# Why is there any vapor above the water surface at atmospheric pressure and room temperature?

If I have a beaker with pure water in it that has a piston right above the surface of the water that maintains a pressure of 1 ATM then below 100° C I would not expect to have any water vapor.

The only reason you would have no water vapor is because you have no air between the piston and water surface for the vapor to be in.

The only coexistence region of water+vapor is at 100° C. This is all in the thermodynamic limit so I am not considering fluctuations.

That is not true. If it were, there would not be any water vapor in the atmosphere at normal temperature and pressures. Water vapor in air is simply gaseous H2O molecules mixed together with the dry air components (primarily nitrogen and oxygen). You are thinking the only way to get water vapor in air is to boil liquid water. Water vapor is also produced by evaporation, which occurs at temperatures below boiling point.

However, now consider the following. Suppose I have a special gas that does not interact or dissolve in water. I introduce some of this special gas between the water surface and the piston and still maintain 1 ATM pressure. Now lets consider the system at 50° C.

I cannot see any difference between the two configurations (with and without the special gas) as far as the water is concerned and thus at 50° C I would not expect any of the water to be in vapor form.

Again, not true. The amount of water vapor in the "special gas" may be different at a given temperature and pressure than the amount in air, but it can still exist nonetheless.

And yet my gut/experience tells me that there is some vapor in the gap between the water surface and the piston.

Trust your gut. If you want to know how much water vapor there is in air without doing calculations, you can check the Psychrometric chart. For example, for a dry bulb temperature of 25° C and a relative humidity of 20%, the amount of water in the air is about 0.002 kilograms H2O per kilogram of dry air.

As for your evaporation comment, I agree there is evaporation and I am asking exactly how it happens at 50° C and 1 ATM and hence what the flaw in my reasoning is.

At the molecular level, evaporation can occur at 50° C and 1 atmosphere because there is a distribution of kinetic energy of the water molecules around the average value that determines temperature. Some have energies (velocities) above the average, others below. Those with high velocity at the surface of the water may have sufficient energy to escape intermolecular attraction forces and enter the air if the air is not saturated with water vapor (e.g, 100% relative humidity). Simultaneously, those water molecules in the air with lower energy condense and fall back to the surface. The process continues until the rate of evaporation equals the rate of condensation, which is when the air is saturated with water vapor.

You know that water boils at less than 100° C at high altitudes, because the air pressure is lower. At 12.3 kPa the boiling point of water is 50° C. That's the partial pressure of the water vapor in air corresponding to the saturation.

As others have said, here's the difference:

Case I: At 50 C and 1 atm, in a system that contains only pure water, all the water will be in a liquid state. That's what phase diagram (a) would show you if the axes were numbered -- you're not on a coexistence line; rather, you're squarely within the liquid region.

Case II: At 50C and 1 atm, in a system that contains water and air, some of the water will be in the gaseous state.

The natural question is then: Why?

The answer: There's no entropy of mixing in Case I, but there is in Case II.

Specifically: Water molecules will move between phases from regions where their chemical potential is higher, to where it's lower, until the chemical potential is uniform, at which point you've reached equilibrium (with respect to the water).

At 50 C & 1 atm, the chemical potential of pure water in the liquid state is lower than that of pure water in the gas state. Thus, in Case I, where we only have the possibility of pure liquid water and pure gaseous water, all the water will stay in the liquid state.

However, let's now consider Case II. Suppose we have some water in the gaseous state. Why doesn't all that gaseous water move to the liquid state, as in Case I? The reason is that the water in the gaseous state is not pure water! It is water mixed with your inert gas. And its chemical potential is lowered (relative to that of pure water vapor) as a result of the increase in entropy associated with this mixing. If the partial pressure of the water is lower than its vapor pressure, it will move from the liquid phase to the gas phase. As this happens, the chemical potential of the water in the gas phase will increase. When the chemical potential of the water in the gas phase (which includes the entropy of mixing) reaches the chemical potential of the liquid water, you have reached equilibrium. At this point, the partial pressure of the water vapor will be equal to its vapor pressure:

Defintion of terms:

$$\mu_{H_2O(g)}^{mix}(T, P_{H_2O(g)})$$ is the chemical potential of water vapor in a gaseous mixture at temperature $$T$$ and partial pressure $$P_{H_2O(g)}$$

$$\mu_{H_2O(g)}^o(T)$$ and $$\mu_{H_2O(l)}^o(T)$$ are the chemical potentials of pure gaseous and liquid water, respectively, at temperature $$T$$ and standard pressure $$P^o$$ (where $$P^o$$ = 1 bar; 1 bar is about 1 atm).

Then:

$$\mu_{H_2O(g)}^{mix}(T, P_{H_2O(g)}) = \mu_{H_2O(g)}^o(T) + RT ln \frac{P_{H_2O(g)}}{P^o},$$

where the logarithmic term is the contribution of the entropy of mixing to the chemical potential. You can see that the lower the partial pressure of the water vapor, the lower the resulting chemical potential.

At equilibrium, the chemical potentials of the water in the pure liquid and mixed gaseous states are equal:

$$\mu_{H_2O(g)}^{mix}(T, P_{H_2O(g)}) = \mu_{H_2O(l)}^o(T)$$

and thus:

$$\mu_{H_2O(g)}^o(T) + RT ln \frac{P_{H_2O(g)}}{P^o} = \mu_{H_2O(l)}^o(T)$$

Consequently:

$$RT ln \frac{P_{H_2O(g)}}{P^o} = \mu_{H_2O(l)}^o(T) - \mu_{H_2O(g)}^o(T)$$

At 50C and 1 atm, the chemical potential of pure liquid water is less than that of pure gaseous water. Hence the RHS is negative. Equilibrium occurs when $$P_{H_2O(g)}$$ is sufficiently low that the LHS is equally negative. This is the vapor pressure of water at that temperature.

N.B.: You specified that your gas didn't dissolve in liquid water. But what if it were a real gas, and did? In that case, wouldn't the presence of the real gas (say, air) also lower the chemical potential of the liquid water? The answer is yes, but the amount by which the chemical potential of the liquid water is lowered by this effect is relatively small.

At a fixed temperature and pressure, as in this situation, water has a particular vapor pressure. Below 100C, that's vapor pressure is less than 1atm and we'll restrict ourselves to that case.

"Vapor pressure" means that, in equilibrium, that will be the partial pressure of water vapor in a gas that's in contact with liquid water.

So in the second case, with the special gas present, there will be a small amount of water vapor in with that special gas. The amount will be determined by the vapor pressure, which is determined by the temperature.

What happens in the first case? The vapor pressure is less than the system pressure. Therefore, the liquid water is pushed "up" with more pressure than the gas above (which is all water vapor by construction) pushes "down": The water will rise up to meet the piston with no space between.