The relative humidity of air equals the ratio of the partial pressure of water in the air to the equilibrium vapor pressure of water at the same temperature times 100%. If the relative humidity of the air is 58% and its temperature is 68 oF , how many moles of water are present in a room measuring 12 ft x 10 ft x 8 ft? The vapor pressure of water at 68 oF is 17.54 torr. The answer is 20, but for some reason I keep getting 15.127moles
Conversion Factor
68F = 20C
The vapor pressure of H2O is equal to 17.54 torr
vapor pressure of air = 58/100 * 17.54 torr = 10.17132 torr
Volume of the room = Length * Breadth * Height
=> 12 ft * 10 ft * 8
=> 960 ft^3
Using the ideal gas equation
PV = nRT
1 ft^3 = 28.3168 L
Volume in Liters = 27184.2 L
Pressure in atm = 10.17132/760 = 0.0133945 atm
Pv = nRT
(0.0133945)(27184.2) = n * 0.0821 * (273+20)
n = 15.1367 moles
This is the corret answer,i am 100% sure of this answer
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