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Reversible heat gain

  1. Nov 11, 2013 #1
    I know that ΔS=qrev/T for any process, but what is qrev here? It can't be identical to the q in ΔU=q+w - if it was, why don't we just write it q? The other problem is that it would mean that ΔS(surroundings)=-ΔS(system) which need not always be true.

    Secondly I want to derive the equation for ΔS(surroundings) in terms of ΔH(system). How do I go about this? Somehow ΔH(surroundings)=qrev seems to be used, but that can't be right - ΔH(system)=q at constant pressure is certainly true but q and qrev are not the same and the system need not necessarily be at constant pressure, and finally this equation gives ΔH(system) rather than ΔH(surroundings)!

    Sorry for the many complaints. Any explanation of qrev and this derivation, with relation to the q meaning heat gain of the system as in ΔU=q+w, would be very appreciated.
  2. jcsd
  3. Nov 11, 2013 #2
    "ΔS=qrev/T for any process" This is not precisely correct. There are an infinite number of processes (i.e., how q and w vary with time) that can take a closed system from one equilibrium state to another equilibrium state. You can calculate q/T (or, more precisely, the integral of dq/T) for each of these processes. However, only the subset of these processes that are reversible ones gives ΔS. All other processes between the two equilibrium states gives a lower value for the integral.
  4. Nov 12, 2013 #3
    What do you mean "only the subset"? Every chemical reaction is reversible - every reaction goes to equilibrium (be it extremely high or low yield and thus termed irreversible, or not). For a certain reaction, we might have a total heat gain of the system q, and an inner energy change, enthalpy change, etc.

    Can you explain the exact relationship between q and qrev? Preferably including any necessary mathematics. How can ΔS(system)=qrev/T and ΔS(surroundings)=-qrev/T yet ΔS(system)=-ΔS(surroundings) is invalid? Must come from some change in the meaning of qrev here. And finally how can we say ΔS(surroundings)=-ΔH(system)/T?
  5. Nov 12, 2013 #4
    The word "reversible" is being used here in two different contexts. Yes, we can control the concentrations of the reactants and products of a reaction so that the reaction can be run either forward or in reverse. In fact, we can even make a chemical reaction proceed along a reversible path so that the products and reactants at any time through the process are only slightly removed from their equilibrium concentrations. The later is close to what we mean by reversibility in the context of entropy change.

    Suppose you have a closed system in which a chemical reaction is occurring spontaneously from an initial equilibrium state in which the reactants are pure and unmixed (and there are no products present) to a final equilibrium state in which the reactants and products have attained equilibrium concentrations. In such a case, there is definitely an increase in entropy for the system, but, if the process had been run adiabatically, both Q and Q/T would be zero. Thus, for this irreversible path between the initial and the final equilibrium states, ΔS>Q/T. However, it is possible to identify other paths between the initial and final equilibrium states of the system, for which not only is Q/T greater than zero, but also Q/T = ΔS. These are limiting reversible paths, for which the reactant and product concentrations are only slightly removed from equilibirum values over the entire course of the process (and reaction). However, it is not possible to identify any process paths for a closed system in which Q/T>ΔS.

    So, q refers to any arbitrary process path between the initial and final equilibrium states, but qrev refers to only those process paths that are reversible in the sense that, at any point along the process, the system is only slightly removed from both mechanical and chemical equilibrium.
  6. Nov 12, 2013 #5

    Andrew Mason

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    The change in entropy of the surroundings is equal and opposite to that of the system only if the system and surroundings are at the same temperature while heat flow is occurring between the system and surroundings. If they are at the same temperature, the process is reversible (by definition). If they are not at the same temperature, the process is not reversible and the total change in entropy will necessarily be greater than zero.

  7. Nov 13, 2013 #6
    These are really excellent questions.

    Consider a closed system that is undergoing a process that takes the system from equilibrium thermodynamic state 1 to equilibrium thermodynamic state 2. At time t = 0, the system is in equilibrium thermodynamic state 1, and at time t = tfinal, the system reaches equilibrium thermodynamic state 2. At any time t during the transition from state 1 to state 2, the system has received a cumulative amount of heat q(t) from the surroundings across the interface between the system and the surroundings, and has done a cumulative amount of work w(t) on the surroundings at the interface between the system and the surroundings. Also, at any time t, the temperature at the interface between the system and the surroundings is T(t), and is the same for both the system and the surroundings. The cumulative amount of heat that the surroundings receives from the system across the interface is -q(t), and the cumulative amount of work that the surroundings does on the system at the interface is -w(t).

    If the path between state 1 and state 2 is a reversible path, the integral of dq/T for the system between its initial and final state is equal in magnitude and opposite in sign to the integral of dq/T for the surroundings between its initial and final state. If the path between state 1 and state 2 is an irreversible path, the same is true. However, if the process is irreversible, it is not possible to identify a single reversible path for both the system and the surroundings that takes them both simultaneously from their initial to their final states. There will be two different reversible paths, and two different changes in entropy, one for the system and one for the surroundings. The sum of these two entropy changes will be greater than zero.

    Here is an illustrative example. Consider a closed, adiabatic, rigid container with a partition, in which an ideal gas is present on one side of the partition and vacuum is present on the other side. At time zero, the partition is broken and the gas fills the container. At the same time, no change occurs in the environment surrounding the container. For this irreversible process, the initial state of the surroundings is identical to the final state of the surroundings. Also, q(t) = 0 and w(t) = 0 for the entire process, and thus ΔU = 0, and thus ΔT =0. Also, the integral of dq/T is equal to zero for both the system and the surroundings. Now, getting the change in entropy of the surroundings is pretty easy, since the initial state is equal to the final state. A reversible process involving no heat flow across the boundary of the surroundings would be necessary to take it from its initial to its final state. However, the heat flow across the boundary of the system required to take it reversibly from its initial state to its final state would be very different. The gas would have to do reversible work on the surroundings, and heat would have to flow from the surroundings to the system. So the heat flow to the system for a reversible path from its initial state to its final state would have to be very different from the reversible path heat flow necessary to bring the surroundings from its initial state to its final state.

    Another way of looking at this is that, if the system is make to transform reversibly from its initial equilibrium state to its final equilibrium state, the final equilibrium state of the surroundings will have to be different from its final equilibrium state along an irreversible path.

  8. Nov 13, 2013 #7
    Ok, thanks so far, I think I have understood somewhat why ΔS(system)=-ΔS(surroundings) is invalid (because qrev in the expressions refers to different reversible paths, one which takes the system to equilibrium and one which takes the surroundings to equilibrium, from their initial states).

    So for your ideal gas example, since no heat is exchanged anywhere, is ΔS(system)=ΔS(surroundings)=ΔS(total)=0 for this case?

    Can you clarify what you mean by "reversible path" and "irreversible path"? The above process is irreversible as you said, but do reversible paths exist for it? (For whom we could have ΔS=qrev/T where qrev is the heat gain of whatever part of the universe (be it system or surroundings) when that part takes this reversible path to reaching equilibrium?) What different paths could be taken for a given process (maybe a reaction, maybe an expansion like this - after all, seems to me reactions must have many possible paths, only one of which is reversible wrt the system and one path of which is reversible wrt the surroundings) and how do we know which are irreversible and which are reversible - and why is there only one reversible path (only one value of qrev) for the system, or independently for surroundings, in the process?

    Lastly, what is the relation to ΔH? Looks to me like, somehow qrev=ΔH(surroundings) where specifically qrev is the heat gain of the surroundings when the surroundings take the reversible path to equilibrium. What's the logic for this?
  9. Nov 13, 2013 #8
    No. ΔS for the surroundings is zero, but not ΔS for the system.
    Yes. A reversible path for the system is one for which it is only slightly out of equilibrium (either mechanical, thermal, chemical or combinations) during its entire transition from its initial to its final equilibrium state. The same goes for the surroundings. Since, in our example, the initial and final equilibrium states for the surroundings are identical, this constitutes a reversible path, and the change in entropy of the surroundings for this path is zero (since q = 0).

    A reversible path to get the system from its initial to its final equilibrium state involves the use of a piston (to allow the gas in the container to expand reversibly and do work on the surroundings) and a constant temperature reservoir in the surroundings (to allow the expansion to occur isothermally). So to get the gas in the chamber from its initial to its final equilibrium state reversibly, we need to allow the gas to interact both thermally and mechanically with the surroundings in a specially designed way (that differs significantly from the path it took during the actual irreversible process).
    As I said, for reversible paths, the system must be maintained close to equilibrium over the entire path (mechanical, thermal, chemical if necessary). My advice to you at this point is to temporarily forget about worrying about chemical reactions for a while. You will learn how this all applies to chemical reactions later in your studies. But right now, you need to solidify your understanding of how these second law basic concepts apply to thermal and mechanical process forcings.
    I don't quite follow specifically what you are referring to here. If a phase change is involved, then heat is given off to the surroundings or taken in from the surroundings essentially at equilibrium at constant temperature and pressure. In such a situation, ΔS=Q/T=ΔH/T for the system. Is this what you had in mind?
    Last edited: Nov 14, 2013
  10. Nov 14, 2013 #9
    Why does it have to be isothermal? Can't we integrate over the changing temperature, from dS=d(qrev)/T? Or is that part of the definition of the path being reversible ...

    As to the values of ΔS, you said that no heat is exchanged with the surroundings in this process, as it is irreversible here. If a reversible process led to the same initial and final results (of the system), there would also have to be an accompanying heat gain of the system, qrev, and thus an entropy change of the system. (Analogous for a reversible process leading to the same initial and final states of the surroundings as this irreversible process does - though this would hold a potentially different qrev, which here would refer to the heat gain of the surroundings during that reversible path.)

    But then how can we apply this to the irreversible process you described? Take the same value of qrev as the reversible-wrt-the-system process would have had? Then this would give us ΔS(system). How do we judge that ΔS(surroundings) is 0 - possibly, from the fact that the initial and the final states are the same, no change is needed at all (if we had a "reversible path" going, it'd be trivial because nothing need occur to bring the initial to the final state) so obviously there is no heat gain? I don't know if there's a more thorough way to explain it.

    This feels quite a lot clearer now.

    Ok so q=ΔH(system) at constant temperature and pressure, sure - but the relation ΔS(surroundings)=-ΔH(system)/T is universal isn't it? In fact it can be used to compare ΔS(universe) to ΔG to show they both determine whether a process (which can in theory be spontaneous both forward and backward) will be spontaneous forward, in the same way. I didn't think these relationships required constant temperature or pressure to hold? And otherwise we need to show that qrev=ΔH(surroundings) more generally, where qrev is the heat gain of the surroundings during the reversible path of the process for which you want to calculate ΔS(surroundings).

    Finally, for the process which occurs that we are taking q for, unless this process is actually happening via the reversible path we cannot say that q(Process)=qrev. Yet q(Process)=ΔH at constant pressure, no? i.e. at constant pressure it is the actual heat change that equals the enthalpy change. So how do we say qrev=ΔH?

    Sorry for the long posts. I feel like I am getting much closer to understanding.
  11. Nov 14, 2013 #10
    Big-Daddy!!!! You're The Man!! You are really catching on to this now, and are able to articulate it very well. You should give yourself a big pat on the back. This stuff is not easy. You were even perceptive enough to ask the following equation:
    You are absolutely correct. We don't need to integrate over a constant temperature. There are an infinite number of reversible paths between the initial equilibrium state and the final equilibrium state that we can integrate over. All of these paths give the exact same value of the integral of dqrev/T. I assume that you are familiar with the Carnot cycle, comprised of four expansion/compression steps, two of which are isothermal and two of which are adiabatic (where the temperature varies). Because it is a cycle, one of the isothermal steps is equivalent to the other three steps (two adiabatic, and one isothermal). So, in our problem, we can get reversibly from the initial equilibrium state to the final equilibrium state using two reversible adiabatic steps (along which the temperature is varying) and an isothermal step. This can be generalized further by connecting up any number of smaller 3-step segments that together take us from the initial equilibrium state to the final equilibrium state. We can make the 3-step segments as small as we want. In this way, we can create an unlimited number of paths with varying temperature, all of which are reversible and give the same integral of dqrev/T.
    I'm having trouble understanding what you are saying here, but ΔS(surroundings)=-ΔH(system)/T is not universal. You may want to think of the surroundings as just another "system" that has undergone a process from its initial state to its final state. This process may have been irreversible, in which case you need to conceive of a reversible path between its initial and final state in order to calculate ΔS of the surroundings. If the process for the surroundings was reversible (even though it may have been irreversible for the system), then you can calculate ΔS from the actual dq/T. In addition to being reversible (for the surroundings), if the pressure at the interface were constant, you could calculate ΔS for the surroundings as the integral of dHsystem/T.
    I am assuming we are talking about qrev for the system here. We can't say that qrev=ΔH unless the path is reversible and the pressure is constant. If we can identify a reversible path between the initial and final states that also has constant pressure, then for that path, qrev=ΔH.

    I'm not sure I've answered your questions. Perhaps you are asking about specific cases in which constant temperature reservoirs are being used in the surroundings, and where mechanical energy is conserved in the surroundings by, say, using weights that are raised and lowered gradually. As I said, I'm having trouble understanding what you are asking. Maybe you can cite a specific example that we can discuss.
  12. Nov 15, 2013 #11
    I see, so there will be an infinite number of reversible paths from the initial to final state, each of which (may) give a different qrev but all of which must share the same ratio qrev/T (at constant T) or more accurately the same value of ∫d(qrev)/T dT which we could obtain by expressing qrev for any temperature as a function of that temperature. So the existence of infinite possible temperatures and infinite possible temperature changes for the reaction to go through enables there to be infinite reversible paths (one at each temperature, if T is constant, or over each range of temperature if T is variable)?

    Hopefully this is pretty much all I'm still having trouble with. So you're saying that we need constant temperature (so that qrev/T is sufficient to calculate ΔS), pressure (so that qP=ΔH(system) where qP is the heat gain of the system in the process, at constant pressure) and that the path taken by the process be reversible (so that qrev=q), for ΔS(surroundings)=-ΔH(system)/T?

    But I thought the equation was just an analogous restatement of equation ΔG=ΔH-TΔS<0 for a forward-process (as ΔS(universe)>0 is true just as ΔG<0 is true, for any process occurring forward). Reversibility is of course a requirement for this equation to apply - that is, the reaction or process can go forward in both directions - but then, since all chemical reactions at constant temperature are, I thought, described by ΔG=ΔH-TΔS, wouldn't that imply that the process must take a reversible path in reality and thus that q=qrev for any chemical reaction?

    I agree, I don't have a basis for confusion except for in the case of chemical equations, because ΔG=ΔH-TΔS may not apply as a criterion for forward-direction spontaneity unless the reaction is reversible (so that if ΔG>0 it is spontaneous backwards and if ΔG<0 it is spontaneous forward). This is not necessarily true for simple expansion systems etc., so to say that ΔS(surroundings)=-ΔH(system)/T for such processes is non-universal is satisfactory. But for chemical reactions, it would seem that all reactions always take a reversible path, so as to write the Gibbs' free energy change equation above. Is this true? And if so, how do we go from there, knowing that q=qrev, to suggesting ΔS(surroundings)=-ΔH(system)/T even at non-constant pressure? (And dS(surroundings)=-dH(system)/T for non-constant P and T.) Take some approximation maybe?

    The reason I think they are analogous is that my book lists ΔG as coming from the same spontaneity principle as entropy of the universe, so that ΔS(universe)=ΔS(system)+ΔS(surroundings)=ΔS(system)-ΔH(system)/T at constant T, which means that ΔG=-TΔS(universe). Or is one (or both) of these two derivations inaccurate unless the condition of constant pressure is also present? Doesn't feel like that would make sense seeing as ΔG is often applied to gaseous phase reactions that change pressure as they occur ... also, are there any other processes to which ΔG is applied except chemical reactions (i.e. processes wherein paths typically taken by isolated systems are reversible)?
  13. Nov 16, 2013 #12
    As I said earlier, I wasn't referring to chemical reactions, but, if you are talking about reversible paths for a closed system, even with chemical reaction present, then, again, there are an infinite number of reversible paths from the initial equilibrium state to the final state, and these can involve temperature changes, pressure changes, and chemical concentration changes. The amount of reversible heat can vary between these paths, but the change in entropy must be the same for all. But, when chemical reactions are involved, you need to carefully control how you specify the way that the path is controlled. Chemical reactions that are occurring at a finite rate of reaction are occurring irreversibly, and you can't let this happen. You have to somehow make sure that, in some way, you control the concentrations along the way so that, at any stage of the transition, the concentrations are only slightly removed from chemical equilibirum.

    No. Even though a chemical reaction can run forward or in reverse, this doesn't mean that chemical reactions all occur reversibly in the thermodynamic sense. We are talking about two different meanings of the word "reversible." Most of the time, chemical reactions take place at a finite rate, and this means irreversibly. Only when the chemical reaction is at equilibrium can we perturb the equilibrium slightly by adding a small amount of reactants or products and say that the resulting change occurred with thermodynamic reversibility.

    When studying the thermodynamics of chemical reactions and chemical equilibrium, it has been found very convenient (and simplifying) to use the Gibbs free energy G as a basis and to work with open systems (rather than closed systems). As you study this more, you will get the idea of why this is. Also, the analysis is usually done at constant temperature and pressure, so that the focus is on the reaction. Once you get the equilibrium constant that way, you can use it in other situations as well.

    Are you familiar with the conceptual device known as the Hoff Equilibrium Box. If you are not, please take the time to read about it. If is a wonderful tool for helping to structure your understanding about the thermodynamics of chemically reacting systems.
  14. Nov 16, 2013 #13
    Thanks. So it seems to me that what we need to do is say that, for any chemical reaction to which the relation dG=dH-TdS applies, the surroundings take a reversible path from the initial to the final state and are at constant pressure. Then we also get the relation that dS(surroundings)=dH(surroundings)/T and we can derive the old relation as dH(surrounding)=-dH(system).

    Even if we have a gaseous phase reaction, e.g. aA + bB = cC + dD, in a container of fixed volume, we are justified in making the assumption that the pressure of the surroundings is constant (even when that of the container evidently is not), and, as we do for all chemical reactions to which we want to apply dG=dH-TdS, that the surroundings take a reversible path to their final state. Is this correct?

    I would like to look into this further, but can't seem to find enough information from Google. Can you suggest a book or internet reference?
  15. Nov 18, 2013 #14
    It is very hard to discuss cases with chemical reaction involved without your having more background in the thermodynamics of multicomponent systems. PF is not the right venue for presenting a tutorial on multicomponent thermodynamics. However, I can provide a reference where you can find what you are looking for, or at least for providing a basis for discussing things further here. Check out Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness. This is not the book that I learned from, but my daughter used it as an undergrad at Penn, and I thought it was a pretty good book. Some chapters you will likely be interested in are:
    10. Systems of Variable Composition. Ideal Behavior.
    15. Chemical Reaction Equilibria (section 15.3 discusses the van't Hoff equilibrium box).

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