Lecture 11 - How to evaluate thermodynamic properties in a...

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Unformatted text preview: How to evaluate thermodynamic properties in a diatomic? Bor n-Oppenh e im e r approximation nuclei remain still while e- move very fast H separation of degrees of freed om H = H nuclear motion + H electronic H electronic n ( r ; R ) = n n ( r ; R ) for each nuclear separation R && the electronic Schrodinger eq. is solved R R r nuclear separation as parameter electronic coordinates U(R) A2 for each electronic state n N H nuclear n ( R ) = n n ( R ) A1 + A A+A R we want to separate the center of mass motion from the internal motions 2 particle system q trans CM 1 particle system, with m=m1+m2 3 2 ( m1 + m2 ) kT 2 = V 2 h for internal (relative) motion as vibrations and rotations 2 particle system 1 particles system with m1 m2 reduced mass = m1+m2 we expand U ( R ) around the Req, considering the motion of a rigid rotor, and small vibrations of a Harmonic Oscillator H nuc .motion = H vibration + H rotation nuc .motion = vibration + rotation H rotation YJM ( , ) = J ( J + 1) hBe YJM ( , ) H vibrationSv ( R - Req ) = h v + 1 Sv ( R - Req ) 2 U ( R ) ; U ( Req ) dU =0 dR Req 2 dU 1 2U d + ( R - Req ) + 2 ( R - Req ) + L dR Req 2 dR Req ( ) 2U d 2 = curvature is a constant dR Req 1 Harmonic Oscillator Req by remembering that we can also separate electronic and nuclear config. we obtain the total molecular q(V,T) H = H tran slatio n + H electro nic + H nuclear + H vibra tio n + H rotation = tra n sl . ( R ) elec . ( r, , ) nucle ar vib. ( R - Req ) JMrot . ( , ) S Y = transla tion + electron ic + nuclea r + vib ration + rotatio n and therefore ( qtransl .qelect .qnuclear qvib.qrot . ) Q(N,V ,T ) = N! N ATTENTION!!!! since the # translation states is enormous H ALW AYS more states than particles! H we ca n use BOLZTMANN STATI STICS for each individual q, we can define a reference energy: translation we integrate (co unt ) all state s f rom 0 electronic 1 = -De 1s t excited st ate energ y = 2 nuclear 1 1 = 0 vibration 1 v = 0 rot ati on 1 J = 0 = 0 1 = h 2 2 1 Do + h = De 2 0 D"o D"e 1 Req We already know q trans CM ( m1 + m2 ) kT 2 = V 2 h De kT 2 3 2 q electronic ( V ,T ) = e 1 q nuclear ( V ,T ) = 1 + e ( 2 ) - kT +L we only have to evaluate qvib and qrot q vib e ( V ,T ) = h 2 - h n - n sum over states n 14 2 4 3 = e n - ( n + 12 ) h q vib e ( V ,T ) = n - e =e vib - h 2 e n ( - h h ) n we recognize the series 1 xj = 1- x j q e 2 ( V ,T ) = - h 1- e - for T , h = 1 and we can use e 2 q ( V ,T ) = e e dn = h Vibrational partition 0 function q h kT vib and since 0 q ( V,T ) ; 2 h vib 2 - h - h n - h the vibrational contribution to the E is given by N Rln q = kT T 2 Evib = kNT 2 e 2 ln - h 1- e T - h - h2 - h ln 1 e - 2 lne Evib = NkT - T T ( ) 2 h = -NkT T 2 Evib = + (1- e + h e - h - h ) ) T Evib h = N 2 + where we used - h =- 1 1- e ( ) T kT 2 h e - h Nh 2 ( Nh e h -1 we define the vibrational temperature as vib h = k v v + the contribution to the E becomes E = Nk v 2 e T -1 ( ) W hat is the population of each vibrational level? for a given state, the relative population is fn = e - h n + 12 ( ) qvib = e - h n + 12 ( ) fn = e - h n + 12 - 12 ( ) ( ( e - h 2 =e - h n + 12 ( ) e + h 2 ( 1- e - h ) 1- e - h - h 1- e ) ) fn = e ( - h n -e - h ( n +1) ) or the fraction of molecules NOT in the ground state -v n -v e 1 fn - f0 = 1 - T 1 - e T n = 0 we can also get Cv (heat capacity @ const. V) E v Cv = = Nk 1 V,N T T for T Evib = Nh 2 Nh 2 + 2 (e e v T v T -1 ) 2 (e Nh h -1 - h kT ) : Nh h = NkT Cv = Nk - h for T 0 Evib = + Nh e 2 kT Cv = -Nk 2 e T 0 ...
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