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 work done by the gas during the part $$CA$$ is

Use first law:

$$Q=\Delta U+W$$

(where W is work done by gas)

For process BC:

Given no heat exchange,

$$Q_{BC}=0$$

Work done on gas = 50 J

So work done by gas

$$W_{BC}=-50J$$

Thus

$$\Delta U_{BC}-50=0$$

$$ΔU_{BC}=50$$

So

$$U_C-U_B=50$$

For process AB:

$$Q_{AB}=40$$

From graph AB is vertical (constant volume), so

$$W_{AB}=0$$

Therefore

$$40=\Delta U_{AB}$$

$$U_B-U_A=40$$

Since

$$U_A=1560$$

$$U_B=1600$$

Then

$$U_C=1650$$

For process CA:

$$Q_{CA}=-60$$

Change in internal energy:

$$\Delta U_{CA}=U_A-U_C$$

$$=1560−1650=−90$$

Now

$$Q_{CA}=\Delta U_{CA}+W_{CA}\\-60=-90+W_{CA}$$

$$W_{CA}=30\text{ J}$$