Increase in entropy principle

If we consider a system to undergo an irreversible process from state 1 to state 2 and a reversible process from state 2 to state 1, then through Clausius inequality

$\int_{1}^{2} \frac{\delta Q_{irrev}}{T} + \int_{2}^{1} \frac{\delta Q_{rev}}{T} \leq 0$

$\int_{1}^{2} \frac{\delta Q_{irrev}}{T} + S_1 - S_2 \leq 0$

$S_2 - S_1 \geq \int_{1}^{2} \frac{\delta Q_{irrev}}{T}$

$\Delta S \geq \int_{1}^{2} \frac{\delta Q_{irrev}}{T}$

Does this mean that the entropy change for a reversible process is greater than that of an irreversible process? I'm convinced I am wrong because my notes say otherwise but isn't Δs the entropy change of a reversible process and ($\int_{1}^{2} \frac{\delta Q_{irrev}}{T}$) the entropy change of an irreversible process?

• "(1to2∫dQirrev/T) the entropy change of an irreversible process" this is wrong. Apr 30, 2016 at 9:19
• If we change (1to2∫dQirrev/T) to (1to2∫dQ/T)arbitrary then we have two cases: ds=(1to2∫dQ/T)arbitrary=(1to2∫dQrev/T) for the reversible case. Thus ds>(1to2∫dQ/T)arbitrary must be the irreversible case and we set it to ds = (1to2∫dQrev/T) + c to combine both into an equation. Is this correct? Apr 30, 2016 at 10:34

Entropy is a property of the system. The change in entropy of a system as it traverses from an initial state(1) to a final state(2) is independent of the path by which the system is taken from state 1 to state 2. The path can be a reversible one, or even irreversible, the change in entropy is always the same as long as the initial and final states are the same. However, in order to calculate the change in entropy $S_{2}-S_{1}$, one has to connect a reversible path between the two states because $\displaystyle S_{2}-S_{1}=\int_{1}^{2}\frac{dQ_{rev}}{T}$, where $dQ_{rev}$ is an infinitesimal amount of heat transferred to the system in a reversible manner at the system temperature T. NOTE: $\displaystyle\int_{1}^{2}\frac{dQ_{irrev}}{T}$ is not the correct formula for calculating the change in entropy. It is always $\displaystyle\int_{1}^{2}\frac{dQ_{rev}}{T}$ irrespective of whether the path is reversible or irreversible.
For an irreversible process, the temperature of the system is typically non-uniform, and when we write $\int{\frac{dQ}{T}}$ for the Clausius inequality, what we really mean is $$\int_1^2{\frac{dQ}{T_I}}$$where $T_I$ is the temperature at the boundary interface (with the surroundings) at which heat transfer is occurring. The $T_I$ is almost always omitted from statements of the Clausius inequality in textbooks, to the endless confusion of students. So the first equation should read:$$\int_1^2{\frac{dQ}{T_I}}+\int_2^1{\frac{dQ_{rev}}{T}}\geq 0$$where, for the reversible path, the interface temperature is equal to the (uniform) system temperature. This leads to:$$\Delta S \geq \int_1^2{\frac{dQ}{T_I}}$$for the reversible path.