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11.Thermodynamics
hard
The heat absorbed by a system in going through the given cyclic process is:
A
$61.6 \mathrm{~J}$
B
$431.2 \mathrm{~J}$
C
$616 \mathrm{~J}$
D
$19.6 \mathrm{~J}$
(JEE MAIN-2024)
Solution
$\Delta \mathrm{U}=0 \text { (Cyclic process) }$
$\Delta \mathrm{Q}=\mathrm{W}=\text { area of } \mathrm{P}-\mathrm{V} \text { curve. }$
$=\pi \times\left(140 \times 10^3 \mathrm{~Pa}\right) \times\left(140 \times 10^{-6} \mathrm{~m}^3\right)$
$\Delta \mathrm{Q}=61.6 \mathrm{~J}$
Standard 11
Physics
