Calculating Entropy of Adsorbed Atoms on a Surface

In summary: Fermions have an integer number of spin states, while bosons have an odd number of spin states. So if we imagine each atom has two spin states, then we could have either 2Nf or 2Nb atoms. If we have an odd number of atoms, then there is no way for them to all be in the same spin state, so they are essentially fermions. If we have an even number of atoms, then there is a 1 in 2 chance of each atom being in the same spin state, so they are essentially bosons.Hi voko,If the atoms were fermions, then I can see...Fermions have an integer number of spin states, while bosons have an odd
  • #1
CAF123
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Homework Statement


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

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, 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.
 
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  • #2
Based on what you know about Argon, would you classify its atoms as bosons or fermions?
 
  • #3
Hi voko,
voko said:
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.
 
  • #4
CAF123 said:
Argon is composed of fermionic constituents so its atoms would be fermions.

Is Helium-4, composed of fermions, a fermion?
 
  • #5
voko said:
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 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
That looks good to me.
 
  • #7
voko said:
That looks good to me.
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
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
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
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
voko said:
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 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
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
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
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
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.
 
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FAQ: Calculating Entropy of Adsorbed Atoms on a Surface

1. What is entropy and how does it relate to combinatorics?

Entropy is a concept in physics that measures the amount of disorder or randomness in a system. In combinatorics, it is used to measure the number of possible arrangements or combinations of elements in a given set.

2. How is entropy calculated in combinatorics?

In combinatorics, entropy is calculated using the formula S = k ln(W), where S is the entropy, k is the Boltzmann constant, and W is the number of possible arrangements or combinations.

3. Can entropy be negative in combinatorics?

No, entropy cannot be negative in combinatorics. It is always a positive value, as it represents the number of possible arrangements or combinations, which cannot be less than zero.

4. What are some real-life applications of entropy and combinatorics?

Entropy and combinatorics have numerous real-life applications, such as in data compression, cryptography, and information theory. They are also used in fields like genetics, linguistics, and computer science.

5. How does the concept of entropy and combinatorics relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. In combinatorics, this can be seen as the number of possible arrangements or combinations increasing as the number of elements in the system increases.

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