GATE 2025 CH – Question 47
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 $\ln P^{sat} = 12 - \frac{4000}{T - 40}$ where $P^{sat}$ 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 (rounded off to the nearest integer).
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Correct answer: 25
Explanation
The mole ratio of water to dry air is $\frac{0.02/18}{1/29} = 0.03222$. The mole fraction of water in the mixture is $\frac{0.03222}{1.03222} = 0.03122$, so the partial pressure is $0.03122$ bar. At the dew point this is the saturation pressure: $\ln 0.03122 = -3.467 = 12 - \frac{4000}{T - 40}$, so $T - 40 = \frac{4000}{15.467} = 258.6$ and $T = 298.6$ K, which is $25.5$ °C. Rounded to the nearest integer this is 25 °C.