One mole of an ideal gas at 300 K in thermal contact with surroundings expands isothermally from 1.0L to 2.0L against a constant pressure of 3.0 atm. In this process, the change in entropy of surroundings (ΔS)inT−1 is (1 L atm = 101.3J)
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 By first law,ΔE=Q+W For isothermal expansion, ΔE=0 ∴Q=−W −Qirrev=Wirrev=pΔV=3(2−1)=3Latm Also,ΔSsurr=QirrevT=(−3×101.3)J300K=−303.9300=−1.013JK−1