High School

Air conditioners not only cool air but also dry it. Suppose that a room in a home measures [tex]5.0 \, \text{m} \times 9.0 \, \text{m} \times 2.4 \, \text{m}[/tex].

If the outdoor temperature is [tex]30^\circ \text{C}[/tex] 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 [tex]30^\circ \text{C}[/tex] is [tex]31.8 \, \text{torr}[/tex].

Answer :

The volume of air in the room. The volume of the room is given by the product of its length, width, and height= 18.015 g/mol.

Calculation-

Volume of air = 5.0 m × 9.0 m × 2.4 m = 108 m^3

Next, we need to find the number of moles of water vapor present in the air at 85% of the vapor pressure of water at 30°C.

The vapor pressure for water at 30°C is 31.8 torr. At 85% of this value, the vapor pressure of water in the air is:

Vapor pressure in air = 0.85 x 31.8 torr = 27.03 torr

We can use the Ideal Gas Law, PV = nRT, to find the number of moles of water vapor present in the air.

P = 27.03 torr

V = 108 m^3

R = 8.314 J/mol·K (The universal gas constant)

T = 303 K (Temperature in kelvins)

n = (PV) / (RT)

Find the mass of water that needs to be removed from the air each time the volume of air in the room is cycled through the air conditioner.

mass of water = n * M(H2O)

Where M(H2O) is the molar mass of water which is 18.015 g/mol. It is important to note that this calculation assumes that air behaves as an ideal gas and that the temperature and pressure remain constant throughout the room. In reality, the temperature and pressure may vary throughout the room and the process of removing water vapor from the air can also affect the temperature and pressure.

to know more about vapor pressure here:

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