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Question

One mole of an ideal gas at 300 K in thermal contact with surrounding 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 surroundings (ΔSsurr) in JK1 is:
(Given:1 Latm=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=ΔUw
As the process is isothermal,so
qsys=0[Pext.ΔV]
=3.0 atm×(2.0L1.0 L)=3.0 Latm=3.0×101.3J.

Putting the relevent values in formula of entropy,
ΔSsurr=qsurrT=qsysT

=3.0×101.3J300K

=1.013 J/K

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