5_60_lecture28

5_60_lecture28 - MIT OpenCourseWare http:/ocw.mit.edu 5.60...

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MIT OpenCourseWare http://ocw.mit.edu 5.60 Thermodynamics & Kinetics Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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5.60 Spring 2008 Lecture #28 page 1 MODEL SYSTEMS Starting with QM energy levels for molecular translation, rotation, & vibration, solve for q and Q, & all the thermodynamics, for these degrees of freedom. The results are the fundamentals of molecular statistical mechanics. We’ll derive the results for a classical model that maps onto QM vibrations. Then we’ll compare to results (given, not derived) for translation and rotation. Double-stranded polymer model Each monomer in one strand interacts with a monomer in the other strand. Interaction energy for each monomer pair is – ε 0 . The strands can “unzip” from one end, rupturing the interactions of the end monomers, then the next ones, then the next, and so on. Each ruptured interaction raises the energy by ε 0 . The three lowest-energy states and the energy levels are illustrated below. μ Configurational energy levels ε conf ª μ n ε 0 0 ª 0 ª 0 ª 0 ª μ 0 ª 0 ª 0 ª 0 ª ε 0 ª 0 ª ª ª ª ª ª ª ª 0 ε 0 2 ε 0 Nondegenerate evenly spaced levels, separated by energy ε 0 : ε = n ε 0 (n = integer)
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5.60 Spring 2008 Lecture #28 page 2 For a very long polymer, there is a large number of levels. Then we can extend the sum over states in q conf to infinity because the highest energies are much bigger than kT anyway (so the corresponding terms in the sum are negligible). q conf = e −ε n kT n 0 kT = 1 + e −ε 0 kT 2 0 e − ε + e − ε kT 3 0 + e − ε kT + " n n =+ 1 e −ε 0 kT −ε 0 kT ) + ( e −ε 0 kT 3 2 3 1 −ε 0 kT + ( e 2 ) + " 1 + x + x + x + " = 1x where x e 1 q conf = 1e −ε 0 kT So q conf takes a very simple closed form. Everything else follows. conf = ( q conf ) N = 1e 1 −ε 0 kT N A =− kTlnQ =− NkTlnq =− NkTln = NkTln 1 conf conf conf 1 −ε 0 kT ( e −ε 0 kT ) 1e = A conf = ( e −ε 0 kT ) A conf scales with N, μ conf = A conf /N μ conf N T,V kTln 1 2 dlnq conf 2 1 −ε 0 kT ε 0 = N ε e −ε 0 kT = N ε 1 ) ( ) kT 0 1e U conf = N k T dT = N k T ( 1e −ε 0 kT e 2 ( −ε 0 kT ) 0 ( e ε 0 kT 1 ) 2 ε 0 kT 2 dU conf e ( ε 0 k T ) ε 0 e ε kT 0 C V conf
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This note was uploaded on 11/27/2011 for the course CHEM 5.43 taught by Professor Timothyf.jamison during the Spring '07 term at MIT.

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5_60_lecture28 - MIT OpenCourseWare http:/ocw.mit.edu 5.60...

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