# Clausius–Clapeyron relation for states of water and vapor

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 $\pu{1 atm}$.

## 3 Answers

The answer may be somewhat counterintuitive. Yes, an inert gas does very much exert pressure on a solid or liquid. The problem is that the chemical potential of the substance in the gas phase is not affected by the inert gas (if the gas is assumed ideal).

Assume you have a substance (liquid or solid) in a rigid isothermal but otherwise empty box. The result is as expected evaporation or sublimation of the substance until the pressure of the vapour has increased to an equilibrium value (the saturation vapor pressure). We assume there is plenty of substance so that some remains in the condensed phase while the rest fills the box as vapour. What drives evaporation/sublimation? The fact that the chemical potential of the substance is (initially) lower in the vapour. Once equilibrium is achieved the chemical potential is the same in both phases.

Now assume an alternate scenario: in addition to the substance (solid or liquid) there is now some inert gas in the box. As suspected this gas exerts a pressure on the liquid and thereby affects its chemical potential (free energy). But the free energy of the substance in the gas phase is unaffected by the inert gas (assuming the gases behave ideally). And here the somewhat counterintuitive result: the substance again evaporates/sublimes until reaching the saturation vapour pressure, but precisely because the inert gas exerts a pressure on the condensed phase and increases its chemical potential in this phase, but does not alter the chemical potential of the vapor, it increases the saturation vapor pressure. The increased vapour pressure raises the chemical potential of the vapour until it matches the heightened potential of the condensed phase.

And the short answer to the question is that the inability of vapor A to 'feel' the total pressure including the inert gas is part of the definition of an ideal or perfect gas, which fails if the pressure is raised sufficiently.

• Nice description. It should be noted that, for the inert gas to work as you describe, it must also be insoluble in the liquid. To the extent it dissolves in the liquid, it would decrease the liquid's mole fraction which, by itself, will decrease the chemical potential of the liquid, thus lowering its vapor pressure. Hence you would have two competing effects from the inert gas, one which raises the chemical potential of the liquid, and one which lowers it. – theorist Sep 29 '20 at 8:31
• @theorist Good point, I guess I should do a calculation to see what Henry's law coefficients and pressures would be necessary for the effect of solubility to become important. – Buck Thorn Sep 29 '20 at 8:57
• @BuckThorn So for the gas phase the chemical potential is function of "partial" and not total pressure? – Anton Mar 8 at 19:41
• @Anton if the gas is assumed ideal, yes. Otherwise a correction factor in the form of an activity coefficient is necessary. – Buck Thorn Mar 9 at 7:17

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.

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 1 atm

It is because water vapor and air mix. If you place a thin foil on the water surface (e.g. saran wrap), the air will work like a piston, pushing on the foil, which pushes on the liquid water and on any molecule that tries to escape.

1. 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.

Yes, and now you could say the water vapor pushes on the liquid so that it does not further evaporate, but it is not a good model. Instead, molecules from the liquid phase are free to evaporate, but the same number of molecules from the gas phase condensate at the same time.

Pressure of the liquid

Where the pressure of the liquid is relevant (and the piston model works no matter whether the pressure is created by a piston, an inert gas or water vapor) is for the inside of the liquid. If the vapor pressure is lower than the pressure of the system, no bubbles form (no boiling occurs). If the vapor pressure is higher, bubbles will form, either displacing the liquid water ("boiling") or increasing the pressure to match the vapor pressure (still no boiling water).