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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 JK−1 is (1L atm=101.3 J)
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 JK−1 is (1L atm=101.3 J)
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Solution
The correct option is C -1.013
Since it is an isothermal process, ΔU=0
dq=−dW=Pext(V2−V1)=3L−atm=3×101.3 J
ΔSsurrounding−3×101.3300 J K−1=−1.013 J K−1
∴ΔSsurr=−1.013 J K−1
Since it is an isothermal process, ΔU=0
dq=−dW=Pext(V2−V1)=3L−atm=3×101.3 J
ΔSsurrounding−3×101.3300 J K−1=−1.013 J K−1
∴ΔSsurr=−1.013 J K−1
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