Question

Air conditioners not only cool air but dry it as well. A room in a home measures 6.0 m×10.0 m×2.2 m. If the outdoor temperature is 30 °C and the partial 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? (Assume that all of the water must be removed from the air.) The vapor pressure for water at 30 °C is 31.8 torr.

Answer 1

`m=3279.85g`

Step-by-step explanation:

Hello,

This example could be analyzed by assuming the ideal gas behavior, thus, recalling the formula:

`PV=nRT`

The explicit unknown are the moles, because we could get mass by including the molecular mass as follows:

`m=(PVM)/(RT)`

Whereas:

`P=0.85*31.8torr*(1atm)/(760torr) =0.03557 atm`

`V=6m*10m*2.2m=132m^3*(1000L)/(1m^3) =1.32x10^5L`

`M=18(gH_2O)/(mol H_2O)`

`R=0.082(atm*L)/(mol*K)`

`T=30+273.15=303.15K`

Now, solving for the mass of water we get:
`m=(0.03557atm*1.32x10^5L*18g/mol)/(0.082(atm*L)/(mol*K)*303.15K ) \\\\m=3279.85g`

Best regards.