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Calculate the work done when the volume changes from 5 m³ to 10 m³ in a non – flow quasi – static process in which pressure is given by :

p = (5V – 5) bar.

Work done,

\begin{aligned}W_{1 \cdot 2} &=\int p \cdot d V=\int_5^{10}(5 V-5) \times 10^5 d V \\&=10^5\left|2.5 V^2-5 V\right|_5^{10}=16250 \ kJ\end{aligned}