# Entropy change in reversible and irreversible processes

$$dS = dq/T$$ for a reversible process and $$dS \gt dq/T$$ for an irreversible process, but why is the entropy change greater in an irreversible process? What difference between the two processes in molecular level is responsible for this change?

Also the statement:

Gibbs free energy is the maximum amount of non expansion work which a system can do reversibly

makes no sense to me.

• Entropy is basically a measure of the information you know about the system. If you lose some information about the system, the entropy of the system increases. A reversible process, unlik an irreversible process, happens so slowly that you can precisely track the motion of all the particles and have complete knowledge about their positions, momentums, etc. Since you don't lose any information the entropy change is zero. Commented Jun 28, 2019 at 4:47
• @thunderbolt How does your comment account for that fact that, between the same two thermodynamic equilibrium end states, the entropy change for an irreversible process is exactly the same as that for a reversible process? And how does your comment address the part of the OP's question about dq/T? Commented Jun 28, 2019 at 16:51

# Background

The form $$d...$$ is a differential. It means an infinitesimal change in something as it undergoes a process. The something of interest here is a thermodynamic state function of a system, its surroundings, or the universe. The change in the universe is the sum of the changes in the system and its surroundings, so only two of the three are independent.

Any change in any thermodynamic state function is always independent of the path taken. Two types of paths are defined: reversible and irreversible. The former is defined as a path where the system and its surroundings are in exact thermodynamic equilibrium at all points during the process. The latter is anything else.

# Entropy

One definition of entropy change is $$dS = \delta q_{rev}/T$$, where $$\delta q$$ is the heat flow that occurs during the process and $$T$$ is the temperature of either the system OR the surroundings at the instant of the infinitesimal change. We include the subscript $$_{rev}$$ on heat flow to acknowledge that, while heat flow is path dependent, entropy change is not. We choose a reversible path because it is typically easier to evaluate than an irreversible path. Writing the expression without the $$_{rev}$$ subscript on heat flow is common but ambiguous.

For a reversible process, the entropy change of the system and the surroundings are equal and opposite at any step. We need only calculate one or the other. The sum is zero. Hence the statement that a reversible process results in $$\Delta S_{universe} = 0$$ always. This means, we only need to know one or the other, not both.

We can and will always evaluate the (infinitesimal) entropy change of a system $$dS_{sys}$$ by choosing a reversible process. How then do we determine entropy change during an irreversible process?

The immediate response back is to clarify that we must also define the exact status of the universe when we propose that a system will undergo an irreversible process. Why? Because the exact nature of the path that is irreversible depends on the exact nature of the differences between the system and the surroundings. By simple analogy, the closer the system and the surroundings are to being at exact thermodynamic equilibrium during the proposed process, the smaller will be the irreversibility of the process that occurs.

The magnitude of the irreversible entropy change for any process depends on the magnitude of the difference between the system and the surroundings from exact thermodynamic equilibrium.

We cannot say that the value (magnitude) of $$dS_{sys}$$ indicates irreversibility by itself. Its value is always evaluated along a reversible path because entropy is a state function. We also cannot say that the value (magnitude) of $$dS_{surr}$$ indicates irreversibility by itself. Otherwise, we could just invert our definitions of system and surroundings and find ourselves in a paradox. We can only determine whether a process is or is not (was or was not) irreversible when we consider $$dS_{univ}$$. In that case, we find $$dS_{univ} > 0$$ indicates a (spontaneous) irreversible process is occurring (has occurred) along the path.

So, if we always determine $$dS_{sys} = \delta q_{rev,sys}/T_{sys}$$, how do we make sense of the statement that "... $$dS > \delta q/T$$ indicates an irreversible process?"

We do so by recognizing that this type of statement is a gross simplification. The true statement is this combination:

A reversible process is one where $$dS_{univ} = 0$$, while $$dS_{univ} > 0$$ indicates a (spontaneous) irreversible process.

In a reversible process, the entropy change of the system and surroundings are equal and opposite. In an irreversible process, we generate extra entropy. We can assign that "extra" irreversible entropy either to the system or to the surroundings. In truth, the irreversible entropy generated belongs to the process, not to a system or the surroundings.

So, where does that irreversible entropy appear at the molecular level? It appears in the fact that the system and the surroundings are at distinctly different statistical configuration levels during the process. By example, imagine a classic example of an ideal gas that fills half of an insulated chamber while the other half is a vacuum. We break an invisible wall between the two half chambers. The resulting process is irreversible. The entropy change of the system can be calculated along a reversible path. The entropy change of the surroundings is zero because no heat flows into or out of the surroundings during the process. The net change of the universe (system + surroundings) is positive. At the molecular level, we anticipated that breaking the wall would result in an irreversible process because the initial system is an ideal gas going into a "surroundings" of a vacuum. The two locations (ideal gas and vacuum) have distinctly different configurational entropy values.

Gradients in the configurational entropy states within a system can be a source of irreversible processes.

Further discussion of entropy is found at these related postings on this forum.

Trying to understand entropy as a novice in thermodynamics

Variation of entropy with volume under constant pressure

Entropy generation of constant volume heat addition process

# Gibb's Energy

As it ends with "... makes no sense to me", your last question shows a need to read further in a textbook. These links may also offer insights.

Does positive total entropy generation means that the entropy generation of the system and the entropy generation of the surroundings are positive too

Why is choosing a suitable thermodynamic potential important?

• This answer was very clear and also very correct! Why does this not have more upvotes. Commented Dec 26, 2020 at 9:55

The detailed development starts out at the molecular scale by demonstrating that the rate of transfer of heat and the rate of transfer of momentum in a typical non-polymeric (small molecule) liquid or a gas experiencing significant temperature- and velocity gradients (as in an irreversible process) can be appropriately quantified using molecular dynamic and statistical thermodynamics considerations in terms two continuum (i.e., slightly larger scale) properties, thermal conductivity k and viscosity $$\eta$$. Bird et al, Transport Phenomena, show specifically how these transport properties can be quantified based on the interactions between molecules (i.e., 6-12 potential). In the case of 1-D heat transfer, for example, the rate of heat flow per unit area (heat flux, q) can be expressed as: $$q=-k\frac{dT}{dx}$$ where the value of k is measured or derived from molecular dynamics and statistical thermodynamics consideration. This is the differential from of the heat conduction equation we all learned in freshman physics. An analogous, although more complicated, equation can be written for a Newtonian (viscous) fluid in terms of viscosity and velocity gradients.

I'm going to stop here before presenting more to see if this initial part of the development is along the lines that you were seeking. I will continue if you feel that it is.