In a room with a volume of 150 m3, air humidity at a temperature of 20 degrees Celsius is 30%.

In a room with a volume of 150 m3, air humidity at a temperature of 20 degrees Celsius is 30%. Determine the mass of steam in the room

Data: V (volume of the indicated room) = 150 m3; t (room temperature) = 20 ºС = 293 K; φ (relative humidity) = 30% = 0.3.

Constants: R (universal gas constant) = 8.31 J / (K * mol); ρн (density of saturated steam) = 17.3 g / m3; Рн (saturated steam pressure) = 2.34 kPa = 2.34 * 10^63 Pa; M (molar mass) = 18 * 10-3 kg / mol.

1 method) Steam mass: m = ρ * V = φ * ρн * V = 0.3 * 17.3 * 150 = 778.5 g.

Method 2) Steam mass: m = P * V * M / (R * T) = 0.3 * 2.34 * 10^3 * 150 * 18 * 10^-3 / (8.31 * 293) = 0.7785 kg = 778.5 g.

Answer: The mass of steam in the indicated room is 778.5 g.