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# Quasistatic and reversible processes

Using theoretically quasi-static and/or reversible processes to stay pretty much at equilibrium. Created by Sal Khan.

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• What is the difference between reversible process and quasi static process according to sal in this video ? •  quasistatic the process is so slow, that the system is always in (or very close to) an equilibrium state
reversible the process goes the other way as soon as you reverse the applied conditions.
Quasistatic and reversible are not the same thing. For example you can mix two gases very slowly (quasistatic) but you can't reverse that easily. The reason is that the entropy between the starting point and the endpoint has increased and that's a main characteristic of an irreversible process.
• You say it is in equilibrium the whole time. That means an infinitesimally small particle is removed each and every time. So basically, the system remains where it was before even after removing all the particles one-by-one. But that doesn't happen when you remove all of them together. I don't get this.
(1 vote) • Think of this as "baby steps". If, instead of using infinitesimally small particles, let's use sand. After removing a single grain, the system would become chaotic, but only very mildly so. After a very short amount of time it would come back to equilibrium. We could take our measurements, and move on to the next grain. Reducing the size of those grains reduces the time it takes to reach equilibrium, such that an infinitesimally small particle will take an infinitesimally short amount of time.
You might need to wrap your head around equilibrium in general to better understand this. The essential point is that the system becomes non-uniform temporarily. The temperature and pressure is not consistent all the way through. Theoretically the average temperature and pressure of all points within the volume will always fall along the line he plotted, but there is no way to measure the pressure at every point simultaneously. It is therefore necessary to wait for the system to come to equilibrium before gathering any meaningful information from it.
• Are microstates also known as INTENSIVE properties ? • So, this is probably a definition issue, but why does a reversible process have to go through infinitesimal changes like a quasi-static process? Doesn't a process which does not lose energy to friction and heat dissipation and which is also able to fully reverse between two states already qualify as a reversible process? Why should it matter if I take one grain of sand away or a big amount of it if by doing the opposite I end up perfectly in the state I started in for both cases?

Is it because the definition of a reversible process is that we need to have 'well-defined' system states to be measured and that the path from initial to final state, like that of a quasi-static process, must be shown? I'm wondering because making a big change, although you will experience a much more dynamic change (oscillations of states etc), you will still end up with equilibrium with time. So say you go from state 1 to state 2 by taking away 1 kg of sand grain by grain which takes a lot of time. Compare this with taing away 1 kg of sand instantaneously and waiting for equilibrium to occur. Both approaches will get from the same state 1 to the same state 2. Now reverse the process by doing the opposite and both will get back to the same initial state. Can the second approach of taking away a large instead of an infinitesimal amount still be called a 'reversible process'?

**UPDATE
Just realized that Sal explained what I wrote in my two comments to this question at the very end of the video (from ). • My take on it is that if you jumped from state A to B quickly (like by removing the large rock off the piston) you would lose a lot of energy to friction, air turbulence, heat, vibration, etc. There is probably also inefficiency in the inner gas having to quickly rearrange itself to equalize pressure and temperature. So I don't know if you can say "non-quasi-static" AND "no heat or energy loss" in this example. I don't think state 2 will be the same if you do this quickly and slowly, which was your assumption.
• Would it be possible for a completely static process, similar to the ones shown in the video? For example, instead of removing a grain of sand at a time, what if you removed an atom or molecule at each time. Would there still be an extremely small period of the system being undefined or is it possible through removing an amount so small that there would be no time when the system is undefined or completely negligible? • Mathematically , if only infinitesimal quantities of pressure is reduced at a time, then the process is actually quasistatic (in the limit the quantity approaches zero). But in reality,, if any non-zero finite amount be removed at a time from "heap of molecules" exerting pressure, it would, (although for a super-smalll interval, be in inequilibrium.
• Why should a reversible reaction be quasi static? Using the rock analogy provided by Sal, if we replace the half rock we evaporated, it will oscillate again and reach the original state. Also, what has loss of energy go to do with the reaction being non-reversible? Does the air inside lose the energy? And even if it does, how does that change the PV graph? • Great question! Let's pretend you have: A + heat <--> B
So if you start with A and heat (energy) you can make B. And now that you have B, you can transform it to make A with heat. What happens now if the heat dissipates (you remove the heat)? Now you do not have one of the ingredients on the left side (heat) and so you cannot reform B. Therefore, the reaction does not become reversible anymore:
A + heat <--- B
• Couldn't one measure the pressure and volume at every instant while the piston is oscillating? One could record the values for every instant and plot those numbers. Why go through all the trouble of reducing pressure grain by grain? • watch the part of video from to and you will get some idea. What we want to do here is to show how we really get from state A to state B on the graph if we kept temperature constant. If we don't do this grain by grain, we wreak havoc on the system undergoing process. Every time you do the process, you will end up with a different pathway from state A to state B due to well... tons of possible macrostates thanks to microstates. If this is not enough, then let me state this bluntly: If you measure pressure and volume at every instant while piston is oscillating, then record and plot their graph, you will get one of the many possible graphs for the same process, even if you do everything exactly the same as before . Going through the trouble of reducing pressure grain by grain tremendously reduces the oscillations, and thus show how Pressure is truely related to volume if temperature is constant(inverse proportion, and well if you know how the graph looks like, you will surely understand why we don't want oscillations to ruin this graph)
• ,if we knew exactly how energy got dissipated in friction,how the energy got transferred in other atoms,and we knew all the other variables,couldn't we at least theoretically reverse the process?sorry for my energy and thanks in advance.
(1 vote) • At , Sal says that there is no loss of energy (in a frictionless world). But how is this possible? The gas in the container applies a force in the direction of the displacement of the piston (up), thus it must be doing positive work. Shouldn't the gas then have less energy?  