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A Big debate: following process is reversible or not

  1. Jul 10, 2016 #1
    There had been a big time debate in my college between chemistry faculties and physics faculties on whether the following process is reversible or not.

    consider that there is some gas (assuming ideal gas) in a chamber with friction less piston of 5kg and 2 kg mass on top of it as shown in the figure. lets say the system is in equilibrium ie pressure of the gas balances 7 kg weight.
    We now remove 2kg block and due to which system gas will expand.
    Will the expansion of gas reversible or irreversible.


    According to physics faculties, this is reversible (assuming ideal gas and friction less piston) since the situation will be just like a spring block system. when the gas will expand then the work done on 5kg piston will be (p(t)-p(atm))*A - mg (here m is mass of piston ie 5kg) and this will be the kinetic energy gained by the 5kg piston. Due to this kinetic energy the piston will further expand to reach at the level where its kinetic energy is zero. The piston will now come down upto the same level from where it started.

    Any opinion is highly appreciated.

    Attached Files:

  2. jcsd
  3. Jul 10, 2016 #2


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    Staff: Mentor

    It looks to me like the scenario has been defined as reversible. Why would it not be?
  4. Jul 10, 2016 #3
    The chemistry faculty was correct and the physics faculty was incorrect. This is definitely a typical example of an irreversible process. I have helped Physics Forums members analyze this basic problem many times in the past. The kinetic energy of the piston does not just vanish into thin air. It is dissipated by viscous stresses occurring within the gas. If it were not for the gas viscosity, the piston would oscillate up and down forever like a mass on a spring. The viscous dissipation of the kinetic energy causes the gas temperature and volume to be higher than they would have been if the gas were made to expand reversibly. We could have done the gas expansion reversibly by, rather than removing the 2 kg mass suddenly, replacing the 2 kg mass with a pile of tiny little masses that totaled 2 kg. We would then very gradually remove the tiny little masses. This would prevent significant viscous dissipation from occurring.

    If you would like me to help you analyze this problem both ways, I would be glad to do so. Or, if you would like me to analyze it myself, I can do that instead. But, of course, it would be better if you gained the experience yourself.

  5. Jul 10, 2016 #4
    For an ideal gas, should we consider loss due viscous forces.
  6. Jul 10, 2016 #5
    Thanks a lot chestermiller for helping
  7. Jul 10, 2016 #6
    Even in the limit of ideal gas behavior, the viscosity is not zero, and approaches a limiting value at low pressures (temperature dependent). (See Bird et al, Transport Phenomena). The piston could not stop oscillating, and the gas could not stop experiencing time- and space-dependent variations of temperature, pressure, and density if the viscosity were zero.
  8. Jul 10, 2016 #7


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    Staff: Mentor

    What am I missing here: is this just that under a non-ideal gas assumption the damping converts the kinetic energy to heat? Typically, reversible isothermal expansion requires the expansion to be done slowly, so there is no kinetic energy in the piston itself. Maybe the problem statement is just muddled?
  9. Jul 10, 2016 #8
    No. Even in the ideal gas limit, a gas still exhibits viscous behavior to bring about the conversion. If the gas were non-ideal, that would not result in damping unless viscous dissipation were also present.
    The problem statement implies (to me) that the mass is removed suddenly so the expansion is allowed to proceed rapidly and spontaneously. As I suggested in my other response, in order for the process to be reversible, the mass would have to be removed gradually (so that the piston is always essentially in equilibrium and the gas pressure is uniform and changing slowly).

    You mentioned isothermal, but the same basic principles apply to adiabatic. Also, for an irreversible process, the term isothermal means only that the boundary of the gas is held at a constant temperature; there can still be substantial spatial (and temporal) variations in temperature within the gas.

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