Counting microstates in Boltzmanns principle

1. May 31, 2013

erm

1. The problem statement, all variables and given/known data
Explain why the number of microstates W in Boltzmanns principle, is

W = ƩNi! / ∏Ni!

when i ideal gasses are mixed at constant volume and temperature. Ni is the number of particles of component i.

2. Relevant equations
S=klnW , where W is the number of microstates and k is Boltzmanns constant.

3. The attempt at a solution
What I am really asking here, is an explanation on how the microstates are counted. It is obviously the sum of all the factorials divided by the product of all the factorials of the different components. What is the logic behind this?

2. May 31, 2013

AGNuke

Here's an example.

Suppose there is a container which is divided in two equal halves. In one half one mole gas is filled and the other half is completely evacuated. So, the gas will only remain in the first half.

Now the membrane separating the halves is removed. The effective volume is doubled. Now we can find the number of microstates that were introduced.

Each gas molecule has two choice, either remain in the first half or migrate to other half. Thus the total number of microstates we're seeing is just the capability of a molecule to choose where it wants to be. Thus it will amount to 2 x 2 x 2 x 2 .... upto NA or $W=2^{N_A}$.

To check its validity, you can find out the entropy using Boltzmann formula AND using conventional formula considering reversible isothermal expansion.

As far as I am concerned, I don't think we can count absolute number of microstates. We can only count the change in their numbers brought about a process.

3. Jun 2, 2013

erm

Thank you for the example. It expains how you can count the microstates in another case, but it really doesn't say anything about the problem I stated (mixing gases at constant volume and temperature). Could someone please explain this?