Chemistry - Chemical Equilibrium Question with Solution | TestHub

${\mathrm{K}}_{\mathrm{P}}=\frac{({\mathrm{n}}_{\mathrm{N}{\mathrm{O}}_2}{)}^2}{{\mathrm{n}}_{{\mathrm{N}}_2{\mathrm{O}}_4}}\times {\left[\frac{\mathrm{P}}{\mathrm{\Sigma }\mathrm{n}}\right]}^1$
For 33% dissociation${\mathrm{K}}_{\mathrm{P}}=\frac{(2\times 0.33{)}^2}{0.67}\times \left[\frac{\mathrm{P}}{1.33}\right]$
For 40% dissociation${\mathrm{K}}_{\mathrm{P}}=\frac{(2\times 0.40{)}^2}{0.60}\times \left[\frac{\mathrm{P}}{1.40}\right]$
So$\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=1.56\simeq 1.60=\frac{8}{5}$.