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Manipulating differentials in thermodynamics

  1. Jan 12, 2015 #1
    Hi everyone,

    Sorry if this has been posted before, but I had a quick question manipulating differentials. This problem is in the context of thermodynamics. We know from the first law that E=E(S,V), and from calculus I know that:
    dE=(∂E/∂S)v dS+(∂E/∂V)s dV sorry if this is hard to read, I'm new to the board and not sure if there is a better way.
    Does it follow from this that (dE/dV)=(∂E/∂S)v (dS/dV)+(∂E/∂V)s

    Thank you in advance for any help.
     
  2. jcsd
  3. Jan 13, 2015 #2
    Yes, this is an application of the chain rule: you are taking the derivative of ##E(S,V)## along a path in ##S,V## space, where you have chosen to use ##V## as the parameter for the path (i.e. the path is given by ##(S(V),V)## for some function ##S(V)##).
     
  4. Jan 13, 2015 #3
    I disagree. In thermodynamics, the state functions are functions of 2 independent variables (at least for a system of constant composition). In this case, E (energy per unit mass) is being expressed as a unique function of the two independent variables S and V (entropy per unit mass and volume per unit mass). So writing dE/dV does not have relevant meaning. Neither does the dS/dV have meaning.

    Chet
     
  5. Jan 13, 2015 #4
    Yes, it's true that state functions typically depend on two or more variables. However, the notation ##\frac{dE}{d\tau}## is meaningful as long as a path ##(S(\tau),V(\tau))## is given as well. In shawnstrausser's example, the notation suggests a path given by ##(S(\tau),V(\tau))=(S(\tau),\tau)## (i.e. parametrized by volume). I believe this notation is standard for derivatives along one-dimensional trajectories (c.f. `Foundations of Differential Geometry' by Kobayashi and Nomizu pg. 4-5, or `Elementary topics in Differential Geometry' by Thorpe). The notation ##\frac{dS}{dV}## is meaningful if ##S## is regarded as a function of 1 variable (i.e. ##S(\tau)\Big|_{\tau = V}##).
     
  6. Jan 14, 2015 #5
    Yes. Actually, I had thought about this over night and, after reconsidering, I had realized that you were right after all. Sorry.

    The example I had though of was expressing the entropy S as a function of temperature T and volume V, and evaluating the differential of dS in terms of the differentials in dT and dV; then assuming that T is known as a function of V.

    Chet
     
  7. Jun 8, 2015 #6
    Thanks to both of you, for your reply to Shawnstrausser. I would like to follow up:

    If I understand correctly, Couchyam affirms that
    (dE/dV)=(∂E/∂S)v (dS/dV)+(∂E/∂V)s
    not from the total differential that Shawnstrausser mentions, but directly from the multivariable chain rule.
     
  8. Jun 8, 2015 #7
    Here's another question from Thermodynamics: work, w, is accomplished when there is a volume change with a fixed pressure. In the text I am following (a geochem text) this is written as dw = P dV (the V actually has an overlying ~ to indicate molar volume). The author wishes dw to be thought of as an "infinitesimal", but doesn't define it and has given no indication of awareness of "non-standard" analysis.

    I can accept dw as a real valued differential that is defined to be a function of the real valued differential dV by the above formula.

    The author goes on to calculate work over two P, T (pressure/Temperature) paths, in a familiar way, but calculates the work by putting a (path) integral sign in front of P dV, apparently co-opting the differential dV as the variable of integration.

    Is there a more standard and justifiable way of setting this calculation up?

    I note that V is, here, considered to be a function V(T,P), and the author mentions the total differential dV. Is that just a red herring? Would this be similar to Shawnstrausser's example in choosing P or T for the path variable, taking the derivative (dV/dT for example) and resorting to an application of the multivariate chain rule?

    I would then think of writing dw/dT = P dV/dT, expand dV/dT using the chain rule, and integrating both sides with respect to T (for part of one path.)
     
  9. Jun 8, 2015 #8
    I assume that the expression you are trying to evaluate has the form
    \begin{equation}
    \int_\gamma PdV
    \end{equation}
    for some path ##\gamma## that is specified by ##(P(\tau),T(\tau))##, where ##\tau## is an arbitrary parameter that runs from ##a## to ##b## say. The above expression is shorthand for
    \begin{equation}
    \int_a^b d\tau \Big[P(\tau)\big(\partial_P V\big|_T\dot{P}+\partial_T V\big|_P\dot{T}\big)\Big]
    \end{equation}
    (mathematically, you are integrating the 1-form ##PdV## over the path ##\gamma##).
    In your case, ##\gamma## is very simple: it's just a line at fixed ##P##, so you can use ##T## as a parameter (as you suggest). However, I think the problem can be solved in a simpler way if you already know the equation of state that relates ##P##, ##V##, and ##T## (since pressure is constant, it can be brought outside the integral).
     
  10. Jun 8, 2015 #9
    Thank you, Couchyam.

    After staring at the problem and the author's treatment for quite a while, I decided that in this simple case (move along an increasing T path, then along an increasing P path, then the opposite, to demonstrate lack of path independence) I could resolve the issue with the substitution rule (basically back to change of variable.)

    As you suggest, the law PV = nRT was reduced to PV = RT, by using "molar volume", hence n=1, and used to make the calculations more explicit. The "total differential" of dV does, in fact, produce the correct forms for substitution and resembles your replacement for dV. Since your formulation seems to avoid differential magic, I prefer it and will study it.
     
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