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Q.

The density of vapour of a substance (X) at 1 atm pressure and 500 K is $0.8 K g / m^{3}$ . 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?
$(R = 0.0821 L . a t m . m o l^{- 1} K^{- 1})$

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answer is 1.52.

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Detailed Solution

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$\frac{r_{x}}{r_{O_{2}}} = \sqrt{\frac{M_{O_{2}}}{M_{X}}} = {(\frac{4}{5})}^{2} = \frac{32}{M_{x}} M_{x} = 50 d x = 0.80 K g / m^{3} V_{m} = \frac{1000}{800} \times 50 = 62.5 L z = \frac{P V_{m}}{R T} = \frac{1 \times 62.5}{0.0821 \times 500} = 1.52$

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