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Rapid adiabatic compression - puzzle

  1. Nov 23, 2007 #1
    I wanna know if in rapid adiabatic compression of a gas, is work done by gas on surrounding (atmosphere) numerically same as work done by surrounding on the gas.
    Please tell if there exists some solid proof of the answers you guys are giving.
  2. jcsd
  3. Nov 24, 2007 #2


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    Didn't quite understand the question; work is being performed on the gas by the atmosphere?
  4. Nov 24, 2007 #3
    The actual question was:
    Consider ‘n’ moles of an ideal monatomic gas placed in a vertical cylinder. The top of the cylinder is closed by a piston of mass M and cross section A. Initially the piston is fixed, and the gas has volume Vo and temperature To. Next, the piston is released, and after several oscillations comes to a stop. Disregarding friction and the heat capacity of the piston and cylinder, find the temperature and volume of the gas at equilibrium. The system is thermally isolated, and the pressure outside the cylinder is P.

    I am not interested in the solution to the question..... my problem is that here in this situation since the changes taking place inside the cylinder are very rapid... so we cannot calculate the work done by the gas inside the cylinder onto the piston.. but we can definitely calculate the work done by Mg and PA on the gas inside the cylinder.
    So can we conclude that the work done by Mg and PA is numerically equal to work done by gas inside the cylinder.
  5. Nov 25, 2007 #4


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    Hi island,
    Why do you think it's not possible to calculate the work done on or by the gas? Yes, it's rapid, but it sounds like you're theorizing that some physical law will not be applicable because the rate of change is large. What would the rate of change have to do with the applicability of thermodynamic laws?
  6. Nov 25, 2007 #5

    Andrew Mason

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    The cylinder does necessarily oscillate for a while - but it is dampened oscillation. The energy of the compression wave in the cylinder gas ends up as heat in the gas. So you don't have to worry about the work done by the gas on the piston during the oscillations. You only have to determine the net work done by the piston on the gas. This is just a matter of applying the ideal gas law (and the first law of thermodynamics).

  7. Nov 25, 2007 #6

    Andrew Mason

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    You can figure it out yourself. W = PdV. For the work done on the atmosphere, you use the pressure of the atmosphere x volume change. For work done by the cylinder gas, use the pressure of the cylinder x volume change.

    In a reversible process, the difference in pressure (inside - outside) is infinitessimal and the piston moves infinitessimally slowly. So the work done by the gas in the cylinder is equal to the work done on the atmosphere. If the process is irreversible, there is a pressure difference or a net force on the piston and this results in the piston acquiring kinetic energy. So the work done by the gas in the cylinder is greater (by the kinetic energy of the piston) than the work done on the atmosphere.

  8. Nov 26, 2007 #7
    but here finally the KE is zero, thus can we say that in this case also:
    the work done by the gas in the cylinder = the work done by the atmosphere.
  9. Nov 26, 2007 #8


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    Yes....so I'm not seeing the problem either. If instead of the atmosphere, you just put a big weight on top of the piston, the resulting scenario is the same.

    Could you explain why you think the word "rapid" makes a difference here? I'm guessing you think it has something to do with kinetic energy, but as you and others pointed out, the initial KE is zero and the final KE is zero and you aren't given any information about viscous damping (if a weight is placed on the piston, there need not even be damping), so there is no calculable effect.

    If the question you are asking is about the effect of that damping, then it still depends on things like insulation, which will determine if the heat eventually goes into the cylinder (keeping it closer to adiabatic) or into the atmosphere (making it eventually isothermal).
    Last edited: Nov 26, 2007
  10. Nov 27, 2007 #9
    yes you caught me on this point. I was a bit worried about the word rapid. And i agree 100% with your answer. But some of my fellow mates were not agreeing to me. Thx for your answer. ..
  11. Dec 1, 2007 #10

    Andrew Mason

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    There some important subtlties here.

    If you expand the gas rapidly more energy goes into the kinetic energy of the gas and piston (kinetic energy is proportional to v^2). That kinetic energy is reduced if the mass of the piston is increased. But it does not disappear. The piston's kinetic energy allows the gas to expand past the point at which its pressure is equal to the outside pressure (when the pressures are equal, the piston is still moving). The piston stops and starts going back the other way because the outside pressure is now greater than the inside gas pressure. It gains kinetic energy in the other direction. Repeat compressions and expansions (oscillations) occur like this. In each oscillation expansion and compression, the gas heats up. Eventually all the kinetic energy is converted to thermal energy of the gas (assuming no friction).

    If you stop the piston system at its initial maximum extension (no kinetic energy), the change in internal energy of the gas is equal to -the work done by the gas ([itex]\Delta U = - P\Delta V[/itex] - first law). The cylinder is thermally insulated so no heat has flowed out of the gas. The result is that more work has been done in the rapid non-reversible expansion than would have been done in the reversible expansion.

    If the process is adiabatic, it means that heat does not flow into or out of the cylinder. But energy can flow into or out of the gas and that energy can be converted to or draw out thermal energy from the gas. This is why I have maintained that we really should not talk about an adiabatic non-reversible process (but I appear to be in a minority on that).

  12. Mar 31, 2008 #11
    For what its worth a 4 stroke engine at 10,000 rpm is is undergoing some pretty rapid adiabatic compression and expansion cylcles. It is the rapidity of the cycle that allows the cycle to be aproximated as adiabatic, because the there is little time for heat to be lost to the walls and cooling system of the engine during the compression and power strokes.
  13. Apr 1, 2008 #12
    Say we had an insulated cylinder, partitioned by a diaphragm with gas under pressure at one end and a vacuum contained by a piston at the other. If the diaphragm is ruptured the gas rapidly expands to fill the complete cylinder. Say its volume has doubled and its pressure has halved. (No change in temperature.) The work done on the surroundings of the cylinder is zero. Now we compress the gas back to its original volume using the piston. The temperature of the gas will rise so it will be necessary to extract heat via a water jacket and radiator to return the gas to its original volume, pressure and temperature. The surroundings will have changed because of the heat absorbed by the water jacket, so the total system (cylinder plus surroundings) is not exactly the same as the one we started with. This is an example of an adiabatic process that is non-reversible. Technically, if I recall correctly, it is called an isenthalpic process (adiabatic + no work exchanged with the surroundings).
  14. Apr 1, 2008 #13
    ya there exists solid proof a ques of this type came in jee and the key idea to the solution was the following:-

    work by gas=work by the weigth+work by external pressure

    if there was a velocity and mass then

    work by gas=work by weigth+work by external pressure+change in K.E
  15. Apr 1, 2008 #14
    hey zarbanx... can you quote the year in which it was asked in Jee pls... this question came in stony brooks university exam...
  16. Apr 3, 2008 #15
    2002 sry for replying late i was busy prep for my xams
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