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Entropy change, simple question

  1. Apr 21, 2008 #1

    lesodk

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    How do i calculate the entropy change for an isobaric process, if i know
    T1, T2, n and Cv ?

    Thanks,
     
    lesodk, Apr 21, 2008
  2. jcsd
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  3. Apr 21, 2008 #2

    astrorob

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    h ttp://en.wikipedia.org/wiki/Isobaric_process

    Try looking there.
     
    astrorob, Apr 21, 2008
  4. Apr 21, 2008 #3

    lesodk

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    There is nothing there about entropy change
     
    lesodk, Apr 21, 2008
  5. Apr 21, 2008 #4

    Hootenanny

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    What is the definition of entropy?
     
    Hootenanny, Apr 21, 2008
  6. Apr 21, 2008 #5

    lesodk

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    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.
     
    lesodk, Apr 21, 2008
  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?
     
    Hootenanny, Apr 21, 2008
  8. Apr 21, 2008 #7

    lesodk

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    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?
     
    lesodk, Apr 21, 2008
  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.
     
    Hootenanny, Apr 21, 2008
  10. Apr 21, 2008 #9

    lesodk

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    thanks alot
     
    lesodk, Apr 21, 2008
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