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Reversible Work and Irreversibility

  1. Jan 9, 2016 #1
    I am trying to grasp the concepts of reversible work, irreversible work and irreversibility.(Last one is the difference between them if i am not mistaken.)

    Let us consider a rigid and evacuated container at volume V. Then, a valve opens and athmospheric air (P0, T0 is filling the tank. The wall of container is thin enough(allowing heat transfer), so eventually the trapped air in the container reaches thermal and mechanical equilibrium. That means air in the container also finally reaches to P0, T0.

    Now, is there any work? I know that gas expansion against vacuum doesn't create work, because there is nothing for you to do work on. Is that the case in here? If atmospheric air fills an evacuated container, it is also gas expansion against vacuum in my opinion. Therefore no work?

    But, there is entropy generation in this case. That means there is exergy destruction and hence, irreversibility. Irreversibility is the difference between reversible and irreversible work. But again, we said there is no work. Was that reversible work, or irreversible? If there is irreversibility can we say that there is no reversible work but some irreversible work.

    I didn't get this concept.
     
  2. jcsd
  3. Jan 9, 2016 #2

    QuantumQuest

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    Gold Member

    You have to keep in mind when the work is positive, regarding the system: when you supply work e.g. you fill some tank with some gas. Now, when system exchanges say heat with its surroundings, then the work is negative. Because there are three variables P, V and T you have also to keep in mind what is changed and what remains constant. In order to have a reversible process, rate of change must be slow. If it is not then the process is irreversible. A simple example usually taught is with a cylinder which is thermally insulated and a piston: when you supply work (i.e. positive work) in a slow manner, you finally reach the end of cylinder. If you release piston, then negative work is done by the system, which pushes the piston backwards to its starting point. Of course in general, processes in nature are irreversible, so entropy increases and a given system does not return to its initial state.
     
  4. Jan 9, 2016 #3
    This is not a very good example because you are not dealing with a closed system (if the container is your system). If the gas that eventually ends up in the tank is your system, then yes, irreversible work was done on this system by the surrounding air in forcing it through the valve.

    To address open systems, you need to study the open system version of the first law (which you may not have gotten to yet in your course).

    If you can provide a better example involving a truly closed system (say gas in a cylinder with a piston experiencing reversible or irreversible expansion), I will be glad to explain in detail about this.

    Chet
     
  5. Jan 9, 2016 #4
    Actually, this was the lecturer's example which confused me on the subject. Before that, I was confident about the work done by gas and irreversibility concepts.

    A rigid and evacuated container's valve is opened and atmospheric air filled it and became in equilibrium with the outside(thermally and mechanically)

    And he went on saying you should think about the reversible work and irreversibility on this case.

    In this case our system is container and it is an open system. I understand that the air rushes in, there is some heat transfer for balance. Entropy is generated and exergy is destroyed. The amount of exergy destroyed is the amount of irreversibility. But what about reversible work?

    I would write down the first law for open systems as Q + minhin = m2u2 I wouldn't include work but he said think about reversible work. So there must be some work, and I'm not sure on how I should include work in this mechanism or what he meant by reversible work.
     
  6. Jan 9, 2016 #5
    The only thing I can think of is that, after gas has entered the container, it gets compressed reversibly by the new gas that is entering the container. This is the opposite of what happens when gas in a container is initially under pressure and then you allow gas to escape through a valve. The gas that remains in the container expands essentially reversibly and adiabatically as it expels gas out of the container. So, whatever gas has remained in the container at any moment of time has experienced an adiabatic reversible expansion. Of course, the temperature and pressure of the gas that is being expelled from the container decreases as time progresses. In the case you are looking at, as the gas in the container is compressed adiabatically and reversibly by the new gas entering the container, its temperature and pressure increase as time progresses.
     
  7. Jan 9, 2016 #6
    Actually, for the valve, you would usually assume that Q = 0.
     
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