One mole of an ideal gas at 300K in thermal contact with surrounding expands isothermally from 1.0Lto2.0L against a constant pressure of 3.0atm. In this process, the change in entropy of surroundings (ΔSsurr) in JK−1 is: (Given:1Latm=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 We know,
From 1st law of thermodynamics qsys=ΔU−w
As the process is isothermal,so qsys=0−[−Pext.ΔV] =3.0atm×(2.0L−1.0L)=3.0Latm=3.0×101.3J.
Putting the relevent values in formula of entropy, ∴ΔSsurr=qsurrT=−qsysT