NotesPartSix - Let us finally solve some problems using...

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1 Let us finally solve some problems using ensembles and partition functions, and their connections to thermodynamics. We will first focus on independent molecules by which is meant that the Hamiltonian of the system of molecules is additive , and the corresponding wavefunction a product : abc a b c HH H H ψ ψψψ =+++ = …… Distinguishable molecules For example, consider a lattice of Argon atoms: a b c d •• They are distinguishable by their respective positions in the lattice. What is the partition function for independent distinguishable molecules? E Ze α β = (use index α to denote system states) Let’s say each molecule has 3 energy levels, and there are two molecules. We will use a j ε to denote the jth eigenstate of molecule a . Then, () ,1 , 2 , 3 ab jk Ej k εε =+ = Now, ( ) ( ) ( ) 11 12 33 123 since aaa bbb Zz z e e eee βε −+ −−− = =++ + ⎛⎞ + + + ⎜⎟ ⎝⎠ If all are really argon atoms, independent distinguishable N Zq = What if we have a gas of identical molecules? The molecules are indistinguishable particles, as they can be anywhere in space.
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2 How would we distinguish between states with energy () 12 21 and ab εε + + , if both particles are of the same species? You cannot. So, in the partition function, you must divide by two to not over count the number of system states! For N particles, with energy ( ) ( ) ( ) 123 abc E α εεε =+++ , divide by ! N , the number of arrangements of permuting 1,2,3,… among a,b,c,… . But, what to do about the system state ( ) ( ) ( ) 112 E . where you do not have N numbers to permute among the N particles? Then, ! N division wouldn’t be correct. But, if number of available configurations is much greater than N ( N Ω>> ) then the chance of obtaining two molecules in same state, i. e., a configuration where ( ) ( ) 11 ε + , is vanishingly small. Thus, N ! division is correct for N . identical ! indistinguishable L.BOLTZMANN N Zz N = >> N When is >> N ? For a particle in box, 3 2 2 2 83 where 62 B mL kT h πε ⎛⎞ Ω= = ⎜⎟ ⎝⎠ for the average particle at temperature T . Combining this with the condition N Ω >> leads to 3 2 2 12 6 B mk T N hV π >> Therefore, Boltzmann behavior is favored by high T, large mass, m , and low density, NV ρ = .
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3 Probabilities: Independent distinguishable : () ( ) ( ) ... ... ab ij E pe Q e e α βε ε β −+ + + == Σ Σ . The probability of finding molecule a in state i regardless of the state of all the other molecules is ( ) ( ) b a j i b a i i aj k i jk ii k ee pp ΣΣ =ΣΣ = Σ Therefore, the probability of finding the “a”th molecule in the ith state is: independent distinguishable i i e p q = Independent indistinguishable The argument is more complicated but the result is the same.
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This note was uploaded on 05/21/2008 for the course CHE 391 taught by Professor Cukier during the Fall '07 term at Michigan State University.

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NotesPartSix - Let us finally solve some problems using...

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