Thermodynamics Ques 16
16. A thermodynamic system is taken through the cycle $A B C D$ as shown in figure. Heat rejected by the gas during the cyclic process is :[2012]
(a) $2 PV$
(b) $4 PV$
(c) $\frac{1}{2} PV$
(d) $PV$
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Answer:
Correct Answer: 16.(a)
Solution:
- (a) $\because$ Internal energy is the state function. $\therefore$ In cyclie process; $\Delta U=0$
According to 1st law of thermodynamics
$\Delta Q=\Delta U+W$
So heat absorbed
$\Delta Q=W=$ Area under the curve
$=-(2 V)(P)=-2 PV$
So heat rejected $=2 PV$
BETA
