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

Solution

Physics sir · 3 steps

Go round the full cycle: net ΔU is zero, so net work equals net heat, and only W on CA is unknown.

1. Cycle rule. The gas returns to A, so Δ U_(cycle) = 0. The first law Q = Δ U + W then gives Wₙₑₜ = Qₙₑₜ. Take heat absorbed and work done by the gas as positive.
2. Known pieces. Qₙₑₜ = 40 + 0 - 60 = -20 J. AB is vertical on the graph, so the volume is constant and W_(AB) = 0. On BC, 50 J of work is done on the gas, so W_(BC) = -50 J.

   W_(AB) + W_(BC) + W_(CA) = Qₙₑₜ ⇒ 0 - 50 + W_(CA) = -20

   Process	Q (J)	W by gas (J)	Δ U (J)
   AB (isochoric)	+40	0	+40
   BC	0	-50	+50
   CA (isobaric expansion)	-60	+30	-90
   Whole cycle	-20	-20	0

3. So. W_(CA) = -20 + 50 = **30 J**. It is positive, which fits the graph: on CA the volume increases, so the gas does work. The given U_A = 1560 J is not needed. The answer is (b).

The idea to keep. For any closed cycle, Δ U = 0, so total heat equals total work. Fix one sign convention first. Then the single unknown work term drops out of one equation.

Answer from the NTA final answer key. Solution by PrepWiser.