What is the effect of atmospheric pressure on the vapour pressure of water?

Question

This question relates to a planned test simulating cold, humid air flowing inside a cold pipe at altitude and measuring the ice that is formed inside the pipe.

I have a value from:

https://courses.lumenlearning.com/physics/chapter/13-6-humidity-evaporation-and-boiling/

Giving the vapour pressure of water at $-10 \ ^\circ \mathrm{C}$ to be $2.60 \times 10^2 \ \mathrm{Pa}$, and at 100% humidity the air will contain water at $2.36 \ \mathrm{g/m^3}$.

I assume that this is at standard atmospheric pressure. What would happen at $20,000 \ \mathrm{feet}$ where atmospheric pressure is $46 \times 10^3 \ \mathrm{Pa}$.

My understanding is that the vapour pressure and water concentration at 100% humidity would not change because the vapour pressure is a property of the water not the air but I would like confirmation.

Answer 1

The vapor pressure $p$ does exhibit some pressure dependence. Intuitively, we can think of a higher pressure as inducing additional strain energy in the condensed phase, making the gas phase more thermodynamically favorable, as Nature tends to minimize the total energy.

For a single phase of condensed matter at constant temperature, according to the Gibbs–Duhem relation, we can always write $$d\mu=\require{cancel} \cancelto{0}{-s\,dT}+v\,dP,$$ where $\mu$ is the chemical potential, $v$ is the molar volume of the condensed phase, and $P$ is the applied pressure. We can also write $$\mu=\mu_0+RT\ln \frac{p}{p_0}$$ where $\mu_0$ is a reference chemical potential, $R$ is the gas constant, $T$ is the temperature, $p$ is the equilibrium partial pressure of the vapor (assumed to be an ideal gas) above the condensed matter, and $p_0$ is a reference pressure. Upon integrating,

$$\Delta\ln \frac{p}{p_0}=\frac{v\Delta P}{RT},$$

which is typically small and often ignored because $v$ is small (1.8×10^-5 m³ for water).

I calculate your new vapor pressure, originally 2.60×10² Pa, to be 2.59×10² Pa—do you concur?

Answer 2

Vapor pressure depends on temperature and is determined by the equilibrium between the water molecules which break loose from the water surface and those that return. If water is placed in a container with air, and the container is then closed; the pressure will go up as some of the water vaporizes. If the container is left open, some of the air (and vapor) molecules will leave, and the pressure will remain constant. In both cases, the vapor contributes some of the pressure in the container.