Lecture05A - Boltzmann Factor (17.1, 2) In Chem 442, we...

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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. j j j H E j ψ = 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: ( 29 ( 29 2 2 2 2 2 , , 8 i x y z x y z h n n n n n n ma ε = + +
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Lecture 5 2 Now, for N atoms (ideal gas has no intermolecular interactions) ( 29 ( 29 3 i 1 Note that each depends on or V= , depends on the number of atoms, N N j i i j j j E a a E E N V E ε = = = 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 . ( 29 (this is called the Boltzmann factor) probability system has energy j j E j E j j p e Ce p E β - - = = 1 1 the partition funciton j j E j j j E j p C e e Q 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 - = ( 29 ( 29 ( 29 , Since: , and Then: , , j E N V j j j E E N V Q e Q Q 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 k T - = - = ( 29 ( 29 ( 29 , / , , , , j B E N V k T j e p N V T Q N V T - =
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Lecture 5 4 Let’s start using Q! Since p j is the prob. of finding the system in energy state E j : ( 29 ( 29 / all states , , , , j B j E k T E j j j j j j j E e E e E E p Q N V T Q N V β - - = = = ( 29 ( 29 , , Sneaky Stuff (Math trick): ln , , 1 , , 1 j j E N V j N V E j j Q Q N V e Q N V Q E e E Q - - = = = - = - , ln N V Q E = -
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Lecture 5 5 ( 29 2 1 lnQ ln 1 Another Trick: T B B k T Q E E T T k T β = = -
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Lecture05A - Boltzmann Factor (17.1, 2) In Chem 442, we...

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