Calculating microstates and entropy

[b100c]

Homework Statement

Two identical brass bars in a chamber with perfect (thermally insulating) vacuum are at respective temperature T hot>T cold. They are brought in contact together so that they touch and make perfect diathermal contact and equilibrate towards a common temperature. We want to know what is the change of entropy to the whole system (cold plus hot bars) according to Boltzmann’s formula S=kln(Ω).

To study that problem, we will consider a simple model. We will assume that the hot bar consists of 4 atoms with 5 quanta of energies that can be distributed arbitrarily among the 4 atoms. For example, three atoms can have one quantum, one atom has two quanta, making a total of (3×1+1×2) quanta=5 quanta. This is one of the many microstates that this system can have with the constraint of 5 quanta in total.

We will assume that the (identical) cold right bar also has 4 atoms, but now with only 1 quantum of energy, as it is colder, that can be distributed arbitrarily among the 4 atoms.

1. Assuming that the atoms are distinguishable, compute the number of microstates initially (before they are brought in contact) for the cold bar.
2. Compute the number of microstates initially for the hot bar.
3. Once they are brought touching together, the energy equilibrates (so that the two bars reach a same/ common temperature). How many quanta of energy exist in the left bar? How many in the right bar?
4. Compute the entropy change for the left bar. Compute the entropy change for the right bar.
5. What is the total entropy change for the whole system? Has it increased or decreased?

Homework Equations

S= kln(Ω)

The Attempt at a Solution

1. Well the cold bar only has one macrostate, which is 1 atom with 1 quantum, and 3 atoms with 0. So this is asking how many permutations are there of 1-0-0-0.
To find the number of microstates (permutations), use the formula
\frac{N!}{{\prod\limits_{k}}M_k!}
with 1 for M₁, 3 for M₂, and 4 for N. This gets the answer 4.

2. Here, I use the same formula from part 1 on the different macrostates, and get 56 microstates in total.

3. I would think that the quanta would even out so that there are 3 quanta in each bar right?

4. I'm confused with this part, I know I use S= kln(Ω), but is Ω the total number of microstates, or the number of microstates for the most likely macrostate?

5. The entropy change for the whole system should just be the sum of the two entropy changes, and I expect it to not decrease by the 2nd law of thermodynamics.

 

[haruspex]

To find the number of microstates (permutations), use the formula
\frac{N!}{{\prod\limits_{k}}M_k!}
with 1 for M₁, 3 for M₂, and 4 for N. This gets the answer 4.

2. Here, I use the same formula from part 1 on the different macrostates, and get 56 microstates in total.

the formula I know for the number of ways of distributing r identical objects into n distinct buckets is $^{n+r-1}C_{r}$. That gives me the same answers.

3. I would think that the quanta would even out so that there are 3 quanta in each bar right?

Yes.

4. I'm confused with this part, I know I use S= kln(Ω), but is Ω the total number of microstates, or the number of microstates for the most likely macrostate?

You know the macrostate, so it's the number for that macrostate.

 

[b100c]

Hi, haruspex thanks for the reply.

You know the macrostate, so it's the number for that macrostate.

How do I know which macrostate to choose, do I just pick the most probable one (the one with the greatest number of microstates)?

 

[haruspex]

Hi, haruspex thanks for the reply.

How do I know which macrostate to choose, do I just pick the most probable one (the one with the greatest number of microstates)?

You are given two macrostates. In the first, the energy is split 5:1, in the second it is 3:3.

 

[b100c]

Oh, okay I think I get it. I was under the impression that a macrostate is a way of distributing the 5 quanta into 4 atoms. And then the microstates are pemutations of that. Like 5-0-0-0 would be one macrostate with 4 microstates, and then 4-1-0-0 would be another macrostate with 12 microstates. This is how I got 56 in part 2, by adding up all the microstates across macrostates. Is this wrong?

 

[haruspex]

Oh, okay I think I get it. I was under the impression that a macrostate is a way of distributing the 5 quanta into 4 atoms. And then the microstates are pemutations of that. Like 5-0-0-0 would be one macrostate with 4 microstates, and then 4-1-0-0 would be another macrostate with 12 microstates. This is how I got 56 in part 2, by adding up all the microstates across macrostates. Is this wrong?

It depends how you choose to define macrostates. If you choose to define a macrostate as "one atom has all 5, but I don't care which" then that would be true. But then you can bundle these macrostates into larger macrostates. These would be macrostates for one bar only, of couse, and you would have to take ordered pairs of such to generate macrostates for the two bar system.
For the purposes of the question, the macrostates of interest are
A. Left bar has 5 quanta and right bar has 1.
B. Each bar has three quanta.