Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Differentials in the context of thermodynamics

  1. Sep 11, 2010 #1
    I just want to make sure I understand differentials in the context of thermal physics. One of the big statements of thermodynamics is the conservation of energy in terms of the state variables [itex]U,T,S,V,P[/itex]:

    dU = T dS - P dV.

    What does this really MEAN though? Is there any way to understand this in a mathematically rigorous way? People keep explaining it to me like this: "Well, what is means is that a SMALL change in energy is equal to a SMALL change in entropy times..." But "small" is a relative term...so what does "small" mean in this context?
  2. jcsd
  3. Sep 12, 2010 #2
    In mathematical terms, every state function is a function of two independent variables, in your case the entropy and the volume. So for example U = U(S, V). Once you have this functional dependance, which can be found for example using the canonical ensembles of statistical mechanics, you define

    [tex]T=\left(\frac{\partial U}{\partial S}\right)_V\qquad\qquad P=-\left(\frac{\partial U}{\partial V}\right)_S[/tex]

    and this way the first principle looks the way you wrote it. You can see it as a first-order Taylor expansion of internal energy:

    [tex]\Delta U=T\Delta S-P\Delta V+O(\Delta S^2+\Delta V^2)[/tex]

    explaining this way the term "small".
    Last edited: Sep 12, 2010
  4. Sep 12, 2010 #3


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In general, when a physics books says that an equality holds for "small" or "infinitesimal" values of the variables, the equality is a Taylor expansion up to some finite order, with all higher order terms dropped. For example, a book might say that f(x)=f(0)+f'(0)x for infinitesimal x. This is stupid for the claim that f(x)=f(0)+f'(0)x+O(x2) in the limit x→0. The "big O" notation is explained here.

    If U is a function of two variables, we can define dU as a function of four variables:

    [tex]dU(S,V,x,y)=\frac{\partial U(S,V)}{\partial S}x+\frac{\partial U(S,V)}{\partial S}y=Tx-Py[/tex]

    and if we call the new variables dS and dV, we have


    where I have used the definitions in Petr's post above. Note that a Taylor expansion of U around (S,V) takes the form

    [tex]U(S',V')=U(S,V)+\frac{\partial U}{\partial S}(S'-S)+\frac{\partial U}{\partial V}(V'-V)+\mathcal O(|(S'-S,V'-V)|^2)[/tex]

    [tex]=U(S,V)+dU(S,V,S'-S,V'-V)+\mathcal O(|(S'-S,V'-V)|^2)[/tex]

    so if we define dS=S'-S, dV=V'-V, then dU(S,V,dS,dV) is the first-order approximation of U(S',V')-U(S,V).
  5. Sep 12, 2010 #4
    Thanks for your help, guys. I'll busy myself wrapping my head around the things you've said. In the meantime, I'd like to ask a few other questions related to thermal physics:

    Suppose you initially have a gas confined, by a partition, to 1/2 of a container (which is isolated from its surroundings). Then you remove the partition and the gas expands to fill the whole container. Certainly, the volume of the gas has changed (it's doubled). But has the temperature of the gas changed? Has the pressure changed? Has the internal energy changed? How do we know?
  6. Sep 12, 2010 #5
    To solve the simple exercise of the "free expansion" you shall consider (Hp: ideal gas):

    -> p, V and T are related by ideal gas law
    -> No energy exchange happens during the expansion
    -> Energy depends only on temperature

    These arguments put a constraint on the final state which allow you to compute it.

  7. Sep 12, 2010 #6
    Ok. Well, then I would think the following: No energy exchange implies [itex]\Delta U = 0[/itex], implying [itex]\Delta T = 0[/itex]. The ideal gas law then tells us [itex]P[/itex] decreases by a factor of two...am I right?

    I guess another general question I could ask is this: We have [itex]dU = \delta Q + \delta W[/itex] (where the latter two are inexact differentials)...for a given system, can we only have [itex]\delta Q \neq 0[/itex] if the system is able to interact with its surroundings in some way?
  8. Sep 12, 2010 #7


    User Avatar
    Homework Helper

    Yes, and yes. And besides, if a system is really unable to interact with its surroundings, [itex]\delta W = 0[/itex] as well. That's just energy conservation: [itex]\mathrm{d} U = \delta Q + \delta W[/itex] means that any change in the internal energy must be a result of some energy entering or leaving the system. And there are only two ways for energy to enter or leave: heat and work.

    Of course, in principle it is possible to set up a system which can exchange energy by heat transfer but not by work (e.g. enclose it in a rigid, thermally conducting box). Or you could set up a system which can exchange energy by work but not by heat transfer (e.g. an insulating piston chamber).
  9. Sep 13, 2010 #8
    Ok, thanks. Again, I appreciate all the help.

    Now, here's a question I'm not even sure makes any sense...if a spring is compressed some amount beyond its equilibrium length and is allowed to decompress by an infinitesimal amount, is there any change in the entropy of the spring?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook