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Kinetic Theory and Elastic Collisions

  1. Mar 2, 2012 #1
    1. The problem statement, all variables and given/known data

    When deriving the Ideal Gas Formula from the Kinetic Theory of Gases, we assumed that the gas molecules made perfectly elastic collisions with the walls of the container. This assumption is not necessary as long as the walls are at the same temperature as the gas. Why?

    3. The attempt at a solution

    My reasoning didn't lead me to the book's conclusion, so I will post my thought process and hope someone takes the time to read it and correct my thinking.

    Since we are talking about the temperature of the wall, I presume we must describe the wall on a microscopic scale. So gas molecules are now colliding with "wall molecules."

    In a perfectly elastic collision between two molecules, kinetic energy is conserved. So if both particles are moving with the same speed and collide, they will both rebound with the same speed they had originally. If they are moving at different speeds, one molecule will impart some energy to the other, and lose speed in that process. But the total kinetic energy of the two molecules will be the same before and after the collision.

    Which leads me to believe that if the wall and gas were at the same temperature, elastic collisions with the wall would not, on average, reduce the speed of the gas molecules. However, if the wall were cooler than the gas, the gas molecules would tend to transfer energy to the wall, losing kinetic energy in the process. Energy would still be conserved, since the wall molecules would gain kinetic energy. But the derivation of the ideal gas law would only work for elastic collisions if the wall and gas were the same temperature.

    In an INELASTIC collision, some kinetic energy will always be lost, even if both molecules have the same speed initially. So even if the wall and gas were at the same temperature, the gas molecule would not rebound with its full initial speed, and the ideal gas law does not follow.

    So that's my reasoning, but it did NOT lead me to the conclusion the book asked for. So where am I going wrong?
     
  2. jcsd
  3. Mar 2, 2012 #2
    Pardon me if I get things wrong. It's been such a long time that I haven't touched my physics.

    In my opinion, one of the most important observations that you've made is that if the temperature of the gas = temperature of the wall, then kinetic energy can be roughly conserved. Nice job! :wink:

    The question, I think, requires you to think more practically. That is, how the theory fits pretty well with some more practical concerns. Ideally the molecule heads to the wall and that molecule would bounce back from the wall. Even in that situation, the wall is so ideal: it is so flat that even a molecule "sees it flat." But we don't need the wall to be too ideal here. Hint: a more relaxed assumption would be one molecule in, one molecule out, and they don't have to be the same molecule.

    You should also at the same time explain yourself why it is one-in-one-out, but not one in and a bunch out or vice versa.

    And in the end, you should also reach to the conclusion that the image of a ball and the wall would be no good. Why? Hint: a ball and a wall is a deterministic case - I mean, you know exactly how both things would behave. Thermodynamic phenomena should NOT be treated as deterministic. It is, at the end of the day, under statistical physics.

    Another issue: why can we apply the-ball-and-the-wall notion so well, such that we have a nice theory (leaving out the fact that gases are not ideal, the ideal gas theory is still a good approximation)? Can you find the correlation between statistical thinking and this deterministic approach?

    P.S.: I don't know if the question requires you to think this far, but I think the beauty of physical ideas can be seen from just a very simple phenomenon.
     
  4. Mar 3, 2012 #3
    I have two problems:

    In an inelastic collision, even if the gas and wall were in thermal equilibrium, wouldn't the gas lose kinetic energy, on average, unless potential energy is being released somehow?

    Also, the book's question seems to imply that the assumption of thermal equilibrium isn't necessary if the collisions are elastic. To me it seems that we would have to assume thermal equilibrium in even if the collisions are elastic. Otherwise, in elastic collisions, the gas molecules would either gain or lose speed after colliding with the wall. Which of course explains heat flow, but means you can't use PV = nRT until the gas and wall reach equilibrium.

    Is this a correct understanding?
     
  5. Mar 4, 2012 #4
    Yes, equilibrium, that's a magic word. But sorry, no, you're still in the-ball-and-the-wall mindset, though it seems that you're heading the correct direction. You have to think of a bunch of balls, each with random behavior, heading to different positions on the wall, each position is rough non-uniformly.
    Get through this, and you're welcome to statistical physics :)

    P.S.: Your understanding of thermodynamic equilibrium is absolutely wrong. This equilibrium, being different from mechanical equilibrium, doesn't apply to one ball and one wall. It is its statistical nature that defines it.
     
    Last edited: Mar 4, 2012
  6. Mar 8, 2012 #5
    I suppose I should check my understanding of thermal equilibrium then.

    Thermal equilibrium is when the average kinetic energies of the molecules in two substances are equal, although some will be higher and some lower, no?

    When any two molecules collide in an elastic collision, the slower one gains kinetic energy and the faster one loses it, but the total amount is conserved.

    So if two substances in thermal equilibrium are in contact and their molecules are undergoing elastic collisions, some molecules in the gas will be losing kinetic energy to the wall molecules, and some will be receiving kinetic energy from the wall molecules. But since both substances have the same average kinetic energy, there will be, statistically, no net flow of energy, or heat. So the momentum of a particle in the gas remain, on average, unchanged after a collision with the wall. From this we can derive the Ideal Gas Law.

    Before I talk about systems that aren't in equilibrium, I'd like to pause and ask if this is correct.
     
  7. Mar 9, 2012 #6
    Yes, it is on average, no net flow, energy flow or mass flow. But I would doubt that it is about kinetic energy. The wall, unlike the ideal gas molecules, has potential energy - and it's definitely not negligible (think of the tight bond among the molecules). Some gas molecules may be held back or stuck in the wall.
     
    Last edited: Mar 9, 2012
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