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Assume we have a cubic box of side length 1m with ideal gas particles inside. We assume the binding energy of a gas molecule to the wall is 1eV. One can make the simplifying assumption that:

  • All states inside the volume have the same energy $0eV$ and the total number of states inside is $N_V=2.45\times10^{28}$
  • In addition the 6 walls have $N_W=3.68\times10^{19}$ bound states all at energy $–1eV$
  • Assume the total number of particles to be either

    • a) $N=3\times10^{20}$
    • b) $N=3\times10^{19}$

Calculate the equilibrium density of particle inside the box (i.e. not on the wall) as a function of temperature using Fermi-Dirac statistics for both cases.

To adress this problem, I assume I need to set up the partition function of the system? $$ Z = \sum_{i} e^{ \frac{-E_i}{k_B T} }$$

For case b), with $N < N_W$ the partition function of the system may look like this, taking into account the combinations of adsorption sites and particles?

$$ Z = \frac{N_W!}{\left(N_W-N\right)! \cdot N!} \left(e^{ \frac{-1eV}{k_B T} }\right)^N $$

Is this the correct and complete partition function of system b)?

How to adress case a), where $N > N_W$?

Any help is appreciated.

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In both cases:

$$Z=\sum_{n=0}^{\min(N,N_W)} F(n) \exp{-\beta (n E_B+(N-n)\cdot0)}$$

Where $F(n)$ is the number of ways $n$ adsorbing states can be occupied. Since the particles themselves are indistinguishable this is $N_w \choose n$.

If $N≥N_W$, the expression that you have is $\sum_n {N_w \choose n} \left(e^{-\beta E_B}\right)^n=(1+e^{-\beta E_B})^{N_W}$.

If $N<N_W$ there is no immediate simplification, but the expression you have written is not correct.

What is asked for is the density function. Note in this case that for each configuration the energy is directly proportional to the amount of bound states. So $\langle E \rangle = c \langle n_\text{bound} \rangle$ for a constant $c$, in this case $c=E_B=-1$ eV.

In general $\langle E \rangle = -\partial_\beta \ln (Z)$. Plug it in for the case (a), you get $\langle n_\text{bound} \rangle=\frac{N_W}{1+e^{-\beta E_B}}$.

Now you want the density in the box, this is $\frac{N-n_\text{bound}}{1m^3}$.

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  • $\begingroup$ Thank you for your kind and structured explanation! So are you saying that case b) with $ Z = \sum_{n=0}^{N} \binom{N_W}{n} \left( e^{ -\beta E_B } \right)^n $ is not analytically solvable? (Remark: In your explanation for case a), I think the minus sign in the exponential term of $ \langle n_\text{bound} \rangle $ must be omitted?) $\endgroup$ – jaruumi6 Jan 5 '16 at 7:54
  • $\begingroup$ I'm not sure what you mean with analytically solvable, the expression is a finite (although huge) sum of analytic functions (in $\beta$ and in $E_B$), so is itself also analytic. I just don't know of any way to put it into a short form like in the first case. The minus in the exponent is correct, though. $\endgroup$ – s.harp Jan 5 '16 at 9:58
  • $\begingroup$ Although I have seen an error in what I have written: I have assumed that in the box itself there is only one state, and the particles act as bosons in how they occupy it. Meaning if I say "there are 100 particles in the box and no particles in the wall" that the configuration is uniquely determined. In reality, since the particles are fermions, you really must have a number $M$ of box states available, and infront of each part of the sum in $Z$ there must also be $M \choose N-n$. $\endgroup$ – s.harp Jan 5 '16 at 10:03

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