One mole of an ideal gas undergoes an isothermal change at temperature 'T' so that its volume ‘V’ is doubled. R is the molar gas constant. Work done by the gas during this change is

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RT log 4

RT log 2

RT log 1

RT log 3

answer is B.

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Detailed Solution

W=nRTlog(v2v1) =RTlog2.

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