Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Why a dust cloud does not obey pV=nRT?

  1. Apr 20, 2016 #1
    Hello all,

    Another very idiotic question (sorry for the idiotic questions today). From what I learned, a perfect gas obbey pV=nRT because the gas molecules collide against each other in an amazingly elastic way; that is, they may collide a trillion times but will still keep their total kinetic energy constant, and redistributes the energy in such way that allows the definition of temperature.

    Now, consider a very cold gas cloud with very low density; then the speed of the molecules and number of collisions is very small. Yet, even when the speed and number of collisions are small, it will still eventually obbey pV=nRT (once enough time has passed to let it settle down into steady state).

    Now, if I build a cloud of jell-o drops, then these collisions are not elastic, so nevermind that. But say I build a very low density cloud of very fine dust specks of the same size of a very rigid solid material, and the speed of the collisions is small enough as to not break the particles apart or melt them or do anything else crazy, then let enough time to pass for it to redistribute the kinetic energy and reach steady state.

    Won't that steady state also be ruled by pV=nRT, even if that is a cloud of solid specks instead of a cloud of gas?
     
  2. jcsd
  3. Apr 20, 2016 #2

    russ_watters

    User Avatar

    Staff: Mentor

    No. Solids have mechanical properties different from gases. Imagine you had a box with a handful of rubber balls in it. Shake it and they bounce around semi randomly. Stop shaking it and they fairly quickly fall to the bottom. They aren't perfectly elastic and the kinetic energy is dissipated as thermal energy. But molecules are perfectly elastic and random because their motion is thermal energy.
     
  4. Apr 20, 2016 #3
    I see. So in order for that to be true I would have to come up with some tiny solid ball that would not dissipate any energy on impact, and as such thing does not exist, then the mechanics are completely different. Thank you, I get it.

    So, for my peace of mind - if you have two clouds of the same size and the same mass, one made of gas and the other made of dust, and both are collapsing into a star. Then the mechanics for that collapse will be different between the two, right? I heard the gas cloud increases temperature due to pV=nRT (less losses for other mechanisms); I guess the dust cloud will increase temperature by some other law, I reckon?
     
  5. Apr 21, 2016 #4
    The dust cloud will obey pV=nRT. With n being pretty small.
     
  6. Apr 21, 2016 #5
    "collapsing into a star"
    These are large populations of either gas molecules or dust particles in a vacuum, bound only by gravity. That is a very different proposition from gas or dust in a box which can be squeezed and measured on a laboratory bench.

    Let's assume the initial condition is an evenly-distributed set of particles in a sphere. These particles each have a gravitational potential energy based on their distance to the cloud's center of mass. As they collapse toward that center of mass they exchange their gravitational potential energy for kinetic energy. Collisions with other particles convert some of that kinetic energy into heat. (That sounds straightforward but since the temperature of a gas is defined by its kinetic energy it's actually a pretty fuzzy explanation.) The end state of this collapse is when the gravitational force (pro-collapse) is balanced by pressure (pro-expansion). The process of stellar collapse results in a hot, dense plasma whether the starting material was a gas or a solid.
    Dust particles are made of more atoms than gas molecules. They have an extra degree of freedom: collisions can cause the dust particles to break apart. This 'spends' some of their kinetic energy to overcome their binding energy (whether that be electrostatic, van der waals, chemical bonds, etc., etc.). This energy is not available to become heat, so the dusty star should reach a lower final temperature as a result of the energy spent breaking dust into plasma.

    That suggests that dust will obey pV=nRT so long as the particles act like perfect gas molecules, while an adjustment term to account for binding energy and fragmentation would have to be introduced once the collisions become too energetic for the particles to survive.
     
  7. Apr 21, 2016 #6
    There is no qualitative difference between a big molecule and a small grain of dust. Only quantitative.
    A lone atom has just 3 degrees of freedom. The 3 translational directions.
    A molecule of 2 atoms has 6 degrees of freedom - still 3 for each atom. But only 3 of these are translations of whole molecule, and only these 3 contribute to pressure.
    The other 3 are movements of atoms relative to each other: 2 are rotations, and the sixth is vibration of atoms along the bond connecting them.
    If a molecule has 3 atoms then it still has 3 degrees of freedom for each atom - and still 3 whole molecule translations, which alone contribute to pressure. The other 6 may be 3 rotations and 3 vibrations, if molecule is not linear, or 2 rotations and 4 vibrations, if it is.
    Et cetera. A grain of dust has a big but finite number of atoms, therefore a big but finite number of degrees of freedom. 3 of which - a small but nonzero fraction - will provide some pressure, while the rest participate in equipartition of energy but contribute no pressure. Most of those being the degrees of freedom for internal vibrations.
     
  8. Apr 22, 2016 #7
    Much better explanation, thank you. I shouldn't have used a term with a specific technical meaning in a more general sense.

    Does that suggest that once the force of impact between dust particles becomes high enough to break them apart that the system would experience a significant increase in the rate of pressure growth?
     
  9. Apr 22, 2016 #8

    ogg

    User Avatar

    The Ideal Gas Law is an approximation and most elementary physical chemistry texts will explain why & why it is a pretty good one for many small gas molecules. So, iirc, using PV=nRT for CO2 will result in ~5% error at STP. (I may be misremembering the gas & T). Few gasses conform to IDG to within 1% (at ~STP, but keep in mind the requirements for it to "hold"). Contrast that to the Law of Conservation of Energy, where there's near exact (local) conformance to the Law. One question to ask yourself is given a cylinder of gas with a height of, say, 100 km - (and say 1 m radius, just to be specific) enclosed in a perfect insulator: will the temperature equilibrate? That is; will T at sea level = T at 100 km? No, "never". Gravity is, in a sense, incompatible with IDG. Finally, an astronomical cloud (gas or dust or both) has a substantial potential energy which results in heating if the cloud collapses - nothing to do with IDG. Just some random factoids to factor into your thinking.
     
  10. Apr 22, 2016 #9

    ogg

    User Avatar

    The Laws of Conservation of Energy and Conservation of Mass (in classical physics) are facets of the Law of Conservation of Mass-Energy (still Classical, but with E=mc² of the early 20th Century (fusion in Sun, radioactive decay, atomic bombs, etc.) added they needed to be combined). With General Relativity, we have the Law of Conservation of Stress-Mass-Energy but most (or, at least "many") models (including the standard lambda-CDM model) assume the vacuum of space-time has a constant energy per space-time volume, and so as the Universe expands, energy is not conserved. But the scales on which this is a significant deviation from the conservation law is cosmological (ie, distance scales of millions of light years). Opinions are divided on this.
     
  11. Apr 22, 2016 #10

    ogg

    User Avatar

    For a cloud to collapse, there has to be some process which allows "scattering" or collisions between the particles making up the cloud. For matter, the electromagnetic interactions allow collapse. "The" reason dark matter was "invented" was because the stars orbiting in our own Milky Way, and in other galaxies exhibited velocities inconsistent with what you'd expect for orbits around a central mass; it required that the mass be "smeared out" to very large distances and in order for that to be possible, whatever that mass was made of, couldn't be doing much colliding (or it would have collapsed). (Collide is another name for "interact", dark matter can't interact much with anything (neither itself nor 'normal' sub-atomic particles)). There's no significant relationship between dark matter and dark energy, two totally different (as far as we know) things, "invented" to explain two totally different "problems" seen in astronomy.
     
  12. Apr 22, 2016 #11
    Very amazing. Whole discussion has been extremely interesting. Thanks all for their insights!

    One last question on this matter of gas cloud and dust cloud. I heard that "temperature" of my old frying pan is vibration of the steel atoms inside (ergo it has no pressure), and that "temperature" of a gas cloud (of how you folks say an ideal gas) is kinetic energy (ergo it has pressure).

    So, if I have a cloud of very fine dust, like say smoke, and then someone tells me it has a temperature of 5,000K... then what is that? If [I have no clue, so no laughs] that 5,000K is a measurement of both kinetic energy and vibration, then is it possible that two dust clouds, both of them at 5,000K, will actually behave quite differently if the particles in one are mostly vibrating while the other they are mostly kinetic?

    If so [no chuckling] then temperature for clouds of dust may not be a very useful measurement, because the actual behaviors may be quite different?
     
  13. Apr 23, 2016 #12
    Wrong. It will become equal.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Why a dust cloud does not obey pV=nRT?
  1. PV=nRT for solids? (Replies: 11)

  2. Pv=nRT derivation (Replies: 7)

Loading...