# Free expansion and adiabatic process

1. Sep 7, 2010

### sita1

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 :

$$\Delta S$$=$$\int$$δq/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.

2. Sep 7, 2010

### Petr Mugver

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.

3. Sep 8, 2010

### sita1

Thanks, that's really a convincing answer.