TS EAMCET 20-July-2022 Shift 2 Solved Paper

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 ‌=−‌

1

2

pV