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Q. A system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is
Question

NTA AbhyasNTA Abhyas 2022

Solution:

In the cyclic process
$\Delta Q=$ Work done $=$ Area inside the closed curve. Treat the circle as an ellipse of area
$\Delta Q=\frac{\pi }{4}\left(P_{2} - P_{1}\right) \, \left(V_{2} - V_{1}\right)$
$\Rightarrow \, \, \, \Delta Q=\frac{\pi }{4}\left(150 - 50\right)\times \left(10\right)^{3}.\left(40 - 20\right)\times \left(10\right)^{- 6}$
$\Rightarrow \, \, \, \Delta Q=\frac{\pi }{2} \, J$