Mathematical definition of thermodynamic reversibility

[studiot]

Many posters here argue against a proposition on the basis that it is 'non mathematical' rather than against the merits or demerits of the proposition itself.

 

So who can offer a mathematical definition of thermodynamic reversibility?

 

The only attempt (not even wikipedia tries http://en.wikipedia.org/wiki/Reversible_process_(thermodynamics ) that I can find is in Callen and that fails counterexamples in both Atkins and Carrington.

 

Remember any proposed definition must be applicable to any conceivable set of circumstances, not just specially constructed ones.

Edited June 16, 2014 by studiot

[Bignose]

Mathematically, a process is reversible if the before and after entropies are the same. A process is irreversible if the final state entropy is greater than the before state.

[studiot]

Given that there are no truly reversible processes in Nature, do you think melting and evaporation are irreversible, if carried out sufficiently slowly?

Neither are isentropic.

[lamironi]

I think melting, freezing, evaporating, dissolving are reversible.

[John Cuthber]

Given that there are no truly reversible processes in Nature, do you think melting and evaporation are irreversible, if carried out sufficiently slowly?

Neither are isentropic.

I don't' think they happen in a closed system so you would have to consider the entropy changes of whatever is driving the change of state as it's part of the system too..

[studiot]

 

I don't' think they happen in a closed system so you would have to consider the entropy changes of whatever is driving the change of state as it's part of the system too..

 

 

Surely that depends upon your specification of the system.

I find that getting this specification is the key to success or failure in thermodynamics.

 

Further I note that entropy has been mentioned twice now.

 

How does one approach that before one has defined 'reversible'?

[John Cuthber]

You are right.

If I look at a hot object, cooling down it is clear that its energy is not conserved and its entropy changes.

That's not a problem with conservation, it's a problem with not choosing the system correctly.

 

For a closed system delta S is zero for a reversible change, but positive for an irreversible one (as Bignose said)

 

When you look at melting ice it is isothermal it's reversible (as long as it's slow) and, it's not isentropic so it looks like it's an exception to the delta S rule.

But the process doesn't happen in a closed system.

A (perfect) thermos flask of ice water doesn't melt.

It will melt if you add heat to it.

But something has to happen to create that heat.

Once you let heat into the thermos flask, it's no longer a closed system and you have to consider the whole system which includes whatever is producing that heat.

For the system as a whole, the change in entropy is positive.and the overall outcome isn't reversible.

[studiot]

Well I don't agree with this interpretation.

 

1) Any heat flow in either the first or second law, by definition, crosses the system boundary. You are suggesting otherwise.

 

2) Bignose's definition should have read for a cyclic process. Therefore we have to melt and refreeze the water. then the entropy change is indeed zero.

 

3) Why only closed systems I asked for any general system, without special conditions.

4) I was looking for a mathematical statement or formulae or calculation I could perform on any process that would have as its outcome "this process is or is not reversible" None of the foregoing discussion shows this.

 

 

This last is a really difficult issue that I have not seen supplied anywhere, as I said at the outset.

 

Edited June 16, 2014 by studiot

[Bignose]

2) Bignose's definition should have read for a cyclic process.

Nope. True for all processes, not just cyclic. You asked for the mathematical definition. I'm sorry that it is complicated. Entropy is not a very intuitive thing, I agree. But that it is the mathematical definition.

Edited June 16, 2014 by Bignose

[studiot]

 

Nope. True for all processes, not just cyclic. You asked for the mathematical definition. I'm sorry that it is complicated. Entropy is not a very intuitive thing, I agree. But that it is the mathematical definition.

 

 

C'mon what you said was neither mathematical, nor correct, and in particular should not involve a quantity called entropy, unless you first define that.

 

Further you have offered no answer to my post#3, that I made in response to your post#2.

 

Here is a non mathematical definition.

 

A thermodynamically reversible process is one that is reversed by an infinitesimal change in the conditions of the surroundings.

 

For instance consider a light piston being pushed out under pressure against a lower resisting pressure.

An infinitesimal change in the surrounding pressure will not reverse the motion of the piston.

 

But if the external pressure is equal to the internal pressure then an infinitesimal change, in either direction to the external pressure will result in the piston moving in or out.

We can also observe that the system is in equilibrium with its surroundings, which is why thermodynamicists often say that reversible means equilibrium.

 

But that is in words, not mathematics.

[Bignose]

C'mon what you said was neither mathematical, nor correct, and in particular should not involve a quantity called entropy, unless you first define that.

 

Further you have offered no answer to my post#3, that I made in response to your post#2.

 

Here is a non mathematical definition.

 

A thermodynamically reversible process is one that is reversed by an infinitesimal change in the conditions of the surroundings.

 

For instance consider a light piston being pushed out under pressure against a lower resisting pressure.

An infinitesimal change in the surrounding pressure will not reverse the motion of the piston.

 

But if the external pressure is equal to the internal pressure then an infinitesimal change, in either direction to the external pressure will result in the piston moving in or out.

We can also observe that the system is in equilibrium with its surroundings, which is why thermodynamicists often say that reversible means equilibrium.

 

But that is in words, not mathematics.

What do want, then? Is this better? Reversible if [math]\Delta S = 0[/math] and Irreversible if [math]\Delta S > 0[/math]. There, now it is in math.

 

If you want to define entropy, I can do that, too, though I suggest you go and take a peek at a good thermodynamics text, because as I wrote above, it is tricky. I'd suggest Smith, Van Ness, and Abbott's Introduction to Chemical Engineering Thermodynamics simply because that is what I learned from (5th edition) and I thought it was good. If you really want to know thermo, though, no better text than Tester and Modell, though it is extremely advanced (don't start with it).

 

Just so that you don't accuse me of not answering you again, entropy is a particular thermodynamic state function. It's calculation can be based on internal energy, enthalpy, pressure, etc. It has a formal definition for heat added reversibly given by

 

[math]dS = \frac{dQ^{rev}}{T}[/math] but with this definition, the above properties can be shown, so it is not just defined for reversible processes. It can be demonstrated to be a state function, so its value is only based upon the current state of the system, not the path on how it got to that state.

 

I didn't think I needed to answer post #3 because I agreed with the answers you have already been given.

 

I have no idea why you think it is incorrect. Please see the above texts. You asked what made a process reversible or irreversible -- zero or positive change in entropy answers that. So it isn't 'not correct'. Or if you think so, please provide some sources, because the definitions I've given above can be found in any decent thermodynamics text (and many no so good ones, too).

 

The answer I gave is directly from the definitions of reversible and irreversible. Again, just because you don't like it, doesn't mean you get to just declare it 'not correct'. If you you really think so, you have some work to do to overturn the standard definitions of thermodynamics as we know them -- and as they have proven themselves supremely successful -- today.

Edited June 17, 2014 by Bignose

[studiot]

 

What do want, then? Is this better? Reversible if and Irreversible if . There, now it is in math.

 

If you want to define entropy, I can do that, too, though I suggest you go and take a peek at a good thermodynamics text, because as I wrote above, it is tricky. I'd suggest Smith, Van Ness, and Abbott's Introduction to Chemical Engineering Thermodynamics simply because that is what I learned from (5th edition) and I thought it was good. If you really want to know thermo, though, no better text than Tester and Modell, though it is extremely advanced (don't start with it).

 

Just so that you don't accuse me of not answering you again, entropy is a particular thermodynamic state function. It's calculation can be based on internal energy, enthalpy, pressure, etc. It has a formal definition for heat added reversibly given by

 

but with this definition, the above properties can be shown, so it is not just defined for reversible processes. It can be demonstrated to be a state function, so its value is only based upon the current state of the system, not the path on how it got to that state.

 

I didn't think I needed to answer post #3 because I agreed with the answers you have already been given.

 

I have no idea why you think it is incorrect. Please see the above texts. You asked what made a process reversible or irreversible -- zero or positive change in entropy answers that. So it isn't 'not correct'. Or if you think so, please provide some sources, because the definitions I've given above can be found in any decent thermodynamics text (and many no so good ones, too).

 

The answer I gave is directly from the definitions of reversible and irreversible. Again, just because you don't like it, doesn't mean you get to just declare it 'not correct'. If you you really think so, you have some work to do to overturn the standard definitions of thermodynamics as we know them -- and as they have proven themselves supremely successful -- today.

 

 

If you were less condescending and more open minded you would realise that all I have stated is purely conventional, and that you yourself are under some misunderstanding.

 

You didn't like the references I gave from full and famous professors at Oxford or Cambridge, and referred me to Van Ness.

 

Well I looked at my old and battered 4th edition and on page 39 found the following standard definition

 

 

A process is reversible when its direction can be reversed by an infinitesimal change in external conditions.

 

How does that compare to your claim that my post #10 attempts to overturn standard definitions?

 

I would say that it is essentially the same as my post#10.

 

So I looked further (in Van Ness) at your claim that the Clausius inequality was not defined by a cyclic process.

 

Well from page 148, he develops the conventional Carnot theory for cyclic processes, leading to a cyclic integral, specifically

 

 

Thus the quantities dQ_rev/T sum to zero for any series of reversible processes that cause a system to undergo a cyclic process

 

The above statement is reasonable and correct.

But it is not what you stated.

 

 

Bignose

Post#2

Mathematically, a process is reversible if the before and after entropies are the same. A process is irreversible if the final state entropy is greater than the before state.

Post#4

Nope. True for all processes, not just cyclic

 

Clearly post#2 claims that all reversible processes have no (zero) entropy change, and post#4 confirms you are asserting this.

 

Compare this with your Van Ness reference when he summaries his 2nd Law discussion

 

 

The change in entropy in any system undergoing a reversible process is found by integration

[math]\Delta S = \int {\frac{{d{Q_{rev}}}}{T}} [/math]

 

 

This in general will not be zero.

Indeed, Van Ness offers various methods of calculation, depending upon the circumstances.

 

Unfortunately Van Ness does not provide a T-S indicator diagram for a Carnot cycle in which there are four reversible processes in series to form a cyclic overall process as per the first quote.

 

I therefore display one here.

 

The T-S diagram has the form of a rectangle with four reversible processes

 

AB in which entropy increases, at constant temperature

BC in which entropy remains constant but temperature falls

CD in which entropy decreasess

DA completes the cycle and raise the temperature, at constant entropy.

 

I know you did not say that entropy cannot decrease, but many believe this so please note that it must decrease somewhere in any cycle for which it's change sums to zero.

The Carnot indicator diagram shows this very clearly.

 

As a matter of interest, the originator of the inequality and cyclic integral, Clausius actually said (translated : Wilson : Cambridge University)

 

 

No cyclic process exists which has as its sole effect the transference of heat from a colder to a hotter body

 

All this is, of course, an interesting digression from the topic.

 

In thermodynamics three conditions (and their opposites) are recognised, reversible, equilibrium and quasi-static.

 

They are all subtly different, but usually coincide.

[Bignose]

If you were less condescending and more open minded

Come on. Don't read context into posts that aren't there. What you take as condescension, I might call being straightforward. E.g. see my very first post. More than that, there is no need to be insulting about it. I have not been deliberately insulting to you.

 

Again, as was pointed out, you cannot just look at the sum total of entropy for one part of the process. In your cycle, you are only looking at the fluid that is going through that cycle. If it comes back to the same state, the change in entropy in that fluid -- entropy being a state function -- will be zero. But that is not the change in entropy of the universe. That has increased. Something was done irreversibly. If it wasn't, I could build a perpetual motion machine of the second kind. Proof is given in Tester and Modell.

[studiot]

Discussion of the 'Entropy of the Universe' is for philosophers and mystics, not classical physicists, so we must agree to differ on the value of such.

 

I would have welcomed some discussion about my topic though, since in the 50 years or so I have been successfully doing thermodynamics I have never come across a purely mathematical definition of reversibility.

[MigL]

Maybe this is a little off topic as it has little to do with thermodynamic aspects, but I always thought you could have a process, reversible or irreversible, where there is no entropy change due to the process involving few particles and proceeding in infinitesimally small steps. Please correct me if I'm wrong and explain.

 

If the above is true, is this not just a case of time reversal symmetry of CPT ?

And I always thought that C, P and CP violation has been confirmed. and so the general feeling is that some process must also violate T symmetry ( for simple systems of course ), and only the product CPT is inviolate.

[studiot]

Thank you for your input, MigL.

 

I'm not sure what part, in any, CPT plays in (particularly classical) thermodynamics.

 

But yes, in classical thermodynamics it is possible to construct theoretical systems/processes where there is no entropy change.

This prevents universal use of the maximum entropy criterion as a 'necessary and sufficient' criterion for equilibrium.

 

This situation is the problem with Callen.

 

An example construct (similar to the recent piston question in homework help) is presented and discussed in Carrington.

 

You mentioned reversibility as though we had a satisfactory (mathematical) definition. Achieving one is the point of this thread. I do not have an answer to pull out of a hat.

 

Do you have any suggestions?

 

One of the difficulties is that folks often rattle of thermodynamic statements without properly distinguishing between the system and the surroundings and changes in each. These variables are not described directly in the laws of thermodynamics, but are connected by the exchange variables across the system boundary. Proper specification of the system boundary and the system process need also to be made to complete a thermodynamic analysis.

 

[John Cuthber]

Entropy has been mathematically perfectly well defined for over a century

s = k ln w

If, for a closed system* the change in entropy is zero then the process is reversible, otherwise it isn't (and if it's negative time is going backwards)

 

 

* (the whole system, not just the bits you choose to look at)

[studiot]

So entropy is undefined, since you have defined neither k nor w, from first principles.

 

Further it is incomplete because I choose to examine non closed systems, and a non universal definition is just that, non universal and therefore certainly non fundamental.

[John Cuthber]

So, "undefined", "first" and "principles" are undefined since you have defined none of them.

 

I think you will find that Boltzmann's constant is quite well defined.

The number of microstates is less well known, but still well enough defined.

 

So, the problem is not the lack of a definition, it's the lack of your background knowledge.

Does this help?

http://en.wikipedia.org/wiki/Boltzmann%27s_entropy_formula

 

 

I suspect that if the system is open, the question of reversibility is undefined so a universal definition will never happen.

Edited June 20, 2014 by John Cuthber

[studiot]

Why are so called experts so condescending?

 

 

Flow system thermodynamics, is 'open'.

 

Do you not know this or is it only those with limited knowledge?

 

At least I am prepared to admit my limitations.

 

An expert should surely be able to explain his words of wisdom and back them up with rational argument, rather than personal attack.

Edited June 20, 2014 by studiot

[John Cuthber]

Why are so called experts so condescending?

 

 

Flow system thermodynamics, is 'open'.

 

Do you not know this or is it only those with limited knowledge?

 

At least I am prepared to admit my limitations.

 

An expert should surely be able to explain his words of wisdom and back them up with rational argument, rather than personal attack.

OK, so who made it personal?

"So entropy is undefined, since you..."

"Further it is incomplete because I..."

Also, re.

"At least I am prepared to admit my limitations."

"Well I don't agree with this interpretation."

 

"Bignose's definition should have read for a cyclic process."

nope. It didn't.

"C'mon what you said was neither mathematical, nor correct,"

Wrong on both counts.

Now, as I said, I'm not sure, but I think that the question of reversibility depends on other factors, notably closedness, so the overall definition needs to include those .

If I'm right in thinking that, then there's no way to address the problem you raise when you say "it is incomplete because I choose to examine non closed systems"

Even if I'm wrong, hopelessly rude and condescending, I backed up the assertions I made as fact, with a wiki page.

I also clearly labeled my beliefs as such, for example, I said "I suspect that if the system is open"

.

Now, would you like to explain the sense in which entropy is undefined as you asserted when you said

"So entropy is undefined, since you have defined neither k nor w, from first principles."?

Obviously, it may not be defined in this thread, but nor are most of the words used here.

Or do you accept that you were simply wrong to make that assertion since the definitions are readily found?

Edited June 21, 2014 by John Cuthber

[studiot]

@JC

Why is it that you post incorrect statements, using incorrect definitions of basic thermodynamic terms in your post #7 and then proceed to mock my 'lack of background knowledge' (your post#19).

 

An open system allows both energy and mass exchange across the system boundary.

 

A closed system allows energy exchange, but prohibits mass exchange.

 

An isolated system prohibits all exchange.

 

So you were using incorrect terminology when using your dewar flask analogy to lambast my sincere attempt to engage in polite discussion rather than fully expose this misuse of terms.

 

The whole purpose of this thread stemmed from my desire to further the cause of avoidance of such misunderstanding of what you call background knowledge and I call first principles, before going on to more advanced and more difficult matters such as entropy.

Edited June 21, 2014 by studiot

[John Cuthber]

This, from wiki, might explain the difference in terminology.

 

"In classical mechanics

In nonrelativistic classical mechanics, a closed system is a physical system which doesn't exchange any matter with its surroundings, and isn't subject to any force whose source is external to the system.A closed system in classical mechanics would be considered an isolated system in thermodynamics."

 

Now, can you address this please

would you like to explain the sense in which entropy is undefined as you asserted when you said
"So entropy is undefined, since you have defined neither k nor w, from first principles."?
Obviously, it may not be defined in this thread, but nor are most of the words used here.
Or do you accept that you were simply wrong to make that assertion since the definitions are readily found?

[studiot]

@JC

 

You made two substantive points, so dealing with them in order;

 

This thread is clearly labelled with the term thermodynamic, so surely it is reasonable to employ thermodynamic terminology.

 

Introducing terms from mechanics is particularly difficult and confusing since within that system perpetual motion is admissible. Note I did not say perpetual motion machines.

Force is an important quantity in mechanics, but very minor to totally unimportant in thermodynamics.

 

So can we please stick to conventional thermodynamic terminology?

 

As to you second point, I noted that you replaced the variable, entropy, with another, enumeration of energy levels along with a constant. Since you did not define your new variable or display a chain of mathematics leading to its relevance I regard the variable entropy as still undefined in this thread.

 

Now I will happily widen the discussion to entropy and the connection between the Boltzmann equation and classical thermodynamics.

 

The first thing to note is that the enumeration variable is not a continuous variable, so there is a mathematical issue in employing it in connection with classical continuous variables, calculus and so forth. Yes it can be overcome but care is needed.

 

The second thing to note is that classical thermodynamics does not attempt to offer calculation of absolute entropy of a system in the same way as statistical mechanics.

The Boltzmann formula refers to system entropy, whereas the Clausius one refers to entropy difference. The distinction is rather like the distinction between voltage and voltage difference.

 

There is a further difficulty employing an extensive state function like entropy in that the definitions, open closed and isolated only make sense for finite systems. Both S and delta S have singularities at zero and infinity.

 

I think that is enough to be going on with if you are serious about a sensible technical discussion.

[John Cuthber]

@JC

 

 

I regard the variable entropy as still undefined in this thread.

 

 

 

The second thing to note is that classical thermodynamics does not attempt to offer calculation of absolute entropy of a system in the same way as statistical mechanics.

The Boltzmann formula refers to system entropy, whereas the Clausius one refers to entropy difference. The distinction is rather like the distinction between voltage and voltage difference.

 

 

Firstly; you might, others don't.

secondly, you are contradicting yourself but why should I care?

You are right there is an analogy between absolute entropy vs entropy changes on one hand and between absolute voltage vs potential difference on the other.

And you can often convert from one to the other.

Imagine an isolated metal sphere.

I can calculate its capacitance (IIRC, it's proportional to the radius).

I can also (in principle) count all the protons and all the electrons and thereby calculate its charge.

Given the capacitance and the charge I can calculate it's potential.

That's an absolute figure.

I can do the same for a second sphere.

I can calculate the potential difference between them.

That's a potential difference calculated as the difference between two absolute values for potential..

 

I can, in principle, do the same with entropy.