The density of the vapour of a substance at `1 atm` pressure and `500 K` is `0.36 kg m^(-3)`. The vapour effuses through a small hole at a rate of `1.33` times faster than oxygen under the same condition.
(`a`) Determine (`i`) the molecular weight, (`ii`) the molar volume (`iii`) the compression factor(`Z`) of the vapour, and (`iv`) which forces among the gas molecules are dominating, the attractive or the repulsive?
(`b`) If the vapour behaves ideally at `100 K`, determine the average translational kinetic energy of a molecule.

Explanation

(`a`) `(r_(vapour))/(r_(O_(2)))=sqrt((M_(O_(2)))/(M_(vapour)))` (Graham's law of diffusion)
or `1.33=sqrt((32)/(M_(vapour)))`
or `M_(vapour)=18.09`
Molar density `=(Molar mass)/(Molar volume)`
or `0.36=(18.09xx10^(-3))/(Molar volume)`
`:. Molar volume= 0.05025 m^(3)` or `50.25 L`
Compressibility factor
`Z=(pV)/(RT)=(1xx50.25)/(0.0821xx500)=1.224`
Repulsive forces dominate because compressibility factor is greater than `1`.
(`b`) Average translational kinetic energy of a molecules is
`(3)/(2)kT=(3)/(2)(R)/(N)T=(3)/(2)xx(8.314)/(6.023xx10^(23))xx1000`
`=2.07xx10^(-20)J`