A sample of an ideal gas is taken through the cyclic process A B C A as shown in figure. It absorbs, 40 J of heat during the part A B , no heat during B C and rejects 60 J of heat during C A . A work of 50 J is done on the gas during the part B C . The internal energy of the gas at A is 1560 J . The work done by the gas during the part C A is

Step-by-step solution

In the part A → B , the volume remains constant. Thus, the work done by the gas is zero. It is given that the heat absorbed by the gas is 40   J . From first law of thermodynamics, Q = Δ U + W ⇒ 40 = Δ U + 0 ⇒ Δ U = 40   J or Δ U = U B - U A = 40   J The increase in internal energy from A to B is 40   J . The internal energy is 1560   J at A , then U B - 1560 = 40   J ⇒ U B = 1600   J In the part B → C , the work done by the gas is