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I Why triatomic gases have internal energy 7RT/2 ?

  1. Nov 14, 2016 #1
    This table is given in my book,

    $$\begin{array}[c!c!c!c!]
    \text{ }&\text{ Transitional }&\text{ Rotational }& \text{ Vibrational} \\
    \hline
    \text{Linear molecules} & 3&2& 3N -5\\
    \hline
    \text{Non-Linear molecules} & 3&3& 3N -6\\
    \hline
    \end{array}$$

    It is also given that each transitional/rotational freedom contributes ##\frac 1 2kT## to internal energy and vibrational freedom do ##kT##.

    So for triatomic linear molecules,
    ##U = {3RT\over 2} + {2RT\over 2} + {2\times (3\times3 - 5 )RT\over 2} = {13 RT\over 2}.##
    ##C_v = {\partial U\over \partial T} = {13R\over2}##

    But ##C_v = 7R/2## is given without any explanation.

    I tried to search on internet and I only find explanations for monoatomic and diatomic gases.
    For triatomic molecules it is simply given "Similarly ##C_v## for triatomic gases is ##7R\over 2##"

    One more thing I don't understand is why we multiply degrees of freedom with gas constant ?
     
    Last edited: Nov 14, 2016
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  3. Nov 14, 2016 #2

    Simon Bridge

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    Depends on how you count the degrees off freedom. How are you reasoning out to get 13/2 there?

    Bear in mind that the equipartition theorem is a rule of thumb not a Law of Nature.
     
  4. Nov 14, 2016 #3
    3 degree of translational freedom contribute ##3RT/2## and 2 degrees of rotational freedom contribute ##2RT/2##.
    No of vibrational degree of freedom is ##3 \times 3-5 = 9 -5 = 4##, Thus total contribution of vibrational degree of freedom is ##4RT##.

    Adding all these I get ##(3/2 + 2/2 + 8/2)RT = 13RT/2## .
     
  5. Nov 15, 2016 #4

    DrClaude

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    For temperatures that are not too high, there is no vibrational excitation and these degrees of freedom do not count for the heat capacity. Hence, a linear triatomic will have ##C_v = 5R/2##. See http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/shegas.html

    Are you sure you are not confusing with ##C_p##?

    Because of the equipartition theorem. Each quadratic degree of freedom has an energy of ##kT/2##.
     
  6. Nov 15, 2016 #5

    DrClaude

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    I would not call it a rule of thumb. If the degrees of freedom were exactly quadratic, then the theorem would hold exactly. It holds approximately because rotational and vibrational degrees of freedom are quadratic only to first order.
     
  7. Nov 15, 2016 #6
    No I don't think so. ##C_p = 9R/2## is given in the book and ##\gamma = {9R/2\over 7R/2} = {9\over 7}##.
    I have no clue why it is so ? neither I can find anywhere else on internet.
     
  8. Nov 15, 2016 #7

    Mapes

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    Contrary to what many introductory textbooks report, the specific heat of a gas is not a constant but rather depends strongly on temperature. Certain degrees of freedom can be "frozen out" at temperatures below a certain threshold and therefore cease to contribute to the count. This is one possible reason why the single value given in your book is less than expected. In any case, the molar heat capacity of no triatomic gas is 7R/2 except possibly at one single temperature.

    I find it mildly fascinating that so many educational materials present a single temperature-independent value for the heat capacity of gases and rarely add the caveat "if the modes described here are fully excited". By similar simplified arguments, you could instruct students that the molar heat capacity of all condensed elements is 3R, for example, and this isn't a bad rule of thumb. But I've never seen this.
     
  9. Nov 15, 2016 #8

    Simon Bridge

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    How did you work out the rotational and vibrational degrees of freedom - in detail?
    If the triatomic molecule is linear ... how many rotational degrees of freedom are there?
    How about if it is not linear - like water?

    It is commonly stated in senior secondary and introductory tertiary texts that monatomic gasses have 3dof per molecule, diatomic have 5, and triatomic+ have 7.
    The gas molecules are usually modelled, at this level, as very small dense spherical masses connected by massless springs.
    The reasoning then goes something like this:
    A single atom is just a ball of mass, it has no vibrational or rotation modes, even though a ball can spin and wobble, because the ball is very small so those modes do not contribute significantly.
    A diatomic molecule may translate (3) rotate end-over-end (1), and vibrate lengthwise (1) for 5dof.
    Spin about the axis does not count because of the the "balls are small" thing. But arn't there 2 axes it could tumble about?
    As soon as you get more than two atoms - things can get tricky.
    For three masses in a line, they can bulk-translate (3), there are 2 linear modes for vibration (2) or is it only along one axis so (1)? - do we count flexing as a vibration mode? Theres rotating end-over-end (1 or 2?), but if it has flexed, then wouldn't an axial rotation become important? In fact, wouldn't an axial rotation tend to accentuate any flexing?
    What I am suggesting is to look through the 7dof as they are reasoned out in the text, or online materials, and compare that reasoning, in detail, with how you got 13. Then you can better approach why they left out some dof that you included.
     
  10. Nov 16, 2016 #9

    DrClaude

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    That doesn't work. If we are considering motional degrees of freedom, then 2 atoms have a total of 2x3 cartesian coordinates = 6 degrees of freedom. Removing 3 degrees of freedom for center-of-mass motion leaves 3 dof, 2 of which correspond to rotation (there are two Euler angles needed to specify the orientation of the molecule in 3D).

    In the case that concerns us, what are important are quadratic degrees of freedom, as these are what enter the equipartition theorem. Translation goes as ##\dot{\mathbf{x}}^2##, so that's 3 qdof, and rotation goes as ##\dot{\mathbf{\omega}}^2##, so 2 qdof. Vibration is treated as a harmonic oscillator, which involves ##\dot{\mathbf{r}}^2## and ##\mathbf{r}^2##, so each vibrational dof corresponds to 2 qdof.

    If the temperature is low enough so that vibrational excitation can be neglected, then for a diatomic molecule there are 5 qdof, ##C_v = \frac{5}{2} R##, otherwise there are ##C_v = \frac{7}{2} R##. I don't see you can get ##C_v = \frac{7}{2} R## for a triatomic molecule, except for the coincidence @Mapes pointed out.
     
  11. Nov 16, 2016 #10
    Ok I found a resource where it is stated that triatomic molecule have 7 dof.
    Reasoning is this :-

    Like a diatomic molecule, a linear triatomic molecule has three translational and only two accessible rotational degrees of freedom
    There are two degrees of freedom for vibrational energy accessible at lower temperatures.
    The total energy of each molecule is 7(1/2)kT = (7/2)kT.

    Is this correct ?
     
  12. Nov 16, 2016 #11

    Simon Bridge

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    That sounds about right - you get now what I was saying about how you count the degrees of freedom?
    The diatomic molecule has 3 translational dof even though there are 2 atoms, because the atoms are not free to be anywhere they like in relation to each other... but DRCaude has the expl. about quadratic dof - just needs to justify restriction.
     
  13. Nov 16, 2016 #12

    DrClaude

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    I don't think so. The bending modes are at a lower frequency than the stretching modes, so you could argue about what "low temperature" means, but there are two degenerate bending modes, so that means Cv = 9/2 R, not 7/2 R.

    The best way to solve it is to look at actual data. You will find the info for CO2 at http://www.engineeringtoolbox.com/carbon-dioxide-d_974.html
    I'll let you convert that yourself to Cv, but you will see that it increases more or less linearly up to about 1000K. There are no plateaus visible, and the 7/2 R point is reach at about 300K and the 9/2 R between 500 and 550K. The high-T limit is about 6.8 R, which is close the 13/2 R you calculated.

    So yes, around room T, CO2 has an effective 7 quadratic dofs, but I would still argue that you can't get that value from theoretical reasoning.

    Lets ask @Chestermiller what he thinks about this.
     
  14. Nov 16, 2016 #13
    Sorry. This is not something I would have spent time thinking about when studying engineering thermodynamics.
     
  15. Nov 16, 2016 #14
    Monoatomic molecules have translational KE only.

    Diatomic and polyatomic molecules have rotational KE as well as translational KE.

    There are three degrees of translational freedom for all molecules two degrees of rotational freedom for diatomic and linear polyatomic molecules and three degrees of rotational KE for non linear polyatomic molecules.

    1.Energy per degree of freedom = 1/2kT from which we can write an expression for the energy for a single molecule.

    2.Multiplying by N gives the energy per mole.

    3. Making T equal to one gives the value Cv. (by definition)

    4. Adding R gives the value of Cp since Cp - Cv = R. In this case R is numerically equal to the work done in raising T by one degree

    1. 2. 3. 4.
    Monoatomic = 3/2kT E = 3/2 NkT Cv = 3/2R Cp = 5/2R
    Diatomic and linear polyatomic = 5/2kT E = 5/2NkT Cv = 5/2R Cp = 7/2R
    Non linear polyatomic = 6/2kT E = 6/2NkT Cv = 6/2R Cp = 8/2R

    Notes . Because of their size monoatomic molecules have negligible moments of inertia and hence zero rotational KE. Diatomic and linear polyatomic have a negligible moment of inertia about the axis joining the atoms and therefore there is one degreee of rotational freedom missing. The analysis can be extended if vibrational energy is to be taken into account.
     
  16. Nov 16, 2016 #15
    @Simon Bridge , @DrClaude , @Chestermiller , @Mapes and @Dadface , Thanks for the help. I think everyone agrees on the conclusion that ##C_v = 5R/2## at low temperatures.
    I will see if I can further find evidence support my claim on Internet.

    @Dadface your display pic is very cute. ##\ddot \smile##.
     
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