# Calculating the boiling point of water

Gold Member
Hi,
I'm looking through wikipedia for a formula to calculate the boiling point of liquids in function of the atmospheric pressure but I didn't find any.
In fact I'm curious what it would be for water on the Moon, Jupiter and so on.
By the way, is the fusion point pressure-dependent? I guess no.
So if a planet has a high pressure at ground level, I could heat up water and put a piece of iron and watching it being liquefied.

Mapes
Homework Helper
Gold Member
By the way, is the fusion point pressure-dependent? I guess no.

The melting temperature is indeed pressure-dependent, but the dependence is small compared to that of the boiling temperature. It's a function of the volume difference between the two phases, which is small and doesn't change much for solids and liquids.

Gold Member
Thank you very much to both.
The melting temperature is indeed pressure-dependent, but the dependence is small compared to that of the boiling temperature. It's a function of the volume difference between the two phases, which is small and doesn't change much for solids and liquids.
Do you have the formula?

Mapes
Homework Helper
Gold Member
It's the same that mgb_phys mentioned, the Clausius-Clapeyron relation. For a phase change to occur (at constant temperature and pressure), the Gibbs free energies must be equal; i.e., $\Delta G=0$ when comparing the two phases. By definition, $G=U+PV-TS$, so $\Delta G=\Delta U+P\Delta V-T\Delta S=0$ and the change in phase change temperature for a given change in pressure is

$$\frac{\partial T}{\partial P}=\frac{\Delta V}{\Delta S}$$

For small changes, we can assume that $\Delta V$ and $\Delta s$ are constant and use the fact that $\Delta S_M=\Delta H_M/T_M$ at the melting temperature ($G=H-TS$ by definition). $\Delta H_M$ is the easily found enthalpy of fusion. So we end up with

$$\Delta T_M\approx\frac{T_M\Delta V}{\Delta H_M}\Delta P$$

Gold Member
Thanks mapes.