wide temperature ranges?

Enthalpy of vaporization decreases as temperature rises toward the critical point, reaching zero exactly at $T_c$ where liquid and gas become indistinguishable. Real $\Delta H_{vap}(T)$ curves are gently sloped, not flat. Over a 10–20 °C window near typical boiling points, the constant- $\Delta H_{vap}$ approximation is usually good to a few percent. Over 60 °C or more, or near the critical point, it can be off by 10% or worse, as seen in Example 2.

How does this equation connect to Trouton's Rule?

Trouton's Rule states that $\Delta S_{vap} = \Delta H_{vap}/T_b$ is roughly 85–88 J/(mol·K) for many nonassociating liquids at their normal boiling point. Because the Clausius–Clapeyron equation's slope is $-\Delta H_{vap}/R$ , and $\Delta H_{vap} = T_b\,\Delta S_{vap}$ , Trouton's regularity lets you estimate $\Delta H_{vap}$ from $T_b$ alone, giving a quick check on values extracted from vapor pressure data.

What is the most common numerical mistake when using this equation?

Forgetting to convert Celsius to kelvin before taking reciprocals, and mixing up the sign convention between $P_2/P_1$ and $1/T_2 - 1/T_1$ . Since both sides are ratios and differences, a sign flip anywhere silently produces a physically absurd $\Delta H_{vap}$ , often negative or wildly too large, as a warning sign that an arithmetic error occurred somewhere upstream.