MOLAR TRANSLATION ENTROPY (SACKUR- TETRODE EQUATION

For a perfect monoatomic gas, the molar partition function is given by      

         Q =  qᴺ 
                 Nǃ

Where N is Avogadro number 

= =  1   [(2πmkT)³/₂  ] ᴺ  ..................... (1)
      Nǃ        h³

Using Stirling's approximation, 

ln Nǃ = Nln N - N = ln Nᴺ − lne = ln Nᴺ − lneᴺ
Nǃ = (N ) ᴺ        .............. (2)
          e

Substituting for Nǃ from equation (2) into equation (1), we get 

Q = (e)ᴺ [(2πmkT) ³/₂ V ] ᴺ ........... (3)
        N          h³

Q = [ (2πmkT)³/₂ Ve ] ᴺ
              Nh³

lnQ = N ln [ (2πmkT)³/₂ Ve ]
                       Nh³

S =  E  + kln  qᴺ  =  + klnQ .............. (4)
        T            Nǃ      T

Substituting for E = 3RT/2 ( trans. K.E) for one molecule. 

S =  R + kNln [(2πmkT)³/₂ Ve ] =  R+ Rln
      2                       Nh³                     2

      [(2πmkT)³/₂ Ve ]
           Nh³

=  Rln e+ Rln [(2πmkT)³/₂ Ve ]
                              Nh³

= Rln [ (2πmkT)³/₂ Ve] = Rln
              Nh³

 [(2πmkT)³/₂ Ve 5/2  ]
          Nh³

Since = M/N, 

S=Rln [ e⁵/₂  (2πk)³/₂ V (TM)³/₂ ]...........(5)
            N⁵/₂ h³

Eqution (5)/is known Sackur- Tetrode equation. Replacing V by RT/P in equation (5) we get 

S = Rln [e⁵/₂ (2πk) ³/₂ RT⁵/₂ M³/₂  ........(6)

Equation (6) is another from of Sackir-Tetrode equation. From equation (5) separating the constants, 

S=2.303 χ R [loɡ  e⁵/₂ (2πk) ³/₂ + loɡ V
(TM)³/₂

Substitute the in c.g.s units and evaluate 

S= 2.303×1.987 [loɡ V +3/2 loɡ T +3/2 loɡ M + loɡ(2.78)⁵/₂ (2χ3.14χ1.38χ10⁻¹⁶)³/₂  ]
            (6.023χ10²³)⁵/₂ (6.626χ10⁻²⁷⁾³

=4.576 [loɡ V+3/2 loɡ T+3/2 loɡ M−2.43]
=−11.12+4.576 [loɡ V +3/2 loɡ T +3/2 loɡ M ] cal k⁻¹ mol⁻¹

Similarly from equation (6) in c.g.s units, 
S=2.303xRln [e⁵/₂ (2πk)³/₂ R + Rln
                          N⁵/₂ h³

        T⁵/₂ M³/₂  ]
             P

S=2.303x1.987[5/2 loɡ T+3/2 loɡ M− 
loɡ P− 0.5065]

S=-2.317+4.576 [5/2 log T+ 3/2 log M-log P] cal k⁻¹ mol ⁻¹

In S.I units, S= 2.303x8.314[5/2 log T+3/2 log M - log P + 8.9998]

= 172.32+19.147[5/2log T + 3/2 log M - log P] J K⁻¹ mol⁻¹


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