Callaga,

You are heading in the right track, I'll try to clarify a few points, from what I understand. First, to clarify there is always the theory and then the actual application of the theory.

A) For part A everything you found is correct, you need all that information to actualy calculate the rate of evaporation for a particular system taking into account pressure, temperature and surface area. Nothing incorrect there.

B) For part B, what it is referring to is how much water needs to evaporate to achieve full vapor pressure. Think of it as this: Pw is the vapor pressure of water at ambient temperature, while Pa is the actual vapor pressure.

For example at 25 Centigrade the vapor pressure of water is supposed to be 23.756 mmHg. If the water temperature is at 20 Celcius the vapor pressure is 17.535. Thus enough water needs to evaporate to make up the difference. Now in relative terms the bigger the difference the higher the rate of evaporation (the ratio you obtained). However, as the two vapor pressures approach each other the rate decreases (the ratio gets smaller).

Also at the same time you are evaporating water, water vapor is condensing, just at a lower rate. Eventually the two pressures are equal and the rate of evaporation remains constant, and it equals the rate of condensation.

Q1) If you increase the temperature of the water, you are just providing more kinetic energy, thus providing more molecules with enough enthalpy to evaporate. Thus your thinking is correct, however, this is where you might run into problems with the equation being used in part a) (theory vs practice).

If the temperature of the water is higher than the air temperature, the ratio is now negative because Pa is bigger than Pw, thus that equation can only be used for normal evaporation of water systems, which are not being heated artificially.

Q2) "Technically" both fusion and vaporization are thermodynamic reactions since they involve the transfer of heat and the breaking of intermolecular bonds. Thus

H_{2}O(l) ---> H_{2}O(g) ΔH_{vap} = 40.8 kJ/mol at 1 atm

You can very easily derive the equation for a system at two different temperatures and arrive at this form of the equation

Ln (k_{1}/k_{2}) = -Ea/R (1/t_{1} - 1/t_{2})

in this form it becomes obious that as temp increases so does the rate.

The important part of the the Arrhenius equation is that the rate of a reaction increases with temperature, however the increase is not linear but exponential. If the temperature difference between the two systems is small enough a ratio can approximate the change, however, over a wider range then the ratio is not valid anymore. ( kind of like the slope in calculus)

As you can see in this equation there is a term Ea (activation energy); increasing the temperature of the water allows more water molecules to obtain enough energy to go over this barrier and thus evaporate.

Valdo