The heat absorbed by a system in going through the given cyclic process is:
JEE MAIN 2024, Diffcult
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$\Delta \mathrm{U}=0 \text { (Cyclic process) }$
$\Delta \mathrm{Q}=\mathrm{W}=\text { area of } \mathrm{P}-\mathrm{V} \text { curve. }$
$=\pi \times\left(140 \times 10^3 \mathrm{~Pa}\right) \times\left(140 \times 10^{-6} \mathrm{~m}^3\right)$
$\Delta \mathrm{Q}=61.6 \mathrm{~J}$
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