Question

Air conditioners not only cool air, but dry it as well. Suppose that a room in a home measures 5.0m×8.0m×2.6m. If the outdoor temperature is 30 ∘C and the vapor pressure of water in the air is 90 % of the vapor pressure of water at this temperature, what mass of water must be removed from the air each time the volume of air in the room is cycled through the air conditioner? The vapor pressure for water at 30 ∘C is 31.8 torr. Express your answer using two significant figures.

Answer 1

Mass of water to be removed = 2.8 kg

Explanation

vapor pressure of water = (90%) * (vapor pressure at 30^oC)

vapor pressure of water = (0.90) * (31.8 torr)

vapor pressure of water = 28.62 torr

vapor pressure of water = 28.62 torr * (1 atm / 760 torr)

vapor pressure of water = 0.03766 atm

volume of room = (length) * (width) * (height)

volume of room = (5.0 m) * (8.0 m) * (2.6 m)

volume of room = 104 m³

volume of room = 104 m³ * (1000 L / m³)

volume of room = 104000 L

moles of water = [(vapor pressure of water) * (volume of room)] / [(R) * (temperature)]

moles of water = [(0.03766 atm) * (104000 L)] / [(0.0821 L-atm/mol-K) * (303 K)]

moles of water = 157.44 mol

mass of water = (moles of water) * (molar mass water)

mass of water = (157.44 mol) * (18.0 g/mol)

mass of water = 2836.33 g

mass of water = 2836.33 g * (1 kg / 1000 g)

mass of water = 2.8 kg