Explanation

(c) Ideal gas equation is given by
$pV = nRT$ …(i)
For oxygen, $p = 1$ atm, $V = 1$ L, $n = {n_{O}}_{2}$
Therefore, Eq. (i) becomes
$\therefore\ \ \ \ \ \ 1 \times 1 = {n_{O}}_{2}RT$
$\Rightarrow \ \ \ \ \ \ \ \ {n_{O}}_{2} = \frac{1}{RT}$
For nitrogen $p =$0.5 atm, $V =$2 L, $n = n_{N}$
$\therefore\ \ \ \ \ \ \ \ \ 0.5 \times 2 = {n_{N}}_{2}RT$
$\Rightarrow \ \ \ \ \ \ \ \ {n_{N}}_{2} = \frac{1}{RT}$
For mixture of gas
$p_{mix}V_{mix} = n_{mix}RT$
Here, $n_{mix} = {n_{O}}_{2} + {n_{N}}_{2}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \frac{p_{mix}V_{mix}}{RT} = \frac{1}{RT} + \frac{1}{RT}$
$\Rightarrow \ \ \ \ \ \ \ \ p_{mix}V_{mix} = 2$ $(V_{mix} = 1)$