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. `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 He 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) 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 in `CH_(4) gt O_2 gt H_(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) lt He lt O_(2) lt CO_(2)` .