Explanation

(a)
$N_{2}O_{4} \rightleftharpoons \ \ \ \ 2{NO}_{2}$
$0.1\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 0$ initialy
$(0.1 - \alpha)\ \ \ \ \ 2\alpha$ at equilibrium
$\frac{0.1 - \alpha}{0.1 + \alpha}p\ \ \frac{2\alpha}{0.1 + \alpha}p\ \ atp_{i}$
$K_{p} = \frac{\left\lbrack NO_{2} \right\rbrack^{2}}{\lbrack N_{2}O_{4}\rbrack}$
$0.14 = \left( \frac{2\alpha \times p}{0.1 + \alpha} \right)^{2} \times \left( \frac{0.1 + \alpha}{0.1 - \alpha} \right)p$
$= \frac{4\alpha^{2}}{(0.1 + \alpha)(0.1 - \alpha)}p$
$= \frac{4\alpha}{0.01 - \alpha^{2}} \times 1$
or $\alpha = 0.018$
Thus, $\left\lbrack NO_{2} \right\rbrack = 2 \times 0.018 = 0.036$ mol