For real gases, van der Waals' equation is written as
`(P+(an^(2))/(V^(2))) (V-nb)= nRT`
where `a` and `b` are van der Waals' constants.
Two sets of gases are:
`(I) O_(2), CO_(2), H_(2)` and `He(II) CH_(4), O_(2)` and `O_(2) and H_(2)`
The gases given in set `I` in increasing order of `b` and gases given in set `II` in decreasing order of `a` are arranged below. Select the correct order from the following:

Options

A. `(I) H_(2) lt He lt O_(2) lt CO_(2) (II) CH_(4) gt O_(2) gt H_(2)`
B. `(I) H_(2) lt O_(2) lt CO_(2) (II) O_(2) gt CH_(4) gt H_(2)`
C. `(I) He lt H_(2) lt CO_(2) lt O_(2) (II) CH_(4) gt H_(2) gt O_(2)`
D. `(I) O_(2) lt He lt H_(2) lt CO_(2) (II) H_(2) gt O_(2) gt CH_(4)`

Answer

A

Explanation

a' is related to intermolecular forces of attraction. Greater the intermolecular forces, higher is the critical temperature and more easily the gas is liquified. Thus, decreasing order of 'a' in II is `CH_(4) gt O_(2) gtH_(2)` (critical temperatures of `CH_(4),O_(2)` and `H_(2)` are 304 K, 154 K and 33 K respectively). 'b' is related to the size of the molecules. Hence, in set I, increasing order of 'b' is `H_(2)ltHeltCO_(2)`.