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Gas Expansion in a vacuum

  1. Oct 14, 2014 #1
    Say that there's a large metal box with nothing but a vacuum inside of it, except for a small bag of compressed gas at the center. If the bag were to suddenly pop, is there a specific rate that this gas would accelerate when expanding to meet the space of the container?

    I think that it might be dependent on the pressure and amount of compressed gas, but I'm not sure, and I haven't had any luck finding an answer.

    Are there variables that I'm missing, and if possible, direct me to a reputable site where I can better study this in depth?
     
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  3. Oct 14, 2014 #2

    e.bar.goum

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    I would think it would be more temperature dependent than anything.

    Consider that an ideal gas has particles that move in a straight line until they hit the walls of the container. Then, if you remove the walls of the container, the particles are just going to keep moving in that straight line velocity, until they hit the walls of the large metal box. Then, the rate of expansion will be given by the average velocity of the gas particles - given by the temperature.
     
  4. Oct 14, 2014 #3

    assume yourself in the bag, and the gas in the bag is just expanding... the bag is expanding with the gas because it may be elastic... but you are not expanding because you are not elastic... rather your body while trying to expand itself just explode and you die...
    so accordingly the popping of the bag depends on its material rather than the velocity...
    :)
     
  5. Oct 14, 2014 #4

    NTW

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    I think I can visualize the expansion imagining the gas initially contained in a small plastic bag that inflates with the expanding gas, without opposing any resistance. The temperature of the expanding gas will remain constant, since the velocity of the gas particles will remain constant too. At first, I thought: well, if it expands, it has to cool down... But then I realized that that expansion does not imply external work, as no force is countering it... It seems to be an isothermal expansion...
     
  6. Oct 14, 2014 #5

    jbriggs444

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    I do not agree that it is isothermal. Certainly, energy is conserved. But some of the thermal energy in the compressed gas will go to bulk kinetic energy of the gas as it expands to fill the container, some parts expanding in one direction and some in another. This bulk kinetic energy will, of course, eventually be converted to thermal energy, leaving the kinetic energy per molecule (and hence the temperature) the same as it was prior to the release.

    In the initial expansion phase, the gas molecules will not be going in straight lines. They will still be colliding with one another. There will be a pressure gradient which is high in the middle of the compressed gas "blob" and low on its exterior. The resulting expansion will be adiabatic.
     
  7. Oct 14, 2014 #6

    NTW

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    If the temperature remains constant, the expansion is -by definition- isothermal. It is also adiabatic, as there is no transfer of energy or matter with the surroundings... It's weird to see those two conditions together, but it's a strange case too, just a thought experiment...
     
  8. Oct 14, 2014 #7

    jbriggs444

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    But does temperature remain constant throughout the scenario? I claim that it does not. By definition, as you point out, this means that the transition from highly compressed, non-turbulent, normal temperature gas to weakly compressed, highly turbulent, low temperature gas is not isothermal. Nor can the subsequent transition from weakly compressed, highly turbulent, low temperature gas to weakly compressed, non-turbulent, normal temperature gas be called "isothermal".
     
  9. Oct 14, 2014 #8

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  10. Oct 14, 2014 #9

    NTW

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    I don't see any reason to believe that the mean velocity of the particles could change during the expansion. As the temperature of a gas changes only when that velocity change, and in our case there is no change, then no variation of temperature will take place...
     
  11. Oct 14, 2014 #10
    I completely missed temperature a a variable, and so I thank all of you, and I'm glad this can generate some form of scientific discussion.

    I'd like to say that on the topic of temperature, the gas is compressed, and under pressure, so when the small balloon pops and the gas expands to fill it's new container, wouldn't the temperature of the gas decrease as volume increases?

    And I wouldn't that decrease in temperature as it expands cause the gas to accelerate at ever slower rates during expansion?
     
  12. Oct 14, 2014 #11
    The expanding gas does no work, as it is expanding into a vacuum. Temperature of the gas will be the same at the beginning and end of the expansion. That is basic thermodynamics.

    If you burst the pressurized balloon in air then of course the temperature of the expanding gas will drop as decrease as it does work to move the air out of the way.
     
  13. Oct 14, 2014 #12

    jbriggs444

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    The mean velocity of the particles does not change. But that does not mean that temperature does not change. The temperature of a small parcel of gas is determined by the mean velocity of the particles relative to the center of mass of that parcel. If the small parcel is moving relative to the center of mass of the entire gas volume, that means that the mean velocity of particles relative to the center of mass of the parcel will be strictly smaller than their mean velocity relative to the center of mass of the entire volume. The temperature of the parcel will be reduced accordingly.

    Every one of the parcels (*), while expanding, will have done positive work on its neighboring parcels -- causing a net increase in total bulk kinetic energy of the entire volume. That is an adiabatic expansion. The energy has to come from somewhere. Temperature is reduced.

    (*) If one treats the entire gas blob as a series of concentric spherical shells, parcels in the outermost shell will not have done positive work on their neighbors. However, an ideal gas is infinitely divisible. We can make the outermost shell as thin as we please, so this effect is negligible.
     
  14. Oct 14, 2014 #13

    jbriggs444

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    That is not correct. The expanding gas does work on itself.
     
  15. Oct 14, 2014 #14

    NTW

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    In the case of an ideal gas, and if no work is done by the expanding gas, then no temperature drop will take place.

    In the case of a real gas, the temperature will drop.
     
  16. Oct 14, 2014 #15

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    The expansion, as assumed in the thought experiment we are discussing, will be symmetrical, the particles being contained within a perfect (in mean terms) spherical shell. Thus, the center of mass of the particles will not change, and will remain where it was throughout the expansion.

    Anyway, the problem of a perfect gas expanding into a vacuum was addressed many years ago, by Joule.

    Below, I copy-paste a fragment of the text in the Wikipedia article 'Joule expansion': http://en.wikipedia.org/wiki/Joule_expansion
    Please note that, in ideal gases, temperature does not change...

    ***************************
    As the system is thermally isolated, it cannot exchange heat with its surroundings. Also, since the system's volume is kept constant, the system does no work on its surroundings.[6] As a result, the change in internal energy ΔU = 0, and because U is a function of temperature only for the ideal gas, we know that Ti = Tf. This implies that PiV0 = Pf(2V0), and thus the pressure halves; i.e. Pf = ½Pi.

     
  17. Oct 14, 2014 #16

    jbriggs444

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    Unless I misread the Joule Expansion article, it describes the end state after equilibrium has been regained. The original post in this thread concerns itself with the intermediate state -- how fast the gas expands when the bag is popped.

    I do not disagree that the final temperature after equilibrium has been regained will (ideally) be identical to the initial temperature of the gas in the bag. What I disagree about is the interim temperature while the gas is expanding but has not yet reached the walls.
     
  18. Oct 14, 2014 #17

    NTW

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    All right...

    We know that the kinetic energy of the molecules of a gas is related to its temperature by v2 = k*T

    Where T is the absolute temperature and k some constant. Now, as the velocity of the particles does not vary during the expansion, the temperature of the gas will remain constant.
     
  19. Oct 14, 2014 #18
    I get you now.

    Something along the line of,
    The expanding front would be doing no work as there is nothing to do work against, only a vacuum.
    Behind the front is where the complication begins.

    At the peripheral the pressure would be zero and at some point as the expansion continues, the pressure along a radius would vary from zero at the outer edge to the initial pressure at the centre, or even at some volume about the centre if the container is not that large. I suppose finding the function of pressure along a radius is what the OP is looking to aquire.
     
  20. Oct 14, 2014 #19

    jbriggs444

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    This ignores the fact that temperature is a local measurement. For a local temperature measurement to be meaningful, the kinetic energy of the molecules in the gas must be measured in a frame of reference in which the gas is locally at rest. Even though the velocity of the molecules in the frame of reference where the total gas volume is at rest is unchanged, the velocity of the molecules in the frame of reference where the local gas parcel is at rest is not unchanged.

    Once the system has returned to equilibrium, such nuances lose importance since all of the parcels are at rest relative to one another. But until that equilibrium is attained the situation is not as simple as you suggest.

    The temperature of the wind whistling past your ears does not depend on the kinetic energy of the air molecules relative to your ear. It depends on their velocity relative to the wind.
     
  21. Oct 14, 2014 #20

    NTW

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    Fascinating concepts... I am left without words...
     
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