Question

You are told to prepare a closed experimental environment (a box) for student mice. The box volume will be 294 litres (about $10f{t}_3$) and the entire air volume will be changed every minute. The relative humidity of the air entering the box is to be controlled at 40 per at ${21}^{\circ }C.$ What weight of $H_{2}O$ (in grams) must be added to the flow of dry air per minute?

[Equilibrium vapour pressure for $H_{2}O$ at ${21}^{\circ }C\simeq 19torr$].[R=0.082 liter atm mole ${}^{-1}$deg${}^{-1},$molecular weight $H_{2}O=18$]

[Write your answer to the nearest integer].

Answer 1

The pressure due to water is the product of vapour pressure and relative humidity.

$P=\frac{19torr\times 0.40}{760torr/atm}=0.01atm$

The mass of water that must be added tol the flow of dry air per minute is

$w=\frac{PVM}{RT}=\frac{0.01atm\times 294L\times 18g/mol}{0.0821L.atm/mol/K\times 294K}=2.2\approx 2g$