An apartment has the dimensions 18 m by 9 m by 6 m. The temperature is 20°C, and the relative humidity is 54 percent. What is the total mass (in kg) of water vapor in the air in the apartment?

Question

Please provide the best equation for calculating this for future reference.

Answer is in kg

Answer 1

Hello Shubbie;

T_kelvin = 20°C + 273.15 = 293.15 K.

The saturation vapor pressure (e) at 20°C: e = 2.3386 kPa ~ 2.3 KPa

The partial pressure of water vapor (P_v) using relative humidity: P_v = (54 / 100) * 2338.6 Pa = 1262.604 Pa ~ 1262.6 Pa

The volume of the apartment: V = 18 m * 9 m * 6 m = 972 m³.

(n) In moles using the ideal gas law: n ≈ Pv * V / (R * T_kelvin) ≈ 1262.6 Pa * 972 m³ / (8.314 J/(mol·K) * 293.15 K) ≈ 0.052 moles.

We need to convert the amount of water vapor to kilograms: mass _{water_vapor} ≈ 0.052 moles * 18.01528 g/mol ≈ 0.94 kg.

So, the total mass of water vapor in the air in the apartment is approximately 0.94 kilograms.