Calculate the partial pressures `N_2` and `H_2` in a mixture of two moles of `N_2` and two moles of `H_2` at STP.

Explanation

`p_(N_(2)) = ("number of moles of" H_(2))/(V) xx RT`
`p_(H_(2)) = ("number of moles of" H_(2))/(V) = xx RT`
Mole fraction of `N_2 = X_(N_(2)) = (2)/(2 +2)`
`= (2)/(4) = 0.5`
`therefore X_(H_(2)) = 0.5 (X_(N_2) + X_(H_2) = 1.0)`
But `P = (RT)/(V)` . For 1 mole `V = 22.4` litres
For 4 moles V = `4 xx 22.4 ` litres and `R = 0.821` lit - atm `K^(-1) mol^(-1)`
`P = (0.0821 xx 273)/(4 xx 22.4) =0. 2501` atm
`p_(N_2) =0.2501 xx 0.5 = 0.1251` atm
`p_(H_2) = 0.2501 xx 0.5 = 0.1251 ` atm