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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 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 ?

A

0.974

B

1.35

C

1.52

D

1.22

Text Solution

Verified by Experts

The correct Answer is:
c
(c) `(r_(x))/(r_(O_(2)))=sqrt(M_(O_(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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