Question

A thermodynamic system is taken through the cycle $ABCD$ as shown in the figure. Heat rejected by the gas during the cycle is

• A. $2\text{PV}$
• B. $4\text{PV}$
• C. $\frac{1}{2}\text{PV}$
• D. $\text{PV}$

Answer 1

Answer: (A)

In a cyclic process $\Delta \text{U}=0$ . In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.


$\therefore \Delta \text{W}=-\text{Area of rectangle ABCD}=-\text{P}\left(2\text{V}\right)=-2\text{PV}$
According to the first law of thermodynamics
$\Delta \text{Q}=\Delta \text{U}+\Delta \text{W}$ or $\Delta \text{Q}=\Delta \text{W }\left(\text{As }\Delta \text{U}=0\right)$
i.e., heat supplied to the system is equal to the work done
So heat absorbed, $\Delta \text{Q}=\Delta \text{W}=-2\text{PV}$
$\therefore$ Heat rejected by the gas = $2PV$