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6.5.Gaseous Systems Composed of Molecules with Internal Motion Assumptions ( ideal Boltzmannian gas ) : 1. Molecules are free particles ( non-interacting). 2. Non-degeneracy (MB stat) : = quantum # for internal DoF
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Internal DoF Molecules : Homopolar molecules (A-A) :
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6.5.A. Monatomic Molecules Let ( All atoms are neutral & in electronic ground state ) Nuclear spin Hyperfine structure : T ~ 10 1 – 10 0 K. Level-splitting treated as degeneracy : Inert gases ( He, Ne, Ar,... ) : Ground state L = S = 0 : = 0 denotes ground state. 0 = 0. L = 0; S 0 :
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L = 0, S 0
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L 0, S 0 Ground state 0 = 0. C V, int = 0 in both limits C V has a maximum. Const of motion: L 2, S 2, J 2, J z S-O: L-S coupling
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6.5.B. Diatomic Molecules Let ( All atoms are neutral & in electronic ground state ) Non-degenerate ground state ( most cases ) g e = 1& j elec (T) = 1 Degenerate ground state ( seldom ) : 1. Orbital angular momentum 0, but spin S = 0 : In the absence of B, depends on | z | doublet ( z = M ) is degenerate ( g e = 2 = j(T) ) C V = 0 2. = 0, S 0 : g e = 2S + 1 = j (T) C V = 0 3. 0 & S 0 : Spin-orbit coupling B eff fine structure
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E.g., NO ( 1/2, 3/2 ) ( splitting of doublets ) : C V has max. for some kT ~
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Vibrational States for diatomic gases Full contribution for T 10 4 K No contribution for T 10 2 K Harmonic oscillations (small amplitude) : “vibrons”
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equipartition value vib DoF frozen out
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Very high T anharmonic effects C vib T( Prob 3.29-30) Mathematica
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Nuclear Spin & Rotational States: Heteropolar Molecules Heteropolar molecules ( AB ) : no exchange effects interaction between nuclear spin & rotational states negligible. From § 6.5.A : C nucl = 0 Molecule ~ rigid rotator with moment of inertia ( bond // z-axis ) = reduced mass r 0 = equilibrium bond length
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<< T room if no H or D present. HCl :~ 15 K N 2, O 2, NO :~ 2-3 K Cl :~ 1/3 K H 2 :~ 85 K D 2 :~ 43 K HD :~ 64 K Euler–Maclaurin formula : B k = Bernoulli numbers B 1 = −1/2, B 2 = 1/6, B 3 = 0, B 4 = −1/30, B 5 = 0, B 6 = 1/42, B 7 = 0, B 8 = −1/30, … ( only l = 0 term survives )
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T >> r : Better approx :set
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B 1 = −1/2, B 2 = 1/6, B 3 = 0, B 4 = −1/30, B 5 = 0, B 6 = 1/42, B 7 = 0, B 8 = −1/30, … Mulholland’s formula ( Evaluated using Mathematica )
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C rot calculated using j rot summed up to l = l max. Thick curve : l max = 20 Thin curve : l max = 30 T << r : T 0 : Mathematica
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CpCp j int indep of V is also indep of V. All contributions from the internal DoF are indep of V. DoF ; f transl: 3 ; 3 rot:2 ; 2 vib:1 ; 2 rot frozen vib frozen rr vv
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Homopolar Molecules High T ( classical region / phase space distinguishable p’cles ; MB ) : homo same as hetero with B = A
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Low T ( quantum region / BE, FD ) : ( Particle exchange ~ spatial inversion ) m m pairings m = m pairings There’re anti-symmetric pairings (m m only ) symmetric pairings (both )&
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Fermion nuclei ( anti-symm. ) : with Boson nuclei ( symm. ) : with where
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Same as classical approach High T ( classical region) :
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Ortho :higher degeneracy Para :lower degeneracy Ratio of ortho-para components : High T : For H 2 : S A = ½ n 3 For D 2 : S A = 1 n 2
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Low T : smallest l term dominates For H 2 (FD) :wholly para as T 0 For D 2 (BE) :wholly ortho as T 0 with l = 0
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Disagree with exp. on H 2 Reason (Dennison) : Transition rate for nuclear spin-flip extremely low ( T ~ 1 yr ) ortho-para ratio not eqm values Lab prep done in room T >> r Let Mathematica FD :wholly para as T 0, C = C even. C even C odd
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6.5.C. Polyatomic Molecules DoF ; f ( linear molecules ) transl: 3 ; 3 rot:3 (2) ; 3 (2) vib:3n 6 (5) ; 2 [3n 6 (5) ] Large moment of inertia r << T of interest. = # of indistinct config in 1 rot. = 2for H 2 O = 3for NH 3 = 12for CH 4 & C 6 H 6 n - atom molecule: I i = principal moments of inertia (Prob. 6.27)
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§ 3.8 : i = normal freq Low i ~ 10 3 K E.g., CO 2 : 1 = 2 = 960 K, 3 = 1990 K, 4 = 3510 K Usually, high i > T disassoc.
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6.6.Chemical Equilibrium Chemical reaction : i = stoichiometric coeff. Let N = # of reactions occured. with N > 0 ( < 0 ) meaning reaction direction is ( ). 0 denotes initial value For a closed, iosothermal, reaction chamber kept at constant pressure, the natural thermodynamic potential is the Gibbs free energy G( T, P, { N i } ).
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At equilibrium : If C is a catalyst, then C C appears at both side of the reaction eq. The equilibrium relation is therefore unaffected. N > 0 ( < 0 ) for ( )
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Hemholtz Free Energy for Ideal Gas int = energy due to internal degrees of freedom. j i (T) = partition function due to int i. i = ground state energy of the i th atomic species.
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Let n 0 = standard # density. Eqm. cond. with P 0 =1 atm. For gas, For solution, n 0 = 1 mole / liter Eqm. cond. : = equilibrium constant where Law of mass action
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Internal Combustion Natural gas combustion : i.e. Combustion : Exhaust : Rapid cooling Actual (non-eqm) exhuast R value is closer to the combustion one. Can be reduced by raising [O 2 ] via reducing [CH 4 ] & using catalyst (Pt,Pd) at exhaust.
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