See image — AITS & Test Series Chemistry Question

💡 Solution & Explanation

u = Q + W = 0 +   ) V – V ( P – dV P – 1 2 ext . ext  CV (T2 – T1) = – Pext.       1 1 2 2 P RT – P RT ; Here Pext = P2  R 2 5 (673–373) = – 673R +       1013 .0 R 373 P2  P2 = 0.3865 M Pa