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Problem about entropy

  1. Jul 8, 2014 #1


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    Combining first and second law of thermodynamics we can get the following equation


    First question: Is this equation available for irreversible process that dS≠dQ/T?

    Second question:If the system temperature Tsys is smaller than the surrounding temperature Tsur, which temperature should we put in the equation? I have this question because sometimes people use Tsur instead of Tsys (e.g. https://www.youtube.com/watch?v=jsoD3oZAAXI&list=WL&index=2 , 19:45) but the equation is supposed to describe changes in the system.
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  3. Jul 8, 2014 #2
    The equation TdS=dU+PdV describes the relationship between the changes in U, S, and V between two differentially separated equilibrium states of a closed system, with one state at T, S, U, P, V, and the neighboring state at T+dT, S+dS, U+dU, P+dP, and V+dV. It is independent of the process path that moves the system from one of these equilibrium states to the other.

    We use a reversible path to measure and quantify these changes, since, only along a reversible path is dS = QdT. For a reversible path, Pext is negligibly different from the pressure of the system P, and Tsur at the interface with the surroundings is negligibly different from Tsys. But, for an irreversible path, you need to use Tsur in calculating the integral of dQ/Tsur and comparing it with ΔS. In this case, ∫dQ/Tsur≤ΔS, with the equality applying to any reversible path, and the less than applying to any irreversible path.

    For more details on this, please see my Physics Forums Blog in my personal PF area.

  4. Jul 9, 2014 #3


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    Thanks. There are several follow up questions:

    1. TdS=dU+PdV is developed by replacing dQ with TdS in the equation dQ=dU+PextdV, is it valid to do so if dQ ≠TdS in an irreversible path? Or if we use Tsur the equality holds?

    2. For irreversible process is it correct to say TdS>dU+PdV because dS>dQ/T ?

    2. For irreversible process ∫dQ/Tsur≤ΔS, what would be the result if we compare ∫dQ/Tsys and ΔS in the case that Tsys≤Tsur (in this case ∫dQ/Tsur≤∫dQ/Tsys, would ΔS still be bigger or the same?
  5. Jul 9, 2014 #4
    You are placing too much emphasis on the path, and not enough emphasis on the end points. The above equation applies to the changes in S,U and V between two closely neighboring equilibrium states of a material, regardless of how the transition between the two equilibrium states was brought about. See my comments below, after my specific answers to the questions.


    No. Irrespective of the path between two differentially neighboring equilibrium states, TdS=dU+PdV
    Please read my Blog over again. In there, I emphasized two important things:
    1. At the interface boundary between the system and the surroundings, Tsys and Tsur are equal to one another. That is, the temperature is continuous at the interface. However,
    2. For an irreversible path, the temperature and pressure within the system are typically not uniform with spatial position (except at the initial and final equilibrium states). That is, T = T(x,y,z) and P = P(x,y,z). So the there is no one single temperature value that you can identify for the system. The Clausius inequality is based on the heat flux at the interface between the system and the surroundings, and the corresponding temperature at this interface.

    In all these questions, you are placing way too much emphasis on the process (whether reversible or irreversible). It is important to think of U, S, and V as equilibrium physical properties of the material comprising the system; they are independent of the process path used to move from one equilibrium state to another. More precisely, if u, s, and v are the internal energy, entropy, volume per unit mass of material, then you can regard these as unique functions of the temperature and pressure of the system at equilibrium: u = u (T,P), s = s(T,P), and v = v(T,P).

    So where does the process path come into play (whether reversible or irreversible)? The process path can come in two ways:

    1. If we want to measure the functions u(T,P) and s(T,P) experimentally, we need to perform a reversible or irreversible process path on the system. In the case of u, we can use both reversible and irreversible paths, and employ the first law: ΔU=Q-∫PsurdV. However, in the case of s, we can only employ reversible paths to make this measurement. We need to dream up a reversible path between the initial and final states of the material, and measure ∫dQ/T for that path.

    2. For actual industrial process calculations, we typically need to consider irreversible or nearly reversible process paths. In such calculations, it is assumed that we already know u(T,P) from laboratory experiments (discussed in item 1. above), so we can use this to calculate, say, the temperature change for the material in the process. If we then know the two end points at the beginning and end of the path, we also know the change in entropy between these states, since we have previously established this from the laboratory measurements of s as a function of T and P.

  6. Jul 9, 2014 #5


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    Thank you for the answer. I have no doubt that the equation is describing equilibrium states from end point to end points. What confuses me is that in a video lecture (youtube.com/watch?v=jsoD3oZAAXI&list=WL, 19:45) the lecturer obvious put TdS>dU+PdV for irreversible process and specially indicate T is Tsur.

    And in applying the first law work done is smaller in irreversible process than in reversible process, we cannot say dW=−pdV, but if we indicate P is Pext (assume constant external pressure for simplicity) we can put dW=−PextdV and so we can say dU=dQ-PextdV. In fact I posted this question because I guess he is doing something like putting Pext in dW=−pdV. Is it correct?
  7. Jul 9, 2014 #6
    With regard to the work W, as I said in my Blog, dW is always given by dW=PextdV for both reversible and irreversible processes (in my Blog, I represent Pext by the parameter PI, where the subscript I represents the interface between the system and the surroundings). Also, for the special case of a reversible process (1) the pressure P within the system is spatially uniform so that (2) Pext=P.

    With regard to the TdS, I will try to review what the guy is saying in the video and make some sense out of it. I'll get back to you.

  8. Jul 9, 2014 #7
    Hi Kevin490.

    I looked over the video, and can understand why you have been confused by what the guy did. I can see what he was trying to convey, but don't like the way he packaged it. He played it very fast and loose with the mathematics, and did not provide additional editorial comments to make the math more precise. I particularly didn't like his use of differential changes, rather than finite changes, in the state functions.

    I have found a development that I think you will find much more understandable, satisfying, and precise. The development is presented in Denbigh, The Principles of Chemical Equilibrium. The development uses finite changes in the state functions. See sections 2.3 and 2.4 in Denbigh, pages 66-70. Pay particular attention to the paragraphs in each section starting, "Consider the special case (a) that the only heat transferred....."

  9. Jul 9, 2014 #8


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    Thanks for looking over the video, I will try to go over the book you suggested. BTW, what's the guy trying to convey in the video? Can he derive the use of Gibbs energy and other type of energies without putting Tds>dU+PdV?
  10. Jul 9, 2014 #9
    He is trying to determine the conditions for a process to proceed spontaneously (a) in a constant volume system in contact with a constant temperature heat reservoir that is held at the same temperature as the initial and final temperatures of the system and (b) in a system in contact with a constant temperature heat reservoir that is held at the same temperature as the initial and final temperatures of the system and in which Pext is held constant at the initial and final pressures of the system. He is trying to show that in case (a), the condition is that the Helmholz free energy has to decrease between the initial and final equilibrium states, and, in case (b), the Gibbs free energy has to decrease between the initial and final equilibrium states.

    Regarding your second question, the inequality is wrong. It is always: TdS=dU+PdV. The confusion arises from using differential changes rather than finite changes. You will be much happier when you read the sections of Denbigh that I referred you to.

  11. Jul 10, 2014 #10
    In my previous response, I was incorrect in saying that TdS=dU+PdV always. This relationship is true for a system in which the chemical composition is not changing. If the concentrations of chemical species within the system change, then the inequality you wrote is valid. Sorry for my confusion.

    It seems that the main idea of the video is to identify criteria for whether a chemical reaction (or set of chemical reactions) will proceed spontaneously (a) in cases in which the surroundings temperature and volume are held constant or (b) cases in which the surroundings temperature and surroundings pressure are held constant.

    Obviously, for a purely thermo-mechanical setup (no chemical reactions) initially at equilibrium in which the external pressure is held constant at the initial system pressure, and the surroundings temperature is held constant at the initial system temperature, there will be no change in the system, and no changes in the state functions. However, if chemical reactions can occur, there can be a change in the system and changes in the state functions. The full equation for the differential changes in a system where mixing or chemical reaction can occur is [tex]dU=TdS-PdV+∑μ_idn_i[/tex]
    where μi and dni are the chemical potential of species i and the change in the number of moles of species i, respectively.

  12. Jul 10, 2014 #11


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    I've just notice the set of equations in p.79 of the book you suggested. Thank you for bring up this point. It's very important to note that chemical reactions results in inequality.

    However, the lecturer in the video doesn't seem like doing the same thing you have mentioned. There is similar treatment in section 2.8, p.82 (try to put first and second law together) but I still need some time to figure out whether it is what the guy wants to convey.

    Another thought is, let's say we agree that TdS=dU+PdV is a state equation and describe reversible and irreversible processes. If we forget this equation temporarily and just consider dQ=dU+PdV and TdS≥dQ as two simple math equations and put them together, we actually get TdS≥dU+PdV. Does it possibly have some physical significance, or is just a meaningless mathematical trick?
  13. Jul 10, 2014 #12


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    I think I start to make some sense of the video now. The book you referred is extremely useful. In the video lecture the guy always use Tsur in the inequality TdS>dU+PdV. In fact Tsur is typically different from the system temperature Tsys.

    In deriving the equation the "surrounding" (in the book p.82 the term "thermostat" is used instead) is included as a bigger isolated system. Let's say the entropy of these bigger system is Si. Therefore dSi≥0, it can be replaced by dS+dSsur≥0 where S is the entropy of the "inner" system and Ssur is the entropy of the "surrounding". As heat is transferred the surrounding loses entropy so it becomes dS-dQ/Tsur≥0 and therefore TsurdS≥dU+PdV.

    In the case that there is no chemical reaction, TdS=dU+PdV is always true as state equation. The T here refers to the "inner" system's temperature. There is no contradiction.
  14. Jul 10, 2014 #13
    I just can't get my mind around doing this for differential changes. The only development that works for me is if the changes between the equilibrium states are finite.

  15. Jul 10, 2014 #14
    I'm really glad that the development in Denbigh has helped you. Another book that I like much more than Denbigh is Smith and Van Ness, Introduction to Chemical Engineering Thermodynamics.

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