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One mole of an ideal gas at 300 K 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 (Ssurr) in JK1is (1 L atm = 101.3 J) Why cant we use ∆S= nR ln(v2/v1) in this. [IIT2016]

One mole of an ideal gas at 300 K 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 (Ssurr) in JK1is (1 L atm = 101.3 J) Why cant we use ∆S= nR ln(v2/v1) in this. [IIT2016]

Grade:12th pass

1 Answers

Vikas TU
14149 Points
3 years ago
From first law of thermodynamics, we realize that change in inner vitality is the contrast between the work done and warm discharged. 
Qsys = U – w 
= 3.0 atm × (2.0 L – 1.0 L) = 3.0 L-atm 
 Ssurr =Qsys/t = - Qsys/t(as vitality is discharged) 
=-3.0 * 101.3J/300k 
= – 1.013 J/K 
 
And here  I hve used for delta that is (change in).

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