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Thermo physics degrees of freedom question!

  1. May 6, 2012 #1
    1. The problem statement, all variables and given/known data

    Consider a sample containing 2.35 mol of an ideal diatomic gas.

    (a) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant volume.

    nCv.=

    (b) Assuming the molecules rotate but do not vibrate, find the total heat capacity of the sample at constant pressure.

    nCp=


    (c) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant volume.

    nCv=

    d) Assuming the molecules both rotate and vibrate, find the total heat capacity of the sample at constant pressure.

    nCp=




    2. Relevant equations



    3. The attempt at a solution

    Since diatomic f= 5 when not considering vibration.

    so for part a)

    nCv= 2.35(5/2)(8.314)=48.9

    b)

    nCp=2.35((5/2)(8.314)+8.314)=68.4

    c) Since Vibrating f=6

    nCv=2.35(3)(8.314)=58.6

    d)

    nCp= 2.35((3)(8.314)+8.314)=78.2


    Is this correct, please let me know If I have made a mistake?

    Thanks!
     
  2. jcsd
  3. May 6, 2012 #2

    ehild

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    Everything is correct, except the vibrational case.

    As the vibration has both kinetic and potential energy, one vibrational mode contributes by RT per mole to the internal energy.

    ehild
     
  4. May 6, 2012 #3

    Andrew Mason

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    I think ehild is correct. I have never really understood why a vibrational mode is equivalent to two degrees of freedom since it seems to me that the potential and kinetic energies involved in vibration are not independent of each other. However, it does seem to be the case and it appears to fit experiment.

    But the fact that the vibrational mode constitutes 2 degrees of freedom does not necessarily mean that that Cv = 7R/2. That is a consequence of the equipartition theorem which does not always apply due to quantum effects. The vibrational mode is quantized. At lower temperatures, the vibrational mode is in the zero energy state. As the temperature increases, the molecule will have enough energy to vibrate in higher energy states. When it reaches very high temperatures, the quantum effects become negligible. It is only at this point that the equipartition theorem applies.

    So you really cannot answer this question properly without more information.

    AM
     
  5. May 7, 2012 #4

    ehild

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    The energy of a vibrating two-atomic molecule is 1/2 mv2 + 1/2 kx2. Imagine that all molecules of the volume of gas are in rest initially, then some initial displacement and velocity is set for every of them, in random. The equipartition principle states that 0.5kT average energy per molecule is assigned to both energy terms. This energy is conserved during vibration.


    The question means that the temperature is high enough for the excitation of the vibration mode. Anyway, the rotation is also quantized and you do not get 5/2R for Cv at low temperatures.

    The rotational energy is entirely kinetic, E=1/2 I ω2. A linear molecule, as any other one has got three principal axes to rotate around, but the moment of inertia is zero when it rotates about the molecular axis: that rotation does not contribute to the average energy.

    In a crystal, there are 3 degrees of freedom per molecule, but they can only vibrate about their equilibrium position (except the translation and rotation of the crystal as a whole) so each molecule has got 3kT average energy at temperatures when all translational vibrational state (acoustic modes) are excited.

    ehild
     
  6. May 7, 2012 #5

    Andrew Mason

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    But without the equipartition theorem, could we not conclude that on average the vibrational energy will be no more potential than kinetic?
    That is true. But there is no diatomic molecule and no temperature for which N>1 vibration modes can occur but where not all higher rotational modes are excited.

    The question says to assume that the molecule is vibrating. This simply means that the temperature has increased so that at least some N>1 vibrational modes are excited. The equipartition theorem does not effectively apply until much higher vibrational modes are excited.

    AM
     
  7. May 7, 2012 #6
    This is an exert from Wiki

    http://en.wikipedia.org/wiki/Degrees_of_freedom_(physics_and_chemistry)#Example:_diatomic_gas

    In three-dimensional space, three degrees of freedom are associated with the movement of a mechanical particle. A diatomic gas molecule thus has 6 degrees of freedom. This set may be decomposed in terms of translations, rotations, and vibrations of the molecule. The center of mass motion of the entire molecule accounts for 3 degrees of freedom. In addition, the molecule has one vibrational mode and two rotational degrees of motion. The rotations occur around the two axes perpendicular to the line between the two atoms. The rotation around the atom-atom bond is not counted. This yields, for a diatomic molecule, a decomposition of:

    3N = 6 = 3+1+2.

    For a general molecule with N > 2 atoms, all 3 rotational degrees of freedom are considered, resulting in the decomposition:

    3 N = 3 + 3 + (3 N - 6)

    So this is saying that it has six degrees of freedom (max),,, is Wiki wrong?
     
  8. May 7, 2012 #7

    Andrew Mason

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    Wikipedia is correct but not complete, in my view.

    A diatomic molecule has 6 degrees of freedom of motion: 3 translational, 2 rotational and 1 vibrational. But the kinetic energy of vibration cannot exist without potential energy of vibration. If you apply the equipartition theorem, you are going to have to split the total kinetic energy 6 ways. But if the kinetic energy of vibration is equal to the translational kinetic energy in each of the x, y and z directions and the rotational kinetic energy on each axis of rotation (which occurs at very high temperatures), you have to then add another equal amount for the potential energy of vibration. So, the molecule behaves as if it has 7 degrees of freedom. But I would say it only has six.

    AM
     
  9. May 7, 2012 #8

    ehild

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    I do not quite understand you. Applying statistical methods to calculate average potential and kinetic energy, it would give the same results at high-temperature limit.
    A diatomic molecule has got one vibrational mode which can be excited to different energy levels, specified by the vibrational quantum number.
    To each rotation, a rotation quantum number belongs and it determines the rotational energy. The molecule can vibrate and rotate at the same time, so its vibrational and rotational quantum numbers can be simultaneously higher than 1.

    You could argue about the rotations, too. The equipartition theorem does not effectively apply until rotation levels of much higher energy are excited. The temperature is not given in the problem. The Equipartition Principle applies when kT is much greater than the energy difference between the energy levels of the molecule.
    The question belongs to Introductory Physics, so it has to be solved by the methods taught at introductory level.


    ehild
     
  10. May 7, 2012 #9

    Andrew Mason

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    I am just saying that you do not apply the equipartition theorem (which deals with the distribution of kinetic energy among independent modes of motion for a system in thermal equilibrium) to conclude that the average kinetic energy and the average potential energy of a vibrating diatomic molecule will be equal.

    For vibration to occur in a diatomic molecule (N>0), the temperature must be well beyond the temperature at which quantum effects for rotation are observed. So, you could have a situation in which the vibrational mode is excited but the Cv of the diatomic gas is more than 5R/2 but less than 7R/2. Really, that is my only point.

    AM
     
  11. May 7, 2012 #10

    ehild

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    I think I might conclude.
    Yes, it is so.

    ehild
     
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