See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

For adiabatic process   U W = P(V2  V1) = 1(20  10) = 10 L atm =  1013 J P H C T    V U C T    P V C H U C    As gas is monoatomic, P 5 C R 2  and V 3 C R 2   5R H 5 2 3 U 3 R 2        5 5 H U 1013 1688 J 3 3       For More Material Join: @JEEAdvanced_2024 AITS-FT-X (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 8     V 2 1 2 1 3 U C T T R T T 1013 2        T2 = 218.8 K