The density of vapour of a substance $(X)$ at 1 atm pressure and 500 K is 0.8 kg/ m $^{3}$ . The vapour effuses through a small hole at a rate of $4 / 5$ times slower than oxygen under the same condition. What is the compressibility factor $(Z)$ of the vapour?

0.974

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1.35

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1.52

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1.22

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Solution

The correct option is C $1.52$
$\frac{r_{x}}{r_{O_{2}}}$ $= \sqrt{\frac{M_{O_{2}}}{M_{x}}}$ $= {(\frac{4}{5})}^{2} = \frac{32}{M_{x}}$

$M_{x} = 50$

$d_{x} = 0.80 k g / m^{3}$

$V_{m} = \frac{1000}{800} \times 50 = 62.5$ L

$Z = \frac{P V_{m}}{R T} = \frac{1 \times 62.50}{0.0821 \times 500} = 1.52$

Hence, te correct option is $C$

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