For the reaction ${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right)$, the relation connecting the degree of dissociation $\left(\alpha \right)$ of ${\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)$ with the equilibrium constant $K_p$ is

$\begin{array}{l}{\mathrm{N}}_2{\mathrm{O}}_4\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NO}}_2\left(\mathrm{g}\right)\\ 1-\alpha 2\alpha ;n_{\text{total }}=1+\alpha \end{array}$

$p_{{\mathrm{N}}_2{\mathrm{O}}_4}=\frac{1-\alpha }{1+\alpha }p$ and $p_{{\mathrm{NO}}_2}=\frac{2\alpha }{1+\alpha }p;K_p=\frac{p_{{\mathrm{NO}}_2}^2}{p_{{\mathrm{N}}_2{\mathrm{O}}_4}}={\left(\frac{2\alpha }{1+\alpha }p\right)}^2\left(\frac{1+\alpha }{1-\alpha }\frac{1}{p}\right)=\frac{4{\alpha }^2}{1-{\alpha }^2}p$

Rearranging, we get $\alpha ={\left(\frac{K_p/p}{4+K_p/p}\right)}^{1/2}$