The air condition inside a closed room is described as follows: temperature = 25^ C , relative humidity = 60 % , and pressure = 104 kPa . If all the water vapour is extracted from the room without altering the temperature, determine the new pressure. The saturation vapour pressure at 25^ C is 3.2 kPa .

Step-by-step solution

Relative humidity is given by the ratio of partial pressure of water vapour to the saturation vapour pressure. RH = P_v P_ sat 0.60 = P_v 3.2 P_v = 1.92 kPa The total pressure of the moist air is the sum of the partial pressure of dry air and the partial pressure of water vapour. P_ total = P_a + P_v 104 = P_a + 1.92 P_a = 102.08 kPa When all the water vapour is extracted at constant temperature and volume, the new pressure in the room is simply the partial pressure of the dry air. New pressure = 102.08 kPa