If for $2{\mathrm{A}}_2\mathrm{B}\left(\mathrm{g}\right) \rightleftharpoons 2{\mathrm{A}}_2\left(\mathrm{g}\right) + {\mathrm{B}}_2\left(\mathrm{g}\right),$

${\mathrm{K}}_{\mathrm{P}}$ = Total Pressure (at equilibrium) and starting the dissociation from 4 mol of ${\mathrm{A}}_2\mathrm{B}$ then

                            $2{\mathrm{A}}_2\mathrm{B} \rightleftharpoons 2{\mathrm{A}}_2 + {\mathrm{B}}_2$           …… (1)

t = 0                        1               0            0

$\mathrm{t}={\mathrm{t}}_{\mathrm{M}}$                   $1 -2\mathrm{\alpha }$       $2\mathrm{\alpha }$          $\mathrm{\alpha }$

Total mole = 1 + $\alpha$

${\mathrm{P}}_{{\mathrm{A}}_2\mathrm{B}} =\left(\frac{1-2\mathrm{\alpha P}}{1+\mathrm{\alpha }}\right) ; {\mathrm{P}}_{{\mathrm{A}}_2}= \left(\frac{2\mathrm{\alpha }}{1 + \mathrm{\alpha }}.\mathrm{P}\right) ; {\mathrm{P}}_{{\mathrm{B}}_2}= \left(\frac{\mathrm{\alpha }}{1 + \mathrm{\alpha }}.\mathrm{P}\right)$

${\mathrm{K}}_{\mathrm{P}}=\frac{{\left({\displaystyle \frac{2\mathrm{\alpha } . \mathrm{P}}{1+\mathrm{\alpha }}}\right)}^2 . \left({\displaystyle \frac{\mathrm{\alpha P}}{1+\mathrm{\alpha }}}\right)}{{\left({\displaystyle \frac{1-2\mathrm{\alpha }}{1+\mathrm{\alpha }}}\right)}^2. {\mathrm{P}}^2}$                             ………..(2)

$\frac{4{\mathrm{\alpha }}^2}{{\left(1+\mathrm{\alpha }\right)}^2}. \frac{\mathrm{\alpha }}{1+\mathrm{\alpha }} = \frac{{\left(1-2\mathrm{\alpha }\right)}^2}{{\left(1+\mathrm{\alpha }\right)}^2}$

$4{\mathrm{\alpha }}^3= \left(1+\mathrm{\alpha }\right) \left(4{\mathrm{\alpha }}^2-4\mathrm{\alpha } + 1\right)$ …….(3)

$4{\mathrm{\alpha }}^3 = 4{\mathrm{\alpha }}^2 + 4{\mathrm{\alpha }}^3 -4\mathrm{\alpha } -4{\mathrm{\alpha }}^2+1+\mathrm{\alpha }$

$1 = 3\mathrm{\alpha }, \mathrm{\alpha } = \frac{1}{3}$

for ${\mathrm{A}}_2\mathrm{B}$ , degree of dissociation=2$\alpha$ = $\frac{2}{3}$