Chemistry - Real Gas Question with Solution | TestHub

ChemistryReal GasMiscellaneous/MixedEasy2 minPYQ_2020

ChemistryEasysingle choice

If density of vapours of a substance of molar mass $18 gm/mole$ at $1 atm$ pressure and $500 K$ is $0.36 kg m^{- 3}$ , then value of $Z$ for the vapours is (Take $R = 0.082 L {atm mole}^{- 1} K^{-}^{1}$ )

Options:

Answer:

Solution:

$V_{r e a l} = \frac{m o l a r m a s s}{d e n s i t y} = \frac{18}{0.36}$
$V_{i d e a l} = \frac{n R T}{P} = \frac{1 \times 0.082 \times 500}{1}$
So, $Z = \frac{V_{r e a l}}{V_{i d e a l}} = \frac{50}{0.082 \times 500} = \frac{50}{41}$

Stream:NTA_ABHYASSubject:ChemistryTopic:Real GasSubtopic:Miscellaneous/Mixed

⏱ 2mℹ️ Source: PYQ_2020

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