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Entropy Equatio for a closed system

  1. Apr 24, 2008 #1
    I must have already been banned for spamming threads.

    But oh well.

    You know how change in entropy [tex]dS[/tex] of a closed system assuming reversibility of the processes = [tex](\frac{dQ}{T})_{rev}=\frac{C_{p}dT}{T}[/tex]

    So when you try to find the actual entropy with respect to temperature, it's:

    [tex]\displaystyle \int S_{T} = S_{0} + \int^{T}_{0} \C_{p}dT/T\[/tex]

    That's understandable. But that's my derivation. The text uses a slightly different formula:

    [tex]S_{T} = S_{0} + \int^{T}_{0} \\frac{C_{p}dT}{T}\ + \sum \frac{\Delta H_{trans}}{T}[/tex]

    So my question 1 is: where did the [tex]\sum \frac{\Delta H_{trans}}{T}[/tex] come from?

    And my question 2 is:
    if we say that heat capacity [tex]C_{p}[/tex] is constant (it's not strictly constant, but with an approximation, we can call it a constant), we have:

    [tex]\int \frac{C_{P}dT}{T} \sim C_{P} \int \frac{dT}{T}=[/tex]
    [tex]\sim C_{P} \ln{T}[/tex]

    Could the top eqn. be used as an approximation (part of an approximation, this is only a part of the equation)

    As you can tell I am not that well-versed in Latex D=


  2. jcsd
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