Question

Five grams of water is placed in a variable volume piston. What is the volume of the pure system when 50% and 75% have been evaporated at (i) 30°C, (ii) 50°C? Use the ideal gas law to model the vapor phase and show that the volume occupied by liquid is negligible compared to the volume occupied by vapor.

Answer 1

PV = nRT

therefore, volume V = nRT/ P

when 50% is evaporated, then mass of vapor = 2.5 g

therefore n = 2.5 /18

so at 30oC, V = 2.5 x 8.2 x 10^-2 x (273+30)/ 18 x 1 = 3.45 L [taking P = 1 atm.]

at 50oC , V = 2.5 x 8.2 x 10^-2 x (273+50)/ 18 x 1 = 3.68 L

when 75% is evaporated, then mass of vapor = 3.75 g

therefore n = 3.75 /18

so at 30oC, V = 3.75 x 8.2 x 10^-2 x (273+30)/ 18 x 1 = 5.18 L [taking P = 1 atm.]

at 50oC , V = 3.75 x 8.2 x 10^-2 x (273+50)/ 18 x 1 = 5.52 L

These volumes are large when compared to volume of liquid, where 5 g of water will occupy merely 5 x 10^-3 L.