Connecting the Statistical Interpretation of Entropy to the Macroscopic
World
or
What we did in Class today (4/9/2004)
These are just copies of the notes I gave you. I’ve annotated them so that you can follow
things I might have skipped or glossed over when I talked about them in class. You don’t
need to be able to regurgitate this derivation, but I do want you to understand it. This
kind of reasoning connects entropy on a microscopic scale to entropy on a macroscopic
scale and links the thermodynamic and statistical interpretations of entropy. On Monday
we’ll continue with the thermodynamic interpretation of entropy and develop the Gibbs
function.
Anyway, we start with the statistical interpretation of entropy
N!
where N is the total number of particles in a
n0!n1!n2!....
collection and n0, n1, n2 .etc are the numbers of those particles in the ground state, first,
excited state, and so on.
S  k ln W  k ln

Using the rules of logarithms,


S  k ln N!k ln( n 0!n1!n 2!...)  k ln N!k  ln n i!  kln N! ln n i! (1)


i
i

now we can simplify this using Stirling’s approximation, which is appropriate as long as
N is a large number (which we always take it to be)
ln N! Nln N  N Stirling’s approximation (Stirling was a preacher’s kid too)
Substituting this into equation (1) above I have



S  kN ln N  N   n i ln n i  n i 




(2)
where I have also used Stirling’s approximation to simplify the expression “ln ni!”

Simplifying equation (2)

 

S  kN ln N  N   n i ln n i   n i  kN ln N  N   n i ln n i  N

 

i
i
i



 kN ln N   n i ln n i 


i
(3)
Okay, now this next part is tricky. Notice that the sum of the numbers of particles in each
state, ni, must equal the total number of particles in the collection N. Therefore we can
reexpress equation (3) as 


S  k n i ln N   n i ln n i 


 i

(4)
We can do this because ln N is a constant independent of the sum, so it doesn’t matter if
we put it inside the sum or not

Anyway, using the rules of logarithms equation (4) can be rewritten
N 
n 
S  k  n i ln   k  n i ln  i 
N 
n i 
i
i
(5)
ni
is simply the probability that a molecule in a collection of N molecules will be
N
in the ith state. Written another way using Boltzman statistics
Now,



 i
ni
e kT
 pi 
; ni  pi N
N
q
(6)
Substituting this into equation (5) gives.
S  kN pi ln pi
(7)
i
Equation (6) also allows us to express ln pi in terms of the partition function, q.

  i kT 
 i

e

 ln 
ln pi  ln 
e kT  ln q  i  ln q
 q  

kT



Using this in equation (7)

 kN


 
S  kN pi  i  ln q kN pi  i   pi ln q
 pii  Nk ln q pi
kT

 i kT  i
 kT i
i
i
This reduces to

S
N i
N
pii  Nk ln q 1 
 Nk ln q

T i
T
where the second equality holds because the sum of the probabilities, pi, has to equal 1
and because the sum of the energies of the states, i, weighted by their probabilities, pi,
equals the average energy, <i>, of particles in the collection.

Finally, the number of particles N multiplied by the average energy of those particles
gives the total energy of particle in the sample- relative to the zero point energy which we
have taken as the bottom of our energy scale (see earlier notes on the derivations of the
partition functions or see me if this last part doesn’t make sense).
So N i  E  E(0)
Finally

S

E  E(0)
 Nkln q
T
(8)
and we now have an expression relating the statistical interpretation of entropy with the
thermodynamic quantity.
We can apply this to a collection of N identical harmonic oscillators which has the
partition function
1
qvib 
1 e

h 0
where 0 is the fundamental vibrational frequency of the molecules
kT
in question. (we showed why this is the vibrational partition function in an earlier
lecture, see your notes for details).
Using the expression for thermodynamic energy that we derived earlier for the energy
above the zero-point energy
E  E(0) 

N q
q ( 1 )
kT
We can derive the following expression for the vibrational entropy of a collection of
harmonic oscillators
 

 kT


S  Nk 
 ln(1 e kT )
 kT 
e 1






Note, for those of you in class on Friday- I
made a sign error in the formula I gave you in
class. This exponent should be positive, not
negative.
where  is just shorthand for h0.
Homework:
1. Using the formulas I have given you, try to derive this expression given above for the
entropy of a collection of oscillators.
2. Calculate the vibrational entropy of 1 mole of CO molecules at 100 K. The
fundamental frequency of CO is 2143 cm-1. (you’ll have to convert this to n0 using the
fact E=h=hc0, c=2.998x1010 cm/s)