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Tds Equation and Entropy

  1. Jan 10, 2013 #1

    I was wondering, with regards to the Tds equation Tds = de + pdv:

    1. All of my textbooks state that integrating this equation, although derived for a reversible process, will give the entropy change regardless of the process or whether or not the process is reversible. However, I don't understand this concept because if the Tds equation was re-derived from first law for a general process, I thought that there would be a Tσ (entropy generation term, zero for reversible) in the equation since ds = δQ/T + σ and entropy generation would be path dependent?

    2. With regards to the reversible Tds equation, I have trouble seeing how this equation is path independent since, for two fixed states, I always thought there was more than one possible work path or pdv expression in which state 1 can be used to move to state 2 and if this were true it would end up giving different s2-s1 values?

    3. I have only found proofs for entropy as a state property for reversible process and argued that by extension it must be a state property for any process but is there a proof that directly show that it is a state property for any process?

    Thanks very much
  2. jcsd
  3. Jan 10, 2013 #2


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    The state variables p, V, E, S etc. are only defined for equilibrium states. In classical thermodynamics, all states proceed from one initial equilibrium state to a final equilibrium state. The entropy difference can be calculated for any path in equilibrium state space to the final state.
  4. Jan 10, 2013 #3
    V, at least, may be defined for a non equilibrium state.
    Last edited: Jan 10, 2013
  5. Jan 10, 2013 #4
    Hi, thanks for the response

    I understand and have accepted the statement but I have trouble understanding how this is reflected, which is why I posed the questions above.

    For instance, I imagine that there is more than one polytropic (or any) process from two equilibrium states for which work can be applied, but from Tds = de + pdv, the integral of the p/Tdv term (taking T constant) would result in different values while e2-e1 doesn't change if the initial and final states are the same. To me this seems to reflect different s2-s1 values despite having different processes from the two states. I have the same confusion regarding the derivation of the Tds equation if entropy generation is added as I see that it would change the form of the Tds equation which many books say is applicable for both reversible or irreversible processes.

    Any help would be appreciated.
    Thanks very much
  6. Jan 10, 2013 #5


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    True, but take in mind that to obtain Delta S you have to integrate de/T, not de. Both p and T will be different in different polytropic processes.
  7. Jan 10, 2013 #6

    So what about a non-insulated piston cylinder system undergoing reversible quasistatic compression such that there is no irreversibility from heat transfer? In this case regardless of the process the temperature would remain the same and the only difference in the Tds equation for different compression processes would be the pdv term?

  8. Jan 10, 2013 #7


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    If T is constant, p is a unique function of V, namely p=nRT/V.
  9. Jan 10, 2013 #8
    Hi, thanks for the response

    So for different polytropic processes to be possible, then the temperature at the end states must not be the same and if it the temperature throughout the process is the same, then the pdv process must satisfy the ideal gas law (assuming an ideal gas is compressed)?

    So this would mean that for different reversible processes with the same end states, the sum of the integral of de/T and p/Tdv will always end up working out to be the same value?

    I'm also wondering if you can take a look at my first question in the OP.

    Thanks very much for your help
  10. Jan 10, 2013 #9


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    Uh, yes, I was assuming an ideal gas here, but only to show that the pressure will be determined by the volume when T is held constant. This will also hold true for other substances.
    The integrals of de/T and p/TdV will not always be the same when T is variable, but their sum will be.
    Regarding your first question you have to distinguish between a process, which may be irreversible, and a path in the state space of equilibrium states used for calculation of the change of S.
    Usually when there is entropy production in an irreversible process, the work is also not equal to -pdV. So there is no contradiction with TdS not being dQ. Think of stirring a viscous liquid which is a form of non-volume work which gives rise to entropy production.
  11. Jan 10, 2013 #10
    Your initial equation TdS = dU + PdV applies to the differential change in the parameters from one equilibrium state to another. It doesn't matter whether you got from the initial state to the final state by a reversible path or an irreversible path. However, the changes described by the equation can not be assumed to apply along the path; they can only be assumed to apply to the overall change from the initial to the final end point. However, if the path is reversible, they will also apply at all points along the path.

    I hope this makes sense.
  12. Jan 11, 2013 #11
    If you would like to learn more about how all this plays out with regard to the contribution of entropy generation over an irreversible path to the total entropy change between the initial and final equilibrium states of a closed system, see my postings on the following PF link:

  13. Jan 12, 2013 #12
    Thanks very much for your responses

    So for any process, reversible or irreversible and regardless of path, I can integrate any path defined by a set of equilibrium states from the same two end points and the integral should work out to be the same? What about the case of a path defined by non-equilibrium states such that integration is not possible?

    I'm still confused as to how the Tσ term plays a role mathematically if the process was irreversible if the first law was δq_irr = de + pdv_irr, is it incorrect to substitute Tσ in as part of δq_irr? What if the system undergoes an irreversible heat transfer with no work?

    I was wondering, when the Tds equation is integrated I thought that it was integrated along the path in which a process takes place whether reversible or irreversible?

    Thanks again
  14. Jan 12, 2013 #13
    To find out how all this can be done for an irreversible path, see my posting #11. In order to find the change in entropy along an irreversible path, you must first simultaneously solve the continuity equation (differential mass balance), equation of motion (differential force balance), and energy balance equation for the system. Then you can take the results of these calculations and plug them into the equation I have presented in the link mentioned in #11 for calculating the entropy change along the reversible path.

    You have to recognize that, in a system undergoing an irreversible process, the key variables, like temperature, pressure, and velocity are not uniform within the system, and vary with spatial position and time. But we definitely have the capabilities for solving for these variations by solving the above differential balance equations. One of the tools that is often used in this regard is computational fluid dynamics. We do not need to throw our arms in the air and simply declare that, because the process is irreversible, there is no useful information we can get about what is going on within the system.

  15. Jan 12, 2013 #14


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    In general non-equilibrium situations, e.g. temperature may not even be defined. How do you want to integrate over it?
  16. Jan 12, 2013 #15


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    That's only the situation engineers are most often confronted with, and they tend to assume local equilibrium. Chemical reactions can be quite irreversible but spatial homogeneous. And finally you can have nasty systems like homogeneous mixtures of gasses with different temperatures like in outer space.
  17. Jan 12, 2013 #16
    Based on the equation presented in the link, the entropy changed is calculated from its original second law formulation applied over a control volume. My question was more related to how entropy can be solved by relating entropy change to the first law in a simple, textbook system where all properties are uniform but not necessarily in equilibrium when undergoing a process. In such a case the entropy generation integral which I expressed σ (or dσ or the differential form) seem to appear in the Tds equations if irreversibilities are present.

    Basically I found that all textbooks say that the equation Tds = de + pdv is valid for between any equilibrium state reversible, irreversible etc. , which I'm confused about since for cases where irreversibilities are present I thought the equation should be Tds = de + pdv + Tdσ.

    So for non-equilibrium situations, this means that the Tds equations are valid in a differential sense only but cannot be used because it's not possible to define a path between two end states and it would be necessary to use the numerical approach mentioned in Chestermiller's post? What about in quasistatic irreversible processes, the integration would be possible?

    Thanks very much for answering my questions thus far.
  18. Jan 12, 2013 #17
    Your comment is correct regarding the fact that I had not included chemical reactions in what I was saying. I was just trying to keep the discussion of a very difficult (and often confusing) subject as simple as possible for the OP. Even so, the development I presented in the link I referred to is a little intense mathematically. The development at that link, as you know, omits diffusion and chemical reactions. The analysis can be extended to these situations via the methodology presented in BSL Transport Phenomena in which effects related to chemical potential are included. However, at this stage, consideration of multicomponent systems are complexities still somewhat in the future for the OP. If he is not able to understand what's involved without multicomponent considerations, he will certainly struggle greatly with multicomponent systems.

  19. Jan 12, 2013 #18
    The material you require to answer all three of your questions was presented by Caratheodory in 1909.

    You will find good discussions of this in books by

    Wilson (1951) Thermodynamics and Statistical Mechanics (Cambridge University Press)

    Pippard (1964) Classical Themodynamics (Oxford University Press)

    Carrington (1994) Basic Thermodynamics (Oxford University Press)

    All three follow and further develop Caratheodory's highly mathematical method and discuss the conditions under which dw and dq may be integrated and prove mathematically that entropy is a state function.

    All three are highly readable, but different, and contain penetrating insights into classical thermodynamics.

    I'm sorry but there are no short form solutions to this. Each proof is many pages long.
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