A closed vessel contains an equal mass of `O_(2) and CH_(4)` gases at `25^(@)C`. What fraction of the total presure is contributed by `CH_(4)` gas?

Explanation

`n_(CH_(4))=(w)/(16),n_(O_(2))=(w)/(32)`
`therefore`Total number of moles`=(w)/(16)+(w)/(32)=(3w)/(32)`
`therefore x_(CH_(4))=(w)/(16)xx(32)/(3w)=(2)/(3)`
`p_(CH_(4))=(2)/(3)` P, thus methane contributes `(2)/(3)`rd of the total pressure.