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 H_(2) Thegases 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 :

For real gases, van der Waals equation is written as (p+(an^(2))/(V^(

1.	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 H_(2) Thegases 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 :
Answer» `(I) H_(2) lt He lt O_(2) lt CO_(2) (II) CH_(4) gt O_(2) gt H_(2)` `(I) H_(2) lt O_(2) lt CO_(2) (II) O_(2) gt CH_(4) gt H_(2)` `(I) He lt H_(2) lt CO_(2) lt O_(2) (II) CH_(4) gt H_(2) gt O_(2)` `(I) O_(2) lt He lt H_(2) lt CO_(2) (II) H_(2) gt O_(2) gt CH_(4)` Solution :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)`.