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(B) For the reaction: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$At $t=0$,moles: $1$ for $N_2O_4$ and $0$ for $NO_2$.At equilibrium,moles: $(1-\alpha)

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$\alpha = \sqrt{\frac{K_p}{4P + K_p}}$

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(B) For the reaction: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$At $t=0$,moles: $1$ for $N_2O_4$ and $0$ for $NO_2$.At equilibrium,moles: $(1-\alpha)$ for $N_2O_4$ and $2\alpha$ for $NO_2$. Total moles $n_f = 1+\alpha$.Mole fractions: $x_{N_2O_4} = \frac{1-\alpha}{1+\alpha}$ and $x_{NO_2} = \frac{2\alpha}{1+\alpha}$.Partial pressures: $P_{N_2O_4} = \left(\frac{1-\alpha}{1+\alpha}\right)P$ and $P_{NO_2} = \left(\frac{2\alpha}{1+\alpha}\right)P$.$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(\frac{2\alpha}{1+\alpha})^2 P^2}{(\frac{1-\alpha}{1+\alpha})P} = \frac{4\alpha^2 P}{(1+\alpha)(1-\alpha)} = \frac{4\alpha^2 P}{1-\alpha^2}$.Rearranging for $\alpha^2$: $K_p(1-\alpha^2) = 4\alpha^2 P \implies K_p = \alpha^2(4P + K_p)$.Thus,$\alpha^2 = \frac{K_p}{4P + K_p}$,which gives $\alpha = \sqrt{\frac{K_p}{4P + K_p}}$.

For the dissociation reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the degree of dissociation $(\alpha)$ in terms of $K_p$ and total equilibrium pressure $P$ is

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