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Name of intermolecular forces of gases

  1. Jan 28, 2012 #1
    For a non-ideal gas expanding in a vacuum, the kinetic energy of each of the molecules does NOT remain consant as the gravitational effect of the other molecules would "slow down" these velocities by deceleration. What are those forces referred to? Are they "Newton" forces or what? I can't remember.

    I don't think it is Van der Waals forces.
     
  2. jcsd
  3. Jan 29, 2012 #2
    ... You're saying that the 'gravitational force' is decelerating the molecules ...
    That's not an intermolecular force.
     
  4. Jan 29, 2012 #3
    "... You're saying that the 'gravitational force' is decelerating the molecules ...
    That's not an intermolecular force."


    No matter what you call it what is the eponym for it? Is it Van der Waals forces, Newtonian forces or what?

    Also, why does an expanding gas (non-ideal, of course) have a propensity to get hotter if it is above a certain "inversion line?" It would seem that no matter where you are with a given molecule in space, all gravitational forces on it would tend to slow it down, not speed it up. Even a rocket launched from Earth at above "escape velocity" continually slows down (although never reaches a nadir.)

    I know it has something to do with enthalpy, whatever that is. As a molecule moves away from other molecules, the gravitational attraction between them lessens and the potential energy of said molecule therefore drops and this is picked up as kinetic energy (which translates into heat) but I cannot figure out the mechanism by which the molecule picks up steam without being "pushed" or "pulled."

    Let us leave relativity out of it for now as we are dealing with markedly subrelativistic speeds.

    H-E-L-P

    stevmg
     
  5. Jan 29, 2012 #4

    Philip Wood

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    Gravitational forces between molecules are utterly negligible. The relevant forces are Van der Waals forces, and they are attractive.

    The rms speed of the molecules is proportional to the kelvin temperature. This is certainly true for an ideal gas, and I believe also for a real gas (for which we don't neglect V der W forces). For a real gas, because of the V der W forces, the molecules' velocities aren't quite constant between one collision and the next, but that doesn't imply that the molecules' rms speeds are different from those for ideal gases at the same temperature.
     
  6. Jan 29, 2012 #5
    Gravity doesn't have an eponym. It's just, gravity.

    What I think you are referring to is the Joule-Thomson effect. It has nothing to do with gravity whatsoever; but the derivation and explanation are included.
     
  7. Jan 30, 2012 #6
    What I think you are referring to is the Joule-Thomson effect. It has nothing to do with gravity whatsoever; but the derivation and explanation are included.

    I read it but don't understand it. I am too old and too stupid to wrap my brain around it. Enough said.
     
  8. Jan 30, 2012 #7
    Well, unless you'd like to expound on what you don't understand, we won't be able to help you.
     
  9. Jan 30, 2012 #8
    Now, let us try again...

    Suppose we have a non-ideal gas in which the molecules do have attractive forces on each other inversely proportional to the square of the distance between them (this not a new concept...) In a freely expanding situation, as the molecules "separate," the attractive forces will diminish. The attractive forces have the propensity to slow down each molecule's speed. As the forces diminish, the "slowing down" effect would tend to diminish. This, however, would not lead to an increase in the speed of the dispersing molecules but would merely reduce the "slowing down" effect. So, how the Dickens would a molecule pick up speed (and consequently temperature) in such a situation? In order for a molecule to pick up speed one would need an accelerating force on it. Where would that come from? The decrease in a decelerating force is not equivalent to an accelerating force.

    Can someone please explain this????

    stevmg
     
  10. Jan 30, 2012 #9

    Philip Wood

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    I've think I've now understood the context. The (thermally insulated) gas cools on expansion into a vacuum as potential energy is gained through work is done against the attractive forces, and kinetic energy is lost. [I believe this is called the Joule effect. The Joule-Thomson effect is temperature change on effusion through a porous plug.] Where did you read that the gas heats up?
     
  11. Jan 30, 2012 #10
    Here is one site which mentions gas increasing in temperature upon expansion (H, He, Ar): http://en.wikipedia.org/wiki/Joule–Thomson_effect. However, this is done with throttling which allows for an increase in linear velocity of molecules under the influence of a (nearly) constant pressure-head behind it. That linear velocity, coupled with a "water-hammer" effect and followup randomization of molecular motion (rather than being straight line which occurs through the throttle) could possibly lead to an increased temperature in small volumes of accelerated gas (much less than the original volume of the gas which exited.)

    Addendum: An ideal gas does not cool on free expansion. I think we agree on that.
     
  12. Jan 30, 2012 #11

    Philip Wood

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    Quite so. Your arguments (and mine) apply to Joule expansion into a vacuum. The J-T effect is something different. And of course we agree that a real gas does not cool on expansion into a vacuum.
     
  13. Jan 30, 2012 #12
    I am a novice at this. Are you stating that the Joule effect is different than the Joule-Thomson effect? Does the J effect apply to a given volume of gas being completely allowed to freely expand whilke the J-T effect applies a given volume of gas passing through a throttle where its flow becomes linear? If so, then we are in complete agreement and I understand what is going on.

    By the way, I have had the darndest times convincing various people that expansion of a gas freely where it does no work implies no cooling. Everyone thinks that is anything expands itmust be cooling. Actually, thermal and other electromagnetic expansion does get more dilute (inverse as the square of the radius) and therefore "cooler" as it expands.
     
  14. Jan 31, 2012 #13

    Philip Wood

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    Stevmg: comments on your post hash12, just to make quite sure we agree...

    First paragraph: that's what I'm saying. The two effects are different.

    Second para: ideal gas expanding into a vacuum: no work and no cooling. Real gas expanding into a vacuum: no external work but, nonetheless, cooling. The internal energy of the gas stays the same, because (we're assuming) no heat enters or leaves the gas, and no external work is done. But internal energy is sum of KE of molecules and PE of molecules arising from intermolecular forces. PE goes up on expansion, so KE must go down (since we're dealing with a case of constant internal energy).
     
  15. Jan 31, 2012 #14
    To Philip Wood, post #13 - Finally! It appears that I do understand the concepts and in simplistic terms. No one else has been able to dumb-it-down for me to my level. Thank you.

    Now, what is the difference between a Joule effect and a Joule-Thomson effect? Is the latter the situation where the expanding gas passes through a throttle (which would imply linear acceleration through a narrow tube) while the former is merely suddenly allowing a gas to expand adiabatically freely without throttling (no linear acceleration through a narrow tube but expanding linear acceleration unencumbered everywhere.)
     
  16. Jan 31, 2012 #15
    The throttling effect would appear to increase the kinetic energy of the exiting gas while, a the same time, decrease the potential and kinetic enegry of the gas in the original chamber. I know that when a CO2 cartridge is allowed to discharge (a throttling phenomenon), the cartridge case feels very cool as compared to originally. The exiting CO2 also feels cool, too, even though its kinetic energy is increased.
     
  17. Jan 31, 2012 #16

    Philip Wood

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    Yes, agree with your distinction. The J-T effect is, I think, not so easy to understand on an intuitive level as 'Joule' expansion into a vacuum. At least I can't summon up an easy way to convince myself as to the circumstances when it will produce cooling and when it will produce heating up.

    The J-T effect is a process where initial enthalpy = final enthalpy. (Enthalpy, H = U + PV.) This can be developed mathematically, and the heating/cooling condition predicted for, say, a Van der Waals gas. It's covered quite nicely in that old war-horse, Heat and Thermodynamics by Mark W Zemansky of blessed memory.
     
  18. Jan 31, 2012 #17
    How about this?

    H, He, Ar heat up under J-T conditions. Could that be due to repulsion rather than intermolecular attraction?

    Repulsion would lead to a decrease in PE with concomitant increase in KE (and therefore, temperature) to maintain enthalpy.

    Why H, He and Ar are the only gases to do that - God only knows.

    stevmg
     
  19. Jan 31, 2012 #18

    Philip Wood

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    If you plot T against P for equi-enthalpy states, then below a certain temperature (the maximum inversion temperature, Tinv max), which depends on the gas, the curve has a maximum. If the gas has an initial pressure less than the pressure at this maximum, then throttling to a lower pressure will produce cooling. Above Tinv max, the enthalpy curves have their maximum at zero pressure, and throttling can't produce cooling. For most gases Tinv max is well above room temperature, so throttling from a suitable starting pressure can produce cooling. But for hydrogen, Tinv max = 202, and for Helium, Tinv max is even lower. So at room temperature throttling won't produce cooling. Argon's a bit of a puzzle because, according to M W Z, Tinv max = 723.

    Maybe it's possible to understand all this intuitively in terms of very simple ideas, but, as I said, I don't. I think my best chance of doing so would be to look at the mathematical treatment in detail and then look to understand each step in simple terms. But I haven't the time just now.
     
    Last edited: Jan 31, 2012
  20. Jan 31, 2012 #19
    Lost me at inversion, etc. I think I have achieved my maximum understanding for my level of knowledge. I have no intuitive concept of inversion and don't think I ever will nor is it worth my time.

    The reason why both the CO2 cartridge AND the escaped CO2 gas to be cooler would be the increase in PE and drop in KE of both to maintain "equi-enthalpy."

    Much thanks for your time in explaining what you did to me!

    stevmg

    P.S. my referrals to water hammers and blowholes are not that far off - trade-off of kinetic energy for potential energy (pressure.)
     
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