The density of vapour of a substance `(X)` at 1 atm pressure and 500 K is `0.8 kg//m^(3)`. The vapour effuse 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 ?

Options

A. 0.974
B. 1.35
C. 1.52
D. 1.22

Answer

C

Explanation

`(r_(x))/( r_(O_(2))) = sqrt((M_(O_(2)))/( M_(x))) = ((4)/( 5))^(2) = ( 32)/(M_(x)) = M_(x) =50`
`d_(x) =0.80 kg //m^(3)`,
`V_(m) = ( 1000)/( 800) xx 50 = 62.5L`
`Z =( PV_(m))/( RT) = (1 xx 62.5 )/( 0.0821 xx500) = 1.52`