# Lecture05A - Boltzmann Factor(17.1 2 In Chem 442 we gave a...

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Lecture 5 1 Boltzmann Factor (17.1, 2) In Chem 442, we gave a brief introduction to temperature and the population of Quantum States. With bulk measurements of ~10 23 molecules, how do we describe the distribution of these molecules in energy levels from our Q.M. solutions? Can we assume all molecules are in the lowest energy state? ˆ From Q.M. we know for eigenstates of a system. jj j HE j ψψ = But now we are dealing with really big (macroscopic) systems. To get some handle, let’s assume that we have an ideal monatomic gas. If we had one atom in a cubic box with side length, a , then: () 2 222 2 ,, 8 ixy z x y z h nnn n n n ma ε =+ + Lecture 5 2 Now, for N atoms (ideal gas has no intermolecular interactions) 3 i 1 Note that each depends on or V= , depends on the number of atoms, N N ji i j Ea a EE N VE εε = = = So, what is the probability that the system of N atoms will be a particular state j? To answer this question, we assume that N is very large (10 23 ) and is in contact with a heat reservoir of infinite capacity at a temp, T. Each system (atom) has some value of N,V,T, but can be in different energy states E j . This collection is called an ensemble . (this is called the Boltzmann factor) probability system has energy j j E j E j j pe Ce p E β == 1 1 the partition funciton j j E j E j pCe eQ C =≡ = ∑∑ Tells us how the system is partitioned into different energy states.

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Lecture 5 3 This gives the important result: j E j e p Q β = () , Since: , and Then: , , j E NV jj j EE N V Q e QQ N V == = Q depends on the energy levels from Q.M. solutions to the system, i.e. microscopic/atomic values. But as we will find, Q will also provide bulk properties of macroscopic collection of these systems. Appended to the end of this lecture is a derivation of the Boltzmann distribution done by Jordan Beck, along with arguments for 1 B 1 , thermal energy of the system, k Boltzmann constant B kT ββ =− = ,/ ,, jB E NV kT j e pNVT QNVT = Lecture 5 4 Let’s start using Q! Since p j is the prob. of finding the system in energy state E j : / all states j E kT E j Ee p QNV −− = ∑∑ , , Sneaky Stuff (Math trick): ln , , 1 1 j j E NV j E j j Q e Q E Q ∂∂ = , ln Q E
Lecture 5 5 () 2 1 lnQ ln 1 Another Trick: T B B kT Q EE TT k T β ⎛⎞ ∂∂ ⎜⎟ == = ⎝⎠ 2 , ln B NV Q Ek T T = Let’s apply this to some systems we’ll study this semester.

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Lecture05A - Boltzmann Factor(17.1 2 In Chem 442 we gave a...

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