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?

  • $\begingroup$ Duplicate of many, including physics.stackexchange.com/questions/60170/… $\endgroup$ – Jon Custer Oct 8 '14 at 14:48
  • $\begingroup$ I have read the link provided prior to posting. It did not answer my question. I am looking for an explanation for why the presence of another ideal gas in the system does not change the observations compared to if the system was homogeneous. $\endgroup$ – Yandle Oct 8 '14 at 14:55
  • $\begingroup$ Nowhere in the prior posting did it say that the applied pressure had to be from any given gas. A (first order) phase change requires a volume change, and that leads to a PV term in the free energies, which affects where the equilibrium temperature is. So, the premise of your question is incorrect - applied pressure indeed impacts phase equilibria. $\endgroup$ – Jon Custer Oct 8 '14 at 16:56
  • $\begingroup$ @JonCuster What I'm wondering is, if I apply a constant 1atm via piston to a block of liquid, the liquid will follow the constant pressure line across the phase diagram when cooled. However, if I have a closed system with 1atm of air (or any ideal gas), cooling the system means that we go along the liquid-vapour saturation line to the triple point. In both cases the system has 1atm of pressure, so how come the air pressure is different than the piston? $\endgroup$ – Yandle Oct 8 '14 at 17:30

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.

  • $\begingroup$ I did not understand why the work is PdV+VdP. For the first term, does this means I am taking the liquid only as my system, the air applies P but no dV (incompressible fluid) so it is zero? I am completely unsure how VdP comes in. $\endgroup$ – Yandle Oct 15 '14 at 5:01
  • $\begingroup$ The differential comes from the energetics of transforming a little liquid to the gas phase (or vice versa) - what is the energy required to do that. Looking at it that way allows you to see that the ideal gas head pressure does not change the equilibrium of the water system at all. The dV is not just of the incompressible fluid, but of the head space as well. $\endgroup$ – Jon Custer Oct 15 '14 at 14:15
  • $\begingroup$ So since the liquid V is constant, the gas V by default is zero? In VdP, is V the volume of water undergoing phase change and dP the pressure difference of liquid/vapour? If so, I'm wondering why dP is zero, since isn't the liquid technically under 1atm of pressure? $\endgroup$ – Yandle Oct 15 '14 at 14:35
  • $\begingroup$ Actually, even if the liquid water is treated as incompressible, adding applied pressure does change the Gibbs free energy of the liquid by vdP. This leads to the Poynting correction. $\endgroup$ – Chet Miller Oct 16 '18 at 12:29

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