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## Q. 2.12

A non – flow reversible process occurs for which $p = 3V²+\frac{1}{V}$, where p is in bar and V in m³. What will be the work done when V changes from 0.5 m³ to 1 m³?

## Verified Solution

Work done,

\begin{aligned}W_{1-2} &=\int p \cdot d V=\int_{0.5}^1\left(3 V^2+\frac{1}{V}\right) d V \\&=\left|V^3+\ln V\right|_{0.5}^1=69.3 \ kJ\end{aligned}