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Homework Help: Quasistatic processes are reversible?

  1. May 28, 2005 #1
    It's probably a silly question.
    I know definitions of quasistatic and reversible process.
    How can we proove that each quasistatic process is reversible? Before it went for me without saying. But now I noticed that I can't proove it.
    In books that I have this conclusion is made after both definitions are given, like it is evident. But how You would proove it? Thank You.
  2. jcsd
  3. May 28, 2005 #2


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    Alright, exactly what are the definitions of "quasistatic" and "reversible"?
    Post them here.
  4. May 28, 2005 #3
    In thermodynamics, a reversible process (or reversible cycle if the process is cyclic) is a process that can be "reversed" by means of infinitesimal changes in some property of the system (Sears and Salinger, 1986) (http://en.wikipedia.org/wiki/Reversible_process).

    In my books is one more reqiurement. After reverse process the environment must be like it did before direct process.

    In thermodynamics a quasistatic process is a process that happens infinitely slowly. In practice, such processes can be approximated by performing them "very slowly" (http://en.wikipedia.org/wiki/Quasistatic_processes).
    (just want to add that each state between initial and final is also equilibrium)

    In some cases, it is important to distinguish between reversible and quasistatic processes. Reversible processes are always quasistatic, but the converse is not always true (http://en.wikipedia.org/wiki/Reversible_process).

    OK. I found that my statement wasn't correct. Not each quasistatic process is reversible (examples are realy evident). But "Reversible processes are always quasistatic". How it may be proven?
    Last edited: May 28, 2005
  5. May 28, 2005 #4
    Hm. Seems like I started to think with my own head :). In general case elastic deformation isn't quasistatic process. But it's reversible!? I'm confused.
  6. May 28, 2005 #5


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    A reversible process must be an infinitesimal slowly process. The idea hidden behind this statement is the mechanism of energy dissipation. Usual rapid processes harness with internal dissipation. This effect causes heat outflow and destruction of work into heat, and so an increasing of universe entropy. An idealized reversible process must be such a slowly process in which there is no chance for internal dissipation (Internal dissipation is proportional to the square of flow velocity). Of course, this kind of processes are impossible in practical industry, but they are a great tool for our calculations.

    I would say that a reversible process must be quasistatic and without friction between mobile parts.

    An idealized elastic behavior must be quasiestatic too. If you deform some material rapidly and leave it free again to return its original state, it could happen two things:

    -The material has reached its original state (for tension=0, deformation=0). Some part of the work employed has been transformed into heat due to internal molecular friction. This heat has been exhausted to the surroundings and so has increased universe entropy.

    -There would be possible to have an hysteresis behavior, in such a way that it has not recovered entirely its original state, but there is some residual deformation. This hysteretic behavior is linked to dissipation effects and fluency, increasing also the universe entropy.

    Remind that every rapid process enhances internal dissipation.
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