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GRE question on entropy

  1. Apr 2, 2017 #1
    1. The problem statement, all variables and given/known data
    It is problem 21 in the attached file.

    2. Relevant equations


    3. The attempt at a solution
    The answer seems to be C. I thought it is D. Can someone explain it to me please?
     

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  3. Apr 2, 2017 #2

    BvU

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    Key word in the problem statement is 'reversible'. For reversible processes entropy is conserved.
     
  4. Apr 2, 2017 #3
    So for a reversible process, the entropy is conserved for both the system and the surroundings (but not for each of them individually)?
     
  5. Apr 2, 2017 #4

    BvU

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    Correct: there is an exchange of energy between the two.
     
  6. Apr 2, 2017 #5
    Thank you so much!
     
  7. Apr 2, 2017 #6

    BvU

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    Is it really clear to you that (A) and (B) are not true ?
     
  8. Apr 2, 2017 #7
    For the gas it is clear that the entropy is constant only during the adiabatic transformation. However I am not sure how can I calculate the entropy of the surroundings
     
  9. Apr 2, 2017 #8

    BvU

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    Nothing happens to the surroundings (if all that occurs is this adiabatic expansion). There is no ##\delta Q##, so no ## \delta S##.
     
  10. Apr 2, 2017 #9
    No, sorry. I meant in the other 2 cases the isothermal and constant volume. How can I know that the change in entropy of the system exactly compensate the one of the surrounding? It is just by the definition of reversible process or there is something more to it?
     
  11. Apr 2, 2017 #10

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    Not much more to it. Reversibility means you can also go through the process in the other direction - in which case ##\delta S## comes out the same value but with opposite sign. That can only be correct if it is zero.
    More through energy conservation plus ##dQ = TdS##.
     
  12. Apr 2, 2017 #11

    vela

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    For a reversible process, when heat is transferred, the temperature difference between the gas and its surroundings is infinitesimal. There has to be some difference in temperature, otherwise heat won't flow, but since the two are essentially at the same temperature, the decrease in entropy of one is equal to the increase of the other.
     
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