Question

Find the mass of the water vapour per cubic metre of air at temperature 300 K and relative humidity 50%. The saturation vapour pressure at 300 K is 3.6 kPa and gas constant R = 8.3 JK^-1mol^-1
1. 15 g
2. 130 g
3. 31 g
4. 13 g

Answer 1

Correct Answer - Option 4 : 13 g

Concept:

Ideal gas equation:

\(pV = nRT = \frac{{{m_{vs}}}}{M}RT\)

Where m_vs = 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 m_v = 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

\({m_{vs}} = \frac{{18 × 3.6 × {{10}^3} × 1}}{{8.3 × 300}}\)

m_vs ≈ 26 g

As the relative humidity is 50% the amount of vapour present in 1 m³ is

m_v = ϕ × m_vs

m_v = 0.5 × 26 = 13 g

∴ Mass of the water vapour per cubic metre of air at temperature 300 K is 13 g