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Entropy expression

  1. May 25, 2014 #1

    ChrisVer

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    Gold Member

    1. The problem statement, all variables and given/known data

    From the 2nd TD law:
    [itex] TdS= d(\rho V) + P dV - \mu d(nV) [/itex]

    find that:
    [itex] S= \frac{V}{T} (\rho+P- \mu n)[/itex]


    2. Relevant equations

    [itex] \frac{dP}{dT}= \frac{P+\rho - \mu n}{T}[/itex]


    3. The attempt at a solution
    [itex] TdS= d(\rho V) + P dV - \mu d(nV) [/itex]

    [itex] TdS= d[(\rho+ P- \mu n) V] - V dP + nV d \mu [/itex]

    or I can write:

    [itex] dS=\frac{1}{T} d[(\rho+ P- \mu n) V] - \frac{V}{T} dP + \frac{nV}{T} d \mu [/itex]

    Now I write that (using the given formula):
    [itex] dP= \frac{P+\rho - \mu n}{T} dT [/itex]

    [itex] dS=\frac{1}{T} d[(\rho+ P- \mu n) V] - V (P+\rho - \mu n) \frac{dT}{T^{2}} + \frac{nV}{T} d \mu [/itex]

    My problem is that this result gives the correct formula I'm looking for the entropy, except for the last [itex] \frac{nV}{T} d \mu [/itex]

    For the last I tried to take:

    [itex] \frac{dS}{dT}= \frac{dS}{d \mu} \frac{d \mu}{dT}= \frac{nV}{T} \frac{d \mu}{dT} [/itex]

    [itex] \frac{d^{2}S}{d(nV)dT}= \frac{1}{T} \frac{d \mu}{dT} [/itex]

    Also:
    [itex]\frac{dS}{d(nV)}= -\frac{ \mu }{T}[/itex]

    [itex]\frac{d^{2}S}{dT d(nV)}= \frac{ \mu }{T^{2}}[/itex]

    So that I have:

    [itex] \frac{1}{T} \frac{d \mu}{dT} =\frac{ \mu }{T^{2}}[/itex]
    That means:
    [itex] d \mu = \frac{ \mu }{T} dT [/itex]

    Inserting in the expression for the entropy at last:

    [itex] dS=\frac{1}{T} d[(\rho+ P- \mu n) V] - V (P+\rho - \mu n) \frac{dT}{T^{2}} + (nV \mu) \frac{dT}{T^{2}} [/itex]

    Which can't be written as:

    [itex] dS= d[(\rho+ P- \mu n) \frac{V}{T}] [/itex]
    due to the last term...

    Any help?
     
  2. jcsd
  3. May 25, 2014 #2
    Start with:

    [tex]dU = TdS - pdV + \mu dN[/tex]
     
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