Q.
A thermodynamic system is taken through the cycle $ABCD$ as shown in figure. Heat rejected by the gas during the cycle is
A) $2\ PV$
B) $4\ PV$
C) $\frac{1}{2}\ PV$
D) $PV$
Explanation
(a) In cyclic process $\mathrm{\Delta}U = 0$
So heat absorbed
$\mathrm{\Delta}Q = W =$ Area under the curve
$= - (2V)(P) = - 2PV$
So heat rejected $= 2PV$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.