Question

Air conditioners not only cool air, but dry it as well. Suppose that a room in a home measures 5.0m×9.0m×2.4m .

If the outdoor temperature is 30 ∘C and the vapor pressure of water in the air is 85 % 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.

Answer 1

I am assuming that all the water vapor is removed from the air as it is cycled through the air conditioner.

Now let’s go to the solution of question.

Convert meter measurements into cm-

5.0m = 500cm

9.0m = 900cm

2.4m = 240cm.

we can apply ideal gas law here-

So, PV=nRT Ideal ,

P = (31.8 Torr/760 Torr) * .85 = 0.0357 atm

V= 500cm * 900 cm * 240 cm= 108,000,000 cm ^3
Convert to Liters: using 1cm^3 = 1 ml = 0.001 L

So, 108,000,000 cm ^3 = 108,000 L
R= 0.08206
T= 273 + 30 = 303 K

Isolate n = PV/RT

n = 0.0357atm x 108000L/0.08206 x 303k

n = 3855.6/24.864

n = 155 mols H20

n = 2791 g H20

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