How much water, in grams, is removed from the air each time the air from the room is cycled through the air conditioner?

Question

Ch10 Q17

Air conditioners dry the air while they cool the air. Suppose an air conditioner is used to cool a room that is 7.62 m×10.88 m×2.31 m when the outside temperature is 37⁢ ∘C. At this temperature, the vapor pressure of water is 47.1 torr. The partial pressure of water in the air is 80.0% of the vapor pressure of water. How much water, in grams, is removed from the air each time the air from the room is cycled through the air conditioner?

mass of H2O:___g

Answer 1

Given data:

T= 37^o C

Volume of room = 7.62 m x 10.88 m x 2.31 m = 191.5 m³.

Pressure of water in air = (80.0/100) x 47.1 torr = 37.68 torr.

To find the mass of water removed from the air, we can apply the ideal gas law PV=nRT.

First we need to make sure that our units are appropriate.

1m³ = 1000 L so V = (191.5 m³) x (1000 L) = 1.915 x 10⁵ L.

1 torr = 1 atm / 760 so P = 37.68 torr / 760 = 0.0496 atm.

K = C + 273 so T = 37^o C + 273 = 310 K.

R is the universal gas constant which is 0.0820574 L atm mol^-1 K.

Now we plug in the above values to the ideal gas law to solve for n.

n = PV/RT = (0.0496 atm) x ( 1.915 x 10⁵ L) / (0.0820574) x (310 K)

n= 373.3 moles H₂O

Now recall that moles = mass/molar mass

So mass H₂O = moles H₂O x molar mass of H₂O , molar mass H₂O = 18 g/mol

Mass H₂O = 373.3 moles x 18 g/mol

Mass H₂O = 6,718.7 g