# Simple Entropy/Multiplicity:

## 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!

## Answers and Replies

anyone tried 'er out?

Tom Mattson
Staff Emeritus
Science Advisor
Gold Member
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. 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?