# How do temperature and pressure influence the rate of evaporation of water in a closed vessel?

I am trying to solve a fairly simple question but I'm kind of stuck on the technicalities:

Imagine that water is evaporating into initially dry air in the closed vessel shown schematically in Fig. 8.1-1(a). The vessel is isothermal at 25 °C, so the water’s vapor pressure is 3.2 kPa. This vessel has 0.8 l of water with 150 cm2 of surface area in a total volume of 19.2 l. After 3 min, the air is five percent saturated. What is the mass transfer coefficient? How long will it take to reach ninety percent saturation?

The answer starts with this: $$N_1=\frac{\text{Vapor concentration}\cdot\text{Air Volume}}{\text{Liquid Area}\cdot\text{Time}}$$

So far this makes sense. The solution then goes on to do this:

$$N_1=\frac{0.05\cdot(\frac{3.2}{101})\cdot(\frac{1\ \mathrm{mol}}{22.4\ \mathrm{liters}})\cdot(\frac{273}{298})(18.4\ \mathrm{liters})}{(150\ \mathrm{cm^2})(180\ \mathrm{sec})}$$

I understand the denominator but the numerator I'm not sure of. Overall there are 18.4/22.4 mol of gas and 5% of that will be water vapor, I get that. But what's with the temperature and pressure adjustment?

$\frac{1}{22.4}\frac{\mathrm{mol}}{\mathrm{L}}$ is only valid for standard conditions. Use the ideal gas law to calculate the concentration for your conditions.

$s \equiv$ standard conditions

$2 \equiv$ your conditions

$p_s V_s = n_s R T_s$

$\frac{p_s V_s}{n_s T_s} = R = const = \frac{p_2 V_2}{n_2 T_2}$

$\frac{n_2}{V_2} = \frac{p_2 T_s n_s}{p_s T_2 V_s}$

$c = \frac{n_2}{V_2} = \frac{3.2 \mathrm{kPa}}{101 \mathrm{kPa}} \frac{273.15 \mathrm{K}}{298.15 \mathrm{K}} \frac{1 \mathrm{mol}}{22.4 \mathrm{L}}$

$N_1 = \dfrac{0.05 \cdot c \cdot V_{Air}}{A \cdot t}$
$N_1 = \dfrac{0.05 \cdot \frac{3.2 \mathrm{kPa}}{101 \mathrm{kPa}} \frac{273.15 \mathrm{K}}{298.15 \mathrm{K}} \frac{1 \mathrm{mol}}{22.4 \mathrm{L}} \cdot 18.4 \mathrm{L}}{150 \mathrm{cm}^2 \cdot 180 \mathrm{s}}$