Physics Asked on December 1, 2020
In this video lecture , at 39:37, The professor describes that if we do the process slowly then the system moves through a set of equilibrium states but why exactly does the speed of doing the process effect reversibility ?
For an analogy, take that video itself, if I play it at 1x or 2x speed by changing the Youtube settings, still the ultimate thing which plays out on my screen would be exactly the same.
You need to keep in mind what a reversible process is. A process is reversible if you can return the system, and the surroundings, to their original states. A system can be always be returned to its original state but only if the process is reversible will the surroundings also return to its original state. For an irreversible process entropy is generated in the system which needs to be transferred to the surroundings in the form of heat resulting in an increase in entropy of the surroundings. That means the surroundings is not returned to its original state.
For example, in the video the instructor talks about carrying out a constant external pressure process so slowly that there is no difference between the pressure in the gas and the external pressure throughout the process, i.e., the gas pressure is always in equilibrium with the external pressure. Moreover, the pressure of the gas is always internally uniform throughout. If you carry out the process too quickly, assuming a massless piston, the pressure of the gas only equals the external pressure at the interface. Internally, however, there will be pressure gradients. Those gradients result in internal motion of the gas and viscous friction heating. In order to return the system to its original state (including its original entropy), heat must be transferred to the surroundings. That heat transfer increases the entropy of the surroundings.
In the case of heat transfer between the system and the surroundings, in order for it to be reversible, the difference between the temperature of the system and surroundings needs to be infinitesimal. That means the heat transfer rate is also infinitesimal causing the process to be carried out very slowly. If the transfer occurs over a finite temperature difference the process is irreversible and entropy is generated. You know that heat only transfers spontaneously from a hotter object to a cooler object. Once the transfer occurs, heat will not spontaneously return from the cooler object to the hotter object. By making the temperature difference infinitesimally small, the process approaches being reversible.
Now for your video analogy. Consider the video as the system, and you the surroundings. True you can run the video forwards and backwards returning the video to its original image. But that doesn't mean the device that played the video has returned to its original state. When it plays it sends out energy in the form of electromagnetic waves that include the visible light you see (not to mention the heat and mechanical friction developed in the device.) That energy does not spontaneously return to the device. An external means is needed. What's more your state, as the surroundings, has also changed having absorbed the electromagnetic energy of the images viewed.
Hope this helps.
Correct answer by Bob D on December 1, 2020
Here is why.
If we are striving to make a process reversible by incrementally moving it through a series of equilibrium states, one thing we must avoid is the development of thermal gradients within the system because their presence signals a non-equilibrium state. We avoid developing thermal gradients by moving the system through those intermediate states so slowly that no thermal gradients can develop- and this takes more time than jamming the system through the changes at whatever rate we choose.
Answered by niels nielsen on December 1, 2020
Running a video faster or slower is fine, but it does not capture what is happening inside a system that is put through a process more rapidly or more slowly. So it may look the same but it is not.
Irreversibility in a rapidly carried out process is manifested in several ways, two of which are of interest here. In a process in which the gas is deformed rapidly, there are viscous stresses developed within the gas that cause some of the usable energy of the gas to be converted to internal energy. The slower the gas is deformed, the less this effect occurs. In addition, if there are large temperature gradients present due to rapid heat transfer, the effects of these large temperature gradients cannot be reversed either. In both cases, there is entropy generated within the gas during the irreversible process, so that either the system or the surroundings (or both) cannot be returned to their original states.
Answered by Chet Miller on December 1, 2020
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