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Problem about entropyby kelvin490
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#1
Jul814, 09:31 PM

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Combining first and second law of thermodynamics we can get the following equation
TdS=dUP_{ext}dV First question: Is this equation available for irreversible process that dS≠dQ/T? Second question:If the system temperature T_{sys} is smaller than the surrounding temperature T_{sur}, which temperature should we put in the equation? I have this question because sometimes people use T_{sur} instead of T_{sys} (e.g. https://www.youtube.com/watch?v=jsoD...ist=WL&index=2 , 19:45) but the equation is supposed to describe changes in the system. 


#2
Jul814, 10:52 PM

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We use a reversible path to measure and quantify these changes, since, only along a reversible path is dS = QdT. For a reversible path, P_{ext} is negligibly different from the pressure of the system P, and T_{sur} at the interface with the surroundings is negligibly different from T_{sys}. But, for an irreversible path, you need to use T_{sur} in calculating the integral of dQ/T_{sur} and comparing it with ΔS. In this case, ∫dQ/T_{sur}≤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. Chet 


#3
Jul914, 01:41 AM

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1. TdS=dU+PdV is developed by replacing dQ with TdS in the equation dQ=dU+P_{ext}dV, is it valid to do so if dQ ≠TdS in an irreversible path? Or if we use T_{sur} the equality holds? 2. For irreversible process is it correct to say TdS>dU+PdV because dS>dQ/T ? 2. For irreversible process ∫dQ/T_{sur}≤ΔS, what would be the result if we compare ∫dQ/T_{sys} and ΔS in the case that T_{sys}≤T_{sur} (in this case ∫dQ/T_{sur}≤∫dQ/T_{sys}, would ΔS still be bigger or the same? 


#4
Jul914, 09:06 AM

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Problem about entropy
1. At the interface boundary between the system and the surroundings, T_{sys} and T_{sur} 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∫P_{sur}dV. 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. Chet 


#5
Jul914, 10:45 AM

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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 P_{ext} (assume constant external pressure for simplicity) we can put dW=−P_{ext}dV and so we can say dU=dQP_{ext}dV. In fact I posted this question because I guess he is doing something like putting Pext in dW=−pdV. Is it correct? 


#6
Jul914, 03:27 PM

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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. Chet 


#7
Jul914, 07:40 PM

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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 6670. Pay particular attention to the paragraphs in each section starting, "Consider the special case (a) that the only heat transferred....." Chet 


#8
Jul914, 08:13 PM

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#9
Jul914, 08:39 PM

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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. Chet 


#10
Jul1014, 07:57 AM

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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 thermomechanical 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=TdSPdV+∑μ_idn_i[/tex] where μ_{i} and dn_{i} are the chemical potential of species i and the change in the number of moles of species i, respectively. Chet 


#11
Jul1014, 09:26 AM

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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? 


#12
Jul1014, 10:41 AM

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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 T_{sur} in the inequality TdS>dU+PdV. In fact T_{sur} is typically different from the system temperature T_{sys}.
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 S_{i}. Therefore dS_{i}≥0, it can be replaced by dS+dS_{sur}≥0 where S is the entropy of the "inner" system and S_{sur} is the entropy of the "surrounding". As heat is transferred the surrounding loses entropy so it becomes dSdQ/T_{sur}≥0 and therefore T_{sur}dS≥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. 


#13
Jul1014, 01:05 PM

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Chet 


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