Vapour Pressure & the Clausius–Clapeyron Equation
For a pure liquid in equilibrium with its vapour, the temperature dependence of the vapour pressure is
well described (over a moderate temperature range) by the integrated Clausius–Clapeyron equation.
Plotting \(\ln P\) versus \(1/T\) gives a straight line whose slope is related to the enthalpy of
vaporization.
Linear forms
\[
\ln P \;=\; -\frac{\Delta_{\mathrm{vap}}H}{R}\,\frac{1}{T} \;+\; B
\]
This is a straight line \(y = m\,x + b\) with the identifications
\[
y=\ln P,\qquad x=\frac{1}{T},\qquad m=-\frac{\Delta_{\mathrm{vap}}H}{R},\qquad b=B.
\]
Eliminating the intercept \(B\) between two temperatures \((T_1,T_2)\) gives the widely used
two-point form:
\[
\ln\!\left(\frac{P_2}{P_1}\right)
\;=\;
-\frac{\Delta_{\mathrm{vap}}H}{R}
\left(
\frac{1}{T_2} - \frac{1}{T_1}
\right).
\]
Units that must be used consistently
- Temperature: Kelvin only (K). Convert from °C by \(T(\mathrm{K})=t(\mathrm{°C})+273.15\).
- Pressure: Any single unit may be used (mmHg, torr, atm, bar, kPa, Pa), but
use the same unit for both \(P_1\) and \(P_2\) when evaluating \(\ln(P_2/P_1)\).
- Gas constant: \(R=8.314462618\ \mathrm{J\ mol^{-1}\ K^{-1}}\).
- Enthalpy units: \(\Delta_{\mathrm{vap}}H\) in \(\mathrm{J\ mol^{-1}}\) (or \(\mathrm{kJ\ mol^{-1}}\) with the
appropriate factor of \(10^3\)).
From natural log to base-10 log (optional)
Some handbooks use base-10 logs. Using \(\ln x = 2.303\,\log_{10} x\) gives
\[
\log_{10} P \;=\; -\frac{\Delta_{\mathrm{vap}}H}{2.303\,R}\,\frac{1}{T} \;+\; B_{10}.
\]
Estimating \(\Delta_{\mathrm{vap}}H\) from two measurements
- Place both pressures in the same unit (e.g., mmHg). Note \(P_1\) at \(T_1\) and \(P_2\) at \(T_2\) (K).
- Compute the logarithmic ratio and the reciprocal-temperature difference:
\[
\ln\!\left(\frac{P_2}{P_1}\right),\qquad
\left(\frac{1}{T_2}-\frac{1}{T_1}\right).
\]
- Solve for the enthalpy:
\[
\Delta_{\mathrm{vap}}H
\;=\;
-\,R\;
\frac{\ln(P_2/P_1)}{\tfrac{1}{T_2}-\tfrac{1}{T_1}}.
\]
Predicting a pressure at a new temperature
If \(\Delta_{\mathrm{vap}}H\) is known (or assumed constant over the range), the pressure at a second
temperature may be predicted:
\[
P_2
\;=\;
P_1\,
\exp\!\Bigg[
-\,\frac{\Delta_{\mathrm{vap}}H}{R}
\left(
\frac{1}{T_2}-\frac{1}{T_1}
\right)
\Bigg].
\]
Graphical interpretation
A plot of \(\ln P\) versus \(1/T\) is linear. The slope equals \(-\Delta_{\mathrm{vap}}H/R\) and the
intercept equals \(B\). A linear fit to multiple \((P,T)\) data pairs therefore yields a more robust
estimate of \(\Delta_{\mathrm{vap}}H\) than any single two-point calculation.
\[
\text{slope} = -\frac{\Delta_{\mathrm{vap}}H}{R}
\qquad\Longrightarrow\qquad
\Delta_{\mathrm{vap}}H = -\,R \times (\text{slope}).
\]
Assumptions and limitations
- Constancy of \(\Delta_{\mathrm{vap}}H\): Treated as temperature-independent over the range used.
Over wider ranges, \(\Delta_{\mathrm{vap}}H\) decreases with \(T\), so the line bends slightly.
- Ideal vapour behaviour: Non-ideality is usually small at modest pressures; near the normal boiling
point and at high \(P\), deviations increase.
- Standard enthalpy: \(\Delta_{\mathrm{vap}}H\) here is not necessarily the standard value
\(\Delta_{\mathrm{vap}}H^\circ\) (which is defined at \(P=1\ \mathrm{bar}\)).
- Units consistency: A common error is mixing pressure units in \(\ln(P_2/P_1)\). Convert first,
then take the logarithm.
- Kelvin only: Insert temperatures in kelvin. Using °C directly will give nonsense.
Quick checklist for calculations
- Convert all temperatures to K.
- Pick one pressure unit (e.g., mmHg) and convert both \(P_1\) and \(P_2\) to that unit.
- Use the two-point equation for single predictions; use a linear fit on \(\ln P\) vs \(1/T\) for multiple data.
- Report \(\Delta_{\mathrm{vap}}H\) with units and sensible significant figures.
