What is the density of wet air with $75 %$ relative humidity at $1$ atm and $300$ K?
[Given: vapour pressure of $H_{2} O$ is $30$ torr and average molar mass of air is $29$ g $m o l^{- 1}$ ]

1.614

g/L

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0.96

g/L

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1.06

g/L

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1.164

g/L

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Solution

The correct option is D $1.164$ g/L
Let us consider the equation

$P_{w e t a i r} = \frac{P_{d}}{R_{d} T} + \frac{P_{v}}{R_{v} T} = \frac{P_{d} M_{d} + P_{v} M_{v}}{R T}$

Given that :
$T =$ temperature $= 300 K$

$M_{d} =$ molar mass of dry air $= 29 g m o l^{- 1}$

$M_{v} =$ molar mass of water vapourr $= 18 g m o l^{- 1}$

$R = 0.821 L a t m m o l^{- 1} K^{- 1}$

$P_{v} = 30 t o r r = 0.0394 a t m$

$P_{d} = P - P_{v} = 1 - 0.0394 = 0.9606 a t m$

where $P_{d}$ partial pressure of dry air

$R_{d} =$ specific gas constant for dry air

$R_{v} =$ Specific gas constant for water vapour

$P_{w e t a i r} = \frac{0.9606 \times 29 + 0.0394 \times 18}{24.63}$

$= 1.164 g L^{- 1}$

Hence, the correct option is $D$

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Similar questions

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

Q. What is the density of wet air with 75% relative humidity at 1 atm and 300 K? Given vapour pressure of

H_{2} O

is 30 torr and average molar mass of air is

29 g m o l^{-}

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 $29 g / m o l$ .