# Vapour pressure equation

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

$\displaystyle \frac{dP}{dT}=\frac{L}{T\Delta V} \ \ \ \ \ (1)$

where ${L}$ is the latent heat for the phase transition and ${\Delta V}$ is the volume change. In order to solve this as a differential equation, we need to know ${L}$ and ${\Delta V}$ as functions of ${P}$ and ${T}$, 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 ${L}$ 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, ${\Delta V}$ becomes essentially just ${V}$, the volume of the ideal gas, at temperature ${T}$ and pressure ${P}$, that is, ${\Delta V=V=nRT/P}$ and if we consider 1 mole so that ${n=1}$, we have

$\displaystyle \frac{dP}{dT}=\frac{L}{TV}=\frac{LP}{RT^{2}} \ \ \ \ \ (2)$

We can then do the integration:

 $\displaystyle \frac{dP}{P}$ $\displaystyle =$ $\displaystyle \frac{L}{RT^{2}}dT\ \ \ \ \ (3)$ $\displaystyle \ln P$ $\displaystyle =$ $\displaystyle -\frac{L}{RT}+\ln K\ \ \ \ \ (4)$ $\displaystyle P$ $\displaystyle =$ $\displaystyle Ke^{-L/RT} \ \ \ \ \ (5)$

where ${K}$ 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 ${50^{\circ}\mbox{ C}}$ and ${100^{\circ}\mbox{ C}}$, we see that the latent heat isn’t quite constant over this range, so we can use the average

 $\displaystyle L$ $\displaystyle =$ $\displaystyle \frac{1}{2}\left(42.92+40.66\right)\times10^{3}\mbox{J mol}^{-1}\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 41.79\times10^{3}\mbox{J mol}^{-1} \ \ \ \ \ (7)$

The gas constant is

$\displaystyle R=8.314\mbox{ J mol}^{-1}\mbox{K}^{-1} \ \ \ \ \ (8)$

Using the given value of ${P=0.1234\times10^{5}\mbox{ Pa}}$ at ${T=50^{\circ}\mbox{ C}=323\mbox{ K}}$ we find

$\displaystyle K=0.1234\times10^{5}e^{41.79\times10^{3}/8.314\times323}=7.07\times10^{10}\mbox{ Pa} \ \ \ \ \ (9)$

With this value of ${K}$, we can use 5 to predict the pressure at ${T=100^{\circ}\mbox{ C}=373\mbox{ K}}$:

 $\displaystyle P\left(373\right)$ $\displaystyle =$ $\displaystyle 7.07\times10^{10}e^{-41.79\times10^{3}/8.314\times373}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 9.93\times10^{4}\mbox{ Pa} \ \ \ \ \ (11)$

The value given in Schroeder’s Figure 5.11 is ${P=1.013\mbox{ bar}=10.13\times10^{4}\mbox{ Pa}}$ (that is, 1 atmosphere, since water boils at ${100^{\circ}\mbox{ C}}$ at sea level) so we’re not too far off.

A plot of ${P}$ versus ${V}$ 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

$\displaystyle T=\frac{L}{R\ln\frac{K}{P}} \ \ \ \ \ (12)$

$\displaystyle T_{c}=\frac{L}{R\ln\frac{K}{P}}-273 \ \ \ \ \ (13)$

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 ${z}$ (m) ${P\left(z\right)}$ (atm) ${T\mbox{ (}^{\circ}\mbox{C)}}$ 1430 0.844 ${95.9}$ 3090 0.693 ${90.6}$ 4420 0.592 ${86.5}$ 8850 0.350 ${73.5}$

The last line is the height of Mt Everest.

We can get an estimate of the dependence of ${T}$ on ${z}$, the height above sea level, as follows. We need to be careful to distinguish the boiling temperature ${T_{w}}$ of the water from the surrounding air temperature ${T_{a}}$. From 2, the change ${\Delta T_{w}}$ as a result of a change ${\Delta P}$ in atmospheric pressure is

$\displaystyle \Delta T_{w}=\frac{RT_{w}^{2}}{LP}\Delta P \ \ \ \ \ (14)$

From the barometric equation, the change in pressure due to an increase in height ${\Delta z}$ is

$\displaystyle \Delta P=-\frac{MgP}{RT_{a}}\Delta z \ \ \ \ \ (15)$

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

$\displaystyle \Delta T_{w}=-\frac{Mg}{L}\frac{T_{w}^{2}}{T_{a}}\Delta z \ \ \ \ \ (16)$

In the limit, we get the differential equation

$\displaystyle \frac{dT_{w}}{dz}=-\frac{Mg}{LT_{a}}T_{w}^{2} \ \ \ \ \ (17)$

which we can integrate to get

 $\displaystyle \frac{1}{T_{w}}$ $\displaystyle =$ $\displaystyle \frac{Mg}{LT_{a}}z+A\ \ \ \ \ (18)$ $\displaystyle T_{w}$ $\displaystyle =$ $\displaystyle \frac{1}{A+\frac{Mg}{LT_{a}}z} \ \ \ \ \ (19)$

for a constant of integration ${A}$. At ${z=0}$, ${T_{w}=373}$ so ${A=1/373=2.68\times10^{-3}\mbox{ K}^{-1}}$. For a molar mass of ${M=0.029\mbox{ kg}}$ and an air temperature of ${T_{a}=10^{\circ}\mbox{C}=283\mbox{ K}}$, we have

 $\displaystyle \frac{Mg}{LT_{a}}$ $\displaystyle =$ $\displaystyle \frac{\left(0.029\right)\left(9.8\right)}{\left(41.79\times10^{3}\right)\left(283\right)}=2.4\times10^{-8}\mbox{m}^{-1}\mbox{K}^{-1}\ \ \ \ \ (20)$ $\displaystyle \frac{Mg}{LT_{a}A}$ $\displaystyle =$ $\displaystyle \frac{2.4\times10^{-8}}{2.68\times10^{-3}}=8.97\times10^{-6}\mbox{m}^{-1} \ \ \ \ \ (21)$

Thus we can write 19 as

 $\displaystyle T_{w}$ $\displaystyle =$ $\displaystyle \frac{1}{A}\left(\frac{1}{1+\frac{Mg}{LT_{a}A}z}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{373}{1+8.97\times10^{-6}z} \ \ \ \ \ (23)$

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

$\displaystyle T_{w}\approx373\left(1-8.97\times10^{-6}z\right) \ \ \ \ \ (24)$

which shows that ${T_{w}}$ is roughly linear in ${z}$. The slope is

$\displaystyle \frac{dT_{w}}{dz}\approx-3.34\times10^{-3}\mbox{K m}^{-1} \ \ \ \ \ (25)$

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

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