The heat Capacity of a Diatomic Gas

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					   Chapter 15



The Heat Capacity of a Diatomic
             Gas
            15.1 Introduction
• Statistical thermodynamics provides deep
  insight into the classical description of a
  MONATOMIC ideal gas.
• In classical thermodynamics, the principle of
  equipartition of energy fails to give the
  observed value of the specific heat capacity for
  diatomic gases.
• The explanation of the above discrepancy was
  considered to be the most important challenge
  in statistical theory.
  15.1 The quantized linear oscillator
• A linear oscillator is a particle constrained to move
  along a straight line and acted on by a restoring
  force      F=-kx2
                  d x
       F = ma= d t = -kx
                m     2


• If displaced from its equilibrium position and
  released, the particle oscillates with simple
  harmonic motion of frequency v, given by         1 K
                                               v
                                                    2   m

Note that the frequency depends on K and m, and is
 independent of the amplitude X.
• Consider an assembly of N one-dimensional
  harmonic oscillators, in which the oscillators
  are loosely coupled so that the energy
  exchange among them is small.

• In classical mechanics, a particle can oscillate
  with any amplitude and energy.

• From quantum mechanics, the single particle
  energy levels are given by
      EJ = (J + ½) hv , J = 0, 1, 2, …..
• The energies are equally spaced and the
  ground state has non-zero energy.
• The internal degrees of freedom include
  vibrations, rotations, and electronic excitations.

• For internal degrees of freedom, Boltzmann
  Statistics applies. The distinguish ability arises
  from the fact that those diatomic molecules
  have different translational energy.

• The states are nondegenerate, i.e. gj = 1
• The partition function of an oscillator
• Introducing the characteristic temperature θ,
  where θ = hv/k




                    2 3
                                    
           e 1  e  e  e  ......
            2T
              
                    T    T   T
                                     
                                    
• The solution for the above eq. is (in class
  derivation)
The distribution function for B-statistics is




                2
• Note that B statistics and M – B statistics have
  the same distribution function, the eq derived
  in chapter 14 for internal energy is also valid
  here.
                 U = NkT2
 since
                       
                                             
               1   e T              1    1 
       U  Nk               Nk         
                2      
                   1 e T             2
                                          eT  1 
                                              
                                    1
                                      0
For T → 0                         
                                eT 1


For


thus
  15.3Vibrational Modes of Diatomic
              Molecules
• The most important application of the above
  result is to the molecules of a diatomic gas




• From classical thermodynamics


                   for a reversible process!
Since




Or

At high temperatures
At low temperature limit

On has

So



                        approaching zero faster
                than the growth of (θ/T)2 as T → 0
Therefore       Cv  0 as T  0




The total energy of a diatomic molecule is made
 up of four contributions that can be separately
 treated:
1. The kinetic energy associated with the
   translational motion
2. The vibrational motion
3. Rotation motion (To be discussed later)

Example: 15.1 a) Calculate the fractional number
   of oscillators in the three lowest quantum
   states (j=0, 1, 2,) for T   4 and forT  
Sol:
J=0
• 15.2) a) For a system of localized
  distinguishable oscillators, Boltzmann
  statistics applies. Show that the entropy S is
  given by                NJ 
             S    N J ln    
                    J         N 

• Solution: according to Boltzmann statistics
                              NJ
                         g
                         n
              W  N!         J
                     J 1 N
                            J
     So
              S  k ln w             1
ln w  ln N !  N J ln g J   ln N J
g J  1  ln 1  0
ln w  ln N !  N J ln 1   ln N J !
ln w  ln N ! ln N !
ln w   ln       J ln 1    J
ln w   ln     J ln  J
   J
ln w    J ln     J ln  J
ln w    J ln   ln  J 
                         
ln w    J ln 
                         
                           
                 J        
                  
 S     j ln     
                 
                 j
         n
                  j 
 S     j ln   
        J 1          

				
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