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Imagine a gas (at room temp. and pressure) enclosed in a thermally insulating spherical container.

At some instant, the container instantly expands symmetrically (radially outward) to, say, twenty times its volume. The gas begins to expand freely (a spontaneous process??) until the first outlying molecules begin to reach the expanded container.

Now the container begins to expand at some rate that nearly matches the spontaneous expansion of the gas, but very gradually slows down, so that the gas expansion rate is slowly decelerated. The pressure on the container is infinitesimally small, but not zero, and that small but steady pressure ultimately brings the gas to a steady state at some fixed new volume.

Does this process approach a reversible process, as we consider the slower and slower deceleration rates?

Edit: Maybe this sub-question is partly the key to the answer: During spontaneous and symmetrical radial expansion, is there any viscosity effect that could slow down the gas? It seems to me that there would be no particular "layer" that would be moving faster or slower than its neighbours, so it's hard to see how viscosity/friction could operate.

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No it doesn't. The expansion of the gas is irreversible, so when then piston finally comes to a rest the entropy will be higher than it was before. There will be no way to restore the initial state without doing work. For example, if you try to push the piston back down again the gas will exert a non-infinitesimal force upon it.

If you want an example of a non-quasi-static but reversible process, why not simply consider an ideal swinging pendulum? It converts gravitational potential into kinetic energy and back again at a finite rate without any losses.

Or if you want an example that involves heat, how about a weight bouncing up and down at the end of an ideal elastic strip? In this case the force is an entropic one, meaning that heat is released as the weight moves downward and is absorbed again as it moves back up.

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  • $\begingroup$ Thanks Nathaniel, but look at the point I added as an "edit" just now. Isn't free symmetrical (spherical) expansion reversible? $\endgroup$
    – swamp thing
    Commented Jun 9, 2014 at 5:28
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    $\begingroup$ No, free expansion is a classic example of an irreversible process. The volume increases but the temperature remains constant, whereas in reversible (quasi-static) expansion the temperature decreases. The irreversibility is not due to friction, it's just due to an increase in the volume accessible to each particle. $\endgroup$
    – N. Virgo
    Commented Jun 9, 2014 at 5:29
  • $\begingroup$ Don't want to try your patience, but... 1. Is it not true to say that free expansion becomes irreversible only after the outward kinetic energy is somehow dissipated into heat? 2. Is there any viscosity/friction operating in the symmetrical expansion case? $\endgroup$
    – swamp thing
    Commented Jun 9, 2014 at 5:34
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    $\begingroup$ @user50166 it's no problem, don't worry! 1. that's quite a difficult and subtle question, but the usual way of looking it is that it's really the expansion itself that's irreversible. In free expansion the temperature remains constant, so there is actually no dissipation into heat as the gas hits the walls of its container and equilibrates. The subtle thing is that if you had a special machine that would reverse the motion of every particle then the gas would return to its initial state. But in practice no such machine is likely ever to exist, so we say expansion is irreversible... $\endgroup$
    – N. Virgo
    Commented Jun 9, 2014 at 6:30
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    $\begingroup$ ... The subtle thing is that for a photon gas, such a machine can be constructed, in the form of a curved mirror. So free expansion of a photon gas is considered reversible, but free expansion of a normal gas isn't, even though in many ways the processes are equivalent. But bear in mind that this is technically true for any irreversible process: if we could precisely reverse the molecules' motion, we could reverse it. (I say all this only for completeness. Your best bet is probably to ignore it most of the time; virtually all thermodynamics texts will just say the expansion is irreversible.) $\endgroup$
    – N. Virgo
    Commented Jun 9, 2014 at 6:32
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Thermodynamics has simple answers to offer regarding reversibility or otherwise of a given process. For a process to be reversible, it must be reversible at every point along the path ie, the system must be in equilibrium, both internally having well defined values for its properties such as temperature, pressure , internal energy etc., and externally with the surroundings as well.

In the given process the gas undergoes a complicated process of expansion. It does not satisfy the requirements of reversibility - the system being in equilibrium at every point along the path. Hence the process as a whole is deemed to be irreversible. Acceleration/decceleration being zero is no criterion for reversibility of the process. Slower and slower decceleration rates of expansion, too, do not ensure reversibility of a process. Hence, the process as a whole is irreversible.

Explanatory note:

It is important to note thermodynamics does not care for the rate(s) at which a process occurs. Therefore, all the elaboration regarding the spherical shape of the container and its instantanious radial expansion followed by a decceleration ending up with a slow steady rate of expansion is of no concern to decide whether the given process is spontaneous (irreversible) or not. Free adiabatic expansion of a gas (whether ideal or not ie whether viscous forces exist or not - assumption of ideal gas simplifies the arguments) is irreversible. Therefore, the answer to the given question is that the process is irreversible. Even the tail process, when the external pressure is infinitesimally small, unless the pressure of the gas is equally infinitesimally small, the process continues to be irreversible! Infact, we can have the expansion going on at a very rapid rate with gas pressure being equal to extenal pressure, the process is deemed to be reversible for an ideal gas!

Given the initial and final equilibrium states A,B of a system, thermodynamics tells us whether it is possible or impossible for the system to go from the A to B, on its own. One criteon used for arring at such a result is to see the change in the entropy of the universe. If it is positive, then we say the process A to B is spontaneous, if negative then the process is not spontaneous (does not occur on its own); if zero the process A to B is said to be reversible.

Radhakrishnamurty Padyala.

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Viscous dissipation of mechanical energy would be the dominant factor in characterizing the irreversibility. This would be readily captured by the tensorial form of the stress-(deformation rate) equation for a Newtonian (viscous) fluid. This describes the deformational behavior of a fluid in both simple shear flow as well as in much more complicated three dimensional deformations. Moreover, in addition to the ordinary shear viscosity, there is a second viscosity parameter that would come into play which describes purely volumetric dilatational effects. This is called the dilatational viscosity (although, for a monatomic gas, the dilatational viscosity is zero).

For more details on the general behavior of Newtonian (viscous) fluid, see Transport Phenomena by Bird, Stewart, and Lightfoot.

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