Quiz #1 Solutions

26 September 1996

Using first principles (plus Stirling's Approximation), evaluate the maximal entropy of any non-degenerate, single-dimensional energy-level system (the harmonic oscillator step-ladder, for instance) of Avogadro's number of particles. (The maximum entropy obtains when you can ignore energy contraints, and any particle can occupy any state. It's like 0, if you like, but it's not necessary to even consider here.)

S = k ln(W) where W = N! / (n₀! n₁! n₂! . . . )

We expect the maximal S when we use the largest W, and the largest W obtains when that denominator = 1. We can expect that to happen when all the n_i are either 0 or 1; that is, no state is multiply occupied.

Not a problem if we don't have to consider energy conservation!

S = k ln(W) = k ln(N!) = k [ N ln(N) - N ] = kN [ ln(N) - 1 ]

S = R [ ln(6.022×10²³) - 1 ] = 8.314 [ 53.75 - 1 ]

S = 446.9 J/mol K = 0.4469 kJ/mol K ... (normal entropies are 1% of that!)

Imagine an equi-spaced energy level ladder (see below), E_j = j (de), where the degeneracy of each of the levels ( j = 0, 1, 2, ... ) is given by

g_j = j+1.
Find W and W* for the case that N=4 and E=3(de). (Remember to include the many distinguishable ways that populations assigned to the various degenerate states of the same energy can be found, but it is much easier to include that count as a multiplicative factor in your W_i calculations than to teach occupation of each degenerate state as a separate W_i! In other words, let W_i mean what it has meant all along, the number of ways of finding the system with a specific energy level occupation scheme, but scale each W_i with a factor accounting for degeneracy correctly.)

j associated degenerate states

...
5 _ _ _ _ _ _
4 _ _ _ _ _ « energy difference between states is (de)
3 _ _ _ _
2 _ _ _
1 _ _
0 _

We've got 4 oscillators but only 3 quanta to play with; the only ways we can arrive at the correct energy are:

E level  Occupancy   Occupancy   Occupancy    Level
          Scheme 1    Scheme 2    Scheme 3  Degeneracies

   3         1           0           0          4
   2         0           1           0          3
   1         0           1           3          2
   0         3           2           1          1

   W'    4!/(3!1!)  4!/(2!1!1!)  4!/(1!3!)

   W'   =    4           12          4

But W' doesn't take the degeneracies into account. For example, scheme 1 puts one oscillator in energy level 3. However it could be in any of the 4 degenerate states at that energy level. So W₁' has undercounted the ways of finding the oscillators by the amount of that degeneracy!

The actual W₁ = 4 W₁' = 16.

Similarly, for scheme 2, the oscillator in energy level 2 might be in any of 3 degenerate states while that in energy level 1 could be in any of 2 degenerate states.

The actual W₂ = (3) (2) W₂' = (6)(12) = 72 = W*

Scheme 3 has 3 oscillators sharing energy level 1 which is doubly degenerate! The various distributions possible in that level will be:

___xxx___  _________  which can only be done 1 way, as 3!/(3!0!),

___x_x___  ____x____  which can be done 3 ways (any of the 3 can be on the right)
                      that is 3!/(2!1!)
____x____  ___x_x___  also 3 possible ways, and finally

_________  ___xxx___  which can only be done 1 way.

That means 1+3+3+1 = 8 ways of scattering the 3 oscillators around the two degenerate states.

The actual W₃ = 8 W₃' = (8)(4) = 32.

W = W₁ + W₂* + W₃ = 16 + 72 + 32 = 120.

That completes the problem. The extra point question was "What are the average n_i?" And we only need to multiply the n_i of each scheme by the weight, W_i / W, of that scheme. Those weights are 16÷120, 72÷120, and 32÷120, or 0.133, 0.6, and 0.267 respectively.

So n₀ = 0.133(3) + 0.6(2) + 0.267(1) = 1.867

and n₁ = 0.133(0) + 0.6(1) + 0.267(3) = 1.401

and n₂ = 0.133(0) + 0.6(1) + 0.267(0) = 0.6

and n₃ = 0.133(1) + 0.6(0) + 0.267(0) = 0.133

definitely declining but not exponentially! The degeneracies see to that.

Last modified 27 September 1996