# [Physics] Increase in entropy principle

entropyreversibilitythermodynamics

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?

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.