Mathematical definition of thermodynamic reversibility

[studiot]

In response to post#25

 

Since both your answers were totally flippant, I see no point discussing this topic further with you.

Edited June 23, 2014 by studiot

[John Cuthber]

I think you gave up discussing it when, in response to me posting that

S= k ln w

and this link

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

 

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

But no doubt others will make up their own minds.

 

And the point remains that the Clausius Entropy and the Boltzmann entropy are essentially the same thing, at least for an ideal gas.

Edited June 23, 2014 by John Cuthber

[studiot]

Once again you are either misunderstanding standard thermodynamic terminology or you are redefining it to terms of your own.

 

 

 

John Cuthber post#7

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

 

This is logical nonsense since including the heat source as part of the system leads to the inescapable fact that the heat transferred from the surrounding to the system equals precisely zero (First Law). This, in turn, leads to the inescapable conclusion that whatever the process, whatever the final system and surroundings the entropy change is always precisely zero (Second Law).

[John Cuthber]

How do you propose to explain the "entropy death" of the universe (which is a closed, isolated system because there's nowhere outside it to transfer energy or mass to or from)?

[studiot]

I did note in post#24 that most of the definitions and much of the maths applies only to finite systems.

 

The question arises is the universe finite or infinite, and, if infinite, how do you treat it?

 

Let us say you calculate a finite entropy 'increase' of S. Can you prove that infinity plus S is greater than infinity?

 

(Sorry I can't present maths symbols from Dundee.)

 

Now let us say you have a system, finite or infinite, and you use S = k ln(w) to calculate the total system entropy.

ln(w) is a function that has no upper bound so as w tends to infinity ln(w) does not tend to a finite limit but increases without bound.

So does an infinite system have infinite entropy?

Can a finite system have infinite entropy? Consider in say the Morse curve where w is a discontinuous independent variable, below a certain level, but continuous above, where there must therefore be an infinite number of energy levels.

Consider also Caratheodory's (hope I spelled it correctly) version of the second law:

For any equilibrium state there are nearby equilibrium states that are inaccessible to the system.

 

I said I was happy to discuss entropy, but I don't see how the above is relevant to reversibility.

[John Cuthber]

"I said I was happy to discuss entropy, but I don't see how the above is relevant to reversibility"

It relates to the nature of change: what is usually referred to as "time's arrow"

 

You have a point about doing arithmetic with infinities.

Lets' do the maths with "some very large bit of the universe"

If that's big enough then only the parts that are near the "surface" can exchange with the outside.

And, as the piece we consider gets bigger those parts will have less and less effect on the overall entropy.

[studiot]

Now that I am back home, let us discuss the following closed, isolated system.

 

Consider an adiabatic cylinder containing a frictionless adiabatic piston separating the cylinder into two chambers, each containing a compressible fluid, so that the piston is in the equilibrium position.

 

Let the piston now be mechanically displaced from its position and then released.

Are the subsequent changes reversible or irreversible and what is the entropy change of the overall system?

Edited June 30, 2014 by studiot

[John Cuthber]

Let's start by checking that we agree what happens

The piston bounces back and to until the fluid's viscosity bring it to a halt (or asymptotically near to a halt).

In doing so, the gas is warmed up.

With a more massive piston, this will happen more slowly

 

As far as I can see, that's mechanical energy degrading to heat.

The entropy change will be positive, and it's a non- reversible system.

The reverse is never seen to happen.

[studiot]

 

46 AM

Let's start by checking that we agree what happens

The piston bounces back and to until the fluid's viscosity bring it to a halt (or asymptotically near to a halt).

 

 

 

 

frictionless adiabatic piston

 

That was meant to mean that there are no dissipative processes.

Edited July 1, 2014 by studiot

[studiot]

This new homework question provides an object lesson in classical four-law thermodynamics as to why you need to separate a system from its surroundings by its boundary and why you need reversibiltiy.

 

http://www.scienceforums.net/topic/84056-thermodynamics-and-gas-law-problem/

 

Of course if you take the view that the whole of the universe is 'the system' then there is only one system in the universe, which is the same for every problem. So bothering to even state the word system, let alone specify it, is pointless.

 

We also seem short of others prepared to help in the thermo homework section?

[John Cuthber]

I apologise for being unable to read your mind.

You didn't specify a non-viscous fluid.

 

Anyway, as the viscosity tends to zero the time taken for the system to come to rest tends to infinity.

As the piston bounces back and to in that case, it repeatedly ends up where it started from and, so the system is reversible.

It's physically impossible and it's not quite the same as the system you specified before- because it's a special case- but it is reversible,

 

Where you say "Of course if you take the view that the whole of the universe is 'the system' then there is only one system in the universe, which is the same for every problem." You have missed the point.

The universe is one clearly isolated system and the laws of thermodynamics tell us that it will suffer an "entropy death" in the long run.

That does not mean that we can not specify other isolated systems.

It's not "the system" it's just "a system"

[studiot]

 

John Cuthber

but it is reversible

 

So you calculated that the entropy change is zero, since that is your criterion for reversibility.

 

Would you mind showing your working?

 

John Cuthber

You have missed the point...........

and the laws of thermodynamics tell us that it will suffer an "entropy death" in the long run.

 

Can you state exactly which law and how?, the above is oft quoted hand waving.

 

Actually you are the one who has missed my point.

 

If you include the surroundings (rest of the universe) in the 'system', then there is room for only one system in all the universe, so every system is identical.

[John Cuthber]

What working do I need to show?

If the piston keeps bouncing back and to forever it keeps returning to exactly the same state that it started in.

Since that state is identical to the initial state, the change in entropy (and everything else) is zero.

 

"Can you state exactly which law and how?"

Yes, I can and I have. For any closed isolated system delta S is always positive.

Since the universe is, or can be treated as closed and isolated, (as discussed at unnecessary length) it will eventually "run down".

 

"If you include the surroundings (rest of the universe) in the 'system', then there is room for only one system in all the universe"

Well then, don't.

Only include those things that are coupled to it.

Then you can make lots of systems (as long as you isolate them suitably)

That doesn't stop you

1) considering other systems or

2) considering the Universe as a whole as a system.

Edited July 4, 2014 by John Cuthber

[studiot]

You are still missing the point.

 

In the development of Thermodynamics as a self consistent discipline we start with definitions propose laws and deduce theorems, much as with Euclid.

Using the second law to define reversibility introduces a circular argument.

 

In classical thermo we start with mechanical quantities such as pressure, volume, work and energy.

 

We propose or deduce (whatever you will) the First Law and the existence of state functions ingeneral and a state function called internal energy.

We discover that pressure and volume form a state function pair for many processes, but sometimes this fails, as with one of the most common substances used in practical thermodynamics, water, since specific volume is not a unique function of temperature.

 

This kit is enough calculate most of the practical requirements in classical thermo. Entropy is not mentioned or needed.

 

But reversibility is required to perform the calculations.

 

For instance this quote from Lewitt : Thermodynamics Applied to Heat Engines"

 

"In article 10 eight thermodynamic processes were defined; any of these processes that can be operated in the reverse direction are known as reversible processes"

 

Lewitt goes on to many pages to analyse adiabtic, isothermal and ther other six processes for reversibility.

 

Similar discussion can be found in Robinson and Dickinson : "Applied Thermodynamics"

 

and of course, that old standby for all engineers Rogers and Mayhew "Engineering Thermodynamics, Work and Heat Transfer"

 

Yes, all introduce and discuss and employ entropy at a much later stage. and none offer a mathematical definition of reversibility.

Edited July 13, 2014 by studiot

[John Cuthber]

OK, what happens to the pistons and fluid system, (in particular, what happens to its reversibility) when you introduce this requirement?

"That was meant to mean that there are no dissipative processes."

[studiot]

It is only reversible when there are no dissipative processes.