An ideal gas is taken through the cycle  A→B→C→A, as shown in figure. If the net heat supplied to the gas in the cycle is 5 J, work done by the gas in the process C→A

given graph has pressure along horizontal axis, volume in meter3 along vertical axiswork done from A to B is WA→B=PdV=10[2-1]=10 J, here volume is increasedwork done from B to C is WB→C=PdV=0, here there is no change in volumework done from C to A is WC→A=?In cyclic process dU=0,( change in internal energy)  from first law of thermodynamics dQ=dU+dWhere dQ=change in heat energy=5JdU=change in inernal energydW=work done  substitute in  first law of thermodynamics  ⇒5=0+[WA→B+WB→C+WC→A] ⇒5=10+0+WC→A ⇒WC→A=5−10 ⇒WC→A=−5 J