Why doesn't the number of $\rm H_2O$ molecules evaporated depend on the number of the other gaseous molecules?

Question

I have a conceptual contradiction. Say we remove a partition between liquid water and gaseous nitrogen at constant volume and temperature. The average number of gaseous water molecules will increase until the partial pressure of water (PP) equals the vapor pressure of water (VP), which only depends on temperature.

$$ VP = PP = \frac{N_{H^2O}}{N_{H^2O}+N_{N^2}}P_{total} = \frac{N_{H^2O}}{N_{H^2O}+N_{N^2}}\frac{RT}{V}(N_{H^2O}+N_{N^2}) = N_{H^2O}\frac{RT}{V} $$

So $N_{H^2O} = VP\frac{V}{RT}$, which doesn't depend on the number of nitrogen molecules. But shouldn't increasing the number of nitrogen molecules put more pressure on the liquid water surface and thus decrease the number of molecules that leave the gas? Thank you

Answer 1

The greater the gas pressure the slower the evaporation rate, due to the gas inhibiting diffusion of water vapor from the water/gas interface. That is, the gas tends to keep the interface saturated inhibiting further evaporation. See https://education.nationalgeographic.org/resource/process-evaporation.

The gas just slows the evaporation rate; eventually the vapor pressure reaches the saturation pressure of the water at its temperature.

If the gas pressure is lowered to less than the saturation pressure of the water at the water temperature, the water flashes to vapor (rapid evaporation).