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JEE Advanced · Chemistry · 6. Thermodynamics (C)

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. A 5.763
  2. B 1.013
  3. C -1.013
  4. D -5.763
Verified Solution

Answer & Solution

Correct Answer

(C) -1.013

Step-by-step Solution

Detailed explanation

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
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