For the gas-phase reaction N2O4(g)rightleftharpoons2NO2(g), take Δ H^°=57.2 kJ mol^-1 and Δ S^°=176 J K^-1mol^-1, both independent of temperature…

First get the equilibrium constant from thermodynamics: $\Delta G^\circ=\Delta H^\circ-T\Delta S^\circ=57200-298\times176=4752\ \mathrm{J\,mol^{-1}}$. Then $K_p=\exp(-\Delta G^\circ/RT)=\exp(-4752/(8.314\times298))=\exp(-1.918)=0.1469$. For $\alpha$ moles dissociating per mole of $\mathrm{N_2O_4}$ the mixture has $1-\alpha$ of $\mathrm{N_2O_4}$ and $2\alpha$ of $\mathrm{NO_2}$, total $1+\alpha$. Therefore $K_p=\dfrac{(2\alpha P/(1+\alpha))^2}{(1-\alpha)P/(1+\alpha)}=\dfrac{4\alpha^2P}{1-\alpha^2}$. Solving, $\alpha^2=K_p/(K_p+4P)=0.1469/(0.1469+2.00)=0.0684$, so $\alpha=0.2616\approx0.26$. At 1 bar the same equation would give 0.19, showing Le Chatelier's effect: lower pressure favours the side with more moles of gas. Note that $K_p$ here is dimensionless because pressures are divided by the 1 bar standard state, and the endothermic sign of $\Delta H^\circ$ means $K_p$ would rise on heating. The sanity check $\alpha<1$ and $\alpha^2<1$ holds.