The system $N_2O_4 \rightleftharpoons 2 NO_2$ maintained in a closed vessel at $60^oC$ & pressure of $5$ atm has an average (i.e. observed) molecular weight of $69$, calculate $K_p$. At what pressure at the same temperature would the observed molecular weight be $(230/3)$?

The average molecular weight M_avg is related to the degree of dissociation alpha by M_avg = M_theoretical / (1 + alpha). For N2O4, M_theoretical = 92. At 5 atm, 69 = 92 / (1 + alpha) implies alpha = 1/3. Kp = (4 * alpha^2 * P) / (1 - alpha^2) = (4 * (1/9) * 5) / (8/9) = 2.5. For the second case, M_avg = 230/3 = 76.67, so 76.67 = 92 / (1 + alpha), giving alpha = 0.2. Using Kp = 2.5 = (4 * 0.2^2 * P) / (1 - 0.2^2), we get P = 15 atm.