Chapter 2
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?
Step-by-Step
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}