# Entropy as a state function?

1. May 11, 2009

### Andrew_

If we consider an isolated system in which a process occurs, then according to the clausius inequality :

$$dS \geq \frac{dQ}{T}$$

Since dQ = 0 , it follows that if the process occurs reversibly dS = 0 and irreversibly dS > 0. But entropy is a state function , how could this possibly be ?

It then occured to me that the universe itself is an isolated system and its entropy is not a state function. Could we then generalize and say that the entropy of an isolated system is not a state function ?

I've found some very brief treatments on this in recent thermodynamic books , but none seems convincing to me. In one book , it is claimed that an irreversible process that corresponds to a reversible process cannot exist in an isolated system since there needs to be the 'uncompensated heat' ( N , according to Clausius ) that must be supplied to the system from the surroundings. Thus , the system is no longer isolated, and the clausius inequality holds no longer for such an irreversible path.

Makes sense ? ... any ideas ?

2. May 11, 2009

### Andy Resnick

Interesting question. My reference is not handy, but I would guess that the entropy is not a state function. Although we can identify an entropy based on micro/macrostates of an equilibrium configuration, the change in entropy as one goes from one equilibrium state to another is, in general, path dependent, IIRC.

Again, I need to double-check my reference (Rational Thermodynamics).

3. May 11, 2009

### Andrew Mason

Entropy is a state function. It is the integral of dQ/T along the reversible path between two states. The change in entropy of a system in a reversible process is 0 only if the path is closed (ie. the system returns to the original state). The entropy change of a system along a reversible path from A to B can be negative or positive. The entropy change of the system + surroundings between any two states can never be negative. It will only be 0 if the system and surroundings return to the original state.

AM

4. May 12, 2009

### Mapes

Yes, entropy is certainly a state function. Andrew (first Andrew), while it may be true that a reversible process could be modeled by an irreversible process followed by heat removal, I don't see what relevance this has. The substitution is for our own convenience; nobody is claiming that thermal energy must be periodically withdrawn from our universe.

If I've misunderstood your question, could you please give the book reference?

5. May 12, 2009

### Andrew_

Andy Resnick, it would be interesting if you provide a rational thermodynamics perpsective because Im myself still new to this field.

Andrew Mason, I agree with you. What you posted is well known , nothing controversial there, but how would you explain the fact that the entropy of the universe - an isolated system - is a path function ?

Mapes, if you assume that a reversible process occurs in an isolated system , then its complementary irreversible path cannot occur without supplying heat/work to the system. Since in general :

$$W_{rev} + N = W_{irr}$$

and N > 0 , it follows that to achieve the irreversible path , the system is no longer isolated. Thus , the clausius inequaity is no longer valid. This is a realistic situation , not a conceptual one , wouldn't you agree ? This could make sense afterall because the clausius inequality is derived on the basis of a non-isolated system which exchanges energy with the surroundings.

This is the treatment I found in http://books.google.com/books?id=RZE2aYsPe0QC" as well in the same author's more recent ( yet unpublished ) book on classical thermodynamics , I just got a copy of the book from one of the co-authors.

Last edited by a moderator: Apr 24, 2017
6. May 12, 2009

### Andrew_

Please note that in generalized thermodynamics , the author introduces something called "calortropy" , which is the usual entropy but for irreversible processes.

He writes , for example, the clausius inequality $$d \Psi \geq 0$$ for isolated systems , where psi = calortropy. Why he chooses a different name for non-equilibrium entropy ... Im yet to find out.

7. May 12, 2009

### Andrew Mason

If you mean that the entropy of the universe is a path dependent function, this is an incorrect statement. The change in entropy of the universe between states A and B is the integral of dQ/T along the reversible path between those two states. There is only one such path.

AM

8. May 12, 2009

### Mapes

No macroscale process is truly reversible. We often assume reversibility to make thermodynamic problems easier to solve, but any proof relying on actual reversibility of an arbitrary process in the known universe is wrong. It's as wacky as assuming a frictionless universe and then saying that something is wrong with our theories of friction when friction is actually observed.

9. May 12, 2009

### atyy

I understood the question something like this. Entropy is a state function that can be assigned to a system in equilibrium. An isolated system is either in equilibrium or not in equilibrium. If it is in equilibrium, then entropy will not increase. If it is not in equilibrium, we cannot assign an entropy to it. So it seems that we cannot see entropy increase in an isolated system? As I understand, the traditional cheat around this is to introduce a "constraint". Consider a thermally isolated box with a wall in the middle that constrains all the gas molecules to be on one side of the box. Each side of the box is in equilibrium, so presumably we can assign an entropy to it. Now remove the constraint, and the gas expands irreversibly to fill the box. When it finally attains equilibrium, we can again assign an entropy to it. So with the help of a "constraint", we can see entropy increase in an "isolated" system. (But if it's "isolated", who removed the "constraint"?) Now, how do we find the entropy change associated with this irreversible adiabatic free expansion? We could find a reversible path that takes the system from the same initial to final states, and integrate dQ/T along it. One such reversible path involves making the system do work and then heating it. (But if you can heat the system, how can it be "isolated"?)

10. May 13, 2009

### Mapes

The "heating" here is virtual, though, yes? Just an imaginary process used to simplify the calculation?

11. May 13, 2009

### Andy Resnick

After further reading, I am more confused than before. Hopefully one day I'll have the time to carefully work through this subject.

First, I have to correct myself: for equilibrium systems and reversible processes, the entropy is a state function: the entropy is a functional only of the current state of the system (no memory).

However, I do not understand how to apply Clausius' expression $\int\frac{dQ}{T}$ to irreversible processes or nonequilibrium conditions. Is 'T' even defined? What does dQ mean for a dynamic process? The more I read, the more I guess that while the entropy can be defined for a given process, the less meaning 'state' has. The Onsager relations appear to be a perburbation approach (using fluxes and forces), but I just don't know enough to say any more than that.

12. May 13, 2009

### Andrew Mason

Q

The process determines the final state. If the process is irreversible, the final state is different than in a reversible process.

For each set of initial and final states there is a reversible path between the two states. The change in entropy is the integral of dQ/T along that path.

Perhaps it is best illustrated with an example. Consider an adiabatic expansion of a gas at pressure P1, temperature T1 and volume V1. It expands adiabatically until its volume doubles. So the process is adiabatic free expansion to double the volume.

The state of the gas is determined by the three variables, P, V and T. If you take the irreversible path of free expansion, with the result that no work is done, there is no change in internal energy (Q=0, W=0). So the final state is P=P1/2, T=T1, V=2V1.

This is a very different state than a reversible adiabatic expansion to double the volume. In this case it would be a process in which the internal and external pressure differ by an infinitessimal amount at all times, so work is done as the gas expands to double the volume. Consequently, the internal energy of the gas is reduced by the amount of work done and, as a result the final state is P<P1/2; T<T1; V=2V1. So the final state of the reversible expansion is different than in the non-reversible path.

To find the change in entropy of the adiabatic free expansion you find a reversible path between the initial and final states. In order to do this, heat must flow into the gas (there is no reversible adiabatic path between the initial and final states) so the integral of dQ/T (entropy change) is greater than 0.

AM

13. May 13, 2009

### Andy Resnick

Is that true? For example, when a protein folds, your statement seems to imply that only one path through configuration space is allowed, but that contradicts experiment. Plus, there is no reversible path under physiological conditions, AFAIK.

Has anyone worked out the entropy for a cycle of Na-K-ATPase, or a motor protein, or anything like that?

Plus, your argument leaves out many other thermodynamic variables such as the magnetic field, particle number, etc. etc.

14. May 13, 2009

### Count Iblis

If the changes in the isolated system are slow enough that they proceed in a quasistatic way, you can still apply thermodynamics by enlarging the thermodynamic state space. Then the entropy is a state function, but the thermodynamic state it depends on is then defined by many more external variables, temperatures, particle numbers etc. etc. E.g. you can think of a gas in an isolated box described by a position dependent temperature and density.

Now, if the changes in the system do not proceed quasistatically, you can still find an equivalent quasistatic process that describes the irreversible transition. Example: Consider a mixture of hydrogen and oxygen in a an isolated container. This gas explodes, but the container is strong enough to contain te explosion. No heat leaks out, and eventually the gas will settle down and it will be in a state that can be described by thermodynamics.

An equivalent quasistatic process can happen using some catalist that lets the oxygen slowly react with hydrogen. You can then interpolate from the initial state to the final state by introducing the number of molecules of hydrogen, oxygen and water. We then have:

dE = T dS - P dV + mu_{O2}dN_{O2} + mu_{H2}dN_{H2}+mu_{H2O}dN_{H2O}

During the process, dE = 0, and dV = 0, so during the chemical reactions we have:

dS = [mu_{O2}dN_{O2} + mu_{H2}dN_{H2}+mu_{H2O}dN_{H2O}]/T

Then the changes in the particle numbers also determine the temperature change, because we know how the internal energy depends on the particle numbers and temperature and the internal energy stays constant.

We can thus integrate dS and dT from the initial state to the fiinal state along the quasistatic trajectory. All this follows from the generalized version of the fundamental thermodynamic relation.

Another example: Free expansion.

As explained above by Andrew Mason, the entropy will increase, and there is a reversible way to realize this process, but that involves an external heat source. So, you let the gas expand adiabatially so that it performs work and then you supply heat from an external source (which can be done in a way that is arbitrary closely to being reversible).

But there is third way which is quasistatic and does not involve an external heat source. We imagine that the volume V of the system is divided in two parts: V = V1 + V2, and all the gas is located in V1. The volume V2 is vacuum. We then split the vacuum section by making a new boundary very close to the boundary between V1 and V2. Then we remove the old boundary so that we get a free expansion of the gas into a slightly larger volume.

Clearly the gas does not perform any work in this process. Because when we remove the old boundary, the new boundary has been fixed in place. The gas expands and then bumps into the new boundary. No enegy is lost from the gas or extracted in the form of work. So, the internal energy of the gas stays what is was.

But if we have increased the volume infinitessimally, then the new state of the gas (when it has settled down) can be computed from the fundamental thermodynamic relation as this relation relates two thermodynamic states that are infinitessimally separated. It doesn't matter if you go from one state to the other in a reversible way or not. So we have:

dE = T dS - P dV

Now, dE = 0, therefore:

dS = P dV/T

Note that we are not saying here that P dV is work performed during the free expansion. All we are using is the fact that if you have two states in thermal equilibrium that differ infinitesimally in S and V by dS and dV, then you have that dE = T dS - P dV. We then choose the two states by considering an infinitessimal free expansion starting from a given state. This is then a quasistatic process that is irreversible.

We can repeatedly insert a extra boundary and remove the old boundary, so this process can be repeated until the volume is increased by a large amount. The temperature remains constant, and we have that P = N k T/V. So, we can integrate the expression for dS to obtain the final entropy.

15. May 13, 2009

### Andrew Mason

I am not sure that this is a reversible or quasistatic process. If you reverse the process, you have to do work on the system to compress the gas. How can a "reversible" process in one direction do no work at all and with an infinitessimal change in conditions, the reverse process requires work?

AM

16. May 14, 2009

### atyy

Count Iblis's solution seems very clean to me. There's still the question of who removes the constraint of an "isolated" system, but I suppose that's done by the expansion of the universe?

17. May 14, 2009

### Andy Resnick

I still don't understand why we have to rely on concepts like "slow enough", "infinitesimally", etc. Do you mean thermodynamics does not apply to explosions, or fractures, or femtosecond processes, ... for example?

18. May 14, 2009

### Andrew Mason

The laws of thermodynamics certainly apply. But they are not reversible processes. No real physical processes are. We study reversible processes because they establish limits on the usefulness of thermodynamic processes.

AM

19. May 14, 2009

### Count Iblis

It is an irreversible quasistatic process in the limit that you change the volume by infinitessimal steps.

It is similar to supplying heat to a system in a quasistatic way. If you supply a large amount of heat Q to a system then the entroopy will increase. But during this process, thermodynamics does not apply. An equivalent quasistatic process is to divide the Q into small chunks of dQ and then after each dQ goes into the system, the entropy increases by dQ/T. The reason why is simply that before you add the dQ and after you add it the gas is in thermal equilibrium (we simply wait long enough before adding the next dQ). We can then take again the relation:

dE = T dS - P dV (1)

and take the dE, dS and dV to be the changes in the state variables. We then have dV = 0, dE = dQ and thus dS = dQ/T.

Of course, this can be a reversible process if the heat is added in a reversible way. But this is not necessary. You can take a gas cylinder and add the heat by some irreversible process, e.g. by rubbing the outside of the cylinder thereby converting work to heat.

Anyway, the point is that the fundamental thermodynamic relation (1):

dE = T dS - P dV

or an appropriately generalized version, can be appied to processes that are quasistatic as it relates state variables that differer by an infinitessimal amount. Whether or not such a process is reversible is not relevant at all here.

If a process is reversible, then T dS must be the heat supplied to a system (which then must also be supplied in a reversible way) and P dV must be the work done by the system.

So, in this sense Eq. (1) is more fundamental than the relation
dS = dQ/T. The usual derivation of (1) involves two arguments. First one argues that in the limit of changing the volume (or other external variables) infinitely slowly, the adiabatic theorem of quantum mechanics implies that the system's entropy will not change. This combined wit the definition of temperature implies (1) in the special case that the system does reversible work.

Then one argues that since (1) involves state variables, the relation should be valid in general. There doesn't need to be any transition from one state to another state at all. You can simply have two systems with the same composition, one with internal energy E, entropy S and volume V and another system with internal energy
E +dE, volume V + dV and entropy S + dS and then dE, dV and dS must satisfy (1).

20. May 14, 2009

### Count Iblis

Thermodynamics doesn't apply in the general case. Because ultimately you have a system that consists of a huge number of particles. You want to describe the situation in a statistical way. Thermodynamics applies when given some external variables, the system has equal probability to be in any of the possible microstates.