See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

For n mole   2 2 n a P V nb nRT V          for negligible volume b  0 PV2 – nRTV + n2a = 0 Quadratic in V thus, 2 2 2 2 nRT n R T 4an P V 2    V has one value at given pressure and temperature thus discriminant = 0 n2R2T2 = 4a n2P P = 42.22 atm.