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Heat capacities definition

  1. Jan 27, 2012 #1
    Hi all,

    I'm working with the heat capacities definition and I have got a confusion. I don't understand why we can express them like

    Cp = T(∂S/∂T)p
    Cv = T(∂S/∂T)v

    I know that Cp=(dQ/dT)p = (∂H/∂T)p with H equal to TdS + VdP and Cv=(dQ/dT)v = (∂U/∂T)v with U equal to TdS + PdV,

    My guess: If I begin for instance with the enthalpy, H, and I constrain it to constant pressure I get just the TdS term, that is the thermodynamic definition of "Heat (Q)" . So I get the definition of the heat capacity if I derive Q respect to T, or TdS respect to T (is the same thing). Doing it, I get:

    dQ/dT = (TdS)/dT = Td^2S/d^2T + dS/dT.

    An expression very different of the definition.

    Can anyone help me?

    Thanks in advance.
     
  2. jcsd
  3. Jan 27, 2012 #2

    Andrew Mason

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    You cannot equate TdS to dQ unless dQ = dU + PdV, and that is true only if the process is reversible. Generally, you have to use the first law: dQ = dU + ∂W.

    I think you mean dU = TdS - PdV. Again, that is only true if it is a reversible process.

    I don't follow what you are doing. Why would it not just be T(dS/dT)? Again, this is true only in a reversible process.

    AM
     
  4. Jan 27, 2012 #3
    ( ∂Q/∂T )p = ( T∂S/∂T )p = T (∂S/∂T)p

    Cp = T (∂S/∂T)p = (∂Q/∂T)p
    Cv = T (∂S/∂T)v = (∂Q/∂T)v
     
  5. Jan 27, 2012 #4
    The problem you have encountered in the italicized step is mathematical.
    (TdS)/dT is T(dS/dT).But in the mentioned step,you have computed the derivative of (T(dS/dT)).
    Read up on differentials.
     
  6. Jan 28, 2012 #5
    Thanks everybody,

    pabloenigma you are right, your advice was very useful!!!
     
  7. Jan 28, 2012 #6
    Thank you
     
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