Question

An open 1.00 m plexiglass box has been sitting for a long time in a room that is maintained at 20.0°C and 30% relative humidity. One tablespoon (15 ml) of water is introduced into the box and the box is then closed and sealed. Once equilibrium is reached, what will be the humidity in the box? Hints: 1) The relative humidity is defined as the partial pressure of the H2O vapor divided by the saturation pressure. 2) For this problem, you may use the ideal gas law.

Answer 1

1. Given data:- volume of plexiglass box = 1.00m² Temperature of air inside box, T= 20° c = 293k Relative humidity (6) - 301 Added amount of water is 15m/2 15x1031 = 15x105 m3 [11-1078 mass of water added, mwa vxfw 5w > 15x106x10, 0.015 kg. ma = 15 gram. Partial Relative humidity (4) Pwa Pressure of water vapour in mixture (ain) [At T = 20e from steam table ] Predaturation pressure of pure water Ps = 0.02337 bar 2 0.02337x15 Pa (N/m2) P = 2331 N/m2 (pa) 30% 2 30 20:32 Pw 2337 pascal Pw = 0.3x2337 N/m2 Pw 2701. 1 N/m2 Assuming lead gas law PNV = mw Rho T mwa 701.11 (R = 261 7/kgk 06153 293 mass of water mo = 0.00888 Kg = 818 gm. mats of watel mo. rapool

Total mass of water vapous after extra addition water will be (mwa mw't mw - 15+5:18 2 20.1@gm. mass of total air applying ideal gas law maz pava 101325 x1 = 1.204 kg. Pa: Atmospheric maz lat 2870 293 = 12048m ressure faa1013250 so Humility in the box a Mwe ma 2 20:18 am 1.204 ng of air = 0.01676 gm gan of air = 16.76 g/kg of air Humidity of the box