A system is taken through a cyclic process represented by a circle as shown. The heat absorbed by the system is
Medium
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(b) In cyclic process $\Delta Q =$ Work done = Area inside the closed curve.
Treat the circle as an ellipse of area $ = \frac{\pi }{4}({P_2} - {P_1})\,({V_2} - {V_1})$
$\Rightarrow \Delta Q = \frac{\pi }{4}\{ (150 - 50) \times {10^3}\} = \frac{\pi }{2}J$
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