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B Naive Doubt About Thermodynamic Equilibrium

  1. May 13, 2017 #1
    This is more of a recurring conceptual doubt that I keep on running into when solving thermodynamics problems. We are taught that variations between extensive state variables in equilibrium are given by the following 'fundamental formula':

    [tex]dE = TdS + \mathbf{J}\cdot{d}\mathbf{x} + \mathbf{\mu}\cdot{d}\mathbf{N}[/tex]

    Now, in equilibrium, we also know that the internal energy [itex]E[/itex] is minimized; thus, first order variations in [itex]E[/itex] with respect to any variable will have to be equal to zero. But then I get weird results like this:

    [tex]\frac{{\partial}E}{{\partial}S} = T = 0[/tex]

    In fact, if I proceed like this I can show that all of the intensive variables will be zero in equilibrium, which clearly makes no sense.

    Where have I gone wrong?
  2. jcsd
  3. May 14, 2017 #2
    In case it doesn't make sense, [itex]\mathbf{J}[/itex] is a generalized force (in the case of gases, this would be [itex]-P[/itex]), and [itex]\mathbf{x}[/itex] is a generalized displacement (in the case of gases, that would be [itex]V[/itex]).
    Last edited: May 14, 2017
  4. May 14, 2017 #3
    Please provide a reference where it says that, at equilibrium, the internal energy is minimized.
  5. May 14, 2017 #4
    I've been reading Mehran Kardar's Statistical Physics of Particles. I remember one problem that led me to this confusion. In the exercises for the first chapter on Thermodynamics, the solution to the first question (given in the book) seems to imply this. Here's the question:

    Surface tension: thermodynamic properties of the interface between two phases are
    described by a state function called the surface tension [itex]\mathcal{S}[/itex]. It is defined in terms of the
    work required to increase the surface area by an amount [itex]dA[/itex] through [itex]dW = \mathcal{S} dA[/itex].
    • (a) By considering the work done against surface tension in an infinitesimal change in radius, show that the pressure inside a spherical drop of water of radius [itex]R[/itex] is larger than outside pressure by [itex]2\mathcal{S}/R[/itex]. What is the air pressure inside a soap bubble of radius [itex]R[/itex]?
    • (b) A water droplet condenses on a solid surface. There are three surface tensions involved, [itex]{\mathcal{S}}_{aw}[/itex] , [itex]{\mathcal{S}}_{sw}[/itex] , and [itex]{\mathcal{S}}_{sa}[/itex] , where [itex]a[/itex], [itex]s[/itex], and [itex]w[/itex] refer to air, solid, and water, respectively. Calculate the angle of contact, and find the condition for the appearance of a water film (complete wetting).

    The solution to part (b) begins with the following:

    In equilibrium, the total energy associated with the three interfaces should be mini-
    mum, and therefore

    [tex]dE = {\mathcal{S}}_{aw} dA_{aw} + {\mathcal{S}}_{as} dA_{as} + {\mathcal{S}}_{ws} dA_{ws} = 0[/tex]
  6. May 14, 2017 #5
    At equilibrium the total differential is zero, that is dE = 0. Not the partial derivatives as you imply in your post.
  7. May 14, 2017 #6
    I don't see how that would work...

    How can [itex]dE[/itex] always be zero? When we can vary [itex]d\mathbf{x}[/itex], [itex]d\mathbf{N}[/itex], etc. in an infinite number of ways, how can we have that [itex]dE[/itex] is zero for all the possible varianions?
  8. May 14, 2017 #7
    What you said is exactly the reason why at equilibrium we conclude that intensive variables have to equal each other. Take the contact of two bodies (1 and 2) through a wall that permits heat transfer and matter transfer of species of type a. Then equilibrium dictates dE=0 which implies (after invoking closure of the number of particles and total energy) that:

    (T1-T2) dS + (μ1a- μ2a) dNa = 0

    For arbitrary changes in dN and dS (as you explained above), the only way to grantee dE=0 is that

    T1 =T2
    μ1a = μ2a.

    The superscripts 1,2 denote the body and the subscript a denotes the chemical species a.

    EDIT: I corrected my post to be fully in the energy representation.
    Last edited: May 14, 2017
  9. May 14, 2017 #8
    I don't know about these surface tension examples, but I do know that, for a gas within a closed system, the criterion for equilibrium is,
    $$dE-TdS+PdV=0\tag{1}$$ This equation shows how the changes in E, S, and V have to be related to maintain thermodynamic equilibrium.

    Here is an example: For an insulated system at constant volume, we have dE=0 and dV=0. Thus, the criterion for equilibrium is dS = 0. This is a maximum for the entropy.

    Another example is this: Since dG=dE-d(TS)+d(PV), if we combine this with Eqn. 1, we obtain: $$dG=-SdT-VdP$$. So, for constant temperature and pressure, the criterion for equilibrium is dG = 0. This is a minimum for the gibbs free energy.

    So, in short, dE = 0 is not a general criterion for equilibrium.
  10. May 15, 2017 #9
    Yes, the deluge of doubts that assail me after assuming this have now convinced me that [itex]dE=0[/itex] cannot be a criterion for equilibrium. I think the underlying reason that I was led into believing so was that (potential) energy minimization seemed to work so well in classical mechanics (although I am now growing weary of that idea, and I I'll just be sticking to the Euler-Lagrange equation in the near future).

    Looking at equation (1), it seems to me that in a general case, we can conclude that the criterion for equilibrium is exactly the equation I started with:

    $$dE = TdS + \mathbf{J}{\cdot}d\mathbf{x} + \mathbf{\mu}{\cdot}d\mathbf{N}$$

    different conditions (constant temperature, constant volume, etc.) can help reduce this to different criterion as you showed. Now, I have just one question, how did you conclude that the [itex]dS=0[/itex] condition meant entropy maximization and not minimization (similarly for [itex]dG[/itex])?
  11. May 15, 2017 #10
    The conditions dE=0 is completely equivalent to dS=0. Minimization of the total energy at a given entropy is equivalent to maximization of the entropy at a given energy. This is proven in Chapter 6 of Herbert Callen text, for example.

    The conditions dG=0, dF=0, dH=0 become handy when there is a contact with reservoirs. For example, dG=0 is useful when in contact with a temperature and pressure reservoirs. And essentially all of these criteria are derived either from dE=0 or dS=0. Notice that in the case of the reservoir, dE (dS) represents the variation in the total energy (entropy) of both the system and reservoir together.

    Depending on the set of axioms you start with:
    1- dS=0 can be proven to correspond to a maximum as Karadar's text does when proves the H-theorem in chapter 3.
    2- dS=0 is postulated to be maximum and from which you prove that energy has to be minimum. This is the approach in Callen's text.
    Both are fine approaches.
  12. May 15, 2017 #11
    In a isolated system, entropy can only increase or stay the same. So, at constant volume, internal energy, and no heat transfer, the final entropy of a closed system must be higher than the initial entropy (for a spontaneous process). So the final thermodynamic equilibrium state must be one of maximum entropy.
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