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?


1 Answer 1


$\frac{1}{22.4}\frac{\mathrm{mol}}{\mathrm{L}}$ 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

$p_s V_s = n_s R T_s$

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

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

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

Insert in your equation

$N_1 = \dfrac{0.05 \cdot c \cdot V_{Air}}{A \cdot t}$

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.