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Adiabatic expansion of real vs ideal gas

  1. Sep 10, 2011 #1
    for the adiabatic expansion of a gas,

    why is the temperature drop smaller for a real gas , vs a bigger drop for an ideal gas

    even though the change in internal energy is greater for a real gas , vs a smaller change for an ideal gas

    assuming starting temperature, and volume change are the same

    is it just due to the offset from the heat capacity?
     
  2. jcsd
  3. Sep 10, 2011 #2

    Andrew Mason

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    Diatomic and polyatomic gases have higher degrees of freedom so their heat capacities (Cv) are higher than an ideal (monatomic) gas.

    AM
     
  4. Sep 11, 2011 #3
    thanksx!
     
  5. Sep 12, 2011 #4

    Philip Wood

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    I've always thought that an ideal gas is one which perfectly obeys Boyle's law (or, in microscopic terms, one in which the volume of molecules is negligible, and for which there are no forces between molecules except during collisions).

    This does not require the gas to be monatomic. I've checked in Jeans, Zemansky, Reif and none of these writers says anything about ideal gases having to be monatomic.
     
  6. Sep 12, 2011 #5

    phyzguy

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    I think the answer to the OPs question is that an ideal gas has completely non-interacting particles, while a real gas has small, long-range interactions between the particles. So as the gas particles move further apart, the potential energy of the interactions between the particles is decreased, which gives the particles a little more kinetic energy, and therefore a slightly higher temperature.
     
  7. Sep 12, 2011 #6

    Philip Wood

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    Surely their moving apart would increase the potential energy. Internal work has been done against the attractive forces, at the expense of mean kinetic energy – which falls, making the overall fall (including that due to the external work) greater. If there is no external work (expansion into a vacuum) there will still be a fall in mean kinetic energy.
     
  8. Sep 12, 2011 #7

    Andrew Mason

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    This is correct.

    This was understood by the quietrain in the original question. His question was why adiabatic expansion did not cause the real gas to have a lower temperature despite having lower internal energy.

    I think adiabatic expansion would have to result in a lower temperature for the real gas if the ideal gas and the real gas were both monatomic gases, for example. So he was talking about real gases that had a higher temperature despite having lost more internal energy ie. non-monatomic gases.

    AM
     
    Last edited: Sep 12, 2011
  9. Sep 12, 2011 #8

    Philip Wood

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    Agreed.
    Yes, agreed, if the distinction the OP wanted to make was between polyatomic and monatomic gases. But as I said in hash 4, this is not, imo, the same thing at all as the distinction between real and ideal gases. One can, I believe, speak without self-contradiction of an ideal polyatomic gas.

    So it's not really Physics I'm arguing about but terminology.
     
  10. Sep 12, 2011 #9

    Andrew Mason

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    Although the term "ideal gas" may just refer to a gas that obeys the ideal gas law PV=nRT, in kinetic theory a true ideal gas consists of point particles with no interactions between particles and no volume occupied by the particles themselves.

    The problem with calling a diatomic or polyatomic gas with non-interacting molecules an "ideal gas" is in determining the specific heat capacity, Cv. The specific heat capacity will depend on vibrational and rotational energies which involves quantum considerations. So when you are talking about the specific heat capacity of an ideal gas you have to be dealing with a monatomic ideal gas.

    AM
     
  11. Sep 12, 2011 #10

    Philip Wood

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    So you wouldn't have collisions between gas molecules in your ideal gas?

    You wouldn't for point particles, would you?

    But writers on kinetic theory (e.g. Jeans) do take into account collisions when deriving the ideal gas kinetic theory formula for pressure: PV = (1/3)Nm crms2.

    In other words, occupying a negligible fraction of the container volume isn't taken to mean that the molecules have to be points, and by the same token, they are not denied enough structure for cv to differ from (3/2)R.
     
    Last edited: Sep 12, 2011
  12. Sep 12, 2011 #11

    Andrew Mason

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    Even in theory it is difficult to model a single diatomic ideal gas. What value do you give for its Cv? 5/2? 7/2? At what temperature? A diatomic gas could follow PV=nRT with a Cv of 5/2 or 7/2 or a Cv that changed from 5/2 to 7/2 over a temperature range of 500K-1000K. Which one is the ideal diatomic gas?

    I don't know how would you compare the change in temperature of this model diatomic ideal gas to a real gas for equivalent adiabatic expansion.

    So it just seems implicit that the OP is asking why a real (diatomic or polyatomic) gas can have a smaller temperature decrease but have a greater internal energy decrease when compared to a (monatomic) ideal gas undergoing the same adiabatic expansion.

    AM
     
  13. Sep 13, 2011 #12

    Philip Wood

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    Agreed. Vibrational modes have big gaps between levels and only come into play at higher temperatures.

    They are all ideal, as long as intermolecular attractive forces are ignored, and molecular volume is negligible compared with container volume (which can be approached by sparseness of molecules rather than by shrinking molecules to points). The ideal requirement is perfectly consistent with quite sophisticated quantum mechanical treatments of individual molecules and doesn't prevent different values of cv at different temperatures.

    Around room temperature for low rmm diatomic gases, for which cv = (5/2)R, the comparison between real and ideal is, I would think, straightforward enough. The real gas would cool more, because of the attractive forces, as we agreed earlier. At temperatures where the vib modes are starting to come into play, and cv for the ideal gas depends on temperature, and so changes during the expansion, the comparison is certainly more complicated.

    As I said before, my point is about terminology. I'm claiming that the distinction between real and ideal has nothing much to do with the distinction between polyatomic and monatomic. What I'm taking issue with is one or two points from your post (hash 9):
    I disagree about the necessity for point molecules in an ideal gas, because then you'd have no collisions between gas molecules (zero probability of collisions) in an ideal gas, which is not what is assumed by authorities such as Jeans and by writers of any thermal physics textbooks I've ever read. I'd be interested in what you have to say about this.

    I also disagree with your last sentence in the post just quoted, for reasons I've explained above.
     
  14. Sep 13, 2011 #13

    Andrew Mason

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    Kinetic theory also assumes that molecules are perfect spheres and collisions are perfectly elastic.
    I agree that the molecules in an ideal gas can have a finite volume so long as the molecular diameter is negligible compared to the average distance between molecules. But it is assumed that all collisions are elastic in terms of translational kinetic energy. This is essential to the derivation of :

    [itex]\frac {1} {2} mv_{rms}^2 = \frac {3} {2} k_B T[/itex]

    One can model an ideal monatomic gas using these criteria. An ideal monatomic gas will necessarily have 3 degrees of freedom. One cannot model a diatomic or polyatomic gas using these criteria because collisions are not elastic in terms of translational kinetic energy.
    I kind of disagree with it myself! I said:
    You can certainly have a diatomic or polyatomic gas that obeys the ideal gas law: PV=nRT. But you must also specify what its heat capacity is for the temperatures in question. You cannot determine what its heat capacity is from the fact that it is an "ideal gas". So what I should have said is: "So you cannot assume anything about the specific heat capacity of an ideal gas at any temperature unless you are dealing with a monatomic ideal gas".

    With that I think we may agree on everything!

    AM
     
  15. Sep 13, 2011 #14

    Philip Wood

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    Well, we're not too far off! I've enjoyed the discussion; thank you for taking the trouble to respond so fully.
     
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