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Q.
$N _{2} O _{4}$ is dissociated to $33 \%$ and $40 \%$ at total pressure $P _{1}$ and $P _{2}$ atm respectively. Then the ratio $P _{1} / P _{2}$ is:
Equilibrium
Solution:
$K _{ p }=\frac{\left( n _{ NO _{2}}\right)^{2}}{ n _{ N _{2} O _{4}}} \times\left[\frac{ P }{\Sigma n }\right]^{1}$
For $33 \%$ dissociation : $K _{ p }=\frac{(2 \times 0.33)^{2}}{0.67} \times\left[\frac{ P }{1.33}\right]$
For $40 \%$ dissociation $: K _{ p }=\frac{(2 \times 0.40)^{2}}{0.60} \times\left[\frac{ P }{1.40}\right]$
$\therefore \frac{ P _{1}}{ P _{2}}=1.56 \approx 1.60=\frac{8}{5}$
