How to Calculate Melting and Boiling Points at Different Pressures

A substance’s melting and boiling points depend on pressure because each transition occurs where two phases are in equilibrium. To calculate a transition temperature at a new pressure, start with a known pressure–temperature point and use data about the phase change. The suitable equation differs for boiling and melting.

The calculations below are useful approximations over a limited pressure and temperature range. For high accuracy, use measured phase-equilibrium data or a trusted phase diagram or property table.

Boiling point: use Clausius–Clapeyron

A liquid boils when its vapor pressure equals the pressure acting on it. Increasing the external pressure therefore generally raises the boiling point; reducing it lowers the boiling point.

If the vapor behaves approximately as an ideal gas, the liquid’s volume is small compared with the vapor’s, and the enthalpy of vaporization is nearly constant over the temperature range, the integrated Clausius–Clapeyron equation is

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

Here, $P_1$ and $P_2$ are absolute pressures, $T_1$ and $T_2$ are temperatures in kelvins, $\Delta H_{\mathrm{vap}}$ is the molar enthalpy of vaporization, and $R=8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}$. Rearranging gives

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

Worked example: water at 80 kPa

Estimate water’s boiling point at $80\ \mathrm{kPa}$, using its normal boiling point, $T_1=373.15\ \mathrm{K}$ at $P_1=101.325\ \mathrm{kPa}$, and $\Delta H_{\mathrm{vap}}\approx40.65\ \mathrm{kJ\,mol^{-1}}$.

Convert the enthalpy to joules per mole so it matches the units of $R$, then substitute:

$$\begin{aligned} \frac{1}{T_2} &=\frac{1}{373.15}-\frac{8.314}{40650}\ln\left(\frac{80}{101.325}\right)\\ &\approx 0.002728\ \mathrm{K^{-1}},\\ T_2&\approx366.5\ \mathrm{K}=93.4\ ^\circ\mathrm{C}. \end{aligned}$$

This is an estimate; steam tables or measured vapor-pressure data give a more precise value. If you have a vapor-pressure table, another direct method is to find the temperature at which the liquid’s vapor pressure equals the external pressure.

Melting point: use the Clapeyron equation

For a solid–liquid transition, the Clapeyron equation gives the slope of the melting curve:

$$\frac{dT}{dP}=\frac{T\,\Delta V_{\mathrm{fus}}}{\Delta H_{\mathrm{fus}}}.$$

Here, $\Delta V_{\mathrm{fus}}=V_{\mathrm{liquid}}-V_{\mathrm{solid}}$ and $\Delta H_{\mathrm{fus}}$ is the enthalpy of fusion. For a small pressure change, treat the quantities on the right as constant and estimate

$$\Delta T\approx\frac{T\,\Delta V_{\mathrm{fus}}}{\Delta H_{\mathrm{fus}}}\,\Delta P.$$

Use consistent molar units: for example, $\Delta V$ in $\mathrm{m^3\,mol^{-1}}$, pressure in pascals, and enthalpy in $\mathrm{J\,mol^{-1}}$. The sign matters. If melting increases volume, pressure tends to raise the melting point; if melting decreases volume, pressure tends to lower it.

Worked example: pressure’s effect on ice

Near $0\ ^\circ\mathrm{C}$ and ordinary pressure, approximate water’s molar volume change on melting as $\Delta V_{\mathrm{fus}}=-1.63\times10^{-6}\ \mathrm{m^3\,mol^{-1}}$, and its enthalpy of fusion as $6.01\times10^3\ \mathrm{J\,mol^{-1}}$. Raising the pressure by one atmosphere, $\Delta P=101325\ \mathrm{Pa}$, gives

$$\Delta T\approx\frac{273.15(-1.63\times10^{-6})}{6.01\times10^3}(101325)\approx-0.0075\ \mathrm{K}.$$

So this approximation predicts that increasing pressure from about one to two atmospheres lowers ice’s melting point by roughly $0.008\ ^\circ\mathrm{C}$. The effect is small at this pressure change. It has this direction because liquid water occupies less volume than the same amount of ice.

When these estimates are not enough

The integrated boiling-point formula assumes a nearly constant vaporization enthalpy and an ideal vapor. It can become inaccurate over a wide temperature range or near a critical point. For melting, the small-change formula is not reliable over a large pressure range if the enthalpy or volume difference changes appreciably. In those cases, use phase-boundary data or integrate the Clapeyron relation with pressure- and temperature-dependent properties.

A phase diagram is often the clearest source of an answer: locate the pressure of interest and read the temperature where the relevant phase boundary crosses it. For precise work, consult experimental data or a validated property model rather than extrapolating these simple equations.

Common mistakes

• Using gauge pressure in the boiling equation. Use absolute pressure, measured relative to a vacuum. Ratios such as $P_2/P_1$ must use the same pressure units.
• Using Celsius in reciprocal-temperature formulas. Convert temperatures to kelvins before calculating; convert back to Celsius only for the final result.
• Assuming melting always shifts in the same direction. The sign depends on whether the liquid has a larger or smaller volume than the solid.
• Treating a rough estimate as exact. Enthalpy and volume changes vary with conditions, and simple equations have limited ranges of validity.