A sample of an ideal gas is expanded 1m³ to 3m³ in a reversible process for which $P = {KV}^{2},$ With K=6 bar/M⁶ then the magnitude of work done (kJ) by the gas is :

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answer is 5200.

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Detailed Solution

$\begin{array}{l} \int dw = - \int P \cdot dv \Rightarrow w_{rev} = - \int 6 \cdot v^{2} dv \\ = - 6 [\frac{v_{2}^{3}}{3} - \frac{v_{1}^{3}}{3}] bar \cdot M^{3} \\ w = - 5200 KJ \end{array}$
So work done by the gas is 5200KJ.

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