The condition of air in a closed room is described as follows. Temperature`=25^(@)C`,relative humidity=60%, pressure=104 kPa. If all the water vapour is removed from the room without changing the tempareture what will be the new pressure?The saturation vapour pressure at `25^(@)C=3.2 kPa`.

Explanation

`T = 25^@C, `
` P = 104 kPa `
` (R_H)=(VP/SVP)`
`[SVP = 3.2 Kpa, (R_H) = 0.6]`
` VP = (0.6 xx 3.2 xx (10^3))`
` = 1.92 xx (10^3) = 2 xx (10^3)`
When vapour pressure removed VP
reduces to zero
Net pressure inside the room now
= ` 104 xx (10^3)-2 xx (10^3)`
` = 102 xx (10^3) = 102 KPa.`