Reversible vs Quasistatic

Question

I have some confusion about the definitions of a quasi-static and a reversible thermodynamic transformation.

As far as I understand, a quasi-static process is one that happens slowly enough for the system to remain in internal equilibrium at all times.

Example:

Consider a cylinder containing some gas in which we push a piston slowly to compress the gas. If the compression is sufficiently slow, the gas molecules have time to react and reach an equilibrium state inside the cylinder, so at any given moment the pressure and temperature of the gas are well defined. That is, the force per unit area is homogeneous across the walls of the container, and the average kinetic energy of any subsystem of gas particles is the same.

Non-example:

If on the other hand we pull up the piston more quickly than the fastest molecules in the gas, there will be a momentary vacuum separating the gas and the piston, which lasts until the gas will rush up to fill the entire volume of the whole space. During this adjustment, the pressure on the lower part of the cylinder is higher than the pressure on the piston (which is zero until the gas molecules reach it), so the pressure isn't well defined.

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Concerning reversibility, I have never seen a rigorous definition. Let me try to formalize what I have understood as best as possible. Let's call a path a function mapping a point in time to a point on the pV diagram: $p(t)=(V(t),P(t))$, defined on some interval of time $t \in [0,t_1]$.

We say a path $p$ connecting a state $A(V_0,P_0)$ to a state $B(V_1,P_1)$ is reversible if $p$ can be traversed in reverse, i.e if the path $p(t_1-t)$ for $t \in [0,t_1]$ is physically possible.

Intuitively, if we take a video of some process transforming a gas from state $A$ to state $B$ and rewind it, then it is possible to do an experiment which brings the gas from state $B$ to state $A$ which will be indistinguishable from the rewinded video.

Question 1: Is this a good definition of reversibility?

Question 2: Furthermore, I don't understand why Wikipedia says "Reversibility refers to performing a reaction continuously at equilibrium" - is this not the definition of quasistatic? Or do they mean a reaction continuously at equilibrium with its surroundings?

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Clear example of an irreversible process

The Joule-Guy-Lussac experiment, in which a barrier enclosing some gas in a subcompartment of a container is suddenly removed, allowing the gas to fill the whole volume of the container. Assuming the walls of the container and the barrier are adiabatic, this is not reversible because the internal energy of the system remains the same throughout the process, but work would be required to compress the gas back to its initial volume. Alternatively, if we took the gas in the state where it has filled the whole volume, and suddenly replaced the barrier, the gas would of course not go back its original state.

Also note that the above example is not quasi-static, since the barrier is removed suddenly.

Answer 1

In order for a process to be regarded as reversible, at a minimum, the system upon which the process is imposed must pass through a continuous sequence of thermodynamic equilibrium states. This is the simplest way of judging whether a process is reversible or not. But why would such a process be considered reversible? Well, passing through a continuous sequence of thermodynamic equilibrium states is just the minimum requirement for the process to be reversible. Such a process is referred to by Moran et al (Fundamentals of Engineering Thermodynamics) as "internally reversible."

But, for total reversibility, the surroundings must also pass through a corresponding/matching set of thermodynamic equilibrium states. If this condition is satisfied, then it is possible to return both the system and its surroundings to their original thermodynamic equilibrium states, without more than negligibly affecting the state of anything else. This is the really stringent requirement for total reversibility.

Note that it is fully possible for a system to pass through a continuous sequence of thermodynamic equilibrium states, without its surroundings undergoing a corresponding/matching set of thermodynamic equilibrium states. Such a process would be considered (internally) reversible in terms of the system, but not for its surroundings, and the overall process would not be considered totally reversible. An example of this would be if you manually caused a gas to adiabatically expand or to contract quasi-statically. The process would be considered (internally) reversible for the gas, but not for your body. Your body (which represents the surroundings) experiences many irreversible interconversions of energy in its muscles which prevent it from passing through a continuous sequence of thermodynamic equilibrium states. So the system itself could be returned to its original thermodynamic equilibrium state, but not your body. However, there are other ways of structuring the surroundings for this example such that the surroundings also experiences a corresponding/matching sequence of thermodynamic equilibrium states.

So, in summary, a good definition of a reversible process for a system (neglecting what is happening in the surroundings) is that the system passes through a continuous sequence of thermodynamic equilibrium states (internally reversible process).

Answer 2

Pippard (in The Elements of Classical Thermodynamics) defines a reversible process as one which may be exactly reversed by an infinitesimal change in the external conditions.

A reversible process has to be quasi-static. A non-quasistatic process, such as your Gay-Lussac experiment (aka Joule expansion) is clearly irreversible – what infinitesimal change in external conditions could possibly take the gas back to its original volume?

But I think that reversibility is a more general requirement than being quasi static. Examples of irreversible changes are heat flowing down a finite temperature gradient, electric charge flowing through a resistor, an object sliding on rough ground, stirring a liquid. It may be that these can all be shown to be non-quasistatic as well as irreversible, but (for me at least) their irreversibility is much clearer to see.

Answer 3

A quasi-static process is in equilibrium at all times, tracing a well-defined trajectory in thermodynamic phase space. If you can follow that trajectory in the opposite direction, the process is called reversible. It is not enough to find another trajectory that brings you back to the initial state!

A necessary condition for reversibility is that the process does not produce entropy (eg it must happen without friction).

Answer 4

Firstly, irreversibility is equivalent to increasing entropy, reversibility is equivalent to constant entropy. Eg an idealised pendulum is a reversible system. An idealised air spring would also do, such as a frictionless piston between two thermally isolated volumes of air in a cylinder, which could oscillate to and fro indefinitely, alternately compressing (and raising the temperature) of the two volumes.

By contrast a quasistatic process might be an air spring that starts with one side compressed and both temperatures the same, and where the piston slowly moves to equalise the pressure while temperatures are kept equal. This increases entropy. The equilibrium here is thermal alone: the pressure is only equalised at the end.

Answer 5

I've always had your same wondering. My own explanation is below.

Quasistatic is clear to everybody: it means that the process dynamics is so slow that the system undergoing the process is always very close to some equilibrium state. Accordingly, the system can move from one point in the diagram P-V (or whatever other diagram) to some "next" infinitesimally close point.

Reversible, in my own idea, is related to the fact that the number of "ways" the system in its microscopical view can take to pass/move from an equilibrium state to the "next" one. If there is just one way for the system to do this step the process is reversible since the microscopical path (or the mapping as you said in your question) is one-to-one; if there are several ways for the system to perform the step the process is not reversible anymore since the microscopical path is no longer biunivocal. In conclusion, to explain reversibility we need to leave the macroscopic view of the classical thermodynamics and switch to consider the system in microscopic terms.

It's not a chance, in my opinion, that adiabatic irreversibile processes are typically associated to entropy changes. Boltzmann definition of entropy is related to the mumber of microscopic states (ways) a system can set up for a given macroscopic state.

Hope I was able to clarify my view (which can be completely wrong, btw)