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Calculate the work done by a gas when its 0.5 mole at $27 ° C$ expands from $5 L$ to $25 L$ reversibly and isothermally.

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$- 2 K J$

$- 0.2 K J$

$- 200 K J$

$- 20 K J$

answer is A.

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Detailed Solution

$w = - 2.303 n R T \log \frac{V_{f}}{V_{i}} = - 2.303 \times 0.5 \times 298 \times \log \frac{25}{5} = - 1997 J = - 2 k J$

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