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Statistical mechanics - characteristics temperature of the HF molecule

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Spectroscopic data (rotational-vibrational lines) show that the hydrogen flouride molecule has a vibrational frequency of 7.8x10^14 rad/sec and a moment of inertia of I=1.35x10^-47 kg.m^2. Find the relevant characteristic temperatures of the HF molecule.


    2. Relevant equations



    3. The attempt at a solution

    I am completely lost on where to start and what formulas to use!
    Please help!
     
  2. jcsd
  3. Sep 18, 2010 #2
    Characteristic temperatures can be figured out easily like this:

    the usual "energy unit" in statistical mechanics is k*T where k is boltzman's constant and T is the temperature.
    Now different kind of motions have different "energy units" for example the energy for vibration is [tex]\hbar \omega [/tex].
    So we shall obtain the characteristic temperature if we make this equal to the "statistical mechanics energy" : [tex]k\cdot T_v=\hbar \omega [/tex]

    And from here you can express T_v.

    Similarly for rotations, the energy of a quantum rotator is hbar^2/(2*I). and you can figure the characteristic temp. for rotations from this...
     
  4. Sep 22, 2010 #3
    thank you heaps!

    the following part says:

    Assuming that HF is a diatomic gas, use the equipartition theorem to predict the specific heat of this gas (quoted as C_v/molecule/k) at T=520K

    Won't there be a contribution to the specific heat from each of the vibrational and rotational parts which don't depend on temperature???
     
  5. Sep 22, 2010 #4
    They do depend on temperature of course :)

    Here what you have to do is figure out the different degrees of freedom. Since HF is a linear diatomic molecule it will have 2 rotational degrees of freedom (you can check this by looking on the molecule) then it will have 3 translational degrees of freedom (it can move in three directions in space) and finally it will have 1 vibrational degree of freedom (it can only vibrate with the atoms approaching eachother, as if on a spring). But the vibrational part is tricky as one vibrational degree of freedom, means that there will be two square terms in the hamiltonian. These means that actually when using the equipartition theorem you will have to take this 2 times.

    So totally there are 3+2+2*1=7 square terms in the hamiltonian. Hence the specific heat is ... and I will leave you to figure that out ;)
     
  6. Sep 22, 2010 #5
    Ok thanks,
    So the internal energy will be (7/2)*k*T so the specific heat will be (7/2)*k which doesn't depend on temperature so why have they given us T=520K??
     
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