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 The Mechanism of Evaporation

The intuitive notion of evaporation is that the heating of a liquid drives the molecules of a liquid into the gas it is in contact with. Likewise the intuitive notion of why warm, dry air causes a liquid to evaporate is that the dry air sucks the molecules out of the liquid. Neither of these notions are correct.

The molecules of a liquid and a gas are aggitated and therefore are diffusing into all directions. This rate of diffusion per unit area depends upon the temperature and density of the substance. If the rate of diffusion from the liquid phase through the interface is greater than the rate of diffusion from the gaseous phase back through that interface then net evaporation is taking place. Note that if air is in contact with water the important density is that of the water vapor in the air and not the air itself.

If the system is closed the buildup of molecules in the gaseous phase would raise the rate of diffusion through the interface back into the liquid until the two rates are equal. At that point net evaporation ceases and the phases are in equilibrium.

However if the more dense gas is constantly blown away from the interface and replaced by less dense gas the evaporation will continue until the liquid is gone.

The flow of molecules through a surface is equal to nv, where n is the molecular density per unit volume and v is the mean molecular velocity in direction perpendicular to the surface. This mean molecular velocity is a function solely of temperature. If the molecular density n is multiplied by the mass per molecule m the result is the mass density and hence the net evaporative mass flow E is given by:

#### E = m[nlvl − ngvg]

where nl and vl are the molecular density and mean perpendicular molecular velocity in the liquid, respectively and ngvg are the corresponding quantities in the gaseous phase. Both the density and the mean velocity are functions of the corresponding temperatures, Tl and Tg.

In an ideal gas the molecular density is given by

#### ng = p/(RTg)

where R is the gas constant and p is the (partial) pressure of the gas.

The average kinetic energy of a molecule, ½mv², is proportional to temperature; i.e.,

#### ½mv² = kTg

where k is Boltzmann's constant. Therefore

#### v = (2kTg/m)½

The mean perpendicular velocity v is proportional to the mean velocity v, say v=rv. Thus

#### n(Tg)v(Tg) = (1/R)(p/Tg)r(2kTg/m)½ = (r/R)(2k/m)½/Tg½

For the liquid phase the molecular density is virtually independent of temperature. Let the molecular density in the liquid phase be denoted as μ. The average kinetic energy is equal to kTl so

#### vl = r(2kTl/m)½

Thus the rate of evaporation is given by

#### E = m[μr(2kTl/m)½ − (r/R)(2k/m)½/Tg½] which reduces to E = r(2km)½[Tl½ − (p/R)/Tg½]

This equation says that when the temperature of the gaseous phase goes up the rate of evaporation goes up because the back flow of molecules from the gas to the liquid decreases. The diffusion of molecules from the gas into the liquid decreases because the molecular density decreases when the temperature goes up. The mean velocity increases when temperature goes up but not by enough to compensate for the decrease in molecular density.