An ideal gas taken through a process ABCA as shown in figure. If the net heat supplied to gas in the cycle is 5 J , then the work done by the gas in process from C to A is

Step-by-step solution

For a cyclic process, the net work done equals the net heat transfer, W_ net = Q_ net , by the first law of thermodynamics since the change in internal energy over a cycle is zero. Given Q_ net = 5 J , it follows that W_ net = 5 J . Process A B is isobaric with P = 10 N/m ^2 , and the volume increases from 1 m ^3 to 2 m ^3 , so the work done is W_ AB = 10(2 - 1) = 10 J . Process B C is isochoric ( V = 0 ), so W_ BC = 0 . Process C A contributes W_ CA to the cycle, and since W_ net = W_ AB + W_ BC + W_ CA , substitu