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How is internal energy U(S,V) a function of temperature?

  1. Jan 28, 2016 #1
    In my chemical thermodynamics class/notes (and other references I've used) it is stated throughout that internal energy U is a function of entropy and volume , i.e. it's "natural" variables are S and V:
    • U = U(S,V)
    I suspect that I must take this "axiomatically" and move on.

    Since U is a state function, it's exact differential will be:
    • dU = (dU/dS) dS + (dU/dV) dV
    Okay, fine until here.

    Then we are introduced to the "fundamental thermodynamic equation" :
    • dU = TdS - pdV
    (BTW I'm not too sure how this comes about, but that's another topic.)

    Comparing these two forms for dU, we readily see that:
    • (dU/dS) = T
    • (dU/dV) = - p
    So as I understand the math above, temperature (as well as pressure) are the partial derivatives of internal energy.

    But then later on in the course I find numerous expressions like (dU/dT) and others where it is implied that internal energy is a function of temperature (and pressure). What am I to make of these expressions? Sure, I understand that temperature actually DOES affect the internal energy of a system, but U has been defined in a way that "excludes" it. I mean there must be a reason that they don't state the dependence explicitly (i.e. U = U(S,V,T,p) ) but I just don't understand it.
    Maybe it has to do with the fact that (dU/dS) = T but then what am I to understand about the form of U from (dU/dT) ? It must mean that whatever the function for U is, it contains the expression for T, so it is a differential equation. For example: If we have
    • U(S,V) = exp(S) + exp(V)
    then
    • T = (dU/dS) = exp(s)
    • p = - (dU/dV) = - exp(V)
    and then all of the following are valid:
    • U(S,V) = exp(S) + exp(V) = T - p = U(T,p)
    • U(S,V) = exp(S) + exp(V) = T + exp(V) = U(T,V)
    • U(S,V) = exp(S) + exp(V) = exp(S) - p = U(S,p)
    So then (dU/dT) & (dU/dp) actually have a meaning:
    • (dU/dT) = exp(V) ( = (dU/dS) )
    • (dU/dp) = exp(S) ( = (dU/dV) )
    But obviously these partial derivatives can be readily expressed by the partial derivatives of U with respect to the natural variables S,V , so what was the meaning of the change of variables in the first place???
    What's going on??? It's things like these that make me hate thermodynamics with a PASSION...
     
    Last edited: Jan 28, 2016
  2. jcsd
  3. Jan 28, 2016 #2

    dU/dT comes about from dividing the dU expression by dT

    So dU = TdS - pdV becomes
    dU/dT = T dS/dT - p dV/dT

    So then the derivatives you worry about are dS/dT and dV/dT

    You don't have to really consider what U is explicitly.
     
  4. Jan 28, 2016 #3
    I agree that most thermodynamics texts mess up this point. The clearest text on this particular point is :
    https://www.amazon.com/Thermodynamics-Introduction-Thermostatistics-Herbert-Callen/dp/0471862568

    Chapter 1-3 cover your question very clearly.

    In summary U=U(S,V) is the fundamental equation of a simple system that is made of one type of matter. It contains all thermodynamic information. Since we are typically interested in changes in U rather than the absolute value of U, we need to compute its total differential. Thus,

    dU= (∂U/∂S)dS + (∂U/∂V)dV

    Let's "define"
    T≡ (∂U/∂S) , P = - (∂U/∂V)
    Then it turns out that this definition matches our intuition about temperature and pressure.

    Then as you showed in your post, you can always eliminate S, V and replace them with T, P and obtain the equation:
    U=U(T,P).
    However, the latter is NOT the fundamental equation and it contains LESS information compared to U=U(S,V) . The reason is that it is a partial differential equation , that is
    U=U(∂U/∂S, ∂U/∂V ).
    A solution of a partial differential equation is alway missing one or two undetermined integration constants. A simple example:

    Suppose U = aT where a is a positive constant (Suppose P is constant for simplicity). Then U=a(dU/dS) , and by integration:

    U= B exp(S/a) where B is an undetermined integration constant.

    In summary U(S,V) has more information compared to U(T,P).

    BUT experimentalists like to deal with T, P not S,V. Simply because T,P are easy to measure and control. Whereas who can measure and control S ?!!! This is the reason why there is interest in thermodynamics to reformulate it around T,P, ... (the intensive variables) without losing the valuable information content in U(S,V,...). The way to do this is called Legendre Transformation which leads to introducing Gibbs , Helmholtz free energies and the enthalpy.
     
  5. Jan 28, 2016 #4
    Wow thanks for the answers! :)
    Both have given me some food for thought.
    I'll try looking into your suggestions and if I'm still having a hard time, I'll ask again!
     
  6. Jan 29, 2016 #5

    vanhees71

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    Thermodynamics can derived from statistical physics. Then all these relations between the various thermodynamic potentials come out very naturally. From the point of view of phenomenological thermodynamics you just have to take the fundamental laws as given. The internal energy, ##U## obeys under changes of the thermodynamic state
    $$\mathrm{d} U=T \mathrm{d} S-p \mathrm{d} V,$$
    which is the 0th-2nd law combined. It's just energy conservation, i.e., the change of the internal energy of the thermodynamic system is due to the change in heat ##T \mathrm{d} S## and mechanical work due to the pressure, ##-p \mathrm{d} V##.

    From this relation you read off
    $$\left (\frac{\partial U}{\partial S} \right )_{V}=T, \quad \left (\frac{\partial U}{\partial V} \right )_{S}=-p,$$
    where the subscript tells you which variable is kept constant in taking the partial derivative. This is very important to remember when switching to another thermal potential, which you always get from ##U## by Legendre transformations.

    E.g., you can introduce the free energy via
    $$F=U-T S.$$
    Then you have
    $$\mathrm{d} F=\mathrm{d} U - T \mathrm{d} S-S \mathrm{d} T=-S \mathrm{d} T - p \mathrm{d} V,$$
    which tells you that the "natural independent variables" for ##F## are ##T## and ##V##, and you have the relations
    $$\left (\frac {\partial F}{\partial T} \right )_V=-S, \quad \left (\frac{\partial F}{\partial V} \right )_{T} = -p.$$
    Via several Legendre transformations you can make any pair of thermodynamical quantities the "natural independent variables".

    In other words, despite the physics behind the 3 (or 4) fundamental laws, all there is to phenomenological thermodynamics are Legendre transformations!
     
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