Reference: Daniel V. Schroeder, *An Introduction to Thermal Physics*, (Addison-Wesley, 2000) – Problems 5.35 – 5.36.

The Clausius-Clapeyron relation is a differential equation relating pressure and temperature on a phase boundary

where is the latent heat for the phase transition and is the volume change. In order to solve this as a differential equation, we need to know and as functions of and , which ordinarily we don’t (or if we do, the relation is too complicated to solve analytically). However, if one of the phases is an ideal gas and we can take constant over the region of interest, then usually the volume of the condensed phase (liquid or solid) is negligible compared to the gas phase. In this case, becomes essentially just , the volume of the ideal gas, at temperature and pressure , that is, and if we consider 1 mole so that , we have

We can then do the integration:

where is a constant. This is known as the *vapour pressure equation*, and gives a measure of how the pressure at which the phase transition occurs varies as the temperature is raised.

As an example, we can consider the steam-water phase transition, for which data are given in Schroeder’s Figure 5.11. Looking at the temperature range between and , we see that the latent heat isn’t quite constant over this range, so we can use the average

The gas constant is

Using the given value of at we find

With this value of , we can use 5 to predict the pressure at :

The value given in Schroeder’s Figure 5.11 is (that is, 1 atmosphere, since water boils at at sea level) so we’re not too far off.

A plot of versus using 5 looks like this:

It has the same general shape as the steam-water phase boundary shown in Schroeder’s Figure 5.11.

We can invert 5 to find the boiling temperature for various altitudes. We get

Or, in centrigrade:

Using the barometric equation, we found the atmospheric pressures at various altitudes. We can use this equation to add a column for the boiling point.

Altitude (m) | (atm) | |

1430 | 0.844 | |

3090 | 0.693 | |

4420 | 0.592 | |

8850 | 0.350 |

The last line is the height of Mt Everest.

We can get an estimate of the dependence of on , the height above sea level, as follows. We need to be careful to distinguish the boiling temperature of the water from the surrounding air temperature . From 2, the change as a result of a change in atmospheric pressure is

From the barometric equation, the change in pressure due to an increase in height is

where is the mass of a mole of air molecules. Inserting 15 into 14, we get

In the limit, we get the differential equation

which we can integrate to get

for a constant of integration . At , so . For a molar mass of and an air temperature of , we have

Thus we can write 19 as

where is in metres. Even if is several thousand metres, the term in the denominator is quite small, so we can approximate it by

which shows that is roughly linear in . The slope is

which is a decrease in the boiling point of about 3.34 K per kilometre.

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