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`) molecular weight, (`ii`) molar volume (`iii`) 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 `1000 K`, determine the average translational kinetic energy of a molecule.

Explanation

According to Graham's law of diffusion
(a) (i) `(r_(V))/(ro^(2)) = sqrt(M_(O_(2))/(M_(V)))`
or `1.33 = 1.33 = sqrt((32)/(M_(V)))`
`M_(V) = 18.1`
(ii0 Molar volume at `500 K`
`(barV)= (mo1 .wt)/("density 1 mole")= (18.1 xx 10^(-3))/(0.36)`
`=50.25 xx 10^(-3) m^(-3)`
(iii) Compression factor
`(Z) = (PbarV)/(RT) = (101325 xx 50.25 xx 10^(-3))/(8.314 xx 500) = 1.225`
`(P = 101325 xx Nm^(-2) =1 atm)`
(iv) Repulsive forces operates among molecules since
`Z gt1`
(b) Average K.E
` = (3)/(2) (RT)/(N) = (3)/(2) kT = (3)/(2) xx 1.38 xx 10^(-23) xx 1000`
`=2.07 xx 10^(-20) J//"molecule"` .