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Q.

A sample of an ideal gas is expanded $1 m^{3}$ to $3 m^{3}$ in a reversible process for which $P = K V^{2}$ , with K = 6 $b a r / m^{6}$ . What is work done by the gas (in kJ)?

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answer is 5200.

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Detailed Solution

$W = - \int P d V = - \int_{1}^{3} 6 V^{2} d V = 2 \times [27 - 1]$
$= 52 b a r - m^{3}$ or 5200 kJ

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