Subtopic: Psychrometry – properties of air-water vapour mixture
Question Statement:
At 60 °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)
Options:
(A) 0.026
(B) 0.032
(C) 0.004
(D) 0.041
Official Answer Key:
A
Step-by-Step Solution & Explanation
Step 1 — partial pressure of water vapour from the definition of relative humidity:
RH = p_v/p_sat ⇒ p_v = 0.20 × 20 = 4 kPa
Step 2 — absolute (specific) humidity. For an ideal gas mixture:
W = (M_w/M_a) × p_v/(P − p_v)
M_w/M_a = 18.02/28.97 = 0.622
W = 0.622 × 4/(101.325 − 4) = 0.622 × 4/97.325 = 0.622 × 0.04110
= 0.0256 ≈ 0.026 kg water/kg dry air
(Common error: dividing by total pressure P instead of the partial pressure of dry air (P − p_v), which gives 0.0246 — close, but conceptually wrong.)
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