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So long as we are in the ideal gas range, the presence of air does not affect the liquid/solid-vapour equilibrium in any way (such as equilibrium vapour pressure, melting point etc.). Is there a reason why this is so? How is 1 atm of air pressure applied to the surface of a liquid "different" than say 1 atm applied by a piston?

Yandle
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1 Answers1

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Thanks for clarifying the question, but I'll restate it here in my words, since comments may disappear:

Given a closed constant volume containing water and a head space, why is the thermodynamics of the water not affected by pressurizing the head space with an ideal gas?

Lets simplify by stating I'll treat the water as an incompressible fluid. Adding applied pressure then does nothing to the Gibbs free energy of the liquid.

Next, the water vapor. Since the head pressure is an ideal gas, it does not interact with, and thus does not affect the Gibbs free energy of the water vapor. The water molecules may get bumped around more with additional ideal gas atoms/molecules, but they still run into each other just as often, and still have the expected velocity distributions.

The only other question is the PV work done to get the water molecules out of the liquid and into the vapor phase. But if you think about it, the ideal gas pressure does not affect this term at all - the differential is $PdV + VdP$. You have a fixed volume to the system, so there is no $PdV$, and the $VdP$ term is solely the pressure difference caused by the water vapor - the ideal gas pressure is static, non-interacting background. From the point of view of the water molecules, it doesn't do anything to the free energies, and thus does not affect the free energy balances between the water phases.

Jon Custer
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