Calculating Entropy of Adsorbed Atoms on a Surface

[CAF123]

Homework Statement

Suppose we put N atoms of argon into a container of volume V at temperature T. Of these N atoms, N_ad stick to the surface, while the remainder N_gas = N - N_ad form an ideal gas inside the container.

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

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

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

Homework Equations

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

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

Many thanks.

 

[voko]

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

 

[CAF123]

Hi voko,

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

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.

 

[voko]

Argon is composed of fermionic constituents so its atoms would be fermions.

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

 

[CAF123]

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

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 N_s positions. There are N_ad atoms to be placed, so Ω= N_s^N_ad. And then we must divide by N_s! if we treat the atoms as indistinguishable.

 

[voko]

That looks good to me.

 

[CAF123]

That looks good to me.

Thanks, but should I not be dividing by N_ad! instead? Since there may be more sites on the surface available than the number N_ad. 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 $\mu_{ad} = \frac{\partial F}{\partial N_{ad}}$

 

[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.

 

[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.

 

[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?

 

[CAF123]

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

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

 

[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.

 

[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.

 

[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 $N_s < N_{ad}$.

 

[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.