Discussion
Air at 1 atmospheric pressure (101.325 kPa) and 30°C with absolute humidity of 0.0218 kg/kg of dry air is flowing in a drying chamber. The saturated vapor pressure of water (\(p_{w}^{0}\), in kPa) is related to temperature (T, in °C) as given below
\[ \ln p_{w}^{0} = 18.6556 - \frac{5217.635}{T + 273} \]
The relative humidity of air (in percentage) is
Answer: D
The relative humidity is about 81.80%.
First the saturation pressure at 30°C from the given equation:
\[ \begin{aligned} \ln p_{w}^{0} &= 18.6556 - \frac{5217.635}{303} \\ &= 18.6556 - 17.2199 = 1.4357 \\ p_{w}^{0} &= 4.20\ \text{kPa} \end{aligned} \]
Then the actual vapour pressure from the absolute humidity, \(H = \dfrac{18}{29}\dfrac{p_{w}}{P - p_{w}}\):
\[ \begin{aligned} p_{w} &= \frac{HP}{0.6207 + H} = \frac{0.0218 \times 101.325}{0.6425} \\ &= 3.438\ \text{kPa} \\ RH &= \frac{3.438}{4.2025} \times 100 = 81.8\% \end{aligned} \]
Using 0.622 for the molar mass ratio gives 81.6%, still closest to 81.80. 62.82, 68.22 and 86.62 do not follow from these numbers.
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