# Fundamental question about thermodynamics

1. Sep 28, 2014

### Sandalwood

Hopefully this is the right place to post this question. This is a very fundamental question on the applicability and limitations of classical equilibrium thermodynamics (CET). I've been learning non-equilibrium thermodynamics (NET), and a few sources mentioned that NET is needed because CET is only limited to reversible processes. In the press release for the 1977 Nobel Prize in Chemistry (went to Prigogine for his work on NET), it's mentioned that CET "cannot be used for the study of irreversible processes but only for reversible processes and transitions between different states of equilibrium." This is something I've also seen mentioned in various sets of lecture notes that I found online. So this leads me to my question. Is it a meaningless question to ask whether a process is reversible or irreversible in the context of CET? Because it seems like one of the assumptions of applying CET is that it only applies to reversible processes, so whenever you apply CET, everything must be a reversible process, even when you have friction and whatnot (whether or not this gives accurate results is a separate issue).

2. Sep 28, 2014

### Staff: Mentor

Even for irreversible processes, you can frequently determine the changes in thermodynamic properties and functions between an initial equilibrium state of the system and a final equilibrium state using CET. But you can not use CET to describe all the states in-between the initial and final equilibrium states. If the process is reversible, CET can also describe all the intermediate states.

If you want to describe all the intermediate states for an irreversible process, you need to model all the transient details that are occurring within the system, and can often accurately do this successfully assuming local thermodynamic equilibrium as a function of location within the system. But you also need to be able to describe the rates of transport processes like momentum transfer (viscous stresses), heat transfer (conduction, convection, radiation), and mass transfer (diffusion) in the system.

Hope this helps.

Chet

3. Sep 29, 2014

### Sandalwood

I read in Reiss' book Methods of Thermodynamics that when you're using a PV diagram (for example), you're assuming that every point on that plane is an equilibrium point. So if you can't use CET to describe states between the initial and final states, then I don't think you can use CET at all. At the very least, I'm not sure it'll give a very accurate answer.

Going back to my original question, if we use the second law to determine whether a process is reversible or irreversible, isn't that kind of a pointless thing to do? Because the impression I get is that CET wasn't designed to handle irreversible processes in the first place.

4. Sep 29, 2014

### Staff: Mentor

This is not correct. For example, if you impose a constant external force per unit area Pext on the piston and allow the system to equilibrate either isothermally (contact with constant temperature reservoir) or adiabatically (insulated), you can determine the final equilibrium state, and thus the changes in all the thermodynamic properties. And you don't need CET to do this.
That's only if you accept your first premise.

Here is a brief tutorial on all this that originally was the contents of my Physics Forums blog, until all the blogs were removed in early September. This should help you better understand what's going on:

FIRST LAW OF THERMODYNAMICS

Suppose that we have a closed system that at initial time ti is in an initial equilibrium state, with internal energy Ui, and at a later time tf, it is in a new equilibrium state with internal energy Uf. The transition from the initial equilibrium state to the final equilibrium state is brought about by imposing a time-dependent heat flow across the interface between the system and the surroundings, and a time-dependent rate of doing work at the interface between the system and the surroundings. Let $\dot{q}(t)$ represent the rate of heat addition across the interface between the system and the surroundings at time t, and let $\dot{w}(t)$ represent the rate at which the system does work on the surroundings at the interface at time t. According to the first law (basically conservation of energy),
$$\Delta U=U_f-U_i=\int_{t_i}^{t_f}{(\dot{q}(t)-\dot{w}(t))dt}=Q-W$$
where Q is the total amount of heat added and W is the total amount of work done by the system on the surroundings at the interface.

The time variation of $\dot{q}(t)$ and $\dot{w}(t)$ between the initial and final states uniquely characterizes the so-called process path. There are an infinite number of possible process paths that can take the system from the initial to the final equilibrium state. The only constraint is that Q-W must be the same for all of them.

If a process path is irreversible, then the temperature and pressure within the system are inhomogeneous (i.e., non-uniform, varying with spatial position), and one cannot define a unique pressure or temperature for the system (except at the initial and the final equilibrium state). However, the pressure and temperature at the interface can be measured and controlled using the surroundings to impose the temperature and pressure boundary conditions that we desire. Thus, TI(t) and PI(t) can be used to impose the process path that we desire. Alternately, and even more fundamentally, we can directly control, by well established methods, the rate of heat flow and the rate of doing work at the interface $\dot{q}(t)$ and $\dot{w}(t)$).

Both for reversible and irreversible process paths, the rate at which the system does work on the surroundings is given by:
$$\dot{w}(t)=P_I(t)\dot{V}(t)$$
where $\dot{V}(t)$ is the rate of change of system volume at time t. However, if the process path is reversible, the pressure P within the system is uniform, and

$P_I(t)=P(t)$ (reversible process path)

Therefore, $\dot{w}(t)=P(t)\dot{V}(t)$ (reversible process path)

Another feature of reversible process paths is that they are carried out very slowly, so that $\dot{q}(t)$ and $\dot{w}(t)$ are both very close to zero over then entire process path. However, the amount of time between the initial equilibrium state and the final equilibrium state (tf-ti) becomes exceedingly large. In this way, Q-W remains constant and finite.

SECOND LAW OF THERMODYNAMICS

In the previous section, we focused on the infinite number of process paths that are capable of taking a closed thermodynamic system from an initial equilibrium state to a final equilibrium state. Each of these process paths is uniquely determined by specifying the heat transfer rate $\dot{q}(t)$ and the rate of doing work $\dot{w}(t)$ as functions of time at the interface between the system and the surroundings. We noted that the cumulative amount of heat transfer and the cumulative amount of work done over an entire process path are given by the two integrals:
$$Q=\int_{t_i}^{t_f}{\dot{q}(t)dt}$$
$$W=\int_{t_i}^{t_f}{\dot{w}(t)dt}$$
In the present section, we will be introducing a third integral of this type (involving the heat transfer rate $\dot{q}(t)$) to provide a basis for establishing a precise mathematical statement of the Second Law of Thermodynamics.

The discovery of the Second Law came about in the 19th century, and involved contributions by many brilliant scientists. There have been many statements of the Second Law over the years, couched in complicated language and multi-word sentences, typically involving heat reservoirs, Carnot engines, and the like. These statements have been a source of unending confusion for students of thermodynamics for over a hundred years. What has been sorely needed is a precise mathematical definition of the Second Law that avoids all the complicated rhetoric. The sad part about all this is that such a precise definition has existed all along. The definition was formulated by Clausius back in the 1800's.

Clausius wondered what would happen if he evaluated the following integral over each of the possible process paths between the initial and final equilibrium states of a closed system:
$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}$$
where TI(t) is the temperature at the interface with the surroundings at time t. He carried out extensive calculations on many systems undergoing a variety of both reversible and irreversible paths and discovered something astonishing. He found that, for any closed system, the values calculated for the integral over all the possible reversible and irreversible paths (between the initial and final equilibrium states) was not arbitrary; instead, there was a unique upper bound (maximum) to the value of the integral. Clausius also found that this result was consistent with all the "word definitions" of the Second Law.

Clearly, if there was an upper bound for this integral, this upper bound had to depend only on the two equilibrium states, and not on the path between them. It must therefore be regarded as a point function of state. Clausius named this point function Entropy.

But how could the value of this point function be determined without evaluating the integral over every possible process path between the initial and final equilibrium states to find the maximum? Clausius made another discovery. He determined that, out of the infinite number of possible process paths, there existed a well-defined subset, each member of which gave the same maximum value for the integral. This subset consisted of what we call today the reversible process paths. So, to determine the change in entropy between two equilibrium states, one must first conceive of a reversible path between the states and then evaluate the integral. Any other process path will give a value for the integral lower than the entropy change.

So, mathematically, we can now state the Second Law as follows:

$$I=\int_{t_i}^{t_f}{\frac{\dot{q}(t)}{T_I(t)}dt}\leq \Delta S=\int_{t_i}^{t_f} {\frac{\dot{q}_{rev}(t)}{T(t)}dt}$$
where $\dot{q}_{rev}(t)$ is the heat transfer rate for any of the reversible paths between the initial and final equilibrium states, and T(t) is the system temperature at time t (which, for a reversible path, is equal to the temperature at the interface with the surroundings). This constitutes a precise mathematical statement of the Second Law of Thermodynamics.

Chet

5. Sep 29, 2014

### Sandalwood

Thanks for the info, Chet. I guess I have a lot of studying to do if I really want to understand thermo!