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Total derivative to partial derivative by division? (Calc./Thermo.)

  1. Apr 27, 2013 #1
    I don't understand the calculus behind this thermodynamics concept:

    S = f(T,P)
    dS = (∂S/∂T)_P*dT + (∂S/∂P)_T*dP
    (∂S/∂T)_V = (∂S/∂T)_P + (∂S/∂P)_T*(∂P/∂T)_V


    Basically, I don't get why and how you get (∂S/∂T) when you divide dS by dT. Also, I don't understand why the constant volume "_V" appears.
     
  2. jcsd
  3. Apr 27, 2013 #2

    Mark44

    Staff: Mentor

    You DON'T get ∂S/∂T by dividing dS by dT.
    Your 2nd equation is the total differential of S, which could be written more simply as
    dS = ∂S/∂T * dT + ∂S/∂P * dP

    Since S is a function of T and P, it's redundant to say (∂S/∂T)_P when writing this partial. P is already treated as a constant when you take the partial of S with respect to T.


    I don't either. S is a function of P and T only, according to your first equation. I know that volume, temperature, and pressure are all related by Boyle's Law or Charles' Law (or Boyle's and Charles' Law), but it's been a very long time since I took physics.
     
  4. Apr 28, 2013 #3
    Thanks! But I was talking about getting from the second equation to the third. Also, it may be redundant but it's my engineering book's convention.
     
  5. Apr 29, 2013 #4

    Mark44

    Staff: Mentor

    You need to explain how V ties into the 2nd equation. To my recollection, the equation is PV = nRT. How that relates to your function S = f(T, P), I don't know.
     
  6. Apr 29, 2013 #5

    DrClaude

    User Avatar

    Staff: Mentor

    The premiss is that, for a given system, you know how to express the entropy as a function of ##T## and ##P##:
    $$
    S=f(T,P)
    $$
    This doesn't mean that entropy is only a function of ##T## and ##P##. It is generally also a function of ##V## and ##N##, for instance. But what is have is an expression for ##S## in terms of ##T## and ##P##.

    From that, you can get the total derivative
    $$
    dS = \left( \frac{\partial S}{\partial T} \right)_P dT + \left( \frac{\partial S}{\partial P} \right)_T dP
    $$
    Now, do something to your system at constant volume and you wish to know how entropy changes with respect to temperature. How can you do that when you don't have an expression for ##S## in terms of ##V##? Either you find a relation between ##P## and ##V## (for instance, if you have an ideal gas, where ##PV=nRT##), or you derive the previous equation with respect to ##T## at constant ##V##. Then you get
    $$
    \left( \frac{\partial S}{\partial T} \right)_V = \left( \frac{\partial S}{\partial T} \right)_P + \left( \frac{\partial S}{\partial P} \right)_T \left( \frac{\partial P}{\partial T} \right)_V
    $$
     
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