Calculate work done when 1 mole of an ideal gas is expanded reversibly from $30 L$ to $60 L$ at a constant temperature of $300 K$ ?

8.78 k J

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- 1.73 k J

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10.73 k J

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- 9.78 k J

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Solution

The correct option is B $- 1.73 k J$
Given,
$T = 300 K$
$V_{1} = 30 L$
$V_{2} = 60 L$

$W_{i s o t h e r m a l} = - 2.303 n R T l o g (\frac{V_{2}}{V_{1}})$

Substituting, we get,
$W_{i s o t h e r m a l} = - 2.303 \times 8.314 \times 300 \times l o g (\frac{60}{30})$
$⟹ W_{i s o t h e r m a l} = - 1729.1 J = - 1.73 k J$

Hence option B is the right answer.

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