Calculate the amount of work done by $2$ mole of an ideal gas at $298 K$ in reversible expansion from $10$ litre to $20$ litre.

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Solution

Amount of work done in reversible isothermal expansion,
$w = - 2.303 n R T log \frac{V_{2}}{V_{1}}$
given, $n = 2, R = 8.314 J K^{- 1} {m o l}^{- 1}, T = 298 K, V_{2} = 20 L$ and $V_{1} = 10 L$
Substituting the values in above equation
$w = - 2.303 \times 2 \times 8.314 \times 298 \times log \frac{20}{10}$
$w = - 2.303 \times 2 \times 8.314 \times 298 \times 0.3010 = - 3434.9 J$
i.e., work done by the system

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