Discussion
A room contains 35 kg of dry air and 0.5 kg of water vapor. The total pressure and temperature of air in the room are 100 kPa and 25°C respectively. Given that the saturation pressure for water at 25°C is 3.17 kPa, the relative humidity of the air in the room is
Answer: D
The relative humidity is about 71 %.
The specific humidity is the mass of vapour per unit mass of dry air:
\[ \begin{aligned} \omega &= \frac{0.5}{35} = 0.01429 \\ &= \frac{0.622\,p_{v}}{p-p_{v}} \end{aligned} \]
Solving for the vapour pressure with p = 100 kPa:
\[ \begin{aligned} p_{v} &= \frac{\omega p}{0.622+\omega} = \frac{1.429}{0.6363} \\ &= 2.245\ \text{kPa} \\ \phi &= \frac{2.245}{3.17} = 0.708 \approx 71\ \% \end{aligned} \]
The relative humidity compares the actual vapour pressure with the saturation pressure at the same temperature, so the 3.17 kPa enters only in the last step. 67 %, 55 % and 83 % do not follow from the specific humidity of 0.0143.
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