Entropy change of an ideal gas that undergoes an irreversible adiabatic (free) expansion?

Question

anonymous · Answer

I think I got the first question, but I have no idea how to answer the second, and how it differs from the first.

anonymous · Answer

I suspect that the entropy change is the same in both cases.

The starting point is dU=TdS-pdV in both cases.

In the isothermal case you can say that dU=0 because for an ideal gas U is a function of T only, and T doesn't change (isothermal expansion)
Thus dS =pdV/T etc

In the second case I think dU=0 again, can you explain why ?