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First Law of Thermodynamics

Work done by ...

Question

Work done by 0.1 mole of a gas at $27^{\circ} C$ to double its volume at constant pressure is ( $R = 2 c a l m o l^{- 1} C^{- 1}$ )

54 cal

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600 cal

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60 cal

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546 cal

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Solution

The correct option is C
60 cal

$W = P △ V = n R △ T = 0.1 \times 2 \times 300 = 60 c a l$

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