Question

E4B.12(a) An open vessel containing water stands in a laboratory measuring 5.0 m x 5,0 m x 3.0 m at 25°C; the vapour pressure of water at this tempera ture is 3.2 kPa. When the system has come to equilibrium, what mass on will be found in the air if there is no ventilation? Repeat the calculation to open vessels containing benzene (vapour pressure 13.1 kPa) and mercury (vapour pressure 0.23 Pa).

Answer 1

biren, Volume (V)-5.0m x 5.0m x 3.0m = she wana 75m3 Temperature (7) -25°C 298 K. we know, Gas Coustaut (R) = 8.314472 Pant3/mol k. for water vapour pressure por pesswe:- P = 3.2 kPa = 8200 Pa, Molar 17 ss218 y had As we know, Ideal gas equation - Pr=nRT water = put 3200 75 8.3144727 298 nwater = 96.8635 moles Mass of water = of water + Molar mass Water = 96-8635+ 18 Mass of = 1743.5 Water for Benzene Pe 13.1 kPa 13100 Pa Molay inass Beuzene = 78 g/mol Benzene 13100 x 75 8314472 X 298 nBenzene = 396.5351 moles Mass of Benzene = 3965351 X 78 230929.74 g Massof Bencene - 30.9297 kg'] for Mercury - P: 0.23 Pa, Molar Mass = 200-59 g/mol Mug 3:314472x298 = 0.00696 mole. Mass of Mercury = 0.00696 X 400-59 mass of Mercurry = 1.3961 g )