A thermodynamic system is taken through the cycle $A B C D$ as shown in figure. Heat rejected by the gas…

A thermodynamic system is taken through the cycle $A B C D$ as shown in figure. Heat rejected by the gas during the cycle is
  1. $2 \mathrm{pV}$
  2. $4 p V$
  3. $\frac{1}{2} p V$
  4. $p V$

Solution

For given cyclic process, $\Delta U=0$
$\therefore Q=W$
Also, $W=-$ area enclosed by the curve
$\begin{aligned}
& =A B \times A D \\
& =(2 p-p)(3 V-V) \\
& =-p \times 2 V
\end{aligned}$
$\therefore$ Heat rejected $=2 p_V$ ^

Asked in: NEET 2012 (Screening)

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