In Van der Waals equation of state for a non-ideal gas,the term that accounts for intermolecular forces is

Dominance of strong repulsive forces among the molecules of the gas ($Z =$ compressibility factor)

When helium gas is allowed to expand into vacuum, a heating effect is observed. The reason for this is (assume $He$ as a non-ideal gas):

For real gases,the relation between $p$,$V$,and $T$ is given by the van der Waals equation:
$(p + \frac{an^2}{V^2})(V - nb) = nRT$
Where $a$ and $b$ are van der Waals constants,$nb$ is approximately equal to the total volume of the molecules of a gas,and $a$ is the measure of the magnitude of intermolecular attraction.
$(i)$ Arrange the following gases in the increasing order of $b$. Give reason: $O_2, CO_2, H_2, He$
$(ii)$ Arrange the following gases in the decreasing order of magnitude of $a$. Give reason: $CH_4, O_2, H_2$

Four gases $P, Q, R$ and $S$ have almost same values of $b$ but their $a$ values ($a, b$ are van der Waals' constants) are in the order $Q < R < S < P$. At a particular temperature among the four gases, the most liquefiable one is

For $1$ mol of a real gas kept at constant temperature $T$,the graph of $PV$ vs $P$ intersects the $PV$ axis at $20$. Hence,the temperature $T$ is ($R$ = universal gas constant).