Question

Assuming that dry air contain $79\%N_{2}and21\%O_2$ by volume. Find the density of moist air at ${25}^{\circ }$C and 1.0atm when the relative humidity is $60\%$. The vapour pressure of water at ${25}^{\circ }$C is 23.76 mm Hg.

• A. $1.33\frac{g}{L}$
• B. $1.1\frac{g}{L}$
• C. $1.19\frac{g}{L}$
• D. $0.96\frac{g}{L}$

Answer 1

The correct option is C

$1.19\frac{g}{L}$

From the given data , we have, vapour pressure of $H_{2}O$ = 0.6 $\times$ 23.76 mm Hg = 14.256 mm Hg
The number of moles of gas in 1L dry air is:
$N_{gas}=\frac{pV}{RT}=\frac{1\times 1}{0.0821\times 298}$ = 0.0409 mol
The number of moles of $N_2=0.0409\times 0.79=0.0323molandmassof{N}_2$ = 0.0323 $\times$ 28=0.9044g