# Vapour pressure

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.41.

Two systems in diffusive equilibrium have equal chemical potentials. We can use this fact to solve the following problem.

We begin with a closed system consisting of a liquid such as water in diffusive equilibrium with its vapour. At the start, only the liquid and its vapour are present. Then we pump in an inert gas (that is, a gas that doesn’t react chemically with the liquid or vapour) to increase the pressure in the container (we’re also assuming that everything is at the same temperature, so no heat flow occurs). Assuming that the inert gas doesn’t dissolve in the liquid, and that the liquid and its vapour remain in diffusive equilibrium, the chemical potentials of the liquid and vapour must change by the same amount: ${d\mu_{\ell}=d\mu_{v}}$. What happens to the partial pressure ${P_{v}}$ of the vapour?

For the vapour, which we’ll assume is an ideal gas, the chemical potential is given by

$\displaystyle \mu_{v}\left(T,P_{v}\right)=\mu_{v}^{\circ}\left(T\right)+kT\ln\frac{P_{v}}{P^{\circ}} \ \ \ \ \ (1)$

where the superscript ‘o’ indicates the value at a reference pressure, usually taken to be 1 bar. If the pressure changes by ${dP_{v}}$, we have

$\displaystyle d\mu_{v}=\frac{kT}{P_{v}}dP_{v} \ \ \ \ \ (2)$

For the liquid, we can use the relation between the Gibbs energy and chemical potential

 $\displaystyle G$ $\displaystyle =$ $\displaystyle N\mu_{\ell} \ \ \ \ \ (3)$

If we combine this with the derivative

$\displaystyle \left(\frac{\partial G}{\partial P}\right)_{T,N}=V \ \ \ \ \ (4)$

we get

 $\displaystyle dG$ $\displaystyle =$ $\displaystyle Nd\mu_{\ell}=V\;dP\ \ \ \ \ (5)$ $\displaystyle d\mu_{\ell}$ $\displaystyle =$ $\displaystyle \frac{V}{N}dP \ \ \ \ \ (6)$

where the pressure ${P}$ is now the pressure on the liquid, which is the sum of the vapour pressure ${P_{v}}$ and the pressure due to the added inert gas. (Note that ${V}$ and ${N}$ refer to the liquid, not the vapour.) Equating 2 and 6, we get a differential equation relating the vapour pressure and total pressure:

 $\displaystyle \frac{kT}{P_{v}}dP_{v}$ $\displaystyle =$ $\displaystyle \frac{V}{N}dP\ \ \ \ \ (7)$ $\displaystyle \frac{dP_{v}}{dP}$ $\displaystyle =$ $\displaystyle \frac{V}{NkT}P_{v} \ \ \ \ \ (8)$

We can solve this as follows:

 $\displaystyle \frac{dP_{v}}{P_{v}}$ $\displaystyle =$ $\displaystyle \frac{V}{NkT}dP\ \ \ \ \ (9)$ $\displaystyle \ln P_{v}+A$ $\displaystyle =$ $\displaystyle \frac{V}{NkT}P+B \ \ \ \ \ (10)$

where ${A}$ and ${B}$ are constants of integration. To determine these constants, we use the fact that when there is no inert gas present, ${P=P_{v}}$. We’ll call this pressure ${P_{v0}}$ to distinguish it from ${P_{v}}$, the latter of which is actually a function of ${P}$. [Schroeder uses the confusing notation ${P_{v}\left(P_{v}\right)}$ to indicate the vapour pressure when there is no inert gas present, which makes it look like a function of a function.] That is, we must have

$\displaystyle \ln P_{v0}+A=\frac{V}{NkT}P_{v0}+B \ \ \ \ \ (11)$

We can choose ${A}$ and ${B}$ so that both sides are zero, giving

 $\displaystyle A$ $\displaystyle =$ $\displaystyle -\ln P_{v0}\ \ \ \ \ (12)$ $\displaystyle B$ $\displaystyle =$ $\displaystyle -\frac{V}{NkT}P_{v0} \ \ \ \ \ (13)$

Substituting these back into 10 we get

 $\displaystyle \ln\frac{P_{v}}{P_{v0}}$ $\displaystyle =$ $\displaystyle \frac{V}{NkT}\left(P-P_{v0}\right)\ \ \ \ \ (14)$ $\displaystyle P_{v}\left(P\right)$ $\displaystyle =$ $\displaystyle P_{v0}e^{\left(P-P_{v0}\right)V/NkT} \ \ \ \ \ (15)$

Thus if ${P>P_{v0}}$, the vapour pressure increases.

As an example, the vapour pressure of water at ${T=25^{\circ}\mbox{ C}}$ is ${3.169\times10^{3}\mbox{ Pa}}$. Suppose we introduce an inert gas at atmospheric pressure (${=1.01\times10^{5}\mbox{ Pa}}$) into the container. For one mole of water, ${V=18.068\times10^{-6}\mbox{ m}^{3}}$, so we have for the new vapour pressure

$\displaystyle P_{v}\left(1.01\times10^{5}\right)=3.169\times10^{3}e^{\left(1.01\times10^{5}-3.169\times10^{3}\right)\left(18.068\times10^{-6}\right)/NkT} \ \ \ \ \ (16)$

Using ${N=6.02\times10^{23}}$, ${k=1.38\times10^{-23}}$ and ${T=298}$, we have

$\displaystyle P_{v}\left(1.01\times10^{5}\right)=3.169\times10^{3}e^{7.14\times10^{-4}} \ \ \ \ \ (17)$

The exponential is very close to 1, so we can expand in a Taylor series:

$\displaystyle e^{7.14\times10^{-4}}\approx1+7.14\times10^{-4} \ \ \ \ \ (18)$

Thus the change in vapour pressure is only about 0.07%.

## 2 thoughts on “Vapour pressure”

1. Nikos

Schroeder says that an increase in the vapour pressure would result in more liquid to evaporate. Why is that? Increasing the vapour pressure due to the presence of the inert gas would increase the chemical potential of the vapour phase. How is this going to result in a change of the chemical potential of the liquid phase (an increase that is) that would result in the evaporation of liquid?

1. sturk

The presence of the inert gas does not directly increase the vapor pressure. It has no direct effect on the chemical potential of the gas phase, since it does not affect the partial pressure of the gas phase in the chamber (the chemical potential of the gas phase depends only on its partial pressure, assuming that the mixture is ideal, which inertness implies). Rather, the introduction of the inert gas increases the total pressure (not vapor pressure) in the chamber, which in turn increases the chemical potential of the liquid phase by an amount (V/N)dP. Now the liquid phase has a higher chemical potential so you have a chemical gradient which is going to cause evaporation of the liquid, i.e. a rise in the vapor pressure.