Question

At \(60\,^{\circ}\text{C}\) dry bulb temperature, saturated water vapour pressure is 20 kPa and relative humidity is 20%. The corresponding absolute humidity of air at atmospheric pressure is nearest to (Take molecular weight of water = 18.02 g.mol\(^{-1}\), molecular weight of air = 28.97 g.mol\(^{-1}\), and atmospheric pressure = 101.325 kPa)

• A. 0.026
• B. 0.032
• C. 0.004
• D. 0.041

Hint:

Get the vapour partial pressure from relative humidity, then use the molecular weight ratio to convert to humidity ratio.

Answer 1

The Correct Option is A

Step 1: Find the actual vapour pressure from relative humidity.
Relative humidity is the ratio of actual to saturated vapour pressure: \( RH = p_v/p_{sat} \).
\( p_v = RH \times p_{sat} = 0.20 \times 20 = 4\ kPa \).

Step 2: Write the absolute humidity formula.
Absolute humidity (humidity ratio) is \( W = \dfrac{M_w}{M_a} \times \dfrac{p_v}{P - p_v} \), where \(P\) is total atmospheric pressure.

Step 3: Substitute the known values.
\( \dfrac{M_w}{M_a} = \dfrac{18.02}{28.97} = 0.622 \).
\( P - p_v = 101.325 - 4 = 97.325\ kPa \).
\( W = 0.622 \times \dfrac{4}{97.325} = 0.622 \times 0.0411 = 0.0256 \).

Final Answer:
The absolute humidity comes out close to 0.026 kg water per kg dry air, matching option (A). \[ \boxed{W \approx 0.026\ kg/kg} \]