See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

Process BC : C B B B B C P P P 1 P 2bar T T 500 250      and   1 BC V m C B U n.C T T    =   2 1.5R 250 500    = –750 R Process CD : U w     1 V m D C n.C T T     ext D C P V V   or,   D C n 1.5R T T    C D D D C nRT nRT P P P         D T 450K   and   1 CD P m D C H n.C . T T      2