1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The gas laws are wrong!

  1. Jun 15, 2010 #1
    Did that get your attention?
    Ok, so I'm mosty kidding, but I have a conundrum for you that exposes one of the weaknesses of the gas laws. Ready?

    In my astronomy class last quarter, we had to do a brief overview of physics, because physics isn't a pre-req, and some people had never taken physics before. At one point, we were given an assignment that involved playing around with a gas law applet online. This "simulator" had a very simple setup. There was a chamber that you could fill with simulated particles of gas, and a readout that displayed pressure and temperature. The chamber also had a plunger that you could push in or out by clicking and dragging, and a heat source that you could turn on or off.

    When the class met and went over the results, the answers were all in accordance with pv=nrt... except for one. One of the questions asked about the effect that moving the plunger had on the temperature. All of the students reported that moving the plunger in heated up the gas, however, many of the students reported that moving the plunger back out did not cool the gas at all! So, how is this possible? How could moving a plunger out sometimes not cool a gas, while moving it in always heats it?

    I got the answer on the drive home.
  2. jcsd
  3. Jun 15, 2010 #2
    When the gas particles collide with the plunger and push it back, assuming that the plunger can only move back by pushing the external atmosphere out of the way, the molecules do work on the plunger. Now all of the the gas particles as a population have less average kinetic energy, which means a lower temperature.
  4. Jun 15, 2010 #3
    True, but the real question is, how is it possible that moving the plunger back could make no difference in the temperature?
  5. Jun 15, 2010 #4
    It does make a difference, just as moving the plunger in. If you pull it back fast (or push it in fast), you will see the effect. The chamber is not thermal-isolated, so when the process is slow, the temperature only changes a little, and sometimes in real life, it changes too little that we human cannot feel anything.
  6. Jun 15, 2010 #5
    First I thought that the computer model was just wrong. But it is also possible that the plunger is supposed to shift to a new position instantly, so that the molecules could not push it back (like in Hirn's experiment). In this case the molecules simply fly over the free space without doing any work, so their internal energy and temperature stay the same.
  7. Jun 15, 2010 #6


    User Avatar
    Science Advisor

    Very simple. Too few particles moving too slow.

    For the plunger to take energy away from the system, the particles must collide with it as the plunger moves back. That allows particles to do work on the plunger.

    If you move the plunger fast enough so that the collisions don't happen, no energy is transfered, and the temperature will remain the same. No violations of gas law happen, because pressure of the new state adjusts to temperature, volume, and number of particles appropriately.

    This will happen in a real gas as well, but you need to move the piston out of the way much faster than speed of sound, as that's roughly the average velocity of the particles.

    Naturally, when you are moving piston in, the collisions are unavoidable, you do work on gas, and heat it up.
  8. Jun 15, 2010 #7
    When V increases, there must be some change in P or T.
    If the plunger is immediately shifted and a free space is left, the gas will go through an adiabatic process, so T will decrease.
  9. Jun 15, 2010 #8
    Did some of them apply external heat while moving the plunger back ?
  10. Jun 15, 2010 #9


    User Avatar
    Science Advisor

    And the gas does work on? In order for T to decrease, gas MUST do work. If the piston is shifted "instantly", no work is done on the piston. So how can T change?
  11. Jun 15, 2010 #10
    K^2 and Lojzek are correct. Many of the students were moving the plunger too fast, faster than the simulated particles (which are much slower than real particles), making a particle-piston interaction impossible. This does not directly conflict with the gas laws but it does make them harder to wield. If you are thinking strictly in terms of pv=nrt, and not in terms of kinetic theory, you may fail to notice that the pressure on the piston by the gas drops to zero temporarily, so no work is done.
  12. Jun 15, 2010 #11
    It actually does. The total energy of the gas must include the "macro kinetic energy" (I'm not sure if I use the right term), i.e. the energy due to the motion of the gas as a whole (the center of mass does move, right?). In most cases, the expanding speed of the gas is about the speed of the piston, which is limited, so the macro kinetic energy is negligible. But in the case the gas expands into vacuum, the speed is large.

    The collision of the gas with the chamber right before the expansion stops will transfer the gas's energy into heat, and so, the gas loses its energy to the chamber. If the chamber is perfectly thermal-isolated (assume that it is kept so that it won't move after the collision), then after the expansion, this energy will transfer back into internal energy, so at the point of equilibrium, T will increase back to its initial value. However during the process, T does change.
  13. Jun 15, 2010 #12


    User Avatar
    Science Advisor

    If we start talking about energy lost to chamber, we'll be nowhere. The walls of the chamber should be assumed to have zero heat capacity and be perfect insulators. That's, by the way, what the simulation would do.

    Yes, the gas will cool as it accelerates towards the piston, briefly. You're right, if it didn't work like that, gas expanding into vacuum wouldn't cool.

    But the initial and final CM velocities are both zero. So after the gas reaches new equilibrium, its final temperature will be the same as initial.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook