Thermodynamics - Result Question 20

21. 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: 21. (a)

Solution:

  1. (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 $=2 PV$



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