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

Answer

C

Explanation

`CO_(2)` is more easily liquefied than `O_(2)` gas. Hence (a) of `CO_(2)` is more than that of `O_(2)`. Also `CH_(4)` is easily liquefied than `H_(2)` and `He`. Hence, a of `CH_(4)` is more than `H_(2)` and `He`
`{:(,He,H_(2),O_(2),CO_(2),CH_(4)),((a),0.434,0.244,1.36,3.59,2.25 "atom 12 mole"^(-2)),((b),0.0237,0.0266,0.0318,0.0427,"0.0428 1 mole"^(-1)):}`
`:.` Order of a `CH_(4) gt O_(2) gt H_(2)`
`:.` Order of `b prop` size and size of `CO_(2) gt O_(2) gt He gt H_(2)`.