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The entropy of mixing thus is |
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S |
= k · ln |
N!
n! · (N n)! |
= k · |
( |
ln N! ln {n! · (N
n)!} |
) |
= k · |
( |
ln N! ln n! ln
(N n)! |
) |
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We now can write down the free enthalpy for a
crystal of N atoms containing n vacancies |
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G(n) |
= |
n · GF |
kT · [ln N! ln n!
ln (N n)!] |
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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. |
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Mathematical approximation: Use the
simplest version of the Stirling
formula |
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Physical approximation, assuming that
there are far fewer vacancies than atoms: |
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n
N n |
» |
n
N |
= |
cV = |
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concentration
of vacancies |
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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. |
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Essentially we need to consider
dS(n)/dn using the Stirling formula |
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dSn
dn |
= k · |
d
dn |
( |
ln N! ln n! ln
(N n)! |
) |
» k · |
d
dn |
( |
N · ln N n · ln n
(N n) · ln (N
n) |
) |
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But we must not yet
use the physical approximation, even so its tempting! With the formula for
taking the derivative of products we obtain
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dSn
dn |
» k · |
( |
( |
( ln n |
n
n |
) |
|
( |
ln ( N n) + |
n N
N n |
) |
· (
1) |
) |
dSn
dn |
» k · |
( |
ln n + 1 ln (N n)
1 |
) |
= k · |
( |
ln n ln (N n) |
) |
= k · ln |
n
N n |
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Now we can use the physical approximation and obtain |
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Putting everything together gives |
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dG(n)
dn |
= 0 = GF
T · |
dSn
dn |
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= GF + kT · ln
cV |
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Reshuffling for cV gives the final
result |
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q.e.d. |
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Here are a few hints and problems in
dealing with faculties and approximations. |
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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. |
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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. |
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The derivative of
u(x) · v(x) is: d/dx(u ·
v) = du/dx · v(x) + dv/dx ·
u(x). |
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The derivative of ln x is:
d/dx(lnx) = 1/x |
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Easy mistake: Don't
forget the inner derivative, it produces an
important minus sign: |
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d
dn |
( |
ln (N n) |
) |
= |
1
N n |
· |
d(N n)
dn |
= |
1
N n |
· (1 ) |
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© H. Föll