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) d=0.36 g kg `m^(-3)`. By Graham's law of diffusion.
`(r_(v))/(r_(O_(2)))=sqrt((M_(O_(2)))/(M_(v)))`, i.e., `1.33=sqrt((32)/(M_(v)))` or `M_(v)=18.09`
(ii) `0.36" kg "m^(-3)=0.36" g "L^(-1)`, i.e., 0.36 g of vapour have volume =1L
`:.` 1 mole, i.e., 18.09 g of vapour will have volume `=(1)/(0.36)xx18.09=50.25" L"`
(iii) Compressibility factor(Z)`=(PV)/(RT)=(1" atm"xx50.25" L")/(0.0821" L atm "K^(-1)mol^(-1)xx500" K")=1.224`
(iv) As `Z gt 1`, repulsive forces would dominate.
(b) `bar(E)=(3)/(2)kT=(3)/(2)xx1.38xx10^(23)" JK"^(-1)"molecule"^(-1)xx1000" K"=2.07xx10^(-20)` J per molecule.