I've seen many questions about this topic on this site, but I still have some issues trying to understand this aspect. Here's what I believe I understood looking at the phase diagram for a pure substance (water):

  1. If you put water in a container with a piston pushing $\pu{1 atm}$ directly on the water with no extra space for inert gases, it will stay 100% liquid as at room temperature and $\pu{1 atm}$ liquid is the stable phase.

  2. If the water is put at room temperature in a closed box where the volume is more than the volume of the water (creating a vacuum), the molecules will evaporate in a number so that the water vapor will be at the equilibrium pressure corresponding to room temperature, which can be calculated by the phase diagram.

I can't understand why the air (and its pressure) doesn't work like a piston, making sure no water is in gas state at $\pu{1 atm}$.

In particular, there was this exercise where a piston with $p = p_\mathrm{tot}$ was pushing down on a mixture of vapor $\ce{A}$ and inert gas above pure liquid $\ce{Ar}$, so that:

\begin{align} p_\mathrm{tot} &= p_\mathrm{inert} + p_\mathrm{a}\\ \text{Equilibrium:}\qquad \ce{A(l, $p_\mathrm{tot}$) &= Inert(g, $p_\mathrm{inert}$) + Vapor(g, $p_\mathrm{a}$)} \end{align}

Why doesn't the vapor $\ce{A}$ 'feel' the $p_\mathrm{tot}$ and get condensed into liquid $\ce{A}$?

Was it because $p_\mathrm{tot}$ is less than the vapour pressure of pure $\ce{A}$ at that temperature? If it was $p_\mathrm{tot} = p_\mathrm{vap}$ ($T = T_\mathrm{exercise}$), then the equilibium would be

$$\ce{A(pure, l, $p_\mathrm{tot}, T$) = Inert(g, $T, p_\mathrm{tot}$)?}$$

This doesn't look like an everyday case to me though where water is in gas form even if the pressure is $\pu{1 atm}$.


Consider the situation from a more microscopic/statistical mechanics perspective: In the liquid water there are quite strong intermolecular forces from all sides, at the water surface the forces are only from "one side" and in the gas the intermolecular forces are negligible.

Remember that the energy is not distributed evenly amongst the individual molecules, but instead some are faster and some are slower. (We shall ignore vibrational modes and such for the sake of simplicity.)

When a certain molecule at the surface of the water has a kinetic energy that is higher than the potential energy from the intermolecular forces; it can overcome the latter and fly away from its neighbors, because the intermolecular forces fall off rapidly, it is essentially moving freely now.

Similarly, a currently free molecule might have a lower than average speed and become "stuck" at the surface of the water, typically transferring some momentum to the body of water.

After a while, a dynamical equilibrium depending on the velocity distribution (i.e. temperature) and the amount of water vapour per volume (partial pressure) will develop. A higher temperature increases the amount of molecules that randomly have a high enough energy to overcome intermolecular forces; whereas the amount of water vapour increases the likelyhood of a water molecules hitting the surface.

That said likelyhood is independent¹ of the pressure of other gasses is not quite easy to see though; for example, one might argue that the mean-free-path is heavily reduced when the ambient pressure is higher. To speak of the vapour pressure however, our system (i.e. slightly moist air and slightly aerated water) must be in a thermodynamic equilibrium². Clearly in equilibrium water molecules are spread evenly across all the space available for the gasses, well mixed with the others constituents of the air.³ (Also clearly, I hope) this is so regardless of the pressure/density that the inert gasses have.

I hope these considerations make clear why only the partial pressure is relevant for evaporation. For very readable explanations of this and similar topics I'd recommend the textbook Thermal Physics by Stephen and Katherine Blundell.

¹ Or almost independent; at very high pressures there are intermolecular forces in the gas complicating things. The ideal gas assumption is very accurate in the troposphere though.

² If you have e.g. a test tube that is open to the atmosphere with a bit of water at the bottom, you would have a gradient with saturated air at the bottom and drier air at the top of the tube. The steepness of that gradient does slightly depend on pressure. Personally considerations like these confused me greatly when first learning about partial pressures and such.

³ Gravity would make for slight gradients, but let's ignore that for now.


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