[sparkle123]

Hello,
I am confused about this: entropy is a state function so Delta S for a reversible change from state 1 to state 2 is the same as Delta S for the irreversible change from state 1 to state 2. Since Delta S of a reversible adiabatic process is zero, shouldn't Delta S for adiabatic free expansion also be zero, if it has the same initial and final states as the reversible adiabatic proces?
For instance, reversible adiabatic expansion from V1 to V2 has Delta S = 0.
Then according to my reasoning, adiabatic free expansion from V1 to V2 should have Delta S.
But NO, Delta S = nRln(V2/V1), says my textbook, and I understand my textbook's derivation of this formula, but my above reasoning seems to contradict it.
Please help me clarify this,
Thanks!

[Jorriss]

I believe there is a subtle difference. Zero heat is transferred, but the process is not reversible, so what is done is a reversible pathway that includes a heat transfer is devised.