Question

Relative humidity of air is ${60}^{\circ }$ and the saturation vapour pressure of water vapour in the air is 3.6 $kPa$. The amount of water vapour present in 2 $L$ air at 300 $K$ is :

• A. 52 $g$
• B. 31.2 $g$
• C. 26 $g$
• D. 5.2 $g$

Answer 1

The correct option is B 31.2 $g$

$PV=\frac{w}{m}RT$ (for vapours of $H_{2}O$)

$P=3.6\times {10}^{3}Pa;V=2L;T=300K$

$\therefore w_{H_{2}O}=\frac{3.6\times {10}^3\times 18\times 2}{8.314\times 300}$

$w_{H_{2}O}=52g$

Since, relative humidity = $60$%, therefore, amount of $H_{2}O=52\times 0.6=31.2g$

Hence, the correct answer is option $\text{B}$.