Question

The density of the vapours of a substance at $1$ atm pressure and $500K$ is $0.36kg{m}^{-3}$. The vapours effuse through a small hole at a rate of $1.33$ times faster than oxygen under the same conditions:

(a) Determine: (i) molecular weight, (ii) molar volume, (iii) compressibility factor $\left(Z\right)$ of the vapours. (iv) Which forces among gas molecules are dominating, the attractive or the repulsive.

(b) If the vapours behave ideally at $1000K$, determine the average translational kinetic energy of a molecule.

Answer 1

(a) (i) $\frac{r_{vapour}}{r_{O_2}}=\sqrt{\frac{M_{O_2}}{M_{vapour}}}$
$1.33=\sqrt{\frac{32}{M_{vapour}}}$
$M_{vapour}=18.1$
(ii) Molar volume $=\frac{Molarmass}{Density}$
$=\frac{18.1}{0.36}=50.25\times {10}^{-3}{m}^{-3}$
(iii) Compressibility factor, $Z=\frac{PV}{RT}=\frac{101325\times 50.25\times {10}^{-3}}{8.314\times 500}$$=1.225$
(iv) $Z>1$ shows that repulsive forces are dominant.

(b) Translational $KE$ per molecule
$=\frac{3}{2}\times \frac{R}{N}\times T$
$=\frac{3}{2}\times \frac{8.314}{6.023\times {10}^{23}}\times 1000$
$=2.07\times {10}^{-20}J$.