Correct Answer - Option 4 : 13 g
Concept:
Ideal gas equation:
\(pV = nRT = \frac{{{m_{vs}}}}{M}RT\)
Where mvs = Mass of saturated vapour, M = Molecular weight of water = 18 g/mol, p = Saturated vapour pressure, R = Universal gas constant, T = Temperature of the air.
Thus, \(m_{vs} = \frac{{MpV}}{{RT}}\)
Relative humidity (ϕ):
It represents water vapour absorbing capacity.
\(ϕ = \frac{{{m_v}}}{{{m_{vs}}}}\)
Where mv = Mass of water vapour present in air when it is unsaturated, mvs = mass of water vapour present in air when it is saturated.
Calculation:
Given:
T = 300 K, p = 3.6 kPa = 3.6 × 103 Pa, R = 8.3 JK-1mol-1, ϕ = 50% = 0.5. M = 18 g/mol, V = 1 m3
\({m_{vs}} = \frac{{18 × 3.6 × {{10}^3} × 1}}{{8.3 × 300}}\)
mvs ≈ 26 g
As the relative humidity is 50% the amount of vapour present in 1 m3 is
mv = ϕ × mvs
mv = 0.5 × 26 = 13 g
∴ Mass of the water vapour per cubic metre of air at temperature 300 K is 13 g