# Entropy and combinatorics

1. May 11, 2014

### CAF123

1. The problem statement, all variables and given/known data
Suppose we put N atoms of argon into a container of volume V at temperature T. Of these N atoms, Nad stick to the surface, while the remainder Ngas = N - Nad form an ideal gas inside the container.

Assume that the atoms on the surface are not able to move and have an energy -eo. Furthermore, let Ns >>1 be the number of sites on the surface the atoms can stick to.

a)Calculate the entropy of the Nad atoms on the surface.

The subscript 'ad' presumably because the atoms are adsorbed onto the surface

2. Relevant equations

# of microstates found by combinatorics and entropy given by Boltzmann's Law.

3. The attempt at a solution
It seems to me that although we have some stated assumptions more is needed to give an answer that matches the one in the answer key. Is more than one atom allowed to occupy a site on the surface? Are the atoms indistinguishable?

From N atoms, Nad are placed on the surface. This can be done in N choose Nad ways. Assuming the atoms are distinguishable, then we multiply this number by Ns!, since there exists that many arrangements of those Nad atoms. If they are indistinguishable, then we do not multiply by Ns!. This is also assuming only one atom can occupy each site.

Many thanks.

2. May 11, 2014

### voko

Based on what you know about Argon, would you classify its atoms as bosons or fermions?

3. May 11, 2014

### CAF123

Hi voko,
Argon is composed of fermionic constituents so its atoms would be fermions. It forms an outer stable octet so the electrons are not all sitting in the ground state. So I think this would imply that the atoms cannot share a site localized on the surface of the container. Also, if we treat the system quantum mechanically, then we cannot put a label on the atoms so they are indistinguishable.

4. May 11, 2014

### voko

Is Helium-4, composed of fermions, a fermion?

5. May 11, 2014

### CAF123

I see your point. Argon (neutral) is composed of an even number of fermions, so the net result is that the atom overall is bosonic. So contrary to what I wrote previously, each atom can occupy any of the sites. So each atom can be put in Ns positions. There are Nad atoms to be placed, so Ω= NsNad. And then we must divide by Ns! if we treat the atoms as indistinguishable.

6. May 11, 2014

### voko

That looks good to me.

7. May 11, 2014

### CAF123

Thanks, but should I not be dividing by Nad! instead? Since there may be more sites on the surface available than the number Nad. Regardless, both lead to an incorrect answer.

I am supposed to use the expression for entropy I obtain here along with the calculation of the free energy to show that the chemical potential satisfies the following equation: $$\mu_{ad} = -e_o - kT \ln \left(\frac{N_s - N_{ad}}{N_{ad}}\right)$$

I calculated the chemical potential using the equation $\mu_{ad} = \frac{\partial F}{\partial N_{ad}}$

8. May 12, 2014

### voko

I was slightly confused by the suffixes, and yes, to take account of the indistinguishability of the atoms, one needs to divide by the number of their permutations, whatever the suffix is :) But both methods leading to the wrong result, hmm. I am afraid I do not have time till much later today, so other contributors are welcome.

9. May 12, 2014

### CAF123

Here is my work: As discussed above, $$\Omega =\frac{ N_s^{N_{ad}}}{N_{ad}!} \rightarrow S = k_B \ln \Omega = k_B \ln \frac{ N_s^{N_{ad}}}{N_{ad}!}$$

Then $$F = E- TS = -N_{ad} e_o - k_B T \ln \frac{ N_s^{N_{ad}}}{N_{ad}!}$$ Expanding the log, using Stirling's approximation to eliminate the factorials, and then computing $\mu_{ad} = \partial F/\partial N_{ad}$ gives $$\mu_{ad} = -e_o - k_BT \ln \left(\frac{N_s}{N_{ad}}\right),$$ so it is close, but not quite correct.

10. May 12, 2014

### voko

The result that you are supposed to demonstrate becomes invalid when $N_s < N_{ad}$. Are there perhaps some additional relations between the two?

11. May 12, 2014

### CAF123

No more information is given in the question, but there is a condition that Ns>>1. I thought initially this was given to validate the use of Stirling's approximation, but then that would require Ns to be large, so I implicitly assumed that. I don't see where I used the condition Ns >> 1.

12. May 13, 2014

### CAF123

It seems weird that the expression becomes invalidated when $N_s < N_{ad}$. Since the atoms are bosons, they can visit any of the sites, so it is not like there is any restriction there. I suppose the more sites there are, the greater probability there will be of a greater energy cost of the system.

If the atoms were fermions, then I can see why intuitively the expression becomes invalidated, but this is not the case here.

13. May 13, 2014

### voko

Well, I cannot see any obvious error in your derivation. And the supposedly correct result, given no stated restriction $N_s > N_{ad}$, is invalid when the latter is not satisfied. So I would suggest that you discuss that with your instructor.

14. May 13, 2014

### CAF123

Hi voko,
I think the problem may be that I treated the sites as if they were quantum states. The Pauli Exclusion principle implies that bosons may occupy the same quantum state, but this does not mean they can occupy the same volume in space.

So if I take $N_s$ sites for the first particle, $N_s - 1$ for the second and so on.. Then dividing by $N_{ad}!$ to take into consideration indistinguishability gives $$\frac{N_s \cdot (N_s - 1)\, ...\, \cdot\, (N_s - (N_{ad} -1))}{N_{ad}!} =\frac{ N_s!}{(N_s - N_{ad})! N_{ad}!}$$ which is exactly the binomial coefficient.

With this, I obtain the correct answer and then I can make sense of why the formula is invalidated when $N_s < N_{ad}$.

15. May 13, 2014

### voko

Well, it is not obvious to me that a "site" means "one atom only", but if does, then indeed that implies the inequality from my previous post, as well as a different structure of microstates. Well done.