ME 083 The Statistical Interpretation Of Entropy Professor David M' Stepp Mechanical Engineering and

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ME 083The Statistical InterpretationOf
EntropyProfessor David M. SteppMechanical
Engineering and Materials Science189 Hudson
Annexdavid.stepp_at_duke.edu549-4329 or 660-5325
http//www.duke.edu/dms1/stepp.htm 26 February
2003
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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)

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• 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

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• 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

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• 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 that n can assume a range of values between
  0 and N

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• 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)

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• 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

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• Example N 4 with n vacant sites and n full
  sites O

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• 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.

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• 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-n1

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• 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-n1)
• N (N-1) (N-2) (N-3) (N-n1) (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.

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• 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!

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• 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!

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• 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
• (abcdef 6 distinct configurations)

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• 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 lt ½ N, and
  becoming smaller than unity for n ½ N
• NCn has a maximum value near n ½ N