Question

The density of vapour of a substance at $1$ at pressure and $500K$ is $0.36Kg$ $m^{-3}$. The vapour effuses through a small hole at a rate of $1.33$ times faster than oxygen under the same condition.
Determine:
(i) mol.wt (ii) molar volume (iii) compression factor(z) of the vapour (iv) which forces among the gas molecules are dominating, the attractive or the repulsive.

Answer 1

By Graham's Law of effusion states rate of effusion is inversely proportional to the molecular weight of the gas.

$\frac{rateofeffuusionofoxygen}{rateofeffusionofvapor}=\surd \frac{molecularweightofvapor}{molecularweightofoxygen}$

hence molecular weight of vapor=$32\times {\left[\frac{3}{4}\right]}^2=18$

molar volume is defined as volume occupied by 1 mole of substance.
$\frac{masof1molofvapor}{densityofvapor}$
$\frac{0.018}{0.36}=0.05m^3$
$\frac{PV}{RT}=Z$
$\frac{1\times 50}{0.0821\times 500}=Z$=1.218

Here Z>1 hence it is a repulsive force.