How do thermodynamic variables behave in reversible and irreversible processes?

Question

I read this definition:

reversible process is one which can be stopped at any stage and reversed so that the system and surroundings are exactly restored to their initial states. An irreversible process is one in which heat is transferred through a finite temperature.

But I just don't get it yet. How can we really differentiate between an irreversible and a reversible process with only measurements or calculations? How do U, H, T, Q and P behave? I only clearly understand that in a reversible process maximum work is done, we are taking more steps to change from one state to the other and entropy must stay constant, the order of the particles of the system can't be restored once it changes, if we had 1 mole it would be hard for $6.022 \times 10^{23}$ atoms to be in the same state as before.

But it is not as clear, for me, how the other variables are affected if we have a reversible or irreversible process.

Answer 1

The difference is this: In a reversible process, when you restore the system to its original state, the surroundings are also restored to their original state (note that the entire cyclic process, including the movement away from the initial state, and the restoration to the initial state, must be reversible). Hence all the state functions and state variables of both the system and surroundings are returned to their original values.

In an irreversible process, when you restore the system to its original state, the surroundings are not restored to their original state. Hence, while the state functions and state variables of the system are returned to their original values, not all the state functions and state variables of the surroundings return to their original values.

I.e., what distinguishes reversible and irreversible processes is not the final state of the system, but the final state of the surroundings. To put it another way, for the same change in state of the system, the difference between a reversible and irreversible process manifests itself through different effects upon the surroundings.

Focusing on the $2$nd law:

In a reversible process, when the system is returned to its original state, $\Delta \text{S}_{system}=0$ (because $\text{S}$ is a state function) and $\Delta \text{S}_{surroundings}=0$ (because the process is reversible). Hence $\Delta \text{S}_{total}=0$.

In an irreversible process, when the system is returned to its original state, $\Delta \text{S}_{system}=0$ (because $\text{S}$ is a state function) but $\Delta \text{S}_{surroundings}>0$. We know that $\Delta \text{S}_{surroundings}>0$ because the $2$nd law requires that, for an irreversible process, $\Delta \text{S}_{total} = \Delta \text{S}_{system}+\Delta \text{S}_{surroundings} > 0 $ .

Answer 2

In thermodynamics, we usually specify the demarcation between the system and the surroundings such that all irreversibility occurs within the confines of the system, while the surroundings are taken to be operated reversibly assuming ideal constant temperature reservoirs and mechanical equipment involving conservative forces (e.g., tiny pebbles added and removed from a piston at different elevations).

Irreversibility within a system is the result of finite rate transport processes including viscous dissipation of mechanical energy to internal energy via finite velocity gradients, heat conduction at finite temperature gradients, and mass diffusion at finite concentration gradients. Another contributor to irreversibility is chemical reaction proceeding at finite reaction rate. All these irreversibilities result in entropy generation within the system, and make it impossible to return both the system and surroundings to their initial states.

Examples of irreversible process involving these mechanisms include rapid expansion or compression of a gas within a cylinder, heat transfer between hot and cold bodies at significantly different temperatures, and mixing of different gases or of the same gas at significantly different concentrations.

Regarding U, H, T, and P, for a reversible process, these are uniform spatially throughout the system. For an irreversible process, if you are willing to accept the idea that, even though the system is not at equilibrium, the internal energy- and enthalpy per unit mass can be defined locally, then these vary with spatial position within the system, and integrate over differential mass to give the total U and H for the system. Similarly, T and P vary with spatial position within the system for an irreversible process.

Answer 3

The following is correct:

the order of the particles of the system can't be restored once it changes, if we had 1 mole it would be hard for 6.022x10^23 atoms to be in the same state as before.

But thermodynamics doesn't really worry about this. Instead, statistical thermodynamics provides a theoretical bridge between the macroscopic and microscopic pictures. In thermodynamics you are looking at a macroscopic system and you ignore the atomic structure of matter. It does not make sense to say that a macroscopic system is ever static when viewed from an atomic perspective, but in thermodynamics this is not a concern, because the macroscopic properties such as U, H, T, and P remain constant, thanks to the statistical behavior of the very large number of particles making up a thermodynamic system. We use the value of those macroscopic properties and the composition of the system to define its state. A transition to another state involves changes in the macroscopic properties and/or the composition. A return to the original state, whether reversible or irreversible, means a return to the original values of U, H, T, P, and S.

Other answers address what it means to have a reversible versus irreversible process in terms of the total change in the entropy of the universe. It is that total entropy change that is zero for a reversible process (not the entropy of the system).