Q158 / 170 · Manipal (MET) 2019 · Physics
A thermodynamic system is taken through the cycle PQRSP process. The net work done by the system is
- A20 J
- Correct answerB$-20 \mathrm{~J}$
- C400 J
- D$-374 \mathrm{~J}$
Solution
$\begin{aligned}
& \text { } W_{P Q}=P \Delta V \\
& =100 \times 10^3(300-100) \times 10^{-6} \\
& =100 \times 10^{-3} \times 200=20 \mathrm{~J} \\
& W_{Q R}=0(\because V \text { is constant }) \\
& W_{R S}=P \Delta V \\
& =200 \times 10^3(100-300) \times 10^{-6} \\
& =-200 \times 10^{-3} \times 200=-40 \mathrm{~J} \\
& W_{S P}=0
\end{aligned}$
Net work done $=20+(-40)=-20 \mathrm{~J}$
Net work done $=20+(-40)=-20 \mathrm{~J}$
Hence, the correct answer is option: 'B'.