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The density of vapour of a substance (X)...

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?

A

`0.974`

B

`1.35`

C

`1.52`

D

`1.22`

Text Solution

Verified by Experts

The correct Answer is:
C
`(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`
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