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From Last Time….

ME 083 The Statistical Interpretation Of Entropy Professor David M. Stepp Mechanical Engineering and Materials Science 189 Hudson Annex david.stepp@duke.edu 549-4329 or 660-5325 http://www.duke.edu/~dms1/stepp.htm 26 February 2003. From Last Time…. The Free Energy of a crystal can be written as

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From Last Time….

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  1. ME 083The Statistical InterpretationOf EntropyProfessor David M. SteppMechanical Engineering and Materials Science189 Hudson Annexdavid.stepp@duke.edu549-4329 or 660-5325http://www.duke.edu/~dms1/stepp.htm26 February 2003

  2. From Last Time…. • The Free Energy of a crystal can be written as • the Free Energy of the perfect crystal (G0) • plus the free energy change necessary to create n defects (n*∆g), which is also internal energy change necessary to create these defects in the crystal (∆H) • minus the entropy increase which arises from the different possible ways in which the defects can be arranged (T∆SC) ∆G = ∆H - T∆SC • The Configurational Entropy of a crystal, ∆SC, is proportional to the number of ways in which the defects can be arranged (W) ∆SC = kB* ln(W)

  3. Remember: ∆SC = kB * ln(W) Configurational Entropy is proportional to the number of ways in which defects can be arranged. Example: Vacancy defect (calculation of W) Imagine a crystal lattice with N sites:

  4. Remember: ∆SC = kB * ln(W) Configurational Entropy is proportional to the number of ways in which defects can be arranged. Example: Vacancy defect (calculation of W) Imagine a crystal lattice with N sites:

  5. One possible complexion N sites and two states (full, vacant) n: Number of sites that are vacant n’: Number of sites that are full Then we have n + n’ = N Here, N = 21, n = 3, and n’ = 18 Note thatn can assume a range of values between 0 and N

  6. The situation where n lattice sites are vacant can be achieved in many alternate ways. Another arrangement (complexion) for our case is: N = 21 n = 3 n’ = 18 • Keep in mind also that we can differentiate between full and empty sites (mathematically); however, all full and all empty sites are equivalent (physically)

  7. In order to determine W (the number of ways in which defects can be arranged), we need to calculate all complexions that are distinguishable. NCn : The number of distinguishable (distinct) configurations of N lattice sites where any n are vacant and the remaining n’ are filled. = W

  8. Example: N = 4 with n vacant sites █ and n’ full sites O

  9. List all possible configurations for N = 4, n = 2 Process: Place █1 on one of the N lattice sites, and create configurations with █2 in each of the remaining (N-1) lattice sites. Now, label each distinct configuration.

  10. Generally: The possible number JN(n) of distinct vacancy placements is obtained by multiplying the number of possible locations of each vacancy. In our last example with N = 4 and n = 2: JN(n) = (# possible █1 locations) * (# possible █2 locations) = (4) * (4-1) = 12 More generally: JN(n) = N * [N-1] * [N-2] * [N-3] *… [N-n+1]

  11. Now, rewrite JN(n) in terms of factorials: N! = N * (N-1) * (N-2) * (N-3) * … (1) JN(n) = N * (N-1) * (N-2) * (N-3) *… (N-n+1) = N * (N-1) * (N-2) * (N-3) *… (N-n+1) * (N-n) *… (1) (N-n) *… (1) = N! (N-n)! But we cannot distinguish (physically) one vacancy from the next; therefore, JN(n)over estimates the number of distinguishable complexions.

  12. Now consider the vacancies alone (i.e., for any set arrangement in a lattice): the possible number of distinct permutations (combinations) of n vacancies is n! In other words, the possible number of permutations of n vacancies (even though these permutations are indistinguishable physically) is n! Example: For n = 3, how many possible distinct permutations exist? 123 213 231 132 312 321 = 6 = 3!

  13. Now, with: • The total possible number of distinct vacancy placements in our lattice • The total possible number of permutations of vacancies within a given placement n! The desired number, NCn, of distinguishable complexions is thus given by dividing JN(n) by n!

  14. Verifying our earlier example: N = 4, n = 2 JN(n) = 12 n! = 2 NCn = 12/2 = 6 Recall all distinct configurations for N = 4, n = 2 (a+b+c+d+e+f = 6 distinct configurations)

  15. Properties of NCn: • Symmetric under interchange of n and n’ NCn = NCn’ NC0 = NCN = 1 The ratio of successive coefficients is initially large, but decreases monotonically with n. Staying larger than unity as long as n < ½ N, and becoming smaller than unity for n ≥ ½ N NCn has a maximum value near n = ½ N

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