x mol $N_2{O}_4$  is taken at $P_1$  atm in a closed vessel & heated. When 75% $N_2{O}_4$  dissociated at equilibrium, total pressure is found to be $P_2$  atm. The relation between $P_1\&P_2$  is

$N_2{O}_4\rightleftharpoons 2N{O}_2$

$x0$

$\underset{\_}{0.75x1.5x}$

$0.25x1.5x$

$\begin{array}{c}Givenxmoles{P}_1\\ 1.75xmoles{P}_2\end{array}$

         $\frac{P_1}{P_2}=\frac{n_1}{n_2}=\frac{x}{1.75x}=\frac{4}{7}$