One mole of an ideal gas at 300   K in thermal contact with its surroundings expands isothermally from…

One mole of an ideal gas at 300 K in thermal contact with its 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 the surroundings ∆Ssurr in J K-1 is:
1 L atm = 101.3 J
  1. 5.763
  2. 1.013
  3. -1.013
  4. -5.763

Solution

From 1st law of thermodynamics,
qsys=∆U-w=0--Pext.∆V
=3.0 atm×2.0 L-1.0 L=3.0 L atm
∴∆Ssurr=qrevsurrT=-qsysT
=-3.0×101.3 J300 K
=-1.013 J/K
Alternate solution:
∆Ssurr=qsurrT=-qsysT=WsysT
∵ For the isothermal process:
∆U=0     qsys=-Wsys
∆Ssurr.=-PextVf-ViT
=-3(2-1)300×101.3
=-1.013  J/K !

Asked in: JEE Advanced 2016 (Paper 1)

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