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I am reading Levine's Physical Chemistry, I know how to arrive to that equation, but I have questions.

First, $\mathrm{d}S=\mathrm{d}Q/T$ for material equilibrium, but if there is a heat flow I understand that there is a reversible process going on, what kind of process can it be? Only a reversible thermal or mechanical process or also a reversible material change?

Second, the book says

$$\mathrm{d}S > \frac{\mathrm{d}Q_\mathrm{irrev}}{T} \qquad\text{and}\qquad \mathrm{d}S = \frac{\mathrm{d}Q_\mathrm{rev}}{T}.$$

Then states the equation $\mathrm{d}S ≥ \mathrm{d}Q/T$ and says $\mathrm{d}S$ equals $\mathrm{d}Q/T$ for a reversible process and $\mathrm{d}S$ is greater than $\mathrm{d}Q/T$ for an irreversible chemical reaction or phase change.

I imagine that the heat flow is different if the process is reversible or irreversible and that the equation $\mathrm{d}S≥\mathrm{d}Q/T$ takes into account that the $\mathrm{d}Q$ is different, is this correct? I think I am a little confused just because there are no more subscripts in $\mathrm{d}S≥\mathrm{d}Q/T$.

Also, I think that the heat flow due to a irreversible process is greater than the heat flow due to a reversible process that goes to the same states, why is that?

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  • $\begingroup$ Here is a tutorial to help you understand the change in entropy for a system experiencing an irreversible process and application of the Clausius inequality: physicsforums.com/insights/grandpa-chets-entropy-recipe $\endgroup$ – Chet Miller Jan 15 at 12:23
  • $\begingroup$ In the inequality, the T is not the system temperature (which can vary from location to location within the system for an irreversible process). The T here is the temperature at the interface between the system and the surroundings where the heat transfer is occurring. $\endgroup$ – Chet Miller Jan 15 at 12:25

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