What is the density of wet air ( in g/L) which have a partial pressure of water of 22.5 torr at 1 atm and $300 K$ ?
$($ Given: vapour pressure of $H_{2} O$ is 30 torr and the average molar mass of air is $29 g)$

1.164 g/L

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2.164 g/L

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3.164 g/L

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4.164 g/L

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Solution

The correct option is A $1.164 g/L$
According to Dalton's law, $\frac{P_{1}}{P_{2}} = \frac{n_{1}}{n_{2}}$
where P and n are the partial pressure and the moles of the gases present
% mole of $H_{2} O$ vapour in air $= \frac{(22.5)}{760} \times 100 = 2.96$
molar mass of wet air $= \frac{29 \times 97.04 + 2.96 \times 18}{100}$
$= \frac{2814.16 + 53.28}{100} = 28.67$
density of wet air $= \frac{P M}{R T} = \frac{1 \times 28.67}{0.0827 \times 300} = 1.164 g/L$

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Q. What is the density of wet air (in

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) that has the partial pressure of water of

22.5

torr at

1

atm and

300 K

(

Given: vapour pressure of

H_{2} O

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torr and the average molar mass of air is

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What is the density of wet air with $75 %$ relative humidity at $1 a t m$ and $300 K$ ? Given : vapour pressure of $H_{2} O$ is $30 t o r r$ and average molar mass of air is $29 g / m o l$ .