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Question

One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0 L to 2.0 L against a constant pressure of 3.0 atm. In this process, the change in entropy of surrounding (ΔSmax) in JK1 is (1L atm=101.3 J)

A
5.763
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B
1.013
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C
-1.013
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D
-5.763
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Solution

The correct option is C -1.013
Since it is an isothermal process, ΔU=0

dq=dW=Pext(V2V1)=3Latm=3×101.3 J

ΔSsurrounding3×101.3300 J K1=1.013 J K1

ΔSsurr=1.013 J K1

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