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Question
One mole of an ideal gas is expanded isothermally and reversibly from 10 L to 15 L at 300 K. Calculate the work done in the process.
Solution
Wmax = – 2.303 × nRT log10 `v_2/v_1`
Wmax = –2.303 × 1 × 8.314 × 300 × log10 `(15/10)`
= – 2.303 × 1 × 8.314 × 300 × log10 1.5
= – 2.303 × 1 × 8.314 × 300 × 0.176
= – 1010.96 J
Wmax = – 1.011 kJ
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