A 15 L container containing 5.6 g of `N_(2)` is fitted with another container of same capacity containing 8.0 g of `O_(2)` at `27^(@)`C . If the valve is opened and if there occurs no change in temperature , then determine the partial pressure of `N_(2)` in the gas mixture .

Explanation

No. of moles of `N_(2)` present in the mixture = `(5.6)/(28) = 0.2`
No. of moles of `O_(2)` present in the mixture = `(8)/(32) = 0.25`
If P be the total pressure of the gas mixture ,then applying equation PV = nRT , we get `P xx (15 + 15) = (0.2 + 0.25) xx 0.082 xx (27 + 273) ` or `30 xx P = 0.45 xx 0.082 xx 300` or P =1.107 atm .
`therefore` Partial pressure of `N_(2)` in the gas mixture = `(0.2)/(0.2 + 0.25) xx 1.107` = 0.492 atm .