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Compression of an ideal gas

  1. Feb 15, 2012 #1
    I really need help on this question, I've tried asking several people but I still don't quite get it.

    The formula for the work done on compressing/expanding an ideal gas is [itex]\int[/itex]-pdV.

    Now first of all - p denotes the internal pressure of the gas right?

    If so, good so far.

    Let us now assume the situation on the picture, where a gas is confined in a cylinder with a movable piston. The gas is not in equilibrium since the external pressure from the atmosphere exceeds the internal (in my example external pressure is twice the internal but it could be anything). Of course the gas will now start to compress.

    My question is what the work done on the gas is and why the external pressure for some reason has no influence on the work done.

    Let me try to do a force analysis so that you can see what I'm thinking and explain, where I go wrong.

    We start noting that the gas will exceed a force on the piston equal to F = Ap, where A is the area of the piston. By newtons third law the piston then exerts a force of -F on the gas.
    Furthermore we see that the external pressure on the piston must exert a force of 2pA and the piston must then exert a force on the atmosphere given by -2pA.

    Let's say that the gas piston moves a distance x and for simplicity assume that the pressure stays the same (NOT realistic, I know)
    Then it is clear that the work done on the gas in the cylinder is indeed -F*x = -pdeltaV.

    *********** BUT SOMETHING IS CLEARLY WRONG. THE PISTON WILL ALSO HAVE GAINED KINETIC ENERGY EQUAL TO THE WORK DONE ON THE GAS! WHERE DOES THIS GO? IT MUST GO TO THE GAS TOO! ***********

    I think fundamentally there is something completely wrong with this way of thinking, but I can't understand it in other ways. Please help me understand it such that it makes sense to think of the work done on the gas as just W = -pdeltaV and not twice that work.
     

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  3. Feb 15, 2012 #2
    In general, the energy you put in you piston manifests itself in the gas as increased pressure (pressure is simply a metric for the kinetic energy of the molecules), or increased temperature.

    If your external pressure is the atmospheric pressure, then it will remain constant throughout the process. If you have 2 pressure vessels, the high pressure will decrease until you reach equilibrium.
     
  4. Feb 15, 2012 #3
    If you consider the piston, there is the work done by the atmospheric pressure L1 (positive), and the work done by the gas L2 (negative). The work done by the gas is less (in absolute value) than that done by the atmospheric pressure. The difference is equal to the kinetic energy of the piston.

    If you consider the gas, there is the work done by the piston on it and that's all. Calculating the actual value may be a little more difficult as the pressure in the gas is not constant and this is not an equilibrium process.
     
  5. Feb 15, 2012 #4

    Andrew Mason

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    Not necessarily. If it is quasi-static then W = ∫PdV where P is the internal pressure of the gas and W is the work done BY the gas (- if work is done on the gas).

    If it is not quasistatic, W will be something other than ∫PdV where P is the internal gas pressure. This is because dynamic energy has to be taken into account, as Nasu points out.

    AM
     
  6. Feb 15, 2012 #5
    oops double post. see next post
     
    Last edited: Feb 15, 2012
  7. Feb 15, 2012 #6
    I still don't quite get it. Let's take it in steps please :) :

    1) Does the atmospheric pressure do a work on the piston or does it not?

    Yes exactly - that's also how I see it. Yet, if this is true the piston does actually receive some kinetic energy. But where does this then go, if it does NOT
    add to the work done on the gas? :(
    Sorry. I forgot to write that we assume the proces to be quasistatic. That still doesn't account for the extra energy as far as I can see, though.
     
    Last edited: Feb 15, 2012
  8. Feb 15, 2012 #7

    Andrew Mason

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    If the gas in the cylinder is being compressed quasistatically, the work is done on the gas, not the piston. The piston acquires no kinetic energy (v is arbitrarily close to 0). The atmosphere does not do work on the piston in such a case. The work is done on the gas and is equal to the work done by the force on the piston (gravity + applied mechanical force) and by the force provided by the atmosphere.

    AM
     
  9. Feb 15, 2012 #8
    hmm okay, I just wish I could see why that is - it would certainly make a lot of things understandable.

    BUT I don't see it. The piston moves down right? Agreed so far. The force from the atmospheric pressure is 2pA right? SO WHY DONT YOU INCLUDE THIS FORCE :((
     
  10. Feb 15, 2012 #9
    The original conditions were that the pressures on the two faces of the piston are different so there is a net force on the piston. This may be the point of confusion.
    The process is not quasistatic. The piston is not in equlibrium but it moves accelerated until the pressures equalize.
     
  11. Feb 15, 2012 #10
    How can be quasistaic in the conditions you described initially?
    Maybe if you have some extra force (friction).
     
  12. Feb 15, 2012 #11
    hmm maybe the whole problem lies in the fact that I'm not certain what quasistatic means.

    1) First I thought it meant that the gas is at rest the whole time so that Pinternal is practically equal to Pexternal throughout the whole proces.

    2)But then my teacher gave me the impression that quasi-static means, that the piston just moves slow enough such that the gas has time to keep a uniform pressure at all times.

    Which of these is the correct. If 2) is the right one, why can't the proces i described be quasistatic?
     
  13. Feb 15, 2012 #12
    Exactly! That's what I've been thinking too. But when I do exercises in thermodynamics and want to find the work done in for instance an adiabatic process, then I just use the formulas. But conditions in those exercises are like the one in my example.
    For example consider a water rocket going off. How is this different from the situation I sketched? Because my teacher just worked through that example today using the equations of adiabatic expansion.

    Im still left with the question of whether quasistatic means that the gas itself has a uniform pressure at all times or also that the force on the piston is infinitely close to equilibrium at all times.
     
  14. Feb 15, 2012 #13

    Andrew Mason

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    In an adiabatic process that is quasistatic, the adiabatic condition applies ([itex]PV^\gamma = K[/itex]). If it is not quasistatic, the condition does not strictly apply although it can be close even for some relatively fast processes (like the compression and expansion of air due to sound).
    The gas expands quickly, and therefore practically adiabatically. In expanding, it does work on the water causing the water to gain kinetic energy as it escapes through the rocket nozzle. The ejection of water through the nozzle imparts momentum (hence kinetic energy) to the rocket (including the water and gas that remains in the rocket). So the gas does work on itself, as well as on the water.

    The adiabatic condition [itex]PV^\gamma = K[/itex] does not strictly apply although it probably not far out.
    Quasistatic processes occur under conditions that are arbitrarily close to equilibrium so it takes an arbitrarily long time to complete. A gas may have virtually uniform pressure during a non-quasistatic process - the gas in a car engine cylinder during the downstroke, for example, will have fairly uniform temperature and pressure, but it is not quasistatic.

    AM
     
  15. Feb 16, 2012 #14
    Okay but consider that rocket again. Just before the cork pops out and the rocket takes of there is an internal pressure much bigger than the external.
    How is this situation then fundamentally different from mine, where the atmosphere does a work on the piston?
     
  16. Feb 16, 2012 #15
    What I'm trying to say is that the rocket just as the cork flies out will have an internal pressure far bigger than the atmospheric. We could for instance say it was 2p, where p is the atmospheric.

    What is then the fundamental difference between the rocket and then situation I sketched - because as far as I understand it, you can treat the rocket propulsion as a quasistatic proces and not the compression of the gas in my cylinder.

    Of course there is a difference in the fact that in my example we have a compression whilst the rocket propulsion is an expansion but of course that wouldn't change anything..
     
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