One mole of N2O4(g) is allowed to reach the equilibrium N2O4(g)rightleftharpoons 2NO2(g) at a fixed temperature. At equilibrium the total pressure is…

With $\alpha=0.40$: moles of $\mathrm{N_2O_4}=0.6$, of $\mathrm{NO_2}=0.8$, total $1.4$. $p(\mathrm{NO_2})=2\times0.8/1.4$ and $p(\mathrm{N_2O_4})=2\times0.6/1.4$, so $K_p=\frac{4\alpha^2P}{1-\alpha^2}=\frac{4(0.16)(2)}{0.84}=1.52\ \mathrm{bar}$. A is correct. At $P=4$ bar, $\alpha=\sqrt{K_p/(K_p+4P)}=\sqrt{1.524/17.524}=0.295<0.30$, so B is correct. The average molar mass is $92/(1+\alpha)=92/1.4=65.7\ \mathrm{g\,mol^{-1}}$, so C is correct. In D, an inert gas at constant volume does not change any partial pressure or concentration, so $Q$ stays equal to $K$ and the equilibrium does not shift; D is incorrect. Answer: A, B, C. Statement D is the standard trap: only a change in volume (equivalently total pressure through compression) can shift this equilibrium, not the addition of an inert gas at constant volume.