Question

on a certain humid day the mole fraction of water vapour in air at 25 degree centigrade is 0.02 87 if total pressure of air is 0.977 bar,calculate relative humidity is vapour pressure of water at 25 degree centigrade is 0.0 313bar​

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.