# What ds>dQ/T mean?

I read the derivation on page 216 over here.

First, it considers an irreversible process between states 1 and 2, followed by a reversible process between states 2 and 1. From my interpretation, equation 8.31 means that the entropy change of a reversible process is greater than that of an irreversible one between the two states. From equation 8.32, did they generalise and change the RHS to a reversible integral?

(We know that $$\delta S = \frac{\delta Q}{T}$$ in the case of equality, hence the inequality must be the case of an irreversible process. Is that how to interpret it?)

Then I read in my class notes that the inequality can be removed by adding an entropy generation term: $$\delta S = \frac{\delta Q}{T} + c$$ where c is the entropy generation term.

For an irreversible process, $$c$$ is positive. Where does entropy generated go, the system or surroundings?

• There is no objective identification which part of the system is the system and which part is the environment. It's the total entropy that grows more than the bound. This extra growth may take place both in the system and in the environment.If the environment has blue and red ink which gets mixed to a purple ink, the entropy of the environment goes up. If the same inks appear within the system, the entropy of the system goes up, even without any heat transfer. Commented May 1, 2016 at 8:09
• Existing answer is great, I just add a note: The first and second laws of thermodynamics are both expressed by equalities by giving due recognition to the irreversible production of entropy. See section 4 of "Thermodynamics in Terms of Gradient Vectors" av8n.com/physics/thermo-forms.htm#sec-extirpate Commented Mar 21, 2023 at 0:45

There is something they forgot to mention in your notes (either from ignorance, or out of omission). The temperature within the system is spatially non-uniform during an irreversible process. So what value of the temperature are you supposed to use in the integral of dq/T? The Clausius inequality calls for the use of the temperature at the boundary interface with the surroundings (where the actual heat transfer is occurring). This requirement is rarely emphasized in textbooks or online sources. So the Clausius inequality should really read: $$\Delta S\geq \left(\int_1^2\frac{\delta q}{T}\right)_{interface}\tag{1}$$The right hand side of this equation is sometimes interpreted as the entropy entering the system from the surroundings. That, combined with the entropy generated within the system by irreveribilities within the system gives rise to the total entropy change of the system.

Eqn. 1 applies to any physical system you desire to use it for.

This is Clausius inequality. The term ds>dQ/T. Holds an impossible reaction which does not obey Clausius statement in 2nd law of thermodynamics. The things in your class thought is about this ds<0. Where you can add c that is entropy generation term. Hope it will help you

I think you have a typo in your question. A reversible process will have a smaller entropy change than an irreversible process. Your interpretation that the equality refers to a reversible process, while the inequality refers to in irreversible process is correct. Looking at the specific equations in that notes document, the integral in 8.31 applies to an irreversible process, while the integral in equation 8.32 is more general, and could apply to either a reversible or irreversible process (reversible only when both sides are equal).

Sometimes it's a bit ambiguous where the distinction between system and surroundings is. A car engine for example is taking in air, and releasing exhaust. The surrounding air could be considered as part of the system. In realistic (irreversible) systems that undergoe a thermodynamic cycle, the extra entropy manifests as heat released to the environment. If this weren't the case, and the system was isolated from the surroundings, then entropy would increase in the system to a maximum, and the system would halt.