See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

2 2 2 an P+ (v -nb) = nRT v n =1 a p+ (v -b) = RT v               If b is negligible For More Material Join: @JEEAdvanced_2024 AITS-CRT-IV (Paper-1)-PCM(Sol.)-JEE(Advanced)/2023 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 9       2 2 a p+ (v) =RT v RT a p = - v v The equation is quadratic in V thus 2 2 +RT ± R T - 4ap V = 2p Since V has one value at given P and T, thus numerical value of discriminant = 0 2 2 2 2 2 2 R T = 4ap R T (0.821) (300) p = = 4a 4×3.592 p = 8 5.277   p = 00042.22