For the equilibrium $\mathrm{A}\mathrm{B}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{A}\left(\mathrm{g}\right)+\mathrm{B}\left(\mathrm{g}\right)$ , at a given temperature $\frac{1}{3}\mathrm{r}\mathrm{d}$ of AB is dissociated then $\frac{\mathrm{P}}{{\mathrm{K}}_{\mathrm{p}}}$ will be numerically equal to

Solution

$\mathrm{\Sigma }\mathrm{n}=\frac{2}{3}+\frac{2}{3}+\frac{1}{3}=\frac{4}{3}$
${\mathrm{K}}_{\mathrm{p}}=\frac{{\mathrm{n}}_{\mathrm{A}}\times {\mathrm{n}}_{\mathrm{B}}}{{\mathrm{n}}_{\mathrm{A}\mathrm{B}}}\times {\left(\frac{\mathrm{P}}{\mathrm{\Sigma }\mathrm{n}}\right)}^{\mathrm{\Delta }{\mathrm{n}}_{\mathrm{g}}}$
$=\left[\frac{\frac{1}{3}\times \frac{1}{3}}{\frac{2}{3}}\right]\times {\left[\frac{\mathrm{P}}{\frac{4}{3}}\right]}^1$
$=\frac{1}{6}\times \frac{3\mathrm{P}}{4}=\frac{\mathrm{P}}{8}$
So, $\frac{\mathrm{P}}{{\mathrm{K}}_{\mathrm{p}}}=8$

Hence, the correct answer is 8.00.