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From this question on Physics SE Thermodynamics of evaporation:

Now imagine the experiment is repeated but instead of vacuum conditions, the water is pressurized with nitrogen at 1 atm. According to the phase diagram of water, liquid is the stable form of water in these conditions. Yet it is commonly observed that the water molecules with the highest kinetic energy will escape and form a gaseous phase. The partial pressure of gaseous water will be equal to the saturation pressure at this temperature.

The accepted answer shows a kinetic perspective. I want to understand things from classical thermodynamics. As even an inert gas is added to the container, there would be some increase of configurational entropy in the gaseous phase compared to the initial conditions. Won't this factor will decrease the chemical potential of the gaseous phase and thus increase the vapour pressure a bit ?.

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If the nitrogen is truly inert, its presence will have no direct effect on the chemical potential of the water in the gas phase. Essentially, the gaseous water is "unaware" of the nitrogen. Consistent with this, in a mixture of ideal gases, the chemical potential of each gas is determined only by the temperature, and its partial pressure. It is indepenent of the partial pressures of the other gases.

However, even an ideal gas can increase the vapor pressure. It doesn't do so by reducing the chemical potential of the gaseous water. Rather, by applying pressure to the liquid water, it increases the chemical potential of the liquid phase, thus increasing the equilibrium vapor pressure. This is a small effect.

Of course, gases aren't inert. As you increase the pressure of the nitrogen, you have two opposing effects on the liquid water: (1) The effect described above, which increases the liquid's chemical potential. (2) With increased pressure, the concentration of nitrogen dissolved in the water will increase, which will lower the water's chemical potential. And of course one also has to account for the non-ideal interactions between the nitrogen and the water vapor.

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  • $\begingroup$ Even if Nitrogen is inert it will surely affect the configurational entropy of the gaseous phase of water? $\endgroup$ – Deepak Arya Jun 5 at 9:38
  • $\begingroup$ If we treat both the nitrogen and the water as ideal gases, then their configurational entropies are each independent of the presence of the other gas. $\endgroup$ – theorist Jun 5 at 10:00
  • $\begingroup$ You're probably thinking: If we have a container in which we have 1 L of ideal gas A on the left side at 1 atm, and 1 L of ideal gas B on the right side at 1 atm, and we remove the partition to allow them to mix, their entropies will increase. Yes, but that's because each now has twice the volume to explore (correspondingly, their partial pressures are halved). But suppose we instead squeeze A and B together into a 1 L space, such that their partial pressures don't change (only the total pressure changes). Then their entropies will be unchanged. $\endgroup$ – theorist Jun 5 at 10:08
  • $\begingroup$ I am trying and the derivative is not cancelling out for a closed container with a fixed amount of inert gas. It is something like 1-ln(x), x = Ratio of moles of water vapour with Inert gas. $\endgroup$ – Deepak Arya Jun 5 at 10:08
  • $\begingroup$ No, I am using xln(x) + yln(y)...x and y mole fraction in the gaseous phase $\endgroup$ – Deepak Arya Jun 5 at 10:18
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At $\pu{46°C}$, the vapor pressure of water in $\pu{0.1 atm}$. It means that if some water is introduced into a vacuum, in an empty container at $\pu{46°C}$, the measured pressure in the container is $\pu{0.1 atm}$. Now if water is heated to $\pu{46°C}$ in an open flask at $\pu{1 atm}$., the composition of the vapor phase is $10$% water vapor, $90$% air. The vapor pressure is the same above the surface of water, whatever the presence of air. The total pressure has to be maintained at 1 atm. of course.

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  • $\begingroup$ There will be a factor of ΔSmix and its derivative will not be zero if we fix the inert gas amount in a closed container. That's why asked. $\endgroup$ – Deepak Arya Jun 5 at 9:35
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    $\begingroup$ -1 This is wrong, putting pressure on water increases it's vapour pressure: Kelvin equation $\endgroup$ – S R Maiti Jun 5 at 9:45

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