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Nilanjan Bhowmick AIR 3, CSIR NET (Earth Science)
Sagar Kumar Biswal posted an Question
December 08, 2020 • 14:37 pm
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30 points
30 points
- IIT JAM
- Physics (PH)
Consider free expansion of one mole of an ideal gas in an adiabatic container from volume v1 to v2. the entropy change of the gas, calculated by considering a r
Consider free expansion of one mole of an ideal gas in an adiabatic container from volume V1 to V2. The entropy change of the gas, calculated by considering a reversible process between the original state (V, T) to the final stale (V2, T) where Tis the temperature of the system, is denoted by AS1. The corresponding change in the entropy of the surrounding is AS2. Which of the following combinations is correct? (A) AS= RIn (V/V2), AS2= - Rn (V1/V2) - Rln (V,/V2), AS>= Rln (V,/V2) (B) AS = (C) AS = Rln (V2/V), AS2 = 0 (D) AS = -RIn (V2/V), AS2-
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Lekhika![best-answer]()
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but mam in a reversible process the total entropy change of the universe should be zero. So I think option B is correct. Although there is no heat exchange between system and surrounding but system is doing some work on surrounding by it's own expansion.
hi sagar, thanks for your comment. but it is a quasi static process.
so then ???
Actually the point you are saying is correct, but here temperature is also constant. So the question itself using two different. I gave answer by taking T=const. also heat cant be exchanged between the system and surrounding so no entropy change should occur in surrounding. That is why i assumed entropy of surrounding to be zero