A bucket full of water is placed in a room at 15^ C with an initial relative humidity of 40 % . The volume of the room is 50 m ^3 . The saturation vapour pressures of water at 15^ C and 20^ C are 1.6 kPa and 2.4 kPa respectively. Determine the mass of water that will evaporate to bring the room to saturation at 15^ C .

Step-by-step solution

The initial vapour pressure of water in the room is given by: P_ i = Relative Humidity Saturation Vapour Pressure P_ i = 0.40 1.6 kPa = 0.64 kPa To bring the room to saturation, the final vapour pressure must be equal to the saturation vapour pressure at 15^ C : P_ f = 1.6 kPa The required change in vapour pressure is: P = P_ f - P_ i = 1.6 - 0.64 = 0.96 kPa = 960 Pa Using the ideal gas equation, PV = nRT = m M RT , the mass of water evaporated is given by: m = P V M RT Substituting the given values ( V = 50 m ^3 ,