A greenhouse contains 216 m3 of air at a temperature of 22°C, and a humidifier in it vaporizes 3.57 L of water?
A greenhouse contains 216 m3 of air at a temperature of 22°C, and a humidifier in it vaporizes 3.57 L of water.
(a) What is the pressure of water vapor in the greenhouse, assuming that none escapes and that the air was originally completely dry? (d of H2O = 1.00 g/mL.)
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For starters, assume that the greenhouse will not become saturated with water vapor. Since 1 kg = 1000 g and 1 meter^3=1000, and the density info, 3.57 liters=3.57 kg. Now, we can determine the kg-moles of water by dividing by 18, which would be about 0.2. Now at 22 deg C, 1 kg-mole of gas occupies about 24 m^3 (you can do the calc for the exact number, if you want, but this is a "neat" quicky estimate). So 0.2 kg would occupy about 4.8 m^3. Since the volume of the greenhouse is unchanged, and as an assumption that the air was initially at 1 atm, there are about 9 kg-mole of air (216/24). With the added vapor, there are 9.2 kg-moles of gas, so the pressure would become 9.2/9 = X mm/760, and X = 1.02x760 or 774 mm Hg. The 14 mm increase is less than the vapor pressure of water, so the answer stands.