How can change in entropy be the same for all processes if the entropy production $\sigma$ is present for irreversible processes?

Question

From the definition of entropy change,

$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{int.rev}$$

From the closed system entropy balance, we have

$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{b}+\sigma $$ where $\sigma$ is the entropy produced within the system, vanishing to zero in the absence of irreversibilities. I don't quite understand how the entropy change between two states is the same for all processes. Is it the case that the entropy transfer in the case of internal irreversibilities present is lower than that of the entropy transfer in an internally reversible process between these same two states, and the entropy production makes up for this difference?

Answer 1

When a system changes from state 1 to state 2, the change in entropy of the system is the same for all processes, because entropy is a state function.

The difference between a reversible and an irreversible process is that, in the latter case, there is net entropy production for the universe, i.e., for the system+surroundings.

Since the entropy change for the system is independent of the process, the difference in entropy production between rev. and irrev. processes is seen in the entropy change of the surroundings.

More broadly, since path differences (which in turn means differences in heat flow and/or work flow) have no effect on the final state of the system, they instead always manifest themselves as differences in the final state of the surroundings.

Answer 2

The way that you describe it is exactly correct. This is the essence of the Clausius Inequality.