The dissociation equilibrium of a gas ${\mathrm{AB}}_2$ can be represented as

$2{\mathrm{AB}}_2\left(\mathrm{g}\right) \rightleftharpoons 2\mathrm{AB}\left(\mathrm{g}\right) + {\mathrm{B}}_2\left(\mathrm{g}\right)$

The degree of dissociation is x and is small compared to 1. The expression relating to the degree of dissociation (x) with equilibrium constant ${\mathrm{K}}_{\mathrm{p}}$ and total pressure P is

                            $2{\mathrm{AB}}_2\left(\mathrm{g}\right) \rightleftharpoons 2\mathrm{AB}\left(\mathrm{g}\right) + {\mathrm{B}}_2\left(\mathrm{g}\right)$

Initial moles              1                      0                 0                    

At equilibrium     $2\left(1-\mathrm{x}\right)$                 2x                 x

Total moles at equilibrium = 2 - 2x + 2x + x = (2 + x)

So,          ${\mathrm{p}}_{{\mathrm{AB}}_2} = \frac{2\left(1-\mathrm{x}\right)\mathrm{p}}{\left(2+\mathrm{x}\right)}, {\mathrm{p}}_{\mathrm{AB}} = \frac{2\mathrm{xp}}{\left(2+\mathrm{x}\right)} {\mathrm{p}}_{{\mathrm{AB}}_2} =\frac{\mathrm{xp}}{\left(2+\mathrm{x}\right)}$

${\mathrm{K}}_{\mathrm{p}} = \frac{{\left({\mathrm{p}}_{\mathrm{AB}}\right)}^2 \left({\mathrm{p}}_{{\mathrm{AB}}_2}\right)}{{\left({\mathrm{p}}_{{\mathrm{AB}}_2}\right)}^2}=\frac{{\left({\displaystyle \frac{2\mathrm{xp}}{2+\mathrm{x}}}\right)}^2\left({\displaystyle \frac{\mathrm{x}}{2+\mathrm{x}}}\right)\mathrm{p} }{\left[\frac{2\left(1-\mathrm{x}\right)\mathrm{p}}{\left(2+\mathrm{x}\right)}\right]}$

$$\begin{gathered} = \frac{{\mathrm{x}}^3\mathrm{p}}{\left(2 + \mathrm{x}\right) {\left(1-\mathrm{x}\right)}^2} = \frac{{\mathrm{x}}^3\mathrm{p}}{2} \\ \left[\because \mathrm{x} << 1 \mathrm{and} 2 \mathrm{So}, \left(1-\mathrm{x}\right) \approx 1 \mathrm{and} \left(2+\mathrm{x}\right) \approx 2.\right] \\ \mathrm{x} = {\left(\frac{2{\mathrm{K}}_{\mathrm{p}}}{\mathrm{p}}\right)}^{1/3} \end{gathered}$$