Consider the pool of mercury diagrammed below:

The pool of mercury is maintained at $25C.$ Suppose the valve leading to the vacuum pump is closed. The system will be at vapor$-$liquid equilibrium, and the rate of vapor molecules striking the pool surface and condensing upon it must be equal to the rate at which liquid molecules evaporate from the pool surface. If we think in terms of gas kinetic theory, we realize that the rate of condensation must be proportional to the rate

of gas molecular collisions $($per unit surface area per time$),$ given by: $r_{\text{c}\text{o}\text{l}}=\frac{1}{4}*N_A\frac{*}{b}ar(u).$

$($a$.)$ Calculate the rate of evaporation of mercury at $25C.$

Not all the colliding molecules will condense, but this calculation can provide us with an upper limit $($maximum$)$ on the condensation rate.

$($b$.)$ Now, the valve to the vacuum pump is opened so that the pump is extracting mercury vapor fast enough so that molecules evaporating from the surface are not able to recondense. What is the maximum rate of evaporation under this condition? What key assumption have you made that is similar to one of the four basic assumptions of CTST$?$

$($c$.)$ What have you assumed about the rate of heat transfer between the air$-$liquid interface and the bulk of the liquid and its magnitude relative to the rate of evaporation into a perfect vacuum? What would happen if this were not true?