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Compressibility Factor

On a certain ...

Question

On a certain humid day, the mole fraction of the water vapour in air at $25^{\circ} C$ is 0.0287. If total pressure of air is 0.977 bar, calculate the partial pressure of water vapour in air and relative humidity if the vapour pressure of water at $25^{\circ} C$ is 0.0313 bar. The partial pressure of water vapour ( $P_{H_{2} O}$ ) and relative humidity ( $R_{H}$ ) respectively are:

A

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B

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C

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D

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Solution

The correct option is A

Here, mole fraction of $H_{2} O (X_{H_{2} O}) = 0.0287, P_{t o t a l}$ = 0.977 bar . Therefore,

$P_{H_{2} O} = X_{H_{2} O} \times P_{t o t a l} = 0.0287 \times 0.977 b a r$ = 0.028 bar.

% Relative humidity = $\frac{P_{H_{2} O} i n a i r}{V a p o u r P r e s s u r e o f H_{2} O} \times 100 = \frac{0.028}{0.313} \times 100$ = 8.94%

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