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Thermo Question: The math behind systems in thermal contact

by E'lir Kramer
Tags: contact, math, systems, thermal, thermo
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E'lir Kramer
#1
Jan14-13, 05:41 PM
P: 66
Hi! I am reading Molecular Driving Forces, 2nd ed., by Dill & Bromberg.

On page 53, example 3.9, we consider why energy exchanges between two systems from the point of view of the 2nd law.

We consider two separate systems. Each has ten particles, and each particle has two possible energy states. System A has total energy [itex]U{a}[/itex] = 2, and [itex]U{b}[/itex] = 4. Thus binomial statistics predicts the multiplicities of these systems:

W([itex]U{a}[/itex]) = [itex]\frac{10!}{8!2!}[/itex] = 45
W([itex]U{b}[/itex]) = [itex]\frac{10!}{6!4!}[/itex] = 210

Now the confusing part, to me, is the math when these two systems come into thermal contact. Then the author asserts that the initial multiplicity is

W([itex]U{total}[/itex]) = [itex]\frac{10!}{8!2!}[/itex][itex]\frac{10!}{6!4!}[/itex]

And that maximum multiplicity is found at

W([itex]U{total}[/itex]) = [itex]\frac{10!}{7!3!}[/itex][itex]\frac{10!}{7!3!}[/itex] = 14,400

But why consider the systems in this way, as opposed to thinking of a new system, with 20 particles, having [itex]U{a+b}[/itex] = 6?

Then we get

W([itex]U{a+b}[/itex]) = [itex]\frac{20!}{14!6!}[/itex] = 38760 ≠ W([itex]U{total}[/itex])

I'm trying to develop a sense of the difference, I suppose, between two systems in thermal contact and one system. After they've equilibrated, how are they not treatable as one system? They clearly aren't, because if they were, then the total multiplicity of that one system must = the total multiplicity of the two systems A and B.
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E'lir Kramer
#2
Jan14-13, 09:39 PM
P: 66
I think I've answered my own question:

The reason there is a difference between two systems in thermal contact and one system has to do with distinguishability.

"Two systems" means that we can independently measure the temperature of each system - or, equivalently, that the two systems have distinguishable particles. In thermal contact without matter transfer, we could, for instance, measure the temperature of the two systems independently. Intuitively, given the same particles of matter and amount energy, if we can distinguish between more macro-states, we expect there to be a smaller number of microstates comprising each macrostate.

The 10+10 systems can of course assume exactly the same energy configurations that the 20/6 state could. The difference is that we can distinguish more of the macrostates arising from various configurations of the 10+10 than that of the 20. For example, a 10/4 - 10/2 split is different from a 10/2 - 10/4 split. In a hypothetical 20/6 system, what we are saying is that all 10/2 - 10/4 splits look the same to us, and in fact those splits look the same as all the 10/3 splits, and all the 10/5 - 10/1 splits.
E'lir Kramer
#3
Jan15-13, 01:59 AM
P: 66
If anyone wants to jump in and confirm or deny this explanation, I'd appreciate it.

Rap
#4
Jan15-13, 10:44 AM
P: 789
Thermo Question: The math behind systems in thermal contact

Quote Quote by E'lir Kramer View Post
If anyone wants to jump in and confirm or deny this explanation, I'd appreciate it.
Basically, I think you are right.

What you are saying seems to be that if you have two containers, before thermal equilibrium, the total entropy is Log(45 x 210) which is, of course, the sum of the separate entropies (=Log(45)+Log(210)=9.15) After equilibrium, they have equal entropies, Log(120) each, and the total entropy is Log(120)+Log(120)=Log(14400)=9.575. Entropy has increased, as expected. What you want to do is consider them as one system with entropy Log(38760)=10.565. You make them one system by removing the partition between them, having one volume containing all the particles.

My problem is that, with real gases, if you have two separated volumes with the same density, different temperatures, they also have entropy equal to the sum of the separate entropies, and if you let them equilibrate, they will have equal entropies, total entropy the sum of the two, and that entropy will be larger than the original. If you then remove the partition, let them mix, the entropy does not change, unlike your scenario.

If the densities are different to begin with, then the entropy will increase when you remove the partition, it all depends on the densities.

The bottom line is that you have assumed that when you remove the partition, the particles still have only two energy states, which is where things differ from the "real" situation.

Entropy is the log of the number of microstates that give a particular macrostate. When the two systems are separated, your macrostate is just the total energies of each system. When they are combined, its just the total energy of the combined system. Given your assumptions, you are correct, but in the real world, you need three variables, energy, density, and (constant) volume, for example, and then things change.


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