# MIT 5 62 - Internal Degrees of Freedom (8 pages)

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## Internal Degrees of Freedom

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## Internal Degrees of Freedom

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Pages:
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School:
Massachusetts Institute of Technology
Course:
5 62 - Physical Chemistry II
##### Physical Chemistry II Documents
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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 classical descriptions and compare results QUANTUM DESCRIPTION Calculate x average kinetic energy in x direction x x P x d x 0 P x kT 1 x kT 1 2 1 x kT 1 2 1 2 1 2 x kT x e 0 x 1 2 x e kT d x kT kT 1 2 1 kT 2 2 1 This is not an accident It is our first glimpse of equipartition of energy 2kT into each independent degree of freedom Integral table for two useful integrals 0 0 x1 2 e ax dx 1 2 2a 3 2 x 1 2 e ax dx a 1 2 dimension length 3 2 dimension length1 2 Since the translational energies in each dimension are uncorrelated average total energy is the sum of the average energy in each direction Also as a consequence of the separability of the Hamiltonian with respect to x y z coordinates total E is sum of Ex E y E z x y z 1 2kT 1 1 3 2kT 2kT 2kT 1 Each degree of freedom contributes 2kT to total energy Agrees with result from ensemble average 5 62 Spring 2008 Lecture 10 Page 2 3 ln Q trans E kT2 P E kT T 2 N V CLASSICAL MECHANICAL DESCRIPTION calculate State of a molecule is described by p momentum q position The energy of N molecule system is p 3N q 3N which is a continuous function of 6N variables q trans e i kT e let s assume this by analogy i quantum V dp dq classical integral over 3 momentum and 3 position coordinates for each particle 2 q trans cl p q kT 2 2 p x p y p z dq x dq y dq z dp x dp y dp z e p dp x e p x 2mkT 2 dp y e p 2y 2mkT 2 2mkT 2 dp z e p z 2mkT 2 3 2 3 V dp x e p x 2mkT V 2 mkT 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 t If the system returns to q t1 p t1 at t2 then it must move to q t1 t p t1 t at t2 t The ergodic hypothesis requires trajectories to visit each cell of phase space of volume h3N not each location in phase space This avoids the problem of self intersecting trajectories 3 2 2 mkT V we see that we need a factor of h3 in the denominator Comparing to q trans qm h 2 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 q trans cl Q trans cl 1 e classical 3 h kT dp dq dimensionless qN 1 trans cl e classical 3N N N h kT N VN 2 mkT dp dq N h 2 3N 2 revised 2 29 08 8 52 AM 5 62 Spring 2008 Lecture 10 Page 3 CONCLUSION Classical and quantum descriptions of translational degrees of freedom yield consistent results QM is operating in classical limit because energies of the translational states are so closely spaced that they can be approximated as continuous as in classical mechanics kT 0 6 kcal mol 1 at 300K 10 20 kcal mol 1 particle in box 3 The average molecular translational energy 2 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 1 Each squared momentum or position term in the energy of a particle contributes 2 kT to the average energy according to this principle translation along each coordinate axis contributes 1 1 2 kT rotation about each principal axis contributes 2 kT and each vibrational mode contributes 1 kT 2 kT each for kinetic and potential energy to the average energy p 3N q 3N kT g e P p e 3N q 3N kT Classical Boltzmann distribution function d p 3N d q 3N N h 3N g is the density of states often denoted as It has units energy 1 1 1 1 3 P d x y z kT kT kT kT 0 2 2 2 2 INTERNAL DEGREES OF FREEDOM FACTORIZATION OF Q ATOMS have one internal degree of freedom ELECTRONIC degree of freedom revised 2 29 08 8 52 AM 5 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 trans int quantum s N M L internal quantum s where int energy from internal degrees of freedom q e i e trans int kT kT i all molecular states all molecular states We do not have to start over qtrans factors out q e trans kT translational states q qtrans qint N q trans q int Q N Q QtransQint Q trans QN trans N Q int Q N int e int kT internal states INTERNAL MOLECULAR PARTITION FUNCTION N q trans q N N int CANONICAL PARTITION FUNCTION CANONICAL TRANSLATIONAL PARTITION FUNCTION 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 motions don t interchange particles Classically Qcl Qtrans cl Qint cl Q trans cl q Ntrans cl N e trans kT dpdq N N h 3 N Q int cl q Nint dp 3N dq 3N e int kT h 3N CONTRIBUTION OF INTERNAL DEGREES OF FREEDOM TO MACROSCOPIC PROPERTIES ln Q E kT2 T N V ln Q transQ int kT2 T N V ln Q trans E kT2 T N V E ln Q int kT2 T N V E trans E int will equipartition be applicable here A kTln Q kTln Q transQ int kTln Q trans kTln Q int A trans Aint Likewise S Strans Sint ln …

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