Calculate ln Omega (k) for N large and k = fN where f is a fraction....

Calculate ln Omega (k) for N large and k = fN where f is a fraction. Assume k and N-k are also large and use Stirling's approximation.

Entropy & 2 nd Law of Thermodynamics

Exam Study. Extremely Urgent.

Expert Solution

Configurational entropy refers to the entropy resulting from imperfect mixing of different atoms in the same site in a crystal.

When you roll a die, there are six possible outcomes, with the following probability

P(X=1) + P(X=2) + ... + P(X=6) = 1

The probability that any particular number will come up is

P(X=1) = P(X=2) = ... = P(X=6) = 1/6

The uncertainty associated with that probability is 0 if one outcome has a probability of 1, and a maximum when the outcomes are equally likely.

Imagine a material with M atoms that can occupy N sites. If the mixing is ideal, the atoms have no interaction with each other, so more than one can occupy the same site. The total number of configurations, , in this system is

= N^M
Copyright c 2009 by C.H. Mak
But, this is slightly incorrect (in what is known as Gibbs' paradox) because there is overcounting of identical configurations like these

With the overcounting removed we have
= N! / (N_p1! * N_p2! ...N_pn!)
where is the probability that a given number of atoms n in a given number of sites N will have a particular configuration. Or
ln = ln N! - ln(N_pi)!
N is always large where moles of material are concerned, so we can simplify this (using Stirling's approximation: ln (N!) = NlnN - N) to
ln = -N _piln_pi
For N atomic sites that can contain fraction X_A A atoms and X_B B atoms,
=
where is the probability that a given number of atoms M in a given number of sites N will have a particular configuration.
Boltzmann defined configurational entropy as
S_{configurational} = k ln (This is engraved on Boltzmann's tomb in Vienna!)
Simplified with Stirling's approximation (and recalling that R = Nk) to
S = - n R (X_Aln X_A + X_Bln X_B)
where n is the number of sites per mole. For example in cordierite there are 4 Al atoms and 5 Si atoms distributed over 9 tetrahedral sites. For a random distribution the entropy is
S = - 9 R (4/9 ln 4/9 + 5/9 ln 5/9) = 51.39 J mol^-1 K^-1

genius_generous answered 3 years ago

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