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)