# Solution to Basic Exercise 2.1-4"Derive the Formula for the Vacancy Equilibrium Concentration"

| First we need to determine the number of possibilites Pn to arrange n vacancies in a crystal of N atoms
|  | This is most easily done by constructing a table and look at the cases n = 1, n = 2, etc. until it becomes obvious what the general law will be
n (= i) | pn = | Comment
1 | N | All N places are available
2 |
N · (N – 1) 2
| N places for the first, only N – 1 places for the second vacancy. Exchanging both vacancies does not change the situation - we have to divide by 2
3 |
N · (N – 1) · (N – 2) 2 · 3
| Exchanging vacancies does not change the microstate, we have to divide by the number of all possible exchanges = 6 = 2 · 3.
Make sure you understand the exchange argument: Here is the detailed reasoning: For vacancy No. 1 on place 1 , you have two possibilities: No. 2 on place 2, No. 3 on place 3 or No 2 on place 3 and No. 3 on place 2. You can do the same thing for No. 2 on place 2 (exchange No. 1 and No. 3) and for No. 3 on place 3., so you have 2 options 3 times = 6 indistinguishable arrangements.
... | ... | and so on
n |
N · (N – 1) · (N – 2) · .. · (N – (n – 1)) 2 · 3 · .... · n
| The obvious law for n vacancies. {1· 2 · 3 · .... · n} of course is simply n!
n |
{N · (N – 1) · (N – 2) · .. · (N – (n – 1))} · {(N – n)!} n! · {(N – n)!}
| Extend the fraction by (N – n)!
n |
N!
n! · (N – n)!
| Final result as used in subchapter 2.1 This is a standard expression in combinatorics and called the binomial coefficient.

| The entropy of mixing thus is
S | = k · ln | N! n! · (N – n)! | = k · | æ è | ln N! – ln {n! · (N – n)!} | ö ø | = k · | æ è | ln N! – ln n! – ln (N – n)! | ö ø
|  | We now can write down the free enthalpy for a crystal of N atoms containing n vacancies
G(n) | = | n · GF – | kT · [ln N! – ln n! – ln (N – n)!]
--- | --- | --- | ---
| Now we need to find the minimum of G(n) by setting dG(n)/dn = 0 and for that we must differentiate factorials. We will not do this directly (how would you do it?), but use suitable approximations as outlined in subchapter 2.1.
|  | Mathematical approximation: Use the simplest version of the Stirling formula
ln x! | » | x · ln x
--- | --- | ---
|  | Physical approximation, assuming that there are far fewer vacancies than atoms:
n | << | N | Þ
--- | --- | --- | ---
n N – n | » | n N | = | cV = |  | concentration of vacancies
--- | --- | --- | --- | --- | --- | ---
| Now all that is left is some trivial math (with some pitfalls, however!). The links lead to an appendix explaining some of the possible problems.
|  | Essentially we need to consider dS(n)/dn using the Stirling formula
dSn dn | = k · | d dn | æ è | ln N! – ln n! – ln (N – n)! | ö ø | » k · | d dn | æ è | N · ln N – n · ln n – (N – n) · ln (N – n) | ö ø
| But we must not yet use the physical approximation, even so its tempting! With the formula for taking the derivative of products we obtain
dSn dn | » k · | æ ç è | æ è | (– ln n – | n n | ö ø | – | æ è | – ln ( N – n) + | n – N N – n | ö ø | · (– 1) | ö ÷ ø
dSn dn | » – k · | æ è | ln n + 1 – ln (N – n) – 1 | ö ø | = – k · | æ è | ln n – ln (N – n) | ö ø | = – k · ln | n N – n
|  | Now we can use the physical approximation and obtain
dSn dn | » | – k · ln cV
--- | --- | ---
| Putting everything together gives
dG(n) dn | = 0 = GF – T · | dSn dn |  | = GF + kT · ln cV
--- | --- | --- | --- | ---
|  | Reshuffling for cV gives the final result
cV | = exp – | GF kT
--- | --- | ---
|  | q.e.d.

| What happens if we use better approximations of the Stirling formula; e.g. ln x! » xln x – x? Lets see:
|  | We start with the equation from above and write it out with the better formula. With the extra terms in red, we obtain
dSn dn | = k · | d dn | æ è | (N · ln N – N) – (n · ln n – n) – [(N – n) · ln (N – n) – (N – n)] | ö ø
|  | After sorting out the signs, we have
dSn dn | = k · | d dn | æ è | N · ln N – N – n · ln n + n – [(N – n) · ln(N – n)] + N – n | ö ø
| Everything in red cancels and we are back to our old equation

Appendix: Mathematical tricks and Pitfalls

| Here are a few hints and problems in dealing with faculties and approximations.
| Having n << N, i.e. n/(N – n) » n/N = cV = concentration of vacancies does not allow us to approximate d/dn{(N – n) · ln (N – n)} by simply doing d/dn{N · lnN} = 0.
|  | This is so because d/dn gives the change of N – n with n and that not only might be large even if n << N, but will be large because N is essentially constant and the only change comes from n.
| The derivative of u(x) · v(x) is: d/dx(u · v) = du/dx · v(x) + dv/dx · u(x).
|  | The derivative of ln x is: d/dx(lnx) = 1/x
| Easy mistake: Don't forget the inner derivative, it produces an important minus sign:
d dn | æ ç è | ln (N – n) | ö ÷ ø | = | 1 N – n | · | d(N – n) dn | = | 1 N – n | · (–1 )

With frame

Internal Energy, Enthalpie, Entropy and Free Enthalpie

Basic Exercise 2-1-4: Derive the Formula for the Vacancy Equilibrium Concentration

Stirlingsche Formel

Exercise 2.1-7 Quick Questions

Internal Energy, Enthalpie, Entropy and Free Enthalpie

© H. Föll (Defects - Script)