Question

On a certain humid day, the mole fraction of water
vapour in air at 25°C is 0.0287. If total pressure of air is
0.977 bar, calculate relative humidity if vapour pressure
of water at 25°C is 0.0313 bar. ​

Answer 1

Answer:

The relative humidity is 90.3%

Explanation:

Given

Vapour pressure of water = 0.0313 bar

Mole fraction of water in air n = 0.0287

Total pressure of air P = 0.977 bar

Pressure due to water

$P_{H_2O}=n\times P$

$\implies P_{H_2O}=0.0287\times 0.977$

$\implies P_{H_2O}=0.028$ bar

Percentage of relative humidity

$=\frac{\text{Pressure of } H_2O\text{ in air}}{\text{Vapour Pressure of }H_2O} \times 100$

$=\frac{0.028}{0.0313} \times 100$

$= 90.3\%$

Hope this helps.