Is it possible to explain evaporation using thermodynamics only? I'm wondering whether it's possible to explain water evaporation using the theory of thermodynamics only, as opposed to the commonly cited statistical mechanics. Indeed, the usually given explanation deals with molecules and their position (near the interphase liquid/gas) and speed, which is a statistical mechanics explanation.
So my question is: "Is it possible to explain water evaporation using only the theory of thermodynamics?" If so, what's the/an explanation?
 A: Yes, it is possible to explain evaporation using thermodynamics only (as opposed to invoking statistical mechanics). The reasoning is the following: 
Assuming we start with the system drop of liquid water on a surface and air (note that it's not in an equilibrium state unless the air is saturated in water), then according to the laws of thermodynamics, equilibrium is reached when the chemical potential of the liquid water matches the one of the air (which is a sum of the chemical potentials of its constitutents). When the air is not saturated in water, its chemical potential is lower than the one of liquid water. 
It is as simple as that. The difference in the chemical potentials of the two different phases drives the motion of substances until it is the same in both regions. This process does not involve boiling because the temperature needs not to be the one of the boiling point. In this way, the laws of thermodynamics (as written in Callen's book) are enough to explain evaporation. There's no need to invoke statistical mechanics and the speed of molecules to explain evaporation; although in the end the two explanations should be in agreement, of course.
As a sidenote, at equilibrium the partial pressure of water vapor is equal to the pressure of liquid water. The difference in both pressures relates to the rate at which the transfer of water from a phase to the other takes place. 
