Lecture14Ch161Mar23 - Summary of last lecture Heat capacity...

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Unformatted text preview: Summary of last lecture Heat capacity of crystals a 1D exact solution Chemistry 161: Statistical Thermodynamics Lecture 14 Thermal Properties of Crystals - the Debye Model Key Concepts and Lessons: a) Debye model of thermodynamics of crystals. b) Concept of Phonons c) Diatomic Ideal gases - partition functions Reading: Reif, Ch10.2 The Debye Model of heat capacity of 3D crystals The 1-dimesional example of lattice dynamics and exact calculation of TD of crystals is useful and very illuminating but it is of course limited because we live in 3D. To that end Peter Debye worked out an approximate formula that works well in most instances and gives good insight into physical behavior of solids. The Debye formula is an approximation for our old friend dispersion relation g ( ) . As before we assume that there is a wave of displacements but now this wave is in 3D: ( i kr r ,t = Ae ( ) t ) We should impose boundary conditions of zero displacement at the boundaries of the crystal. Such conditions determine possible values of k: kx = n Lx x kx = kz = Ly Lz ny nz Where n are of course integer numbers and Lx,y,z are spatial dimensions of our crystal. The counting of standing waves in k space is easy: V 3 kx k y kz = nx n y nz The Debye Model of heat capacity of 3D crystals Converting to polar coordinate in k-space and taking into account that we are interested only in positive n's (i.e. 1/8 of the space) we get the number of normal modes between k and k+dk as: Vk 2 dk g(k )dk = 2 2 Now we have to relate and k. This is done using the relation (approximately!) where u is speed of sound. In other words: =2 = uk / 2 g ( ) 4 V d = u3 2 d This is (almost) the desired result. To make it more precise we note that there are 3 sound waves: 2 transverse and 1 longitudinal and their speed of sound may differ. Therefore a better approximation would be: g () 2 1 12 V d = 3 + 3 4 V 2d = ut ul u3 2 d The Debye Model of heat capacity of 3D crystals (Cont) Now we define the limiting frequency (called Debye frequency) 1D case with max ) so that: D D (pretty much as we've seen in g( )d = 3N; i.e. 0 D 3N = 4 V 1/ 3 u This relation tells us that total number of normal modes should be 3N. Then presented in a simple form: g( ) can be g ( )d = 9N 3 D 2 d > D 0 D 0 Now we define the Debye temperature D = 3 h k D and change variables to /T x= h kT to get: CV = 9Nk T D D 0 (e x 4ex x 1 ) 2 dx The Debye Model of heat capacity of 3D crystals (Cont) The integral can be calculated approximately and in asymptotics. When extended up to infinity and: T D then integration can be CV = AT 3 In the opposite case when T D exponentials can be expanded (because x is always small) CV (prove that: 20pt) 3Nk for soft molecular crystals lies in the range 100-200K while for several metals it is in the range of 300-500K. For diamond D is 1860K so that this solid should described as low-temperature case where QM is crucial. D The Debye Model of heat capacity of 3D crystals (Cont) The Debye function can be defined as: 1/ x D(x) = 3x 3 0 ( e 1) y y 4e y 2 dy Then: T D CV = 3NkD The concept of phonons The picture of normal modes suggests that we can present total vibrational energy of a crystal as: E= 3N h j =1 j 1 = nj + 2 3N j =1 h j n j + E0 It is as if we have independent particles with energies per particle and with occupancy numbers Since there are no limits on n these particles are bosons. Now we can write the basic formula for occupancy numbers of these particles (called phonons) nj = e 1 h kT 1 E = E0 + 0 With density of states g we get as before: g( )h d e h kT 1 Why is this a useful concept? The reason is that in more complicated situations where behavior of electrons depends on the state of nuclear lattice this factor can be treated as interaction of electrons with phonons as if they were real particles and that helps a lot in the analysis of complex situations where anharmonicity of lattice vibrations plays an important role. Diatomic Gases Now we turn our attention to the discussion of diatomic gases. In this case molecules possess additional degrees of freedom (rotation and vibration change of distance between atoms and their orientation). Here we note that electronic motions are much faster than nuclear ones. This means that nuclear motions can be viewed as motion in an effective potential which is comprised of direct Coulomb interactions between nuclei which are repulsive of course - and electronic energy of the whole molecule as a function of distance R between the nuclei. This is called the Born-Oppenheimer approximation. Assuming the ground state for electrons (good assumptions at room and higher T) we get the motion of nuclei in an effective potential with a deep minimum around U min = U ( R0 ) . Now we can consider small deviation from equilibrium and get the nuclear motion to occur in the potential U R1 , R2 = U ( R1 where d 2U ( R) = dR 2 ( ) R2 ) = 2 (R 1 R2 ) 2 The Quantum Hamiltonian of a diatomic molecule is then: p2 p2 ^ R H diatomic = 1 + 2 + 2m1 2m2 2 1 ( R2 ) 2 where m1 and m2 are nuclear masses and the center of mass of the molecule: ^ p1,2 = m1,2 R1,2 are operators of momenta for nuclei 1 and 2. Now introduc R= m1 R1 + m2 R2 m1 + m2 Diatomic gases (Cont) we get: R1 = R + R2 = R m2 ( R1 R2 R2 ) ) m1 ( R1 m1 + m2 m1 + m2 ^ H diatomic = where d = R1 R2 2 m1 + m2 ( 2 pc.m ) + pv2 2 + 2 ( R1 R2 = ) 2 1 1 1 2 m1 + m2 R 2 + d 2 + d 2 2 2 = m1m2 ( ) is distance between nuclei and m1 + m2 is so-called reduced mass of the molecule (do not confuse with chemical potential!!!!!). Now we see that our diatomic molecule participates in three independent motions: 1) Motion of its center of mass as a whole, effective mass m1 + m2 2) Nuclear vibration, effective mass and frequency = 3) Rotation around common center 2of mass, inertia moment is 2 2 m1 ( R1 R ) + m2 ( R2 R ) = ( R1 R2 )0 = d02 where subscript 0 means that we take this distance at equilibrium. Diatomic gases (Cont) Contributions to CPF from different degrees of freedom are independent and therefore CPF factorizes into: Qdiatomic = ( qtransl qvibqrot qel qnucl N! ) N Where qtransl is contribution due to translational motion of center of mass, qvib is contribution due to the vibrational motions described earlier, qrot is contribution due to the rotational motions of the molecule as a whole; qel is a contribution to electronic degrees of freedom and qnucl is due to nuclear degrees of freedom; the latter two terms account mostly for degeneracies of ground states of electrons and nuclei because thermal motion is not likely to excite these degrees of freedom. The HFE is then: A = Atransl + Arot + Avib + Ael + Anucl Where we already derived the translational contribution: Atransl = NkT ln 2 ( m + m ) kT 1 2 3/ 2 h 2 Ve N This term includes the N! denominator as usual. Diatomic Gases - Vibrational PF The vibrational partition function of a diatomic molecule is easi;y derived because it is just a standard PF function for a Harmonic Oscillator which we derived earlier: qvib = ( n +1/ 2) e n=0 kT = 1 e 2 kT e kT We denote as usual a characteristic vibrational temperature as Avib = Nk v v = kB / 2 + NkT ln 1 e v T And for energy and contribution to heat capacity we get in a straightforward way: Evib = NkT 2 ln qvib = Nk T 2 v 2 + e v v v T 1 Ev T = cvib = Nk N v e v T 2 T e T 1 How important is the TD contribution from vibrational motion? H 2 : 6100 ; N 2 : 3340 ; O2 : 2230 ; NO : 2690 ; HCl : 4140 ; These numbers are determined spectroscopically in vibrational adsorption experiments - in the IR region around 1000cm-1. Apparently at room temperature of 300K all vibrations of covalent bonds remain frozen - vibrationall motions do not contribute to heat capacity and contribute only zero energy to total energy. Rotational PF of Diatomic Ideal gas heteroatomic molecules Now consider another essential degree of freedom rigid rotor rotations. The quantum number describing this motion is angular momentum (Maximal) J and square of angular momentum is: M2 = 2 J J +1 2 ( ) ) Energy associated with this motion can be evaluated in analogy with classical mechanics: J M2 = = 2J J J + 1) 2I ( Where I = R1 ( R2 ) 2 is moment of inertia of the molecule Each rotational level is 2J+1 times degenerate to account for possible z-projections of the angular Momentum for the molecule. We can also introduce the `'rotational'' temperature r = 2 2Ik And get the rotational PF for a diatomic molecule: qrot = J =0 ( 2J + 1)e r J ( J +1) T Rotational PF - high T hetero- and homoatomic molecules Now we note that in contrast to vibrations are usually very low (the highest for H2 is about 85.3K, r more massive molecules, e.g. O2 have much lower values of around 2K or so). That means that rotational PF can be presented as an integral: qrot = 0 ( 2J + 1)e r J ( J +1) T dJ = 0 d J ( J + 1) e { } r J ( J +1) T = T r = 8 IkT h2 2 T and the contribution to heat capacity: crot Nk In case of homoatomic molecules rotation by 1800 does not change anything. This means that we are overcounting qrot in reality it should be divided by 2, i.e. in this case: Erot = NkT 2 ln qrot NkT q hom oatomic rot T 4 2 IkT = = 2 r h2 This will become very important when we consider chemical reactions. Low T rotation - para and ortho hydrogen In the quantum case the situation is more complex. We will carry out the discussion in case of 0 molecules consisting of spin nuclei (the most important example hydrogen molecule with r = 64 K ). Here we note that nuclei act as fermions (spin !!!!). Therefore total wavefunction (WF) of the two nuclei molecule should be antisymmetric with respect to exchange of the nuclei. Now the total WF of two nuclei must be antisymmetric with respect to nuclei exchange (because they are fermions)! ( ( 1 , 2, R1 , R2 = ) ( 1 2 , 1, R2 , R1 ) The total WF consists of coordinate part and spin part because two are `'independent'': 1 , 2, 1 R , R2 = ) ( , 2, ) (R ,R ) 1 2 Now comes real FUN. The total spin state for two nuclei can be either 1 or 0. In the first case the spin ; ; 2 1/ 2 ( + ) WF can be composed of three symmetric wave functions 1/ 2 ) and in the case of total spin 0 only one antisymmetric nuclear WF 2 ( Now: the total WF must be antisymmetric. Therefore if spin WF is symmetric (when total spin is 1) the coordinate WF must be antisymmetric and vice versa. Antisymmetric coordinate WF corresponds to odd values of angular momentum quantum numbers J. The symmetric spin WF is three times degenerate (because with spin 1 there are 2J+1=3 `'directions''). Low T rotation - para and ortho hydrogen qrot,nucl = 3 odd J (2J + 1)exp( r J(J + 1) / T + even J ) (2J + 1)exp( r J(J + 1) / T ) If r T the discreteness (integer character) of J becomes irrelevant and we get odd J ( 2J + 1) exp ( ( ) ( r J ( J + 1) / T ) ) even J ( 2J + 1) exp ( r J ( J + 1) / T ) 1 2J + 1 exp 2 all J r J ( J + 1) / T = T 2 r qrot ,nucl = 4 T 2 r The factor is purely due to all possible orientations of two nuclear spins. So at high T the result is trivial nuclear and rotations degrees of freedom again factorize and we get the same result. 4 = 2I + 1 = 2 ( ) 2 1 +1 2 2 Low T rotation - para and ortho hydrogen However the picture is different when we consider T (640 for for H2). In this case we cannot replace r sums by integrals. Hydrogen with total nucl spin of 0 is called para-hydrogen and with total nucl spin of 1 is called ortho-hydrogen. N ortho = N para 3 odd J ( 2 J + 1) e r J ( J +1) / T even J ( 2 J + 1) e r J ( J +1) / T A really striking result: of orto- and para H depends on Temperature! Indeed since J=0 is an even state at very low T para- will dominate while at high T its fraction will drop to because the sum over even and odd angular momenta are the same! Since fraction of ortho and para are easily measurable by NMR this results allows a direct test of very deep principle of Quantum theory - Pauli principle itself! (In reality this experiment requires long time because interconversion between ortho- and para forms of hydrogen molecules is very slow). At shorter time the gas will present a mixture of ortho And para forms in fixed concentrations. However their thermal properties are different at low T: ortho ortho Arot = kT ln qrot 2 0 r 2kT ln 3 A para rot = kT ln q para rot Again that would have consequences for the reactivity of ortho- and para- hydrogen Next lecture: 1) Partition Functions of Polyatomic Ideal gases (Reif 9.12, McQuarrie 6.1-6.3. 2) Chemical Equilibrium (Reif 8.7,8.10. McQuarrie Ch9) ...
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This note was uploaded on 05/04/2010 for the course CHEM 161 taught by Professor Shaklovich during the Spring '10 term at Harvard.

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