[clinz63]

Let's say a have a piston with adiabatic walls which contains a gas. The piston has two stops. If I release the first stop against vacuum and it stops at the second stop, that means the gas has expanded against vacuum so work is zero. Heat is also zero since the walls are adiabatic. However, the state of the system changed. There is a different volume, pressure and temperature. Since internal energy is also a state function, it means it has also changed. So the summary is:
q = 0, w = 0, ΔU≠0
How?
Does this mean there some value for w since it is impossible for q. Irreversible work? Free expansion work is not equal to zero? Or is it that p, T and V changed in some way that U is the same?

[Vital Y]

Ideal gases obey Boyle's law so, (PV)_intial = (PV)_final this holds because temperature of the system hasn't changed. Since PV,T are the same, the state of the system hasn't changed. Hence ΔU = q + w = 0

[clinz63]

I didn't hold T constant.

[mjc123]

Why should T change? No heat goes in or out, no work is done, so what should cause a change in T? Real gases expanding adiabatically cool down because they do work against the intermolecular forces. And real pistons aren't massless or frictionless. But I assume we're talking ideal here.

[clinz63]

Okay okay. Thanks!

[Enthalpy]

Either you had a counter-force on the piston to limit its speed, and this has received work, or the piston accelerated, stored the work as kinetic energy, and released it in the shock.