Question

On a certain humid day, the mole fraction of the water vapour in air at ${25}^{\circ }C$ is 0.0287. If total pressure of air is 0.977 bar, calculate the partial pressure of water vapour in air and relative humidity if the vapour pressure of water at ${25}^{\circ }C$ is 0.0313 bar. The partial pressure of water vapour ($P_{H_{2}O}$) and relative humidity ($R_H$) respectively are:

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Here, mole fraction of $H_{2}O\left(X_{H_{2}O}\right)=0.0287,P_{total}$ = 0.977 bar . Therefore,

$P_{H_{2}O}=X_{H_{2}O}\times P_{total}=0.0287\times 0.977bar$ = 0.028 bar.

% Relative humidity = $\frac{P_{H_{2}O}inair}{VapourPressureof{H}_{2}O}\times 100=\frac{0.028}{0.313}\times 100$ = 8.94%