\(\Delta U = 0\)

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwiase.

\(\therefore \,\,\,\Delta W =  - Area\,of\,rectangle\,ABCD =  - P\left( {2V} \right)\)

\( =  - 2PV\)

According to first law of thermodynamics

\(\Delta Q = \Delta u + \Delta W\,or\,\Delta Q = \Delta W\,\,\left( {As\,\Delta u = 0} \right)\)

\(i.e.,\) heat supplied to the system is equal to the work done

So heat absorbed,\(\Delta Q = \Delta W =  - 2PV\)

\(\therefore \) Heat rejected by the gas  \( = 2PV\)