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## Q. 2.13

A fluid is contained in a cylinder behind a piston of specific volume 0.2 m³/ kg at 5 bar pressure. The fluid is compressed reversibly to a pressure of 0.5 bar according to the law : $p v^{1.5}$ = C, where C is a constant. Calculate the work done by the fluid on the piston.

## Verified Solution

$v_2=v_1\left(\frac{p_1}{p_2}\right)^{\frac{1}{1.5}}=0.2\left(\frac{5}{0.5}\right)^{\frac{1}{1.5}}=0.92832 \ m ^3 / kg$

Work done,

\begin{aligned}w_{1-2} &=\int\limits_{v_1}^{v_2} p . d v=C \int\limits_{0.2}^{0.92832} \frac{d v}{v^{1.5}}=C\left[\frac{-v^{-0.5}}{0.5}\right]_{0.2}^{0.92832} \\ \\&=2 C\left[0.2^{-0.5}-0.92832^{-0.5}\right]=2.39635 C\end{aligned}

Now

\begin{aligned}C &=p_1 v_1^{1.5}=5 \times 10^5 \times 0.2^{1.5}=44.72 \times 10^3 \\ \\w_{1-2} &=2.39635 \times 44.72 \times 10^3=107.168 \ kJ / kg\end{aligned}