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 JK1is (1 L atm = 101.3 J) Why cant we use ∆S= nR ln(v2/v1) in this. [IIT2016]
qwerty , 7 Years ago
Grade 12th pass
1 Answers
Vikas TU
Last Activity: 7 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).
Provide a better Answer & Earn Cool Goodies
Enter text here...
LIVE ONLINE CLASSES
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Ask a Doubt
Get your questions answered by the expert for free