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GATE 2025 Chemical Engineering, Q47 · Mass Transfer

The correct answer is 23 to 27.

Question

Consider moist air with absolute humidity of 0.02 (kg moisture)/(kg dry air) at 1 bar pressure. The vapour pressure of water is given by the equation ln(Psat)=12−ln(P_{sat}) = 12 − 4000/(T−40), where PsatP_{sat} is in bar and T is in K. The molecular weight of water and dry air are 18 kg/kmol and 29 kg/kmol, respectively. The dew temperature of the moist air is _____ °C°C (rounded off to the nearest integer).

Official answer: 23 to 27

Why

GATE 2025 official key: 23 to 27°C.27°C. From humidity, pw=0.02×29/(18+0.02×29)≈0.0312p_{w} = 0.02 \times 29/(18+0.02 \times 29) \approx 0.0312 bar. Solving ln(0.0312) = 12 − 4000/(T−40) gives T ≈298.6\approx 298.6 K ≈25.6°C,\approx 25.6°C, within range.

Explanation cross-checked for consistency. Final answer matches the official IIT answer key.

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