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I Heat and Entropy Increase Confusion

  1. Jul 9, 2016 #1
    This may be a matter of me being confused by the definition of heat. However, I view heat as the energy passed between systems of different temperatures.

    My problem is the following:

    By the principle of minimum energy/max entropy, in an isolated system (and therefore fixed internal energy). the entropy of the system should tend to maximize.

    I thought of a very simple example. I have a bunch of charged positive particles all initially pretty close together. As time goes on they will separate, and their potential energies will drop. However, they will all have gained kinetic ( thermal) energy and hence the temperature will have increased and the entropy of the system will have also clearly increased. (Fast movement and particles now clearly in a state of higher disorder). However, change in entropy is defined as q/T. However, where could this heat q have come from? I understand the obvious that the potential energy of the charged particles was converted into kinetic, but this comes from the work done on the particles does it not? There is no outside system to transfer heat?
     
  2. jcsd
  3. Jul 9, 2016 #2
    You certainly chose a very complicated system as an example to frame your main question: If an irreversible process occurs within a closed system that is both rigid and insulated (so that it cannot exchange heat or work with its surroundings, and the internal energy is constant), how can the entropy of the system increase if entropy change is q/T, and not heat transfer is allowed? Is this basically your question?

    A better example would be: You have a rigid insulated chamber split into two halves by a panel, with a gas at high pressure in one half and the same gas at a lower pressure in the other half and you poke a hole in the panel and allow the system to re-equilibrate. How can the entropy of the system increase if the entropy change is q/T, and no heat transfer is allowed?
     
  4. Jul 9, 2016 #3
    Okay in this case there could be some heat exchange among the two gases if they are of uneven temperature, this would obviously result in increased entropy. However this is somewhat different from my scenario where this no heat exchange even within the system, but solely potential energy being converted to kinetic.
     
  5. Jul 9, 2016 #4
    What if the gases are at the same temperature? The entropy would certainly increase in that case too, right?
    Your problem is with your understanding of how to determine the entropy change.

    The entropy change of a closed system is not equal to the integral of dq/T unless the process is reversible. Here is my recipe for determining the change in entropy for a closed system that has experienced an irreversible change between two thermodynamic equilibrium states:

    1. This is the most important step. Totally forget about the actual irreversible path between the initial and final equilibrium states. Focus only on the two end states.

    2. Devise a reversible path between the initial and final thermodynamic equilibrium end states. This reversible path does not have to bear any resemblance whatsoever to actual irreversible path that the system experienced. For example, even if the actual path was adiabatic, the chosen reversible path does not have to be adiabatic. You can even separate different portions of the system, and subject each of them to a reversible path individually. All reversible paths will give exactly the same value for the change in entropy of the system (according to the equation in step 3), so choose one that is convenient.

    3. For the chosen reversible path, calculate the integral of dq/T. This will be the change in entropy of the system between the initial and final equilibrium states.

    Can you think of a reversible path to calculate the change in entropy for the system involving the two chambers of gas with different pressures?
     
  6. Jul 9, 2016 #5
    So that makes sense to me, given that Entropy is a state function. However, why does that definition only apply for reversible processes?

    Also I can't really think of one. Honestly I'm not too sure about what actually makes a reversible process reversible. I know it has something to do with infetesimal temperature differences, but I don't get how this is different from a finite temperature difference. Could you possible explain this, and such a process for your example?
     
  7. Jul 9, 2016 #6
    If doesn't apply only to reversible processes. It applies to any process that takes the system from the initial equilibrium state to the final equilibrium state, including the irreversible process of interest. The entropy change is the same for all of these. To understand why we need to use a reversible path to calculate the entropy change, see my Physics Forums Insights article: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

    This is also covered in my Physics Forums Insights article. A reversible path consists of a continuous sequence of thermodynamic equilibrium states.

    To get the change in entropy for the two-chamber gas pressure problem, you first need to use the first law of thermodynamics to determine the final thermodynamic equilibrium state of the system for the irreversible process. There is no heat transfer and no work done by the rigid container on the gas in the process. So, the final internal energy of the combined gases is the same as the initial internal energy. This tells you that the final temperature is the same as the initial temperature. Once you know the final temperature, you can use the ideal gas law to determine the final pressure. This then gives you your starting point for applying the 3 step procedure to get the entropy change.
     
  8. Jul 9, 2016 #7
    Alright, so would I be using a PV diagram here? Also if that's the final state, what is the initial? Aren't there technically two?

    edit: Isn't simply using a piston to compress a gas reversible? However, along this path there is still no heat transfer..?
     
    Last edited: Jul 9, 2016
  9. Jul 9, 2016 #8

    Twigg

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    First, entropy can be transferred from surroundings (as Q/T) or produced in the system (during an irreversible process). In English, the change in entropy of a system is equal to the entropy transferred into the system (the sum of all Q/T's) plus any entropy generated by irreversibilities. In math, ##\delta S = \sum \frac{\delta Q}{T} + \sigma##. What makes a reversible process reversible is that the entropy generated (##\sigma##) equals 0. The second law says that entropy can't be destroyed, so ##\sigma ## is never less than 0. This means you can't reverse a process that generates entropy, because to do so would destroy entropy. You are correct in saying that the entropy in this system is coming from the work done by the electrical field. To be a little more precise, the entropy is generated by the inefficient work done by the electric field. The stored potential energy spontaneously decays into uncontrollable kinetic energy, and by Earnshaw's theorem there's no way to restore that potential energy once you let it go without doing a whole lot of work to gather up those particles while keeping the second law happy. Hope that helps!
     
  10. Jul 9, 2016 #9
    That did help, as did the other responses by Chester! So the way to quantify what sigma is would be to use his process, got it. However, I'm still trying to understand. Say I did have two states, how would I run an integral for this "pretend" reversible process? Would it be through utilizing a PV diagram or?
     
  11. Jul 9, 2016 #10

    Twigg

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    Gold Member

    You can do it with a PV diagram, but that's not the only way to do it. By doing the process integral on a PV diagram, you are effectively treating it as a chain of infinitesimal constant volume and constant pressure processes, for example constant volume hearing and isobaric expansion/compression. You can calculate and integrate dQ/T for each, rewriting the integrated in terms of V or p respectively. There's no reason you couldn't do isothermal processes and isobaric processes instead. It just changes the path integral you do from one diagram to another.
     
  12. Jul 9, 2016 #11
    Alright, I guess I'm just struggling to figure out how to develop this process. Say I did know the initial pressures of both sides of gas, and I know know the final pressure of the combined gases. As stated, the temperature remained constant. So how would I develop some reversible path integral with this information?
     
  13. Jul 10, 2016 #12
    Here's one way. You take the two original gases and separate them, putting each of them into a cylinder with a piston. You then compress the lower pressure gas isothermally and reversibly until its pressure is equal to the final pressure from the irreversible process, and you expand the higher pressure gas isothermally and reversibly until its pressure is equal to the final pressure from the irreversible process. Then each gas will be in the same final state it had in the irreversible process. You do this isothermally because the final temperature in the irreversible process was the same as the initial temperature. You calculate the entropy change of each gas in each of these separate reversible processes and add them up. This will be the entropy change for the gases in the irreversible process.
     
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