I'm struggling to understand this concept of reversible process as a transformation from a initial state i ( generic ) to a final state f, but done through small changes/steps. What does it mean that the system is always in equilibrium with itself if the changes are very small? Wouldn't it be the same for "big" changes?
If you apply a sudden big change to a gas, the ideal gas law (or other equation of state) no longer describes the relationship between pressure, volume, and temperature globally. For one thing, the pressure and temperature are no longer spatially uniform within the gas (as pressure waves and temperature waves are traveling through the gas). So what value do you use for the pressure in calculating the work? Secondly, there are viscous forces present within the gas (which are not significant when the gas is deforming slowly), such that the forces acting on the boundary of the gas (where work is being done) depend not only on the volume, but also on the rate of change of volume.
It is still possible to describe the behavior of the gas, the forces at the boundary, and the work done if one solves the partial differential equations for heat and momentum transfer (Navier Stokes equation and differential thermal energy balance) and one assumes that, locally, the conditions satisfy the equation of state. But this is typically beyond what thermodynamics alone can handle, and requires the use of computational fluid dynamics (CFD). All these complexities go away however if the deformation of the gas is slow (i.e., involves small changes).
Thermodynamics applies to a system that is in equilibrium (there is a separate subject of non-equilibrium thermodynamics). That is quantities like pressure, entropy, free energy etc are only well defined for a system that is in equilibrium.
The problem is that if a system is at equilibrium then it isn't changing because by definition equilibrium means the system is in a steady state i.e. not changing with time.
So we have a problem because:
if a system is in equilibrium it can't change
if it isn't in equilibrium then our usual thermodynamic laws don't apply.
There is a contradiction here, so how can we analyse systems that change with time?
The solution is that provided the change to the system is slow the system will never be very far from equilibrium. Suppose we change the volume by a very small amount. This will shift the system out of equilibrium, but because the change is slow the system will come back into equilibrium very quickly. So quickly, in fact, that to a very good approximation we can treat the system as always having been in equilibrium.
And this is what we mean when we talk about a reversible process being made up of a series of infinitesimal steps. We imagine some change, e.g. increasing the volume, as being made up of a sum of tiny changes, and each of those tiny changes takes place slowly enough that the system quickly equilibrates again. So our system remains very close to equilibrium all the way through the change.
Thermodynamics of equilibrium is simpler to deal with than its non-equilibrium counterpart, so considering quasi-static processes is useful, since that allows us to use the thermodynamics of equilibrium also when (slow) change takes place.
If changes are fast, even basic concepts can fail to apply. For instance, when the valve between a container with gas and an evacuated one is open, pressure is not defined during the sudden process of expansion. That's a strong statement: it means that it doesn't make sense to attribute a pressure to the system during this process. As a consequence, a line connecting the initial ($i$) and final ($f$) states in, e.g., a PV plot cannot be drawn, because the system didn't go through states of this space between $i$ and $f$, which obviously prevents you from doing many of the usual calculations.
So, when possible, it's often helpful to approximate a given transformation by a quasi-static one, which is done by considering it took place at small steps.
protected by Qmechanic♦ Dec 20 '17 at 12:43
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