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Statistical Interpretation of Entropy in Crystalline Structures

This lecture presented by Professor David M. Stepp focuses on the statistical interpretation of entropy in the context of crystallography. It explores the free energy of crystals, including the implications of defects and configurational entropy. Through examples like vacancy defects, the discussion details how to calculate the number of ways in which defects can be arranged (W). Key concepts include distinguishing between full and vacant lattice sites, the process for determining distinct configurations, and the relationships between free energy changes, enthalpy, and entropy.

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Statistical Interpretation of Entropy in Crystalline Structures

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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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