Free expansion and adiabatic process

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In an adiabatic process, the change in entropy (ΔS) is zero because heat transfer (Q) is absent. However, in a free expansion, although both work (W) and heat (Q) are zero, the process is irreversible, leading to a non-zero change in entropy. The formula for ΔS is only applicable to reversible processes, which is why ΔS cannot be zero for free expansion. To accurately calculate the entropy change in free expansion, one must connect the initial and final states through a reversible path. This distinction clarifies the misunderstanding regarding the application of the entropy formula in different types of processes.
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I hope someone in here can help me out of this problem.
I know that for calculating the change in entropy of a system (\Delta S) we can use this formula :

\Delta S=\intδq/T
Well the problem is here, when we calculating ΔS for an adiabatic process we know Q= 0 so ΔS=0 in this process, but why we just can't let Q be equal to zero in a free expansion (while we know that in free expansion W=Q=0) and get to the same result (ΔS=0) for the free expansion process?
I should add that I know ΔS≠ 0 for free expansion process but, you know I just don't know where the problem is in my argument.
 
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The formula you wrote is only valid for a reversible process. And the free expansion is not a reversible process. To calculate the difference in entropy in this case, connect the initial and final state by any reversible path, for which generally you won't have dQ = dL = 0, and use your formula on this path.
 
Thanks, that's really a convincing answer.
 
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