Initial pressure of N2O5 $\begin{aligned}
& =\frac{\frac{37.8}{108} \times 0.082 \times 500}{1}=14.35 \text { bar } \
& 2 \mathrm{N}_2 \mathrm{O}_5 \rightleftharpoons 2 \mathrm{N}_2 \mathrm{O}4+\mathrm{O}2
\end{aligned}\begin{array}{lll}\mathrm{t}=0 & 14.35 \ \mathrm{t}=\mathrm{eq} & 14.35-2 \mathrm{P} & 2 \mathrm{P}\end{array}\begin{aligned} & \mathrm{P}{\text {Total }} \text { at eqb }=14.35+\mathrm{P}=18.65 \ & \mathrm{P}=4.3 \ & \mathrm{P}{\mathrm{N}_2 \mathrm{O}5}=5.75 \text { bar } \ & \mathrm{P}{\mathrm{N}_2 \mathrm{O}4}=8.6 \mathrm{bar} \ & \mathrm{P}{\mathrm{O}2}=4.3 \text { bar } \ & \mathrm{k}{\mathrm{p}}=\frac{(8.6)^2 \times(4.3)}{(5.75)^2}=9.619=\mathrm{x} \times 10^{-2} \ & \mathrm{x}=961.9 \approx 962\end{aligned}$