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DE-TdS+PdV ? 0

  1. Jul 7, 2012 #1
    My textbook says that dE-TdS+PdV<0 for all irreversible processes. However, for reversible processes, the author says that dE-TdS+PdV = 0, and in fact this equation can be applied to irreversible processes because E is a state function. I'm confused -- this seems to be a contradiction.
  2. jcsd
  3. Jul 7, 2012 #2

    Philip Wood

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    dE - TdS + pdV = 0 for all infinitesimal changes, reversible or irreversible, because E, T, S, p, V are all functions of state, and depend only on the starting state and finishing state, not on the process of getting from one to the other, provided the change is infinitesimal.

    It is also true (first law of thermodynamics) that, in any process, dE - Q + W = 0, in which Q is the heat input and W is the work done by the system.

    In a reversible process, W = pdV and Q = TdS. In an irreversible process, W < pdV and Q < TdS. [An extreme case of an irreversible change would be the expansion of a thermally isolated gas into a vacuum. Here pdV has a positive value but W is zero. TdS has a positive value equal to pdV, but Q is zero. Thus both equations give dE = 0, which is correct because the system is thermally isolated and does no work.]

    [This is a corrected version of the earlier post. Note the sign convention I'm using for W. Should this post be re-written for the opposite sign convention?]
    Last edited: Jul 7, 2012
  4. Jul 10, 2012 #3
    Any clue as to why Dr. C.H.P. Lupis uses this (dE-TdS+PdV < 0) as a jumping off point for studying criterion for equilibrium? Does this inequality ever have a context and a meaning? Thanks!
  5. Jul 10, 2012 #4
    Can you possibly scan or alternatively quote the passage where he makes the "dE - TdS + PdV < 0" claim?

    Are you sure he doesn't simply state that "Q - TdS < 0" and that you suppose the other inequality follows from this (it doesn't!)?
  6. Jul 11, 2012 #5
    Careful here, p is not constant

    [tex]q = TdS = \int\limits_{{v_1}}^{{v_2}} {pdv} = RT\int\limits_{{v_1}}^{{v_2}} {\frac{{dv}}{v}} = RT{\log _e}\left[ {\frac{{{v_2}}}{{{v_1}}}} \right][/tex]
  7. Jul 11, 2012 #6
    Studiot, I believe what you're saying is incorrect. Wood is correct in saying dQ = 0. The infinitesimal version of your statement is dQ = RT dV/V.

    Firstly I hope it's clear that dQ should be equal to zero, since Wood is talking about an adiabatic expansion.

    Secondly, I believe your calculation is incorrect since you assume that dQ = TdS which is not true in an irreversible process such as an adiabatic expansion.
  8. Jul 11, 2012 #7
    Let us suppose you expand a gas into a vacuum, you can't just declare the process to be adiabatic (although in this case it is). Nor can you continue the process indefinitely as far as the equations are concerned. You have to state a final volume.

    Philip Wood is basically descibing Joules experiment where he connected a cylinder (volume V1) of gas under pressure P to an evacuated cylinder (volume V2).

    When he allowed the gas to freely expand to fill the second cylinder he found that there was no temperature change.

    This we know to mean that the internal energy of an ideal gas depends upon temperature only.

    So no temperature change means no change internal energy.

    However there was an entropy change.

    Furthermore if we now recompress the gas back to the first cylinder we have to do work and we find that it heats up, unless we do it slowly and extract an equivalent amount of heat such that

    q = ∫pdv

    Then the gas ends up in its original condition and the recompression is reversible.
    Thus we can use it to calculate the work done, the heat exchanged and since Internal Energy and Entropy are a state functions we can use it to calculate the entropy change for the original expansion.

    Finding a reversible path between two states of a state function to calculate the Δ for some irreversible process between the same two states is a powerful technique in thermodynamics.

    Does this explain the proceedings?
    Last edited: Jul 11, 2012
  9. Jul 11, 2012 #8
    Studiot, I agree with almost all you say, but not with the part that counts (i.e. the part which I disagreed upon in my last post): you can't use the reversible path to calculate the work done or the heat done, since these are path-dependent quantities.

    I'm not sure what exactly you're disagreeing on here: your post #5 implied that dQ for a free expansion is not zero, while it is actually equal to zero (and I tried to point out where you went wrong in your calculations). What are you disagreeing with here?
  10. Jul 11, 2012 #9
    Yes you are right I didn't make it clear first time, I was just concerned to note that since p is not constant there is a problem with saying the work = pdV. What p?

    I showed a very shortform explanation to calculate the entropy change. Obviously it was too short.

  11. Jul 11, 2012 #10
    Well your calculation to calculate the entropy change is correct you should just leave out he first equality "q = TdS = ..." and start with "TdS = ...", then it's all good.

    As for your other statement
    I'm not sure whether you're talking about a reversible process or an irreversible process. In the first case, pressure is a continuous function so for a really small change we can assume p is constant (and for "really" an infinitesimal change, pressure is constant, but of course changes are never "really" infinitesimal). In case you're talking about an irreversible process, like an adiabatic expansion, in that case I'm not sure but I suppose if the expansion is "infinitesimal" (or just really really small), we can again assume p is constant. But this time I'm not really sure since it's not clear to me whether p is a continuous function for irreversible processes. It might very well not be, since at least during an irreversible change p is not even defined, so there's no real reasoning to conclude that it should be continuous, except for perhaps "small disturbances lead to small changes". In that case you're right that PdV is not well defined. Note though that no one is saying that dW = PdV for an irreversible process, rather dW < PdV (for adiabatic expansion: dW = 0), but indeed, the problem whether PdV is well-defined remains.
  12. Jul 11, 2012 #11
    Thank you for pointing out the lack of clarity in my original statement, hopefully that is now cleared up.

    As to the issue of pressure I had thought we were discussing a particular well specified system, not a general case.

    As such the pressure is well defined at all times, it is just not constant. In fact it starts out at theoretically zero.

    For the general case, of course there are many irreversible processes that happen under constant pressure. Some are chemical reactions.
  13. Jul 11, 2012 #12
    When you talk about "a particular well specified system", are you referring to the adiabatic expansion? I would say that in a non-infinitesimal adiabatic expansion pressure is certainly not defined; there would be a pressure gradient (so okay, it's defined locally, but I don't think that's what you were talking about).
  14. Jul 11, 2012 #13
    The original process was described as the expansion of a thermally isolated gas into a vacuum.

    Now either this expansion is into a limitless vacuum like outer space in which case the pressure is at all times zero.

    Or this is a description of the Joules experiment as I already noted.
    In this case the the volume of the second chamber is always known as is the temperature and the total mass of gas is fixed so at any instant we can calculate the pressure from a mass balance between the two chambers.

    The pressure in the supply chamber is less certain that the pressure in the receiving chamber

    I am, of course, assuming a perfect gas.
  15. Jul 11, 2012 #14
    All the situations described are to me synonymous for "free expansion of gas into vacuum" (and I'm imagining there is a surrounding wall so the expansion is not limitless). Maybe we got a small terminology problem? Anyway, assuming we're talking about free expansion: during the expansion, surely there is no homogeneous pressure in the system?
  16. Jul 11, 2012 #15
    No, of course there is no homogenous pressure in the system, if the system is the chamber of pressurised gas.

    The pressure is provided by the surroundings or in this case the lack of them since it is a vacuum.
  17. Jul 11, 2012 #16
    I'm afraid I don't understand what you're saying.
  18. Jul 11, 2012 #17
    Are you familiar with the Joule experiment?
  19. Jul 11, 2012 #18
    To make sure I googled. It's the free expansion we were talking about, yes. I understood you as noting that the pressure is not constant. I noted that not only is it not constant, it's not even well-defined. Unless you're talking about the reversible path between the same two points? In which case there is no problem and we were just talking past each other the last few posts.
  20. Jul 11, 2012 #19
    It is often useful to imagine the expansion as taking place against either a frictionless massless piston or a membrane that are opposed by a defineable pressure. that way this can be used to calculate the work done. this defineable pressure is the pressure in the expression

    First law work is always a quantity transferred between the system and the surroundings. It is never done within a system - that could not change the internal energy as per the first law.
    As such there are two ways of calculating it. Work done on the system and work done on the surroundings. Sometimes one is easier to calculate than the other.

    It's signing off time here but if you want to continue tomorrow we can do that.

    go well
  21. Jul 11, 2012 #20
    Yes, you're talking about the reversible path, the concept is familiar to me, it was just not clear to me you were talking about the reversible path, I thought you were talking about the actual, irreversible path. No need to continue, I think.
  22. Jul 12, 2012 #21

    Andrew Mason

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    You are missing at ΔU in there. This should be:

    [tex]q_{rev} = \int TdS = \Delta U + \int\limits_{{V_1}}^{{V_2}} {PdV} [/tex]

    If T is constant, then

    [tex]q_{rev} = T\Delta S = \Delta U + \int\limits_{{V_1}}^{{V_2}} {PdV}[/tex]

    If the substance is an ideal gas then for an isothermal process ΔU = 0, so:

    [tex]q_{rev} = T\Delta S = RT\int\limits_{{V_1}}^{{V_2}} {\frac{{dV}}{V}} = RT{\ln}\left[ {\frac{{{V_2}}}{{{V_1}}}} \right][/tex]

  23. Jul 12, 2012 #22

    Thank you for your comment, Andrew.
    I did actually mention that in my explanation.

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