GATE 2026 CH – Question 38
Vapor pressure of water at various temperature is given in the table.
| Temperature (K) | 284 | 289 | 294 | 299 | 310 |
|---|---|---|---|---|---|
| Vapor pressure (kPa) | 1.28 | 1.80 | 2.50 | 3.40 | 6.40 |
An air and water vapor mixture at 100 kPa with a relative humidity of 20% has a dry-bulb temperature of 310 K. Assume latent heat of vaporization for water is 44 kJ mol$^{-1}$ and specific heat capacity of the air and water vapor mixture is 0.035 kJ mol$^{-1}$K$^{-1}$. Which one of the following is the closest to its wet-bulb temperature (in K)?
Practise this question in The GATE Grind →
Show answer and explanation
Correct answer: (B) 294
Explanation
The vapour pressure in the air is $0.2 \times 6.4 = 1.28$ kPa, so the molar humidity is $Y = \frac{1.28}{100 - 1.28} = 0.01297$. At the wet bulb temperature the heat balance is $C(T - T_w) = \lambda (Y_w - Y)$ with $Y_w = \frac{p_w}{100 - p_w}$ (per mole of dry air, in the same units). Try $T_w = 294$ K: $p_w = 2.5$ kPa, $Y_w = 0.02564$, so the right side is $44 \times 0.01267 = 0.558$ kJ and the left side is $0.035 \times 16 = 0.560$ kJ. They agree, so the wet bulb temperature is about 294 K. At 289 K the sides are 0.735 and 0.236, and at 299 K they are 0.385 and 0.978, so neither fits.