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 temperature, what will be the new pressure ? The saturation vapour pressure at 25°C = 32 kPa.

The condition of air in a closed room is described as follows. Tempera

1.	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 temperature, what will be the new pressure ? The saturation vapour pressure at 25°C = 32 kPa.
Answer» T = 25°C P = 104 KPa RH =VP/SVP [SVP = 3.2 KPa, RH = 0.6] VP = 0.6 × 3.2 × 10³ = 1.92 × 10³ ≈ 2 × 10³ When vapours are removed VP reduces to zero Net pressure inside the room now = 104 × 10³ – 2 × 10³ = 102 × 10³ = 102 KPa