I've managed to get myself thoroughly confused in thinking about the vapor pressure of a substance at a particular temperature $T$, and how it relates to the pressure of the substance if it is in a phase equilibrium at that same temperature $T$.

Consider two sealed, rigid containers. All three containers are in thermal equilibrium with a large temperature reservoir at (say) 298 K.

  • We evacuate Container A completely, and place into it a sample of liquid water. Some of the water boils away until the pressure of the water vapor attains the appropriate pressure on the phase boundary in the P-T diagram. (We assume that the sample of water is appropriately sized so that some water is left in each phase once it reaches equilibrium.) Once equilibrium is reached, the water vapor has reached a pressure $P_A$.

  • Into Container B (which is of the same size) we also place an indentical sample of water. However, Container B also contains dry air at $P_\text{air} = 1$ atm. The water evaporates, mixing with the air, until it reaches a particular vapor pressure. (Again, assume that the sample of water is appropriately sized so that some water remains unevaporated in equilibrium.) Once equilibrium is reached, the water vapor has partial pressure $P_B$.

Does $P_A = P_B$ exactly? Or, to put it another way, does the value of $P_\text{air}$ affect the partial pressure of the water vapor in equilibrium? If $P_A \neq P_B$ precisely, is there some limit or approximation under which $P_A = P_B$ exactly? If so, what is it?

I haven't been able to find a definitive answer in my searching online, and it seems plausible to me that the presence of an external source of pressure on the water (namely the air in container B) might affect the phase transition properties of the water.


The way to understand the effect of the pressure of an inert gas (oxygen here assumed to be inert) on the vapor pressure of a liquid is to consider the effect of the extra pressure on the chemical potential of the liquid: $$\left(\frac{\partial \mu}{\partial P} \right)_T=V_m$$

The additional pressure compresses the liquid and thereby increases its chemical potential, resulting in an increased vapor pressure. However, since $V_{m,~l}$ is relatively small, the effect on the vapor pressure is also small except at very high pressures. The effect of air at atmospheric pressure and $\pu{298 K}$ on the vapour pressure of water ($\pu{3.2 kPa}$ in the absence of inert gas) is negligible, an increase of $\pu{2.3 Pa}$.

So $P_A \neq P_B$, but they become nearly identical when $PV_{m,l}\ll RT$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.