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Homework Help: Entropy, reversible process

  1. Aug 17, 2010 #1
    1. The problem statement, all variables and given/known data
    An ideal gas is taken from an initial temperature Ti to a higher final temperature Tf along two different reversible paths: Path A is at constant pressure; Path B is at constant volume. The relation between the entropy changes of the gas for these paths is
    a) delta S(A) > delta S(B)
    b) delta S(A) = delta S(B)
    c) delta S(A) < delta S(B)

    2. Relevant equations

    delta S = delta Qr / T
    Qr = heat transferred to system while the system is going along a reversible path

    3. The attempt at a solution
    This is one of those checkpoint questions in the chapter and the answer is given as choice a (delta S(A) > delta S(B)).

    I'm confused though because in this book, it says that entropy is a state variable and as such, it only depends on the endpoints and is therefore independent of the actual path taken from A to B. But here, we're taking two different paths and yet we're getting that the change in entropy going from one path is different than when we take the other path.

    I think the answer should be choice b (delta S(A) = delta S(B)).

    It would seem that if you're only dependent on the endpoints, then regardless of the path taken, if you're going from A to B in multiple ways, that the entropy should be the same for all cases.

    Where am I going wrong in my thought process? Thanks a lot ahead of time.
  2. jcsd
  3. Aug 17, 2010 #2


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    Remind me -- does state include pressure and volume too? Or does the state involve temperature alone?
  4. Aug 17, 2010 #3


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    The final states are different. Though they end at the same temperature, the two paths end at different pressures and volumes.
  5. Aug 17, 2010 #4
    Ah, I think I see. So if we a system taking two paths to some final state, then that final state is the same for that system only if the pressure, volume, and temperature are all the same?

    And if this is true, then the entropy should be the same as well?

    Thanks a lot for your quick response.
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