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`) (`i`) Using Graham's law of diffusion,
`(r_(1))/(r_(2))=sqrt((M_(2))/(M_(1)))`
We get
`(1.33)/(1)=((32 g mol^(-1))/(M_(1)))^(1//2)`
`impliesM_(1)=(32g mol^(-1))/((1.33)^(2))=18.09 g mol^(-1)`
(`ii`) The molar volume of the vapour is
`V_(m)=(M_(1))/(rho)=(18.09 g mol^(-1))/(0.36 kg m^(-3))`
`=(18.09 g mol^(-1))/(0.36 g dm^(-3))=50.25 dm^(3) mol^(-1)`
(`iii`) The compression factor is
`Z=(V_(m,real))/(V_(m,ideal))=(V_(m,real))/(RT//P)=(pV_(m,real))/(RT)`
`=((101.325 kPa)(50.25 dm^(3) mol^(-1)))/((8.314 JK^(-1)mol^(-1))(500 K))`
`=1.22`
(`iv`) Since `Z gt 1`, the repulsive forces dominate among the gaseous molecules.
(`b`) The average translational kinetic energy of a molecule is `(3)/(2) kT`
`Ke=((3)/(2))(1.38xx10^(-23)JK^(-1))(1000 K)`
`=2.07xx10^(-20)J`