# Homework Help: Basic Statistical Thermodynamics

1. Nov 30, 2007

### the keck

[SOLVED] Basic Statistical Thermodynamics

1. The problem statement, all variables and given/known data

Two distinguishable particles are to be distributed among nondegenerate energy levels 0,e,2e,3e.... such that the total energy is U = 2e

If a distinguishable particle with zero energy is added to the system show that the entropy of the assembly is increased by a factor of 1.63 when compared to the entropy of the original assembly.

2. Relevant equations
S=k*ln(W), where k is Boltzmann's Constant and W is thermodynamic probability

3. The attempt at a solution

I know that without the particle added, W is three. And so a factor of 1.63 means that W must equal to 6 i.e. Solving ln(x)/ln(3) = 1.63 => x = 6

However, I do not know how adding an extra particle would produce three more states.

Regards,
The Keck

2. Nov 30, 2007

### George Jones

Staff Emeritus
That's not what I get. Could you write down your three configurations?

3. Nov 30, 2007

### nrqed

george, I do agree with his/her result. Which number do you not agree with? # initial configurations or 6 final?

4. Nov 30, 2007

### George Jones

Staff Emeritus
With the initial configuration of 3 possible states.

Since the particles are distinguishable, aren't there two separate states where each particle has energy e?

5. Nov 30, 2007

### nrqed

I think that this is a single state. There is no way to distinguish "particle 1 having energy e and particle 2 having energy e" from the reverse situation even if the aprticles are distinguishable. I think this is a single state.

to the OP: I do get 6 states for the case of 3 particles.

6. Nov 30, 2007

### siddharth

I agree. That's what I get as well.

Last edited: Nov 30, 2007
7. Nov 30, 2007

### George Jones

Staff Emeritus
It seems that I had talked myself into thinking that order matters. Thanks.

Writing things a different way, suppose one particle is green and one particle is yellow. Green with energy e and yellow with energy e is that same state as yellow with energy e and green with energy e.

I also see the six states for the configuration with three distinguishable particles, but I have a bit of a problem with the wording "a distinguishable particle with zero energy is added to the system".

Start out with a 2-particle system consisting of one green particle and one yellow particle. Add a third particle. Since the particles are distinguishable, we know which particle was added, say a blue particle. Under this interpretation, the blue particle has to have energy zero, and there still are only three states.

My interpretation of the wording probably is non-standard, or maybe even, as before, outright wrong.

8. Nov 30, 2007

### nrqed

Right. If I close my eyes and you switch them, I won't be able to tell if they ahve been switched after I open my eyes.
I see what you are saying. Good point. The question is a bit ambiguous because of that. It depends if the particles interact or not. One other interpretation is the following: the third particle has no energy when it is added (so the total energy remains 2E) but since it may interact with the others (which is the assumption I made and which is not clear from the question, I grant you that!), once it is added teh energy may be redistributed among all three particles. If there is no energy exchanged, I agree that the number of states remains 3. If energy may be redistributed among all three particles, the number of states increases to 6.

9. Nov 30, 2007

### the keck

Thanks for your help, George and nrqed. I also took the assumption that the particles cannot interact and so the third particle only has energy zero, leading to only three states.

I admit the question is definitely ambiguous, but that seems to be one of the those things which seem to occur more often in Statistical Physics questions than in other types of physics questions.

Thanks again

Regards,
The Keck