A gas mixture consisting of `O_(2) and N_(2)` gases gas a volume of 5 L at `25^(@)C`. In themixture, if the mass of `O_(2)` gas is twice that of `N_(2)` gas, then which one of them will have a greater contribution to the total pressure of the mixture.

Explanation

In the mixture, `n_(O_(2))=(2w)/(32)=(w)/(16) mol and n_(N_(2))=(w)/(28)mol`.
`therefore`Total number of moles `(n)=(w)/(16)+(w)/(28)=(11w)/(112)mol`
Hence, `x_(N_(2))=(w)/(28)xx(112)/(11w)=(4)/(11) and x_(O_(2))=(w)/(16)xx(112)/(11w)=(7)/(11)`.
Thus, `x_(O_(2)) gt x_(N_(2)) gt So, p_(O_(2)) gt p_(N_(2)) (because p_(i)=x_(i)P)`.