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Entropy change, simple question

  1. Apr 21, 2008 #1
    How do i calculate the entropy change for an isobaric process, if i know
    T1, T2, n and Cv ?

    Thanks,
     
  2. jcsd
  3. Apr 21, 2008 #2
    h ttp://en.wikipedia.org/wiki/Isobaric_process

    Try looking there.
     
  4. Apr 21, 2008 #3
    There is nothing there about entropy change
     
  5. Apr 21, 2008 #4

    Hootenanny

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    What is the definition of entropy?
     
  6. Apr 21, 2008 #5
    entropy change = dS = int( dQ/T )
    where you integrate from state 1, to state 2.
    The forumula only holds for reversible processes and i don't know if thats the case in my question.
     
  7. Apr 21, 2008 #6

    Hootenanny

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    Correct, you can now use this in conjunction with the information which astrorob provided to calculate the entropy change.
    Indeed, the formula does only hold for a reversible process, but it may be useful to note that since entropy is a state function it is independent of the path taken and depends only on the initial and final states. Therefore, for any process, even an irreversible one, we can chose a reversible path between the initial and final states that allows us to calculate the change in entropy.

    Does that make sense?
     
  8. Apr 21, 2008 #7
    so i shoud insert the value for dQ = nCvdT and by that getting
    dS = int( (n Cv dT)/T) = n Cv ( (1/T) dT ) = nCv ln(T2/T1) ?
    Is that correct, even though the process is not reversible?
     
  9. Apr 21, 2008 #8

    Hootenanny

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    The only thing that concerns me is the Cv. Since the process is isobaric, you need Cp. Apart from that your expression is correct.
     
  10. Apr 21, 2008 #9
    thanks alot
     
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