Question

Given the value of their van der Waals' constant, arrange the following gases in the order of their expected liquefaction pressure at a temperature T. T is below the critical point of all the gases.

Gas	$CH_4$	$Kr$	$N_2$	$Cl_2$
$a(atmL^2.mol^{-1})$	2283	2439	1.408	6.579

A) $CH_4 < Kr < N_2 < Cl_2$
B) $N_2 < CH_4 < Kr < Cl_2$
C) $Cl_2 < Kr < CH_4 < N_2$
D) $Cl_2 < N_2 < Kr < CH_4$

Answer 1

Hint: The pressure needed for the liquefaction of a real gas has an inverse relationship with the Van der Waals constant ‘a’. So, we first arrange the given real gases in the decreasing order of the Van der Waals constant ‘a’. Then we can arrange these gases in the increasing order of the pressure needed for the liquefaction of a real gas.

Complete answer:
Real gases show the deviation from the ideal gas behavior. The properties of the real gas are different from the properties of an ideal gas. To quantify the properties of real gas, we can use two parameters that are the Van der Waals constants ‘a’ and ‘b’.
The parameter Van der Waals constant 'a' gives indirect measure of the force of attraction present between different gas molecules. When a gas has a high value of the Van der Waals constant ‘a’, we can easily liquify the gas. So, we need less pressure for the liquefaction of this gas.
When a gas has low value of the Van der Waals constant ‘a’, we cannot easily liquify the gas. So, we will need more pressure for the liquefaction of this gas.
We can write the increasing order of the Van der Waals constant 'a' as $N_2 < CH_4 < Kr < Cl_2$. We can also write the decreasing order of the liquification pressure as $N_2 > CH_4 > Kr > Cl_2$.

Hence, the correct answer is option C.

Note: The behaviour of a real gas is different from that of the ideal gas. This is due to presence of intermolecular forces and the fact that the volume of gas molecules cannot be neglected in comparison to the volume of the container. For this purpose, the ideal gas equation is replaced with the real gas equation.