See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

  m m x P V y RT V          Yashpatil TG~ @bohring_bot For More Material Join: @JEEAdvanced_2024 Rankers Academy JEE AITS-PT-I-PCM(Sol.)-JEE(Main)/2024 FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com 16 m m RT x P V y V     At critical point 2 2 m m dP d P 0 and 0 dV dV   Or   2 2 m m x RT V V y   … (1)   2 2 3 3 m m m d P 2RT 2x 0 dV V V y     or   3 3 m m x RT V V y   … (2) Dividing (2) by (1) m m V V y y 0     Which is NOT possible. So, gas has no critical point.  Answer is 33.