1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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.

  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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook