Single species
At the surface of a liquid, molecules are constantly being ejected from the liquid surface and gas molecules rejoining it. When those two rates are equal there's no net change and the system is in equilibrium.
The rate at which liquid water will eject a molecule is roughly independent of pressure and is based solely on temperature. The temperature determines how much kinetic energy the molecules have while the pressure just determines how tightly packed they are and they're already very tightly packed so the density doesn't change much. There are always water molecules right on the surface because of this high density.
The rate at which gaseous water molecules rejoin the water depends on how often one hits the surface which depends on the density. By the ideal gas law, the density is related to pressure and temperature, so if we've already set the temperature, it only depends on pressure. Thus the higher the pressure, the higher the density, the more gaseous water molecules there are to hit the surface and join it.
Locally, the density and pressure of the gaseous water molecules will increase until the rate of rejoining is equal to the rate of evaporation.
Multi gas species
Other gasses can feel free to bombard the surface of the water, compressing it a little bit, and increasing the pressure significantly. However, this doesn't increase the rate at which water molecules escape, so the vapor will still equilibrate at the same partial pressure.
Multi liquid species
This is where Raoult's law comes in. On the liquid surface there are now two species of molecules but the rate at which molecules are ejected is still the same. Now those ejections must be split between species, and according to Raoult's law, it's based on the concentration. If 30% of the liquid's molecules are water then 30% of the ejections will be water, so the rate of ejection is reduced to 30%. So in equilibrium the rate of rejoining must also be 30% so there must be 30% as many collisions thus 30% the density and thus 30% the vapor pressure.
Non Equilibrium
Examining the gas directly above the surface would show a partial pressure of water that approaches the vapor pressure of water for that temperature very quickly. These water molecules would then diffuse through the nitrogen which would slowly lower the local partial pressure if it was not being replenished by additional evaporation from the surface. This evaporation requires energy to overcome the latent heat and as such it will lower the local temperature of the water. Water is a much better heat conductor than air so this heat is drawn from the water.
In some evaporation processes the slow diffusion of the gas dominates, in others, the temperature drops until the diffusion process dominates. This is why liquid nitrogen gets and stays so cold, and why wet bulb temperatures are lower than dry bulb temperatures.
If you'd like to learn more about the steady state solution I would recommend reading about relative/absolute humidity, condensation etc.
If you'd like to learn more about the rate limiting effect look into gas diffusion.
If you'd like to learn more about the evaporation condensation process learning about statistical thermodynamics would help.
You have a 2 liter rigid container, featuring 1 liter of liquid water and, above it, one liter of a mixture of air and water vapor all at 1 atm. The temperature is 20 C, and the partial pressure of the water vapor in the head space is the equilibrium vapor pressure, so that the system is at equilibrium. This is the initial thermdynamic equilibrium state of the system. Start out by determining the partial pressure of air in the head space, the mass of air in the container, and the mass of water. In the ensuing calculations, it is permissible to assume that the partial pressure of the water vapor in the head space is equal to the equilibrium vapor pressure at the liquid temperature and the air is not soluble in the liquid water.
Vapor pressure of water at 20 C = 17.5 torr = 0.023 atm
Partial pressure of air in container at 20 C = 0.977 atm
From ideal gas law, moles of water vapor in head space = 0.00096
Mass of water in head space = 0.017 grams
Total mass of water in container = 1000.017 grams
Moles of air in head space = 0.04066
Mass of air in head space = 1.18 grams
Now you raise the temperature of the system to 50 C and let it equilibrate. What is the partial pressure of water vapor in the head space, and the mass split between the liquid water and water vapor? And what is the partial pressure of the air in the head space? What is the total pressure.
NOW FOR 50 C
Vapor pressure of water at 50 C = 92.5 torr = 0.121 atm
From ideal gas law, mass density of water vapor in head space = 0.0822 g/l = 0.0000822 g/cc
Let x = mass of water in vapor phase
Mass of water in liquid phase = 1000.017-x
Volume of water in container (cc) = $(1000.017-x)+\frac{x}{0.0000822}=2000$
Solving for x : x = 0.0822 grams
Liquid water remaining = 999.93 grams
Volume of vapor = 1.00007
From ideal gas law, partial pressure of air = 1.077 atm
Total pressure = 1.197 atm.
Now try 100 C, 150 C, 200 C, etc.
Best Answer
Metastable water vapor exists as supercooled vapor or in supersaturated air e.g. in the absence of cloud condensation nuclei.
At room temperature and a pressure of 1 atm supersaturated air exists, but I doubt that it is possible to prepare pure supercooled water vapor at this conditions.
The usual phase diagram of water describes water as a pure substance, not as an n-ary mixture with gases. According to this phase diagram, water is liquid at $20\mathrm{°C}$ and $1~\mathrm{atm}$ ($1013.25~\mathrm{hPa}$).
However, when you follow the line $T = 20\mathrm{°C}$, you will find an intersection with the curve that separates the vapor and liquid phase. The pressure at this point ($23~\mathrm{hPa}$) is called the vapor pressure at $20\mathrm{°C}$. This is approximately the same as the partial pressure of water in saturated air at the same temperature. So you can think of air molecules as mere spectators that do not contribute to the "water pressure".
Conclusion: The answer would be no under the additional assumption that the air is saturated with water vapor. However if air is supersatured with water vapor then the water vapor would be in a mestastable state and the answer would be yes.