Result

The Clausius-Clapeyron equation

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Statement

Treating liquid–vapour equilibrium as a special case of gibbs-equilibrium-constant's own relation (with the pure liquid's activity conventionally taken as \(1\), so the "equilibrium constant" is simply the vapour pressure itself), vapour pressure varies with temperature according to \(\ln\dfrac{P_2}{P_1}=-\dfrac{\Delta H_{\text{vap}}}{R}\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\), valid so long as the enthalpy of vaporisation \(\Delta H_{\text{vap}}\) is approximately constant over the temperature range considered.

Why it matters

This is a direct, elegant extension of the same machinery just derived in gibbs-equilibrium-constant, applied to a physical phase transition rather than a chemical reaction — strong evidence that "chemical equilibrium" and "phase equilibrium" are two instances of the identical underlying thermodynamic logic, not separate topics requiring separate foundations. Practically, it allows a substance's vapour pressure (and hence its boiling point) to be predicted at any temperature or pressure from a single known reference point plus \(\Delta H_{\text{vap}}\) — the quantitative basis for everyday observations such as water boiling at a lower temperature at high altitude.

Hypotheses

\(\Delta H_{\text{vap}}\) (and \(\Delta S_{\text{vap}}\)) are approximately constant across the temperature range being considered.This is an approximation: heat capacities of the liquid and vapour phases generally differ somewhat, so \(\Delta H_{\text{vap}}\) does have some genuine temperature dependence of its own (illustrated directly in Worked examples, where water's \(\Delta H_{\text{vap}}\) at its normal boiling point, \(40.7\,\text{kJ/mol}\), differs measurably from its value at \(298\,\text{K}\), \(44.0\,\text{kJ/mol}\), already used in hess-law). The approximation is good over a modest temperature range and degrades over a wide one. The vapour phase behaves as an ideal gas, and the liquid's molar volume is negligible compared to the vapour's.This is the assumption underlying the alternative, more general derivation route via the Clapeyron equation (\(dP/dT=\Delta H/(T\Delta V)\), Discussion), where \(\Delta V\approx V_{\text{vapour}}=RT/P\) once the liquid's much smaller molar volume is neglected; it is not needed for the equilibrium-constant-based derivation used as the primary route here (Proof), but is required for the differential (Clapeyron) form to reduce cleanly to the same integrated result.

Proof

\[ \Delta G^\circ_{\text{vap}} = -RT\ln\frac{P_{\text{vap}}}{P^\circ} \]

Applying gibbs-equilibrium-constant's own \(\Delta G^\circ=-RT\ln K\) relation directly to the phase-equilibrium "reaction" liquid \(\rightleftharpoons\) vapour, with the pure liquid's activity conventionally set to \(1\) (it does not appear in the equilibrium expression), leaving vapour pressure alone playing the role of \(K\). A

\[ \ln\frac{P_{\text{vap}}}{P^\circ} = -\frac{\Delta H_{\text{vap}}}{RT} + \frac{\Delta S_{\text{vap}}}{R} \]

Substituting \(\Delta G^\circ=\Delta H_{\text{vap}}-T\Delta S_{\text{vap}}\) (gibbs-free-energy) into Step 1 and rearranging. A

\[ \ln\frac{P_2}{P^\circ} - \ln\frac{P_1}{P^\circ} = \ln\frac{P_2}{P_1} = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right) \]

Writing Step 2 at two different temperatures \(T_1,T_2\) and subtracting: assuming \(\Delta H_{\text{vap}}\) and \(\Delta S_{\text{vap}}\) are approximately constant over the range (Hypotheses), the \(\Delta S_{\text{vap}}/R\) term is identical at both temperatures and cancels exactly, leaving the integrated Clausius-Clapeyron relation. A

Result

\[ \ln\frac{P_2}{P_1} = -\frac{\Delta H_{\text{vap}}}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right), \qquad \ln P = -\frac{\Delta H_{\text{vap}}}{RT} + \text{constant} \]

Reading. Vapour pressure and absolute temperature are related through a single, testable equation, derived directly from the same equilibrium-constant machinery already established for chemical reactions; a plot of \(\ln P\) against \(1/T\) is predicted to be a straight line of slope \(-\Delta H_{\text{vap}}/R\), a standard experimental technique for measuring \(\Delta H_{\text{vap}}\) from vapour-pressure data alone.

Scope. Most accurate over a modest temperature range where \(\Delta H_{\text{vap}}\)'s own temperature dependence (Hypotheses) can be safely neglected; applies equally, with the appropriate enthalpy of transition, to sublimation (solid\(\rightleftharpoons\)vapour) as well as ordinary vaporisation.

Corollaries & converses

Given one known (temperature, vapour pressure) reference point and \(\Delta H_{\text{vap}}\), the equation predicts the boiling point at any other pressure — the quantitative explanation for why water boils at a measurably lower temperature at high altitude, where atmospheric pressure is reduced (Worked examples).
Measuring vapour pressure at several known temperatures and plotting \(\ln P\) against \(1/T\) gives \(\Delta H_{\text{vap}}\) directly from the slope, without needing any calorimetric measurement at all — an independent, purely pressure-based method for determining a quantity otherwise obtained via calorimetry.
Converse: if measured \((\ln P, 1/T)\) data from a real substance deviates noticeably from a straight line, this signals that \(\Delta H_{\text{vap}}\) is not, in fact, constant across the measured range (Hypotheses' known limitation), rather than an error in the underlying derivation.

Fails without

Apply the equation across a very wide temperature range where \(\Delta H_{\text{vap}}\)'s own temperature dependence becomes significant (violate Hypotheses): the predicted \(\ln P\)-versus-\(1/T\) line would noticeably curve rather than remain straight, and predictions far from the reference temperature become progressively less accurate.
Apply the vapour-pressure equation to a phase transition where the liquid's molar volume is not negligible compared to the vapour's (an assumption specifically needed for the alternative Clapeyron-equation derivation route, not the primary route used here): for transitions where both phases have comparable density (e.g. some solid-solid transitions), the full Clapeyron equation \(dP/dT=\Delta H/(T\Delta V)\), without the vapour-ideal-gas simplification, must be used instead.

Common errors

Using Celsius temperatures directly rather than converting to absolute (kelvin) temperature, the same fundamental error already flagged for the ideal gas law (ideal-gas-law's Common errors) — the \(1/T\) terms here require an absolute temperature scale for the same underlying reason.
Using a \(\Delta H_{\text{vap}}\) value measured at one reference temperature (e.g. \(298\,\text{K}\)) when the calculation actually spans a temperature range far from that reference, without acknowledging the resulting approximation (Hypotheses).
Mixing up the sign or order of \(T_1\) and \(T_2\) (or \(P_1\) and \(P_2\)) in the integrated formula, which flips the predicted direction of the vapour-pressure change.

Discussion

The relation traces to Benoît Paul Émile Clapeyron's 1834 general equation for phase equilibria (\(dP/dT=\Delta H/(T\Delta V)\), the same Clapeyron already credited in ideal-gas-law with first assembling the combined ideal gas law), later refined and given its now-standard vapour-pressure form by Rudolf Clausius (the same Clausius who introduced entropy itself, entropy-second-law) once the vapour-ideal-gas approximation was applied to simplify Clapeyron's more general equation.

The equilibrium-constant-based derivation used here (Steps 1–3) is a somewhat less traditional but genuinely rigorous route to the same result as the classical Clapeyron-equation derivation; both are entirely equivalent once the vapour-ideal-gas and liquid-negligible-volume approximations (Hypotheses, second postulate) are applied, and the choice between them is largely one of pedagogical convenience — the route used here has the advantage of making explicit that phase equilibrium and chemical equilibrium share an identical thermodynamic foundation.

Common misconception: that boiling point is a fixed, universal property of a substance rather than a pressure-dependent one. A substance's "boiling point" quoted without qualification conventionally means the normal boiling point (at exactly \(1\,\text{atm}\)); at any other pressure, the temperature at which boiling occurs shifts according to exactly this equation, sometimes substantially (Worked examples, high-altitude cooking being a familiar everyday consequence).

Worked examples

\[ \text{Water: } T_1=373.15\,\text{K (normal bp)},\ P_1=1.000\,\text{atm},\ \Delta H_{\text{vap}}=40.7\,\text{kJ/mol (value at normal bp)} \]

Predict the vapour pressure at \(T_2=363.15\,\text{K}\) (\(90^\circ\text{C}\)): \(\ln(P_2/P_1)=-\dfrac{40{,}700}{8.314}\left(\dfrac{1}{363.15}-\dfrac{1}{373.15}\right)\), giving \(P_2\approx0.697\,\text{atm}\). A

\[ \text{Comparison with the experimentally tabulated vapour pressure of water at } 90^\circ\text{C}\approx0.692\,\text{atm} \]

The predicted value (\(0.697\,\text{atm}\)) matches the tabulated experimental value closely, confirming both the equation and the standard \(\Delta H_{\text{vap}}\) value used; the small residual discrepancy is consistent with \(\Delta H_{\text{vap}}\)'s own mild temperature dependence (Hypotheses), not an error in the calculation. A

\[ P(\text{H}_2\text{O}, 90^\circ\text{C}) \approx 0.697\,\text{atm}\ (\text{matches tabulated }\approx0.692\,\text{atm}) \]

Reading. A single reference point (water's well-known normal boiling point) plus a single tabulated enthalpy of vaporisation predicts water's vapour pressure at a different temperature with good accuracy, entirely from thermodynamic first principles.

Scope. The same calculation, run in reverse, predicts the boiling point at any specified pressure (Problems).

Further example. Ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) has a normal boiling point of \(351.4\,\text{K}\) (\(78.3^\circ\text{C}\)) at \(1.000\,\text{atm}\). Given that its molar enthalpy of vaporisation at this temperature is \(\Delta H_{\text{vap}} = 38.6\,\text{kJ/mol}\), calculate the vapour pressure of ethanol at \(298.15\,\text{K}\) (\(25^\circ\text{C}\)).

\[ T_1=351.4\,\text{K},\ P_1=1.000\,\text{atm},\ T_2=298.15\,\text{K},\ \Delta H_{\text{vap}}=38{,}600\,\text{J/mol} \]

Identify the known reference state (normal boiling point) and the target temperature. Ensure all units are consistent by converting \(\Delta H_{\text{vap}}\) to J/mol. A

\[ \ln\left(\dfrac{P_2}{P_1}\right)=-\dfrac{38{,}600}{8.314}\left(\dfrac{1}{298.15}-\dfrac{1}{351.4}\right) \]

Substitute the values into the integrated Clausius-Clapeyron equation. The term \(\left(\dfrac{1}{T_2}-\dfrac{1}{T_1}\right)\) is negative because \(T_2 < T_1\), which will result in a positive value for \(\ln(P_2/P_1)\) only if the sign convention is handled carefully; here, since \(T_2 < T_1\), \(1/T_2 > 1/T_1\), so the bracket is positive, and the negative sign in front makes the RHS negative, implying \(P_2 < P_1\). A

\[ \dfrac{1}{298.15} \approx 0.0033540,\quad \dfrac{1}{351.4} \approx 0.0028458 \]

Calculate the reciprocal temperatures to sufficient precision to avoid rounding errors in the difference. A

\[ \ln\left(\dfrac{P_2}{1.000}\right)=-\dfrac{38{,}600}{8.314}(0.0033540-0.0028458) = -4642.8 \times 0.0005082 \approx -2.359 \]

Compute the product of the enthalpy term and the temperature difference. The result is the natural logarithm of the pressure ratio. A

\[ P_2 = 1.000 \times e^{-2.359} \approx 0.0945\,\text{atm} \]

Exponentiate to solve for \(P_2\). The value \(0.0945\,\text{atm}\) is approximately \(71.3\,\text{mmHg}\), which is consistent with the known vapour pressure of ethanol at room temperature (typically cited around \(59\,\text{mmHg}\) or \(0.078\,\text{atm}\); the discrepancy arises because \(\Delta H_{\text{vap}}\) varies with temperature, and using the boiling point value over a large \(\Delta T\) introduces error, but it is a reasonable first-order estimate). A

\[ P(\text{ethanol}, 25^\circ\text{C}) \approx 0.095\,\text{atm} \]

Reading. This calculation demonstrates how the Clausius-Clapeyron equation allows estimation of vapour pressure at a temperature far from the reference point, provided the enthalpy of vaporisation is assumed constant.

Scope. Accuracy decreases as the temperature difference \(\Delta T\) increases, due to the temperature dependence of \(\Delta H_{\text{vap}}\) and the non-ideality of the vapour phase at higher pressures.

Problems

Using the same reference point and \(\Delta H_{\text{vap}}\) as Worked example 1, predict the temperature at which water boils at \(0.700\,\text{atm}\) (roughly the atmospheric pressure at a moderately high mountain altitude).
Solution
Rearranging the Result for \(T_2\): \(\dfrac{1}{T_2}=\dfrac{1}{T_1}-\dfrac{R\ln(P_2/P_1)}{\Delta H_{\text{vap}}}=\dfrac{1}{373.15}-\dfrac{(8.314)\ln(0.700/1.000)}{40{,}700}\), giving \(T_2\approx363.3\,\text{K}\approx90.1^\circ\text{C}\) — noticeably below water's normal \(100^\circ\text{C}\) boiling point, the quantitative basis for the familiar observation that water boils at a lower temperature at high altitude.
A substance has vapour pressure \(0.500\,\text{atm}\) at \(350\,\text{K}\) and \(1.000\,\text{atm}\) at \(380\,\text{K}\). Determine its enthalpy of vaporisation.
Solution
Rearranging the Result for \(\Delta H_{\text{vap}}\): \(\Delta H_{\text{vap}}=\dfrac{-R\ln(P_2/P_1)}{(1/T_2-1/T_1)}=\dfrac{-(8.314)\ln(1.000/0.500)}{(1/380-1/350)}\approx25{,}600\,\text{J/mol}\approx25.6\,\text{kJ/mol}\).
Explain why using water's \(\Delta H_{\text{vap}}=44.0\,\text{kJ/mol}\) value (measured at \(298\,\text{K}\), already used in hess-law) instead of the \(40.7\,\text{kJ/mol}\) value used in Worked example 1 would give a slightly different, though still reasonably accurate, prediction for water's vapour pressure at \(90^\circ\text{C}\).
Solution
\(\Delta H_{\text{vap}}\) is not perfectly constant with temperature (Hypotheses); the \(44.0\,\text{kJ/mol}\) value describes vaporisation specifically near \(298\,\text{K}\), while \(40.7\,\text{kJ/mol}\) describes vaporisation specifically near water's normal boiling point (\(373\,\text{K}\)), a temperature much closer to the \(90^\circ\text{C}\) prediction target. Using the value from further away (\(298\,\text{K}\)) as if it applied uniformly across the entire range down to \(90^\circ\text{C}\) introduces exactly the kind of approximation error the Result's scope note warns about, even though the qualitative prediction would remain reasonably close either way.
Explain, using the derivation in Proof, why the Clausius-Clapeyron equation can be viewed as a special case of the more general \(\Delta G^\circ=-RT\ln K\) relation (gibbs-equilibrium-constant), rather than as an entirely separate physical law.
Solution
Step 1 applies gibbs-equilibrium-constant's relation directly to the liquid-vapour phase change, treated formally as an "equilibrium reaction" with the pure liquid's activity conventionally fixed at \(1\); vapour pressure then plays exactly the role of the equilibrium constant \(K\) in that more general relation. Every subsequent step (Steps 2–3) follows purely algebraically from substituting \(\Delta G^\circ=\Delta H-T\Delta S\) (gibbs-free-energy) and comparing two temperatures — no new physical postulate beyond what was already established for chemical equilibrium is introduced anywhere in the derivation.
A volatile organic compound has a vapour pressure of \(12.0\,\text{kPa}\) at \(298\,\text{K}\) and \(50.0\,\text{kPa}\) at \(323\,\text{K}\). Calculate the molar enthalpy of vaporisation \(\Delta H_{\text{vap}}\) for this compound in \(\text{kJ/mol}\).
Solution
Rearranging the Clausius-Clapeyron equation to solve for \(\Delta H_{\text{vap}}\): \[ \Delta H_{\text{vap}} = \dfrac{-R \ln(P_2/P_1)}{\left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)} \] Substitute the given values (\(R = 8.314\,\text{J mol}^{-1}\text{K}^{-1}\)): \[ \ln\left(\dfrac{50.0}{12.0}\right) = \ln(4.1667) \approx 1.4271 \] \[ \dfrac{1}{323} - \dfrac{1}{298} = 0.0030960 - 0.0033557 = -0.0002597\,\text{K}^{-1} \] \[ \Delta H_{\text{vap}} = \dfrac{-(8.314)(1.4271)}{-0.0002597} = \dfrac{-11.865}{-0.0002597} \approx 45{,}687\,\text{J/mol} \] Converting to \(\text{kJ/mol}\): \[ \Delta H_{\text{vap}} \approx 45.7\,\text{kJ/mol} \]