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Water (or any liquid/solid for that matter) has a saturation vapour pressure (SVP) that follows the curve given by Clausius-Clapeyron. At room temperature the SVP of water is about $0.023$ bar, so in the atmosphere at $20$°C, the fraction of water vapour in the air above the sea (in equilibrium) will be $2.3\%$. If we had vacuum or 10 bar ambient pressure the partial pressure of water vapour would still be $0.023$ bar, right? What I'm confused about is, why the ambient pressure of dry air (or any ambient gas other than water vapour) is not relevant to this? Water vapour will accumulate in the gas from the liquid phase till it reaches the SVP. Is there no interaction with the other molecules in the air? I hope it's clear what I mean.

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This is Dalton's law of partial pressures. In the low density regime where all the gases can be treated as "ideal" the total pressure is simply the sum of all the pressures that the different components of gas (water vapour, nitrogen, oxygen, argon etc.) would exert if the other components were not there. In other words the the molecules of the gases don't see each other. This is not exactly true of course, or the gas would never be able to relax to thermal equilibrium.

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  • $\begingroup$ Ok, I see were your "ideal" gas comes into play where interactions are neglected, but I don't quite see how this applies to the liquid and gas phase mixture. Don't you have to consider the system liquid-gas together and not just treat it as an ideal gas? I'm wondering why the liquid water doesn't "care" about the other constituents making up the total pressure in the gas phase above. $\endgroup$
    – Diger
    Commented Jan 7, 2021 at 15:55
  • $\begingroup$ It cares when it comes to boiling. The liquid can only turn into vapour on its surface, as the pressure inside the liquid is the total pressure, water vapour plus all the rest. That's why a liquid can only boil (turn to vapour throughout its volume) when the vapour pressure is greater than or equal to the total atmospheric pressure. $\endgroup$
    – mike stone
    Commented Jan 7, 2021 at 17:44
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If we had vacuum or 10 bar ambient pressure the partial pressure of water vapour would still be 0.023 bar, right? What I'm confused about is, why the ambient pressure of dry air (or any ambient gas other than water vapour) is not relevant to this?

Assuming validity of the Clausius-Clapeyron equation, ambient pressure of the air may influence the equilibrium pressure of water vapor. This is because higher ambient pressure means liquid is more compressed so it has lower specific volume and latent specific heat can be changed by that as well. But liquid water is only weakly compressible, any effect of external pressure in ranges 1-10bar on any water property seems to be very weak so all this is usually neglected and for these ranges of pressure, we can use the approximation of constant specific volume.

High enough pressure is however needed to maintain part of water in liquid phase so that equilibrium between liquid and vapor can exist. If the pressure was decreased too low, the liquid would start to evaporate from inside - it would start to boil and eventually turn into all vapor or (if in closed vessel with enough liquid) find a new equilibrium between liquid and vapor.

Other gases such as air also don't interact with vapor or liquid water much so the evaporation process isn't affected by atmospheric gases.

This could change if the other gas was too dense or more reactive so its interaction with the water vapor or solubility in water was stronger. I would expect strongly soluble gases such as $\text{SO}_2$ or $\text{HCl}$ to change equilibrium between liquid and its vapor more visibly. For example, the solution can be hygroscopic and thus decrease the partial pressure of water in the gas phase.

This paper [1] studies partial pressure water dependence on concentration of $\text{H}_2\text{SO}_4$ in the water liquid phase and it says there is a dependence. Intuitively, latent heat $L$ in the C-C equation should be very sensitive to concentration of other polar molecules if there is enough of them (so that major part of water molecules are affected by them).

[1] Crawford H. Greenewalt, Partial Pressure of Water Out of Aqueous Solutions of Sulfuric Acid., Ind. Eng. Chem. 1925, 17, 5, 522–523, https://doi.org/10.1021/ie50185a036

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  • $\begingroup$ You probably meant to say "find a new equilibrium at HIGHER pressure", or? $\endgroup$
    – Diger
    Commented Jan 8, 2021 at 12:24
  • $\begingroup$ That was not a good formulation, fixed. $\endgroup$ Commented Jan 8, 2021 at 12:30

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