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((MO_(2))/(M_(x)))=((4)/(5))^(2)=(32)/(M_(x))`
`M_(x)=50`
`d_(x)=0.80kg//m^(3)`
`V_(m)=(1000)/(800)xx50=62.5L`
`Z=(PV_(m))/(RT)=(1xx62.5)/(0.0821xx500)=1.52`