Question

What is the density of wet air with 75% relative humidity at 1 atm and 300 K? Given :vapour pressure of $H_{2}O$ is 30 torr and average molar mass of air is 29 g $mo{l}^{-1}$.

• A. 2.614 g/L
• B. 0.96 g/L
• C. 1.06 g/L
• D. 1.164 g/L

Answer 1

The correct option is D

1.164 g/L

% Relative Humidity = $\frac{PartialPressureof{H}_{2}O}{VapourPressureof{H}_{2}O}\times 100$

$75=\frac{P_{H_{2}O}}{30}\times 100$

$P_{H_{2}O}=22.5torr$

% of $H_{2}O$ in air = $\frac{22.5}{760}\times 100\Rightarrow 2.96$

Molar mass of wet air = $\frac{29\times 97.04+2.96\times 18}{100}\Rightarrow \frac{2814.16+53.28}{100}\Rightarrow 28.67$

Density of wet air = $\frac{PM}{RT}=\frac{1\times 28.67}{0.0821\times 300}\Rightarrow 1.164\frac{g}{L}$