Question

If the volume of most air with 50% RH is isothermally reduced to half its original volume then relative humidity of moist air becomes

1. 25%
2. 60%
3. 70%
4. 100%

Answer 1

Correct Answer - Option 4 : 100%

Concept:

Relative humidity is given by

\(ϕ = \frac{{{P_v}}}{{{P_{vs}}}}\)

where P _v is pressure of water vapour and P_vs is pressure of water vapour at saturation point.

For isothermal process:

PV = constant

Calculation:

Given:

ϕ₁ = 50%, P_v1

Since volume is reduced to half so the pressure has become twice.

P_v2 = 2P_v1

\(\frac{{{ϕ _2}}}{{{ϕ _1}}} = \frac{{{P_{{v_2}}}}}{{{P_{v1\;}}}}\)

\(\frac{{{ϕ _2}}}{{{ϕ _1}}} = 2\)

ϕ₂ = 2ϕ₁

ϕ₂ = 2 × 50 ⇒ 100%.