Set-I: `O_(2),CO_(2),H_(2) and He, ` Set-II: `CH_(4),O_(2) and H_(2)`. The gases in set-I in increasing order of 'b' and gases given in set-II in decreasing order of 'a', are arranged below, where 'a' and 'b' are van der waals constants. Selection the correct order from the following-

Options

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

Answer

B

Explanation

A gas with strong intermolecular forces of attraction has a large value of 'a', and a gas has a large value of 'b' if its molecules are big is size.
The increasing order of sizes of `H_(2),He,O_(2) and CO_(2)` colecules is `H_(2) lt He lt O_(2) lt CO_(2)`. so, the increasing order of 'b' values for these gases is `H_(2) lt He lt O_(2) lt CO_(2).`
`CH_(4),O_(2)` and `H_(2)` are all non-polar molecules. only intermolecular forces of attraction that act in `CH_(4),O_(2) and H_(2)` gases are London forces. the strength of London forces increases with molecular size. so, the increasing order of intermolecular forces in `CH_(4),O_(2) and H_(2)` gases will be: `CH_(4) gt O_(2) gt H_(2)`. Again, the stronger the intermolecular forces of attraction in a gas, the larger the value of 'a' for the gas. therefore, the decreasing order of 'b' values for these gases will be: `CH_(4) gt O_(2) gt H_(2)`.