A sample of an ideal gas is taken through the cyclic process ABCA as shown in figure below. It absorbs 60J of heat during the part AB and rejects 80 J of heat during CA . There is no heat exchanged during the process BC . A work of 40 J is done on the gas during the part BC . If the internal energy of the gas at A is 1450 J , then the work done by the gas during the part CA is:

Step-by-step solution

The process is a cycle ABCA. We are given the heat exchange Q and work done W for each part of the cycle. For process AB: Q_ AB = 60 J. Since W_ AB = 0 (as it is a vertical line on the P-V diagram, volume is constant), the change in internal energy is U_ AB = Q_ AB - W_ AB = 60 - 0 = 60 J. Thus, U_B - U_A = 60 J. For process BC: Q_ BC = 0 J. Work is done on the gas, so W_ BC = -40 J. The change in internal energy is U_ BC = Q_ BC - W_ BC = 0 - (-40) = 40 J. Thus, U_C - U_B = 40 J. For process CA: Q_ CA = -80 J (hea