What is the total mass (in kg) of water vapor in the air in the apartment?

Question

An apartment has the dimensions 19 m by 10 m by 6 m. The temperature is 20°C, and the relative humidity is 50 percent. What is the total mass (in kg) of water vapor in the air in the apartment?

Answer 1

First, we need the Saturation Density for water in air at 20ºC. We can modify the Ideal Gas Law to give us density:

m/V = PM/RT

Density = (Saturation Vapor Pressure • Molar Mass of Water) / (Ideal Gas Constant • Kelvin Temperature)

P = Saturation Vapor Pressure for Water at 20ºC = 2338.8 Pa

(from a table of water vapor pressures, usually approximated or determined experimentally)

M = 18.0146 g/mol

R = 8.314462 (m^3 • Pa)/(K•mol)

T = 293.15 K (20ºC + 273.15)

All units will cancel out except g/m^3, which is our density of water at 100% saturation.

m/V = (2338.8 Pa • 18.0146 g/mol) / (8.314462 m^3•Pa/K•mol) = 42,132.55 / 2437.38 = 17.286 g/m^3

The room in question measures 19m x 10m x 6m = 1140 m^3

We multiply the Saturation Density of Water at 20C by the volume of the room to get the mass of water at 100% humidity.

1140 m^3 x 17.286 g/m^3 = 19,706.04 g = 19.706 kg

Finally, the questions asks for the mass of water at 50% humidity, so we multiply the saturation mass by 50%.

19.706 kg x 0.5 = 9.853 kg of water at 20ºC and 50% humidity