In cyclic process, ∆U=0 So, by first law, ‌∆Q=∆U+∆W ‌∆Q=∆W As cycle given is clockwise so, work done is positive. Hence, ∆Q is positive that means heat is absorbed. Now, ∆Q=∆W= Area of cycle =‌
1
2
×(2V−V)×(2p−p)=‌
1
2
pV Hence, heat released =−‌ Heat absorbed ‌=−‌