# Quasistatic and Reversible processes

I'm having trouble understanding the quasistatic process concept. I understand that for any process we have well defined initial and final states and the problem is in specifying the path, and the path is important as we need to know it in order to calculate work or heat because they vary depending on the path.

But I don't quite understand how implementing the process slowly can make the system in equilibrium at all points during the process, in other words, how does the system change but in the same time it remains in equilibrium? Also I read in my textbook that if a piston compresses a gas very fast, that results in a higher pressure region near the piston's surface and hence the pressure is not uniform in the gas.

I just find all of that confusing and I want to understand why the concept of quasistatic processes is important and what problem would there be in the theory of thermodynamics if we don't define processes this way? ( we cant apply integration if there is not a set of points or a path for the process?

And if we want a path we have to assume that at any point the pressure(or any property) is uniform and we can only assume that if the process is done extremely slowly so that its almost not happening and the system isn't changing!).

But I don't quite understand how implementing the process slowly can make the system in equilibrium at all points during the process, in other words, how does the system change but in the same time it remains in equilibrium?

The quasistatic process is not any more real than the ideal gas laws are. Apologies if you know this already, but to me, and I do sincerely apologise if I have misunderstood you, you seem to be asking a question that asks how it works in real life. It doesn't. It is an idealisation, so that we can take it that the gas, in theory, has time to continually equilibrate to the changing conditions. We also, for ease of calculation, ignore other aspects, such as the friction of the piston against the cylinder. This is work done when pushing on the piston, which we assume it is negligible.

As Jackl points out, integration of tiny little steps is involved.

Quasistatic ( I suppose you could say the clue is in the name), is usually good enough in practice, in that to avoid it, you would need to push the piston so fast that the gas fails to respond at the same rate. That is, the piston would exceed the speed of sound in the gas.

What problem would there be in the theory of thermodynamics if we don't define processes this way? (we can't apply integration if there is not a set of points or a path for the process?

As far as I know, if we did not make this reasonable approximation to real life, we would get caught up in having to account for factors that have no real bearing on the teaching of thermodynamics.

In other words, we would be obliged to, unlike the ideal gas law, obtain the particular function, for each case, that relates volume to pressure. In cases such as free expansion, tbere is no defined path or function and we can use the quasistatic process to replicate the initial and final states, and allow an estimate of work done, heat and entropy changes using a reversable processes

• Good, except for the last sentence: if the process is not quasi-static, the system is unlikely to have a well-defined pressure at all (due to pressure gradients within the system), which means there is no such function that relates volume to pressure. Dec 9, 2016 at 0:53
• @march Such as path unknown free expansion processes, so quasi static allows us to do a reversible calculation, with the same initial and final states?? Sorry, I am still at the stage of one notebook dedicated solely to keep the zillions of equations and their twists and turns in. I never studied TD before, it's much broader in scope than I had assumed. If I am wrong in my question in the first line above, that's OK ,please ignore it, I will keep slogging at it. Thank you.
– user108787
Dec 9, 2016 at 1:53
• @CountTo10, The point of march's statement is that during irreversible processes, there can be internal viscous gradient at job; so you can't define a unique pressure for the system. That's one of the reasons why reversible processes keep track of its path and the system can be returned along the same path since it doesn't lose the information of the trajectory. Irreversible processes due to the viscous gradient lose the information of the intermediate states, so it can't re-trace the same path even though the system can be returned to the initial state.
– user36790
Dec 9, 2016 at 4:00
• @ MAFIA36790, can you elaborate more on the "viscous" gradient? Dec 10, 2016 at 13:04

The quasistatic hypothesis is what make you use the equations, such as Gas Laws, in every point of your transition (in every point in space and in every moment in time), because you know that in every point that gas is in equilibrium. This is necessary especially when you have to integrate them and therefore you need that the equation you are integrating is valid in the whole interval of integration. This idea of making changements happen slowly implies that the equations are still true at least approximately, because changements are small.