Advanced Thermodynamics   Quiz #2

  1. We showed in class that the Harmonic Oscillator partition function, zHO, was simply 1/(1-e-a/T) where a=hv/k.

    Imagine an isomerization reaction, A=B, involving a molecular vibration which changes in the way diagrammed below. (We'll pretend that nothing else in the molecule changes except this vibration.) In particular, vA=2vB and B's zero point level lies 4hvB above that of A.

    etc. on up...etc. on up...etc. on up...
    106
    945
    84
    733
    6   ---(case c below)---2
    521
    40
    31
    2
    10
    0
    Energy units (in hvB)Quantum levels of AQuantum Levels of B


    Predict KP = PB/PA = [B]/[A] for the following cases:

    1. T = 0 K

This doesn't take any calculation. At T=0 only the ground state of the system is occupied; that is A's zero point level given that both A and B are merged states open to the system. So at T=0, there isn't any B and KP=0.

  • T is infinitely large.

    • This too takes no calculations. At indefinitely high T, all of the states of the system are equally occupied (since there is no energy penalty associated with having all the energy there is). Since there are twice as many states in B's manifold as A's, KP=2.

  • kT = 6.46426 hvB (above A's zero point level).

    • Ah...now we have to calculate something.

    If there was no energy difference between A and B, then KP=zBHO/zAHO. But every term in the zBHO sum over states has 4hvB added to its state energy as a result of the endothermicity of the isomerization. But fortunately e-(4hvB+nBhvB)/kT = e-4hvB/kT × e-nBhv/kT and only the latter factor has to stay inside the sum. Thus zBHO × e-4hvB/kt replaces zBHO in the expression for KP.

    For simplicity, let y = e-hvB/kT = e-1/6.46426 = 0.856675 at the chosen average energy. Also
    1. zAHO = 1/(1-y2) = 3.75787
    2. zBHO = 1/(1-y) = 6.997715
    3. e-4hvB/kT = y4 = 0.538597
    making KP = 6.997715×0.538597/3.75787 = 1 (Would you have guessed the breakeven point to be there? Obviously it involves the tails above kT!)


  • The 3rd Law of Thermodynamics demands that S arrow right 0 as T arrow right 0 (ignoring the "residual entropy" of imperfect crystals). Boltzmann has no problem with that since W=1 at absolute zero (only one way in which to configuration a perfect crystal), hence S=klnW vanishes there. However, our derived version of statistical entropy,

    S = k ln(Z) + E/T

    seems it might have a problem with that last term! We hope that E approaches zero much faster than does T itself so the ratio vanishes just as the first term will when Z, the sum over available states, goes to zero.

    Given the zHO above and our derived

    E = + kT2 [ d lnZ / dT ]V

    satisfy yourself (and me) that the 3rd Law holds for harmonic oscillators.

    Hint: Your algebra (and calculus) is simplified using zHO = 1/(1-e-a/T) where, as above, a=hv/k. And, of course, e-a/T falls off to zero faster than any power of T as T gets small.

    • So we must determine if EHO/T behaves itself (vanishes) as T disappears to zero, and the statistical
    E / T = + kT [ d ln(Z) / dT ]V
    is of cold comfort since that differential might be hiding 1/T2 or worse!

    So we must evaluate the differential. In doing so, we can dismiss the difference between Z and z since Z=zN for HO, and since the ln is involved, that N superscript can be taken completely outside the differntial:
    E / T = +NkT [ d ln(z) / dT ]V
    That same trick will simplify the derivation for zHO=(1-e-a/T)-1 since
    [ d ln(1-e-a/T)-1 / dT ] = (-1) [ d ln(1-e-a/T) / dT ]
    But now we actually have to slug our way through the differentiation:

    (-1) [ d ln(1-e-a/T) / dT ] =
    (-1) (1-e-a/T)-1 [ d (1-e-a/T) / dT ] =
    (-1) (1-e-a/T)-1 (-e-a/T) [ d (-a/T) / dT ] =
    (-1) (1-e-a/T)-1 (-e-a/T) (-a) [ d T-1 / dT ] =
    (-1) (1-e-a/T)-1 (-e-a/T) (-a) (-1) T-2 =

    + a / [ T2 (e+a/T-1) ]
    And that makes
    EHO / T = + aNk / [ T (e+a/T-1) ]

    where that exponenial e+a/T gets large much faster than T vanishes, and the combination then kills EHO/T to zero as T approaches zero.
    Last modified 7 October 1997.