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MIT 5 62 - Internal Degrees of Freedom

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MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.5.62 Lecture #10: Quantum vs. Classical. qtrans. Equipartition. Internal Degrees of Freedom. GOAL: Calculate average translational energy via quantum and classicaldescriptions and compare results. –QUANTUM DESCRIPTION — Calculate — average kinetic energy in x-directionxε εx = ∫∞ εxP (εx )dεx0 kT P (εx )=(πkT)−1/2 εx −1/2e−εx ∞ ⎛ 1 ⎞1/2 1/2e−εx kTdεx =εx ⎝⎜ πkT⎠⎟ ∫0 εx ⎛ 1 ⎞1/2 kT (πkT)1/2 = 1 kTεx = ⎝⎜ πkT⎠⎟ 2 2 This is not an accident. It is our first glimpse of “equipartition” of energy, 12kT into each independent degree of freedom. Integral table for two useful integrals x1/2e−axdx = π1/2 ∫∞ 2a3/ 2 (dimension: length3/2 )0 ∫0 ∞ x−1/ 2e−axdx =(π a)1/2 (dimension: length1/2 ) Since the translational energies in each dimension are uncorrelated, average total energyis the sum of the average energy in each direction. Also, as a consequence of theseparability of the Hamiltonian with respect to x, y, z coordinates, total E is sum of Ex + Ey + Ez. ε – = 1 2kT + 12kT + 12kT = 23kT Each degree of freedom contributes 12kT to total energy. Agrees with result from ensemble average – – – ε + ε + ε x y z  5.62 Spring 2008 Lecture #10, Page 2 ⎛∂lnQtrans ⎞ 3ε = E = kT2 ⎝⎜ ∂T ⎠⎟ = ∑ PαEα = 2 kT N ,V α CLASSICAL MECHANICAL DESCRIPTION — calculate – ε State of a molecule is described by p momentum, q position. The energy of~ ~N–molecule system is ε(p3N, q3N), which is a continuous function of 6N variables.~ ~ q trans ∑ e− ε i kT let's assume this by analogy ∫∫ e−ε( p~,q~)/kTd p d q= ~ ~ i quantum ⇒ classical — integral over 3momentum and 3 positioncoordinates for each particle = px2 + py2 + pz2 ↓ ( )qtrans,cl = ∫∫∫ dqxdqydqz ∫∫∫ dpxdpydpze−p2 / 2mkT= V ∫ dpxe−px2 /2mkT ∫ dpye−py2 /2mkT ∫ dpze−pz2 /2mkT = V⎣⎢⎡ −∞ dpxepx2 /2mkT ⎦⎥⎤ ( )3 /2 ∫∞ 3 = V 2πmkT [qtrans,cl. should be dimensionless, but this has units m3l6t–3 = h3] The need to divide by a factor of h3 was recognized by Boltzmann even before Planck “invented”h. It is impossible for a trajectory in phase space to intersect itself. [Why? Because the classical Hamiltonian is a function of q,p. If the system is at q(t1), p(t1) at t1 then the H tells us the values of q and p at t1 + δt. If the system returns to q(t1), p(t1) at t2, then it must move to q(t1 + δt), p(t1 + δt) at t2 + δt.] The ergodic hypothesis requires trajectories to visit “each cell” of phase space ofvolume h3N, not “each location” in phase space. This avoids the problem of self-intersectingtrajectories. ⎛ 2πmkT ⎞ 3 /2 Comparing to qtrans,qm = ⎝⎜ h2 ⎠⎟ V , we see that we need a factor of h3 in the denominator of qtrans,cl.. In fact, it turns out that to construct a partition function from a classical Hamiltonian,this h–3 factor is required, so our assumption above must be corrected: = h13 ∫ ∫ e−εclassical kTd p d q (dimensionless) qtrans,cl ~ ~ ⎛ 2πmkT ⎞ = qtrans,cl1 ⎡⎣∫ ∫ e−εclassical kTdp dq⎤⎦ = N! ⎝⎜ h2 ⎠⎟=Qtrans,cl NN! N!h3N N VN 3N/2 revised 2/29/08 8:52 AM5.62 Spring 2008 Lecture #10, Page 3 CONCLUSION: Classical and quantum descriptions of translational degrees of freedom yieldconsistent results. QM is operating in classical limit because energies of the translational statesare so closely spaced that they can be approximated as continuous, as in classical mechanics. kT = 0.6 kcal mol–1 at 300K ∆ε = 10–20 kcal mol–1 (particle in box) The average molecular translational energy, 23 kT, is independent of the kind of molecule, or more precisely, independent of the energy level spacings because these spacings are ~ 1020 times smaller than kT. Spacings don't change much from He, to C6H6, to DNA. If T were decreased to – 10–19 K, then He, C6H6, and DNA would have different Etrans. CLASSICAL EQUIPARTITION OF ENERGY PRINCIPLE Each squared momentum or position term in the energy of a particle contributes 21 kT to the average energy; according to this principle, translation along each coordinate axis contributes1 1 2 kT; rotation about each principal axis contributes 2 kT; and each vibrational mode contributes kT (21 kT each for kinetic and potential energy) to the average energy. g(ε)e 3N ,q~3N ) −ε(p3N ,q3N ) kT−ε (p kT Classical Boltzmann ~ distribution function( ) =P ε∫∫ e ~ ~ d p~3Nd q~3N N!h3N g(ε) is the density of states, often denoted as ρ(ε). It has units (energy)–1. ε = ∫0 ∞εP ε( )dε = εx + εy + εz = 1 1 1 3kT + kT + kT = kT2 2 2 2 INTERNAL DEGREES OF FREEDOM — FACTORIZATION OF Q ATOMS — have one internal degree of freedomELECTRONIC degree of freedom revised 2/29/08 8:52 AM5.62 Spring 2008 Lecture #10, Page 4 MOLECULES — have other degrees of freedom ELECTRONIC, VIBRATION, AND ROTATION, each of which contributes to total energy and to other macroscopic properties. Nuclear hyperfine? [Nuclear spin degeneracy factors. LATER.] Internal energy adds to translational energy to get total energy ε = εtrans + εint quantum #'s internal quantum #'sN,M,L where εint = energy from internal degrees of freedom q = ∑ e−εi kT ( ) kT = ∑ e− εtrans +εint i all molecular states all molecular states We do not have to start over. qtrans factors out. kT )(∑ e−εint kT )q =(∑ e−εtrans translational states internal states q = qtrans • qint ← INTERNAL MOLECULAR PARTITION FUNCTION ⎞Q =(qtransqint )N ⎛ qNtrans N= N! ⎝⎜ N! ⎠⎟ qintQ = QtransQint CANONICAL PARTITION FUNCTION Qtrans = Qtrans N N! CANONICAL TRANSLATIONAL PARTITION FUNCTION Qint = QintN CANONICAL INTERNAL PARTITION FUNCTION revised 2/29/08 8:52 AM 5.62 Spring 2008 Lecture #10, Page 5 NOTE: N! was included with Qtrans. This is because it's the translational motion that causes the positions of identical particles to be interchanged, requiring the factor of N! The internal motionsdon't interchange particles. Classically Qcl = Qtrans,cl Qint,cl Qtrans,cl = qtrans,cl N N! = [ e− ε trans /kT∫ dpdq]N N!h3 N Qint,cl = qint N = d p3N ∫ d q3N e−εint /kT h3N CONTRIBUTION OF INTERNAL DEGREES OF FREEDOM TO MACROSCOPIC PROPERTIES ⎛∂lnQ ⎞


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MIT 5 62 - Internal Degrees of Freedom

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