# 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. Commented 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
Commented 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
Commented Dec 9, 2016 at 4:00
• @ MAFIA36790, can you elaborate more on the "viscous" gradient? Commented 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.

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?

Thermodynamics (despite the name) is the study of systems in equilibrium. The set of all possible equilibrium states is called the state space. If you imagine a high-dimensional space, with a dimension for each thermodynamic variable (e.g., pressure, temperature, volume, etc), then the state space is a surface in that space determined by the constraints imposed by equation(s) of state. The ideal gas is a good example: consider an ideal gas with fixed particle number. We have a three-dimensional space with axes for pressure, temperature, and volume. But the equilibrium states of an ideal gas are only those values of $$(p, V, T)$$ that satisfy $$pV = N k T$$. The state space is therefore a two-dimensional surface. Almost all points in $$(p, V, T)$$ space do not represent an ideal gas. We could add another thermodynamic variable as a fourth dimension, e.g., the internal energy, $$U$$. But there is another equation of state: $$U = \frac{3}{2} N kT$$ (for a monatomic gas), so the state space remains two-dimensional.

A quasistatic process is a transition from one equibrium state to another equilibrium state where the system remains in equilibrium at all times. Therefore, a quasistatic process is represented as a curve on the state space, i.e., a continuous sequence of equilibrium states. This is obviously an idealization: such a transition would have to occur infinitely slowly. But, it is meant to represent the infinitesimal limit of making a very small change to the system, allowing it to return to equilibrium, and then repeating.

On the state space, we can apply the tools of differential geometry to calculate changes in the thermodynamic variables along a given curve (just as one does, e.g, on the curved spacetime of Einstein's general relativity). The basic tool for performing such analysis is the thermodynamic identity, which, in its simplest form, is: $$dU = T dS - p dV$$ This is a relationship between covectors (or, differential forms) that is valid at every point on the state space. While the equation(s) of state determine what the state space is, the thermodynamic identity provides the fundamental relationship between thermodynamic variables on the state space. (Operationally, we can think of covectors as objects that can be integrated over a path).

All of this analysis is performed using quasistatic transitions/processes on a state space.

So, you might ask, why do we need to call them "quasistatic transitions"? Can't we just call them thermodynamic transitions, since they seem to be all there is? Well, for physical systems they are definitively not the only kind of transitions. As mentioned above, they are an idealization that is impossible to realize in practice. Most realistic thermodynamic transitions, from one equilibrium state to another equilibrium state, occur completely out of equilibrium, with only the initial and final states represented on the state space. Examples include: rapid expansion/compression of a gas, mixing of two gases, expansion of a gas into a vacuum, heating a pot of water on a stove (even slowly: the heating sets up a temperature gradient such that the water is always out of equilibrium). Unless we want to restrict the theory to only idealized, impossible processes, we must at least recognize that much of thermodynamics occurs completely "off of" the state space.

The preceding paragraph does not mean that quasistatic processes are not useful, however. Because each equilibrium state is represented on the state space, we could take a non-quasistatic transition (represented by just the distinct initial and final points on the state space) and consider a quasistatic transition (curve) that connects these two points. For any calculations involving changes in state variables (functions that depend only on the position on the state space, e.g., internal energy, entropy, and all thermodynamic variables) the result will be the same for the quasistatic and non-quasistatic process. A typical example is calculating the change in entropy when a hot cup of water and a cold cup of water are allowed to come to equilibrium with each other. The system is never in equilibrium while their temperatures are changing. But, because entropy is a state variable, we can use any quasistatic process that would connect the initial and final states, e.g., separately, and slowly, changing the temperatures of each cup using hot plates (to that final equilibrium temperature required by conservation of energy). In that way, the entropy change of each cup is can be calculated using the thermodynamic identity: $$\Delta S = \int dU/T = \int c m dT/T$$.

So, in summary: (1) quasistatic processes and the thermodynamic identity allow for calculations involving the integration over a path on the state space; (2) often these calculations can be used as a proxy for a process that is not quasistatic, but has the same initial and final states; (3) there would be no "problem ... in the theory of thermodynamics" if we consider other transitions, it's just that there are not formal tools for making calculations for them.