See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

102  403 4 3 0 0 V V 10 Lt 2P P   V4  V3 = 5 Lt Work done = ar(102)  ar(034) = 2 1 0 4 3 0 1 1 (V V ) 2P (V V ) P 2 2      D  = 1+x S1 S2 Screen O t = 12 mm   = 2.5 d P F 2f f  AITS-FT-IX-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 6 = 0 0 0 1 10P (5P )