Question

1 kg of moist air of RH 70% at 21°C is cooled at a constant pressure of 1 bar to 5°C. The vapor pressure at 21°C and 5°C are 0.025 bar and 0.0087 bar. The percentage of water vapor that condenses into water, at 5°C is
1. 66%
2. 85%
3. 51%
4. 17%

Answer 1

Correct Answer - Option 1 : 66%

Concept:

To find the percentage of water vapor that condenses into water, \(\frac{{{\omega _1} - {\omega _2}}}{{{\omega _1}}} \times 100\)

We know that specific humidity ω = \(\frac{{0.622{p_v}}}{{{p_t} - {p_v}}}\) [where p_t = total pressure of dry air and water vapor, p_v = partial pressure of water vapor]

Calculation:

GIven:

At 21℃ ⇒ Pv = 0.025 bar, P = 1 bar

At 5℃ ⇒ Pv = 0.0087 bar, P = 1 bar

Specific humidity at 21℃ 

ω₁ = \(\frac{{0.622 \times 0.025}}{{1 - 0.025}}\)

And Specific humidity at 5℃ 

ω₂ = \(\frac{{0.622 \times 0.0087}}{{1 - 0.0087}}\)

So the percentage of water vapour that condenses into water-

= \(\frac{{{\omega _1} - {\omega _2}}}{{{\omega _1}}} \times 100\) = \(\frac{{\frac{{0.025}}{{1 - 0.025}} - \frac{{0.0087}}{{1 - 0.0087}}}}{{\frac{{0.025}}{{1 - 0.025}}}} \times 100\) ≈ 66%