An ideal gas is taken through cycle A→B→C−A, as shown in figure. If the net heat supplied to the gas inthe cycle is 5J, the work done by the gas in the process C→A is
-5 J
It is a cyclic process (Δv=0) from 1st law of Thermodynamics,
ΔQ=ΔU+ΔW⇒ΔQ=ΔW5=wAB+wBC+wCA5=10(2−1)+0+wCA⇒wCA=−5J