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Question

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) in T1 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(21)=3L atm
Also, ΔSsurr=QirrevT=(3×101.3)J300K=303.9300=1.013 JK1

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