The differential of the entropy dS is generally defined as $dS = \frac{{d'Q}}{T} + {d_i}S$, where 1st and 2nd terms are the contributions from the heat transfer to the system and the disorder created in the system by the spontaneity (I actually don't clearly understand what this means). The textbook said the latter term ${d_i}S $ is zero (positive) for a reversible (irreversible) process. It looks that the entropy differential is defined for both the reversible and irreversible process.

My question is this: how is "defining the state variable" possible for the irreversible process? As far as I know, the meaning of the state variables such as S (entropy), T (temperature), P (pressure), etc. are not clearly defined if the state is not equilibrium. So, I hardly accept the concept of dS in the irreversible process. Could you help to clarify my confusion?

And in the derivation of the fundamental equation of the thermodynamics $TS = U - YX - \mathop \sum \limits_{j = 1}^v {\mu _j}{N_j}$, the starting point is that $S = S\left( {U,X,{N_i}} \right)$. Why is S a function of like this? Here, U: internal energy, X: generalized displacement (ex: volume V), ${{N_i}}$: a number of particles in i-th species.


Your question is very perceptive. The equation requires you to accept the idea that entropy can be conceived of even for a system that is not at thermodynamic equilibrium. In the equation you have written, the T in the denominator of the first term is not the system temperature, because the temperature in a system experiencing an irreversible change does not have to be uniform. This temperature is the temperature at the boundary of the system (with the surroundings) where the heat transfer d'Q is occurring. Therefore, this term represents transport of entropy across the boundary. The second term in the equation represents entropy generation within the system, typically produced by viscous dissipation, dissipation of temperature gradients due to conduction, and dissipation of concentration gradients due to diffusion. For a reversible path between the same initial and final thermodynamic equilibrium states of the system, the second term is negligible, and, in the first term, the system temperature is uniform (and matches the boundary temperature), and, here again, it represents transport across the system boundary. However, since the second term is zero for a reversible process, all the entropy change in the reversible process occurs by transport across the boundary, and the first term is greater for the reversible process than for the irreversible process. This is basically the Clausius inequality.

  • 1
    $\begingroup$ +1 To add to the answer, assigning temperature to points on system's boundary, when the system is not in thermal equilibrium, requires that we make the additional assumption of "local thermodynamic equilibrium" (inscc.utah.edu/~tgarrett/5140/Notes/LTE_and_Kirchoffs.pdf). $\endgroup$ – Deep Apr 1 '17 at 3:43

1st and 2nd terms are the contributions from the heat transfer to the system and the disorder created in the system by the spontaneity

The second term captures entropy produced within the system. Examples would be free expansion of a gas (a process that is however not quasistatic, which means the state variables aren't well-defined at all times and thus no infinitesimal thermodynamic description is possible) or a real-world piston that isn't frictionless.

For a process to be reversible, we need a way to undo any changes in entropy. That's possible if they are due to heat transfer, [heat transfer due to a finite temperature differential is of course irreversible!] but in case of free expansion, you cannot put the genie (microscopic system trajectories) back into the bottle (the original smaller phase space volume available for system evolution), or in the second example, have the piston absorb the heat it produced through friction.

My question is this: how is "defining the state variable" possible for the irreversible process?

For state variables to be well-defined, the process needs to be quasistatic, ie slow enough that the system is (approximately) in equilibrium at all times. But as seen above, a quasistatic process that has well-defined entropy at all times can still be irreversible.

Why is S a function of like this?

In principle, $S$ is a function on thermodynamic phase-space that can be described using any parametrization you like. The traditional choice of using a certain set of extensive variables is a convention that is motivated on physical grounds eg by entropy being extremal at equilibrium when its natural variables remain constant; similar reasoning applies to other thermodynamic potentials.

For comparison, think about Hamiltonian mechanics: If you're so inclined, you can use a canonical transformation to have position and momentum exchange their roles, or even do a non-canonical transformation if you promote the minus in Hamilton's equations to a tensor (the symplectic product). You could do similar things in thermodynamics, we just generally don't because we like the traditional parametrization.

  • $\begingroup$ if the internally generated entropy is known then there is no need for a quasistatic and/or reversible process to calculate the entropy. For example, the Joule heat (rate) is completely irreversible, $\dot q = I^2R$ and internally it generates $\dot \sigma = \frac{I^2R}{T}$ entropy per unit time. If this is added to the external heat exchange the process entropy can be calculated. $\endgroup$ – hyportnex Mar 31 '17 at 11:57
  • $\begingroup$ @hyportnex: what I was getting at was that if the process is not quasistatic, some state variable (such as $T$ occurring in your infinitesimal formula) may become ill-defined (or at the very least need to be promoted to fields) $\endgroup$ – Christoph Mar 31 '17 at 13:43
  • $\begingroup$ @Christoph: Thanks for giving me a detailed comment, But I still don't clearly understand one thing: a quasistatic process that is irreversible? I mean my textbook said that the reversible process is the process which is always infinitesimally close to the equilibrium and it can be always reversed without changing the thermodynamic state of the universe. But you mentioned that "a quasistatic process that has well-defined entropy at all times can still be irreversible". Is there any difference between "infinitesimally close to the equilibrium" and "quasistatic"? $\endgroup$ – Donggyu Jang Apr 1 '17 at 3:25
  • $\begingroup$ quasistatic implies the ability to plot the process as a trajectory in thermodynamic phase space, and irreversible implies that you can follow the trajectory in the opposite direction; as an illustration, consider an isolated system of two chambers with a separator, one of them filled with gas; now, let the gas expand into both chambers by (a) 'instantaneously' removing the separator (b) puncturing the separator so the expansion happens (ideally) molecule by molecule (c) slowly moving the separator without friction (d) slowly moving the separator with friction $\endgroup$ – Christoph Apr 1 '17 at 10:36
  • $\begingroup$ process (a) is neither quasistatic nor reversible, process (b) is not reversible, but each subsystem changes its state quasistatically (whereas the combined system does not as it is not in global equilibrium), process (c) is quasistatic and reversible, and process (d) is quasistatic, but not reversible $\endgroup$ – Christoph Apr 1 '17 at 10:40

First of all, a reversible process is not necessarily necessarily a stationary one. All it means is that there exists another possible transformation that can bring the new state back to the initial state. With that in mind, your reasoning would mean we would be unable to define state variables during any process.

To quickly understand why we can still define the these "state variables" consider a particle moving in one dimension. The state variables are x,p (position and momentum), however is the particle is not stationary we can still talk about it's value. We write it as a function of time, (define it for "snapshots" in time). Similar logic is valid for the thermodynamic state variables.

What I believe your textbook is trying to tell you about irreversible vs. reversible processes is that for a reversible process we add the minimum amount of entropy possible ($d_i S = 0$). This then allows another minimal process to bring the state back to the initial one. An irreversible process ($d_i S > 0$) will add more then the minimum, thus any process intended to reverse it would not be able to get rid of that extra amount added. This means the system can never return to it's initial state afterwards (hence the name irreversible).

To answer your second question, we will go back again to a single particle moving in one-dimension. Notice I could have equivalently fully described it's state by it's position and velocity. The are different sets of variables that can fully describe your state. The entropy of the system depends on the systems state, $U,X,N_i$ is just one of the possible sets that can be used to describe the state. (It just works best for what we are trying to do in this circumstance so we choose it). You can think of it kinda like the basis of plane, sometimes polar coordinates are easier, other times cartesian are.

  • $\begingroup$ Hello. Could you tell me more about "a reversible process is not necessarily a stationary one"? Yes, any process is a change of the state, so the state can not be completely stationary when the process is being done. However, as far as I know (I've learned from the textbook), the reversible process is the process in which the system is very close to the equilibrium all the time, so I think the state variables including S can be clearly defined. However, in the irreversible process, the system deviates from the equlibrium, so how we can define these? how do they get physical meanings in $\endgroup$ – Donggyu Jang Apr 1 '17 at 3:49
  • $\begingroup$ a non-equilibrium state? Hm...may be the entropy S is an exceptional one which can be defined even for the irreversible process? Even if it was true, I would hardly accept this concept if there is no consistancy between other state variables and S. I think I need a little more explanation. $\endgroup$ – Donggyu Jang Apr 1 '17 at 3:53

Please note the state variable is different from the stationary variable. The state variable means the parameter that defines the state of a system. For example, let's say Temperature. During any process, it may get changed over time; it need not to be constant. Even if the temperature is transient during the process, yet it is a state variable that indicates the hotness of system. For further clarity, we may perceive it as the temperature at an instant.

The temperature i.e. hotness does not depend on how it achieved i.e. irrespective of whether reversible or irreversible process. However, the value of temperature may be different for different processes.

Similar logic is applicable for the entropy, etc.

Additionally, it is to clarify that the reversible process is a fake term. Any process is actually irreversible. However where the extent of irreversibility is very small, we consider as reversible. That means it is a limiting case.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.