# Entropy in irreversible adiabtic process

We know that, $$dS=\dfrac{\delta Q_{rev}}{T}$$ If you have an irreversible adiabatic process between two thermodynamic equilibrium end states of a system, there exists no possible reversible adiabatic process between these same two end states. So to get the entropy change for the irreversible adiabatic process, you need to devise an alternative reversible path between the same two end states, and this reversible path will not be adiabatic.

But , we know that between two themodynamic states there exists many reversible paths becauseof which the value of $$δQ_{rev}$$ can vary beacuse of heat being a path function which will then lead to different values of entropy for the same irreversible adiabatic process

Where am i going wrong

• Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
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Commented Jun 3 at 8:36
• If two equilibrium states can be connected by one reversible path then all such reversible paths connecting the same two states have the same entropy change. It is like work in a "potential" field: the work per charge depends on the start and end points and is independent of the connecting path. In fact, the "points" representing the equilibrium states constitute a potential field irrespective how you get to reach one, equilibrium is equilibrium. Commented Jun 3 at 8:43
• The heat flow during different reversible processes will be different, but so will the temperatures as a function of "position" along the paths, and these two changes (in $\delta Q$ and $T$) "cancel" to make the entropy change the same along any two reversible paths joining the same two equilibrium states. In mathematical terms, $1/T$ is known as an integrating factor that turns the inexact $\delta Q$ into an exact differential. Commented Jun 3 at 16:17

But , we know that between two themodynamic states there exists many reversible paths becauseof which the value of $$δQ_{rev}$$ can vary beacuse of heat being a path function which will then lead to different values of entropy for the same irreversible adiabatic process
$$\Delta S=\int_1^2\frac{\delta Q_{rev}}{T}$$