For following equilibrium: ${\mathrm{N}}_2{\mathrm{O}}_4( \mathrm{g})\rightleftharpoons 2{\mathrm{NO}}_2( \mathrm{g})$ ; ${\mathrm{K}}_{\mathrm{P}}=\frac{{\mathrm{P}}_{{\mathrm{NO}}_2}^2}{{\mathrm{P}}_{{\mathrm{N}}_2{\mathrm{O}}_4}}$; Graph between $P_{\text{total }}Vs$ $\alpha$ is given. Where $P_{total}$ = Total pressure at equilibrium a ® degree of dissociation (a << 1) 

 

$$\begin{gathered} \underset{{\mathrm{a}}_{ 0}-{\mathrm{a}}_0\alpha }{{\mathrm{N}}_2{\mathrm{O}}_4( \mathrm{g})} \rightleftharpoons \underset{ 2{\mathrm{a}}_0\alpha }{2 {\mathrm{NO}}_2(\mathrm{g} ) } \\ {\mathrm{K}}_{\mathrm{P}}=\frac{4{\alpha }^2}{1-{\alpha }^2}·{\mathrm{P}}_{\mathrm{t}}\approx 4{\alpha }^2·{\mathrm{P}}_{\mathrm{t}} \text{ if }\alpha <<1 \end{gathered}$$

$$\begin{gathered} \left(1\right) {\mathrm{K}}_{\mathrm{P}}=4\times (0.3{)}^2\times 100=36 \mathrm{atm} \\ { \mathrm{K}}_{\mathrm{P}}=4{\alpha }_2^2{\mathrm{P}}_2 \\ 4\times (0.3{)}^2\times 100=4(0.4{)}^2{\mathrm{P}}_2 \end{gathered}$$

$\left(2\right) {\mathrm{P}}_2=\frac{900}{16}=56.25 \mathrm{atm}$

$\left(3\right) Q_P=4{\alpha }^2. P_t$ so if $P_t$ increases then $\alpha$ decreases to reset $Q_P=K_P.$

$\left(4\right)$ As reaction is endothermic in forward direction so on increasing temperature, it favour endothermic step i.e. $\alpha$ increases.