One mole of an ideal gas at 300K 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 surroundings ΔSsurr in JK−1 is(1L atm = 101.3 J)

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5.763

1.013

-1.013

-5.763

answer is C.

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Detailed Solution

Isothermal irreversible reactionw=−Pv2−v1=−3(2−1)=−3L. atm ΔSsyst =QT and ΔSsurr =−ΔSsysQ=−w=−(−3) lit . atm ΔSsurr =−3×101.3300=−1.013JK−1

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