What is the mass of the water vapor in kg

Question

An apartment has the dimensions 11 m by 6 m by 4 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

There's a number to look up, after which this becomes an ideal gas law problem.

The equilibrium (saturation) vapor pressure of water at 20 °C is

2.3388 × 10³ Pa (source: CRC Handbook of Chemistry and Physics)

The volume of the apartment is

V = 11 m ×  6 m ×  4 m = 264 m³

Using the ideal gas law (PV = nRT) to calculate,

the maximum capacity of water (in mol) in 264 cubic meters of air is

n = PV/RT where

P = 2.3388 × 10³ Pa

V = 264 m³

R = m³-Pa/mol-K

T = 20 + 273 = 293 K

n = [(2.3388 × 10³)(264)] / [(8.314)(293)] = 253.466 mol water

253.466 mol × 18 g/mol = 4.56 × 10³ g = 4.56 kg of water

At 50% relative humidity there is half this much.

½(4.56) = 2.28 kg

Answer

The total mass (in kg) of water vapor in the air in the apartment is 2.28 kg