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 

Internal DoF   Molecules : Homopolar molecules (A-A) :

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 : 

  L = 0, S  0 

 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

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

E.g., NO (  1/2, 3/2 ) ( splitting of  doublets ) :  C V has max. for some kT ~ 

Vibrational States for diatomic gases  Full contribution for T  10 4 K No contribution for T  10 2 K Harmonic oscillations (small amplitude) : “vibrons”

 equipartition value  vib DoF frozen out

Very high T  anharmonic effects C vib  T( Prob ) Mathematica

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 

<< 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 )

T >>  r :  Better approx :set  

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 ) 

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

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 rr vv

Homopolar Molecules High T ( classical region / phase space distinguishable p’cles ; MB ) :  homo same as hetero with B = A

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 )&

Fermion nuclei (  anti-symm. ) : with  Boson nuclei (  symm. ) : with  where

  Same as classical approach High T ( classical region) :

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

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

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

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)

§ 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.

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 } ).

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  (  )

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.

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

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.