The heat absorbed by a system in going through the given cyclic process is :
The heat absorbed by a system in going through the given cyclic process is :

- $19.6 \mathrm{~J}$
- $61.6 \mathrm{~J}$
- $616 \mathrm{~J}$
- $431.2 \mathrm{~J}$
Solution
$\begin{aligned} & \Delta \mathrm{U}=0(\text { Cyclic process }) \\ & \Delta \mathrm{Q}=\mathrm{W}=\text { area of } \mathrm{P}-\mathrm{V} \text { curve. } \\ & =\pi \times\left(140 \times 10^3 \mathrm{~Pa}\right) \times\left(140 \times 10^{-6} \mathrm{~m}^3\right) \\ & \Delta \mathrm{Q}=61.6 \mathrm{~J}\end{aligned}$
Asked in: JEE Main 2024 (05 Apr Shift 1)
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