How Does Adding Particles Affect Entropy in a Multi-Level Energy System?

[HalfManHalfAmazing]

Homework Statement

Two distinguishable particles are to be distributed between 3 energy levels: 0, 1e and 2e such that the total energy level E = 2e.

a) What is the entropy of the system?
b) if a third distinguishable particle with zero energy were added to the system show that the entropy increases by a factor of 1.63.
c) Reevaulate a) with undistinguishable particles. does b) still hold if this time the added particle is undistinguistable?

Homework Equations

$$S = k\ln\Omega$$
Various methods of determining multiplicity...

The Attempt at a Solution

Okay. First up this system has 3 possible states per particle and we've got 2 particles. We want the multiplicity of the macrostate 2e. Thus we need to determine the number of microstates that this macrostate has. Looks to me like we have the following states: 0/2, 2/0, 1/1, 1/1. Thus 4 microstates? so we plug 4 into the omega from the entropy formula and we're done?

Next, adding the third 0 energy level particle confuses me. It won't change the number of combinations required to reach 2e, will it? or do we do something like this: 0/2/0 0/0/2 2/0/0 0/1/1, 1/1/0, 1/0/1? Also, what formula could i use to make this a general expression such as what is the entropy of n particles in n states? I know it involves factorials...

Thanks!

 

[HalfManHalfAmazing]

anyone tried 'er out?

 

[quantumdude]

You're right on both counts. Adding the 3rd particle changes the number of combinations in just the way you surmised.

As for n particles in n states, I'll leave that for someone else.

 

[HalfManHalfAmazing]

I find a multiplicity of 6 for b. in which case the when i check the factor of increase i only find a factor of 1.29, not 1.63. So that can't be right?