1. May 16, 2013

burakumin

Hello

Can we consider the sentence “Adiabatic process cannot decrease entropy” (found for example here) to be true in any circumstances ? Here I do not want to focus on any particular situation. Hence the system is not necessarily a (perfect) gas; the transformation is not necessarily quasi-static; neither the system nor the environment are necessarily in equilibrium during and at the end of the transformation.

If true and considering this statement is not presented as a thermodynamic principle, how can we deduce it from the two principles? In particular I think we cannot say that the “exchanged entropy” is something like ∫δQ/T as T is not necessarily a defined entity because temperature might be heterogeneous.

Thanks

Last edited: May 16, 2013
2. May 16, 2013

DrDu

Maybe you should have a look at the work of Caratheodory.
http://en.wikipedia.org/wiki/Second_Law_of_Thermodynamics#Principle_of_Carath.C3.A9odory
He derived the existence of entropy from the statement that not all states in the neighbourhood of a given state can be reached in adiabatic processes.

3. May 16, 2013

burakumin

Hi DrDru,

So this mean you do agree with the fact this statement is always true ?

(edit)

The related article on adiabatic accessibility apparenlty agrees that entropy can be defined as a quantity that increases (or remains constant) between adiabatically accessible states so I guess the statement is true.

I already knew Lieb and Yngvason's method but I was not sure the definition they use for adiabatic process was fully equivalent to usual ones.

Thanks

Last edited: May 16, 2013
4. May 17, 2013

Staff: Mentor

If you are referring to a closed system, δQ and T refer to the heat flow at the interface (boundary area) between the system and the surroundings, and T is both defined and measurable at the boundary. δQ and T refer to what is happening during the irreversible (or reversible) change. However, when applying the Clausius inequality, you need to compare ∫δQ/T with the entropy change between an initial and a final equilibrium states of the system.

5. May 17, 2013

Jano L.

burakumin, those are very good questions! You will not have trouble finding process where the statement is true - in all reversible processes it is true, as the entropy is constant; and in all processes in which the reservoir is in equilibrium state, it is again true, because of the Clausius theorem.

However, as you rightly suspect, these are not all possible processes. Consider the following example.

We have a mixture of oxygen and hydrogen in a vessel of initial volume $V_1$ with adiabatic walls, closed above by a movable piston and surrounded by reservoir at pressure $p_0$. The gases can react, either slowly or violently (explosion), depending on how we arrange the reaction.

We let the reaction proceed A) slowly, B) violently, in both cases the volume will expand, and we will freeze the volume at value $V_2$.

In case A), the work done on the reservoir is $p_0\Delta V$ and the entropy of the gas does not change.

In case B), the work done will be higher, because the piston will move so fast that it will work on a gas with higher local pressure $p_{ext} > p_0$.

In the end, the gases have the same volume $V_2$, but the energy of the gas in case B) is lower.

As the entropy is an increasing function of volume and energy, the final entropy in B) is lower than in A), which means it is lower than at the beginning.

So it seems that the Caratheodory principle cannot be extended to all adiabatic processes; it is true only for quasi-static ones.

6. May 18, 2013

burakumin

Hi Chestermiller. I do refer to a closed system (I'm not sure you can call a process adiabatic if it exchanges matter). My question does includes cases where the temperature is not global or not even definable (given that it's apparently possible). So things ∫δQ/T may be undefinable. And this is precisely the purpose of my question. Can I assert that, even in these "chaotic" cases and considering that my system σ is closed from its environment ε and that the process ε applies to σ is adiabatic, the entropy of σ cannot decrease (as well as the entropy of ε as ε and σ have symmetric properties here) ?

Hi Jano L.

I'm sorry but I'm not sure I agree with you about your example. I think there is a trap with "we will freeze the volume at value $V_2$". How will you ? I would say that:
- Either you let the system evolve naturally and in that case you will have $V_2^{(A)} < V_2^{(B)}$.
- Or you really constrain the volume by choosing $V_2^{(A)} = V_2^{(B)}$ and in that case you must provide some energy to the system to do so (or equivalently you must reduce the energy it sends to its environment)

7. May 18, 2013

Staff: Mentor

The reaction between hydrogen and oxygen is highly irreversible, and this does not depend on the reaction rate. Process A cannot be regarded as quasi-static, because the reaction is not maintained close to equilibrium. There is a major increase in entropy in both scenarios because of the entropy change due to reaction. Just consider the case where the volume does not change and the process is adiabatic. In the beginning you have hydrogen and oxygen, and in the end you have water. This spontaneous process in an isolated system takes place with a big increase in entropy. So in both scenarios, the change in entropy is positive, although, in scenario B, it may be a little less of an increase than in scenario A. The conclusion is that, in both scenarios, the Clausius inequality is satisfied.

Incidentally, the entropy is a function not only of temperature and pressure, but also of species concentrations.

Chet

8. May 18, 2013

Staff: Mentor

By definition, a closed system does not exchange matter.

As I said in my earlier post, ∫dQ/T is always definable, since, irrespective of how the temperature and conditions vary within the system, the heat flux at the boundary and the temperature at the boundary are always available for measurement. If there is no heat transfer between a closed system and the surroundings, the Clausius inequality is always satisfied.

9. May 18, 2013

burakumin

I know what it is a closed system. Sorry but my remark was perhaps not clear (and I'm not an English native speaker). What I meant was: given that I'd only mention adiabatic process, I thought it was obvious that the system was necessarily closed, because I'm not sure it's correct to speak about an adiabatic process acting on a open system anyway.

You were refering to cases where "initial and a final equilibrium states of the system." and I do not want to restrict myself to those cases only.

Furthermore, just a small remark about the "∫dQ/T is always definable", this requires that temperature is (possibly locally) definable, right ? But Mr Denker says here :

10. May 18, 2013

Staff: Mentor

It actually is correct to speak about an adiabatic process acting on an open system. For both open- and closed systems, dQ represents the heat flow at the boundary associated with conductive and/or radiative heat transfer. This does not include sensible heat (i.e., internal energy) carried into and out of the system by mass entering or leaving the system. This is included separately in the heat balance for an open system.

As I said in previous posts, ∫dQ/T in Clausius inequality refers to the heat flux and the temperature exclusively at the boundary of a closed system. The temperature at the boundary is accessible, and is not mysteriously undefinable. If you want to, you can measure the temperature at the boundary as a function of time during the process. The temperature T in this integral does not refer to the temperature at any other location within the system, except at the boundary.

If you are willing to accept the concept that the entropy and internal energy are well-defined even during non-equilibrium situations in which irreversible changes are occurring within a system, then the Clausius inequality applies to these situations as well. I happen to be one of those who subscribe to this concept. However, there are many members of Physics Forums who assert that the thermodynamic functions can only be defined for equilibrium situations.

11. May 20, 2013

Jano L.

The final volume can be restricted by installing some locking mechanism to the cylinder, which will arrest the motion whenever the piston reaches position with volume $V_2$. This does not require any further work, but even if it did, it does not seem to be a problem, since only the transfer of heat is forbidden.

Indeed, ordinarily the reaction is done in such a way that it is not reversible (combustion with release of water to atmosphere). But it can be made reversible if the system is in equilibrium and the parameters of the system are changed slowly. For example, we can use membrane that is penetrable to hydrogen, but not oxygen, and slowly change the volume (and pressure) of the container.

However, now I realize that the example with chemical reaction is quite complicated, as there will be three different gases, and it is not clear that one could really set the process in such a way that would achieve decrease of entropy.

I tried to think of a simpler example, but so far with no success. I thought that the fact that the system can do more work in adiabatic irreversible processes than in adiabatic reversible process with same final volume will be sufficient to show decrease of entropy, but I was wrong - the other variables play important role.

I still think that it may be possible to decrease the entropy in adiabatic process, if it involves irreversible process in reservoir. The common formulations of 2nd law do not seem to be as strong as to prohibit drain of the system entropy to reservoir.

12. May 20, 2013

Jano L.

I am afraid that this is not general enough to answer burakumin's question.

Why do you think the Clausius inequality holds in non-equilibrium process?

True, if the system is contained in a metal vessel, the boundary has temperature.

But the temperature in the Clausius inequality is actually the temperature of a large reservoir that is maintained at equilibrium and in contact with the system. It may be sometimes replaced by the temperature of a part of a system on the boundary, if the heat transfer is quasi-static, but I think not in general.

In our case there is no reservoir, and no heat flux normal to the boundary. It is not clear whether the Clausius inequality is applicable to the heat transfer occurring in the boundary itself, since the boundary is quite thin and not a great reservoir maintained at equilibrium.

Internal energy may be defined for non-equilibrium, but I think for the entropy and thermodynamic temperature there is no apparent way to do so - their definition is based on equilibrium states and their quasi-static changes. Do you know other definition which works for non-equilibrium states as well?

13. May 20, 2013

Staff: Mentor

14. May 21, 2013

DrDu

The reaction proceeding slowly does not mean that it is reversible. As you said in a latter post, you will need semipermeable membranes to make the process reversible. I see no reason why the work done on these membranes should sum up to $p_0 \Delta V$.

15. May 21, 2013

Jano L.

Chestermiller, your view is expressed very clearly, and it might be right. However, you are phrasing the Clausius inequality differently from what I learned. You use the temperature of the system $T_s$ on the surface to state the inequality

$$\Delta S \geq \int \frac{dQ}{T_s}. ~(1)$$

But as far as I know, the Clausius derivation implies only

$$\Delta S \geq \int \frac{dQ}{T_r}, ~(2)$$

where $T_r$ is the temperature of the reservoir in thermal contact with the system.

Often the difference is immaterial, mainly for quasi-static heat transfer, but sometimes it isn't. For example, if $T_r > T_s$, the formula (1) predicts higher increase of entropy of the system (and the whole super-system as well) than the formula (2). Can you say why do you prefer (1) over (2) ?

The form (2) is particularly interesting, because it allows the system to get so far from equilibrium that it does not have temperature at all, not even locally (consider hot ionized gas). It thus seems more general.

16. May 21, 2013

Jano L.

The basic idea was that if the process is slow, the pressure outside the container will be maintained at $p_0$. Then the volume work on the atmosphere is given by $p_0 \Delta V$. But now I realize that the example had another flaw: it is not clear that the reaction can be made such that B) will end up with decreased entropy (the presence of three chemical species makes it hard).

17. May 21, 2013

Staff: Mentor

The temperature and the normal component of heat flux at the interface between the system and the surroundings are both continuous. At the boundary, the temperature of the system is equal to the temperature of the surroundings and the normal component of heat flux vector within the system at the interface is equal to the normal component of the heat flux vector within the surroundings at the interface. This applies both to quasi-static heat transfer as well as to highly irreversible heat transfer. For example, when we solve unsteady state heat transfer problems (and other more complicated systems), we always consider the temperature and normal component of heat flux to be continuous at the boundary. When we do this, we get the right answer.

The system does not know whether it is in contact with a reservoir, or with just other material that comprises the surroundings. Therefore, at the boundary, if there were a reservoir present, Ts=Tr. The reservoir is merely considered a part of the surroundings. Also continuous at the interface between the system and the surroundings is the traction vector, which is equal to the stress tensor dotted with the unit normal to the interface.

I hope this helps.

Chet

18. May 21, 2013

Jano L.

I agree, the formula (1) is quite plausible. In most situations it probably will work well, although it seems more restrictive than (2).

EDIT:
Yes, that is a good reasoning when the system actually is in thermal contact with the surroundings.

But the situation proposed in OP is somewhat out of the scope of the continuum theory. Imagine a system within a container made of hypothetical solid walls that are not subject to heat transfer. They are not part of the system, they are just boundary condition - the system cannot get into the wall. The system can interact with the outside only by performing/accepting work, by displacing some parts of the container (piston, propeller) or by changing the external electric/magnetic field.

Now imagine that some process is set up that begins at equilibrium state A, involves non-equilibrium states of the system and reservoir (explosion, sudden vibration of the piston, fields, etc...), and finally ends in equilibrium state B.

Then we do not have any temperature or heat differential to use in (1). As there is no heat reservoir either, the original analysis based on reduced heats is not valid and the Clausius inequality (2) has no meaning.

It seems that the Clausius inequality is not sufficiently general to address the original question in its generality.

I realize that without example, all this is bit hypothetical, but nevertheless, I think that it is interesting to think about this possibility.

Last edited: May 21, 2013
19. May 21, 2013

Staff: Mentor

I totally disagree. The continuum form of the Clausius Inequality is perfectly well suited to the irreversible scenario you describe. The problem with the usual form of the Clausius inequality is that it was stated in a manner that is not sufficiently precise mathematically, and this deficiency has been an endless source of confusion and misunderstanding by physicists, engineers, and chemists. The continuum formulation removes this deficiency. The continuum form of the Clausius Inequality is:

$$\Delta{S}≥ \int\int \frac{\mathbf q \cdot \mathbf n}{T}dAdt$$

where q is the heat flux at the boundary, T is the temperature at the boundary, n is an inwardly directed unit normal at the boundary, t is process time during the transition from state A to state B, and A is the boundary area of the system.
For the case you describe, q = 0 at the interface between the system and the surroundings for all times (adiabatic process), so that ΔS≥0. Therefore, the entropy must increase between state A and state B for an adiabatic process on a closed system.

Last edited: May 21, 2013
20. May 21, 2013

Jano L.

I understand your position. Do you have any other support for the validity of the above formula? Is there a derivation, or do you regard it an axiom?

21. May 21, 2013

Staff: Mentor

Both. I regard it as an axiom, but also, in the link provided in post #13, I discuss the derivation not only of the inequality relationship, but also of the missing terms that can change it from and inequality to an equality, even for an irreversible process. Here is the starting point for the derivation:

In the "bible of Chemical Engineering", Transport Phenomena by Bird, Stewart, and Lightfoot, there is a discussion of entropy generation on page 372 of the 2nd Edition (the most recent edition). It basically shows how to calculate the transient change in entropy within a system in which an irreversible process is occurring. It also spells out explicitly and in detail the local rate of change of entropy with respect to time at each location within the system. The development is very compelling. By using this formulation, you don't need to consider a reversible path in determining the change in entropy from equilibrium state A to equilibrium state B. But, instead, you would need to solve the transient differential equations of heat transfer and fluid flow, and from this solution you could calculate the entropy change. I think this is the kind of development that most students of thermo crave, and it can greatly contribute to one's understanding of entropy generation.

The development in BSL presents the derivation of the microscopic entropy balance in the form of of a partial differential equation (basically, the continuum mechanics approach). If one integrates this partial differential equation over a moving control volume that forms the boundary of a closed system, applies the divergence theorem, and applies the Reynolds transport theorem, one obtains the macroscopic entropy balance equation. I discuss all this in another post within the same thread. In the macroscopic balance for the change in entropy along an irreversible path between two equilibrium states, there are two positive definite terms that are integrated over the control volume. These represent viscous dissipation, and dissipation from irreversible heat conduction. If these terms are omitted, one ends up with the Clausius inequality in the integrated form that I presented.

22. May 22, 2013

DrDu

Chestermiller, so you basically plea to use some non-equilibrium thermodynamics approach as that found in your engineering book. The point is that all non-equilibrium thermodynamics have a limited range of applicability, too, and this range is often much harder to make precise than that of equilibrium thermodynamics. Chemical reactions can leave rather easily the range of linear non-equilibrium thermodynamics and there are but few general statements on entropy production beyond non-linear approximation (e.g. Prigogines criterion). Highly diluted gasses like in the outer atmosphere or the heliosphere of the sun are also examples where you can't even describe the system in terms of a single temperature.

23. May 22, 2013

Staff: Mentor

Well, yes, in order for this to apply, you must be able to treat the substances within the closed system as continuua, in which case the mean free path of the particles must be small compared to the physical dimensions of the system. I don't regard this as much of a limitation for most practical situations.

Even if you are unhappy with the entropy generation treatment that I presented, I still stand by the integrated form of the Clausius inequality that I gave. I regard this both as a variational calculus statement of the second law of thermodynamics as well as an operational definition of entropy.

(Incidentally, I should mention that the equations I provided did not include the effects of chemical reaction or diffusion. These are handled in the BSL development in a later chapter).

Chet

Last edited: May 22, 2013
24. May 26, 2013

burakumin

This is clear. Thanks to you, to Jano L. and to DrDru for this informative conversation.

25. May 26, 2013