Thermodynamic Process in a Thermally Insulated Container Suppose we have a specific amount of gas in a thermally insulated container having a frictionless-massless piston. The piston initially is in the middle, dividing the container into equal two chambers, one of which has the gas while the other is vacuum. Now the piston is released. Then the gas, being inside pressure must expand quickly and go through an adiabatic process to expand upto the full volume of the container. My question is: whether this thermodynamic expansion is a reversible process or not? Why is it reversible or irreversible? 
Again, suppose, somehow, we get back to the initial state of the system. Then, instead of releasing the piston, we just hold the piston and allow the force that holds the piston to slowly decrease so that the piston moves slowly. This causes a slow change in the expansion of the gas, not as fast as the one before. As much as I know, this slow process is mainly an isothermal process. Since, the container have ideal gas, the change of internal energy of the gas is $0$, which means:
$$
dE_{int}=0
$$
Now, according to the first law of thermodynamics, we know that:
$$
dQ=dW+dE_{int}
$$
Now, since $dE_{int}=0$, we find that for the isothermal process:
$$
dQ=dW
$$
Therefore, the heat applied to the gas is the amount of work done by the gas. But as mentioned, the container is thermally insulated, therefore, the gas is not allowed to receive any heat from outside the system, that indicates $dQ=0$. So, finally $dW=0$ is bound to be true. 
But as we can see, the gas here actually expands, then the work done by the gas cannot be $0J$ anyhow in this case. Since, another probable equation to measure $dW$ is
$$
dQ=pdV
$$
the gas have to do a work to expand itself. Then what's the mystery here?
 A: The first process you describe is irreversible.  Why?  Because there is no process you can devise for returning the system to its original state without resulting a change in the surroundings.
The second process you describe is not isothermal.  So, in this process, the internal energy decreases as a result of the work done by the gas.  This process can be carried out reversibly by replacing your manual control with a more energy-conserving method (such as raising tiny weights to different elevations).
A: You might be better able to understand the seemingly paradoxical situation by asking yourself a few questions.

*

*How is the second situation different from the first?

*Why can you find two ways to calculate the change in Heat?

Alternatively if know about the second law of thermodynamics you can use Entropy to understand why this process is not reversible. Specifically when a process is reservable in relation to it's change in entropy and how you can see the entropy in the beginning and end of the process. If the sometimes weird definitions of entropy in the classical view of thermodynamics trip you up then try to use the statistical understanding of entropy.
If that does not bring you to an answer I recommend checking out the Wikipedia page on adiabatic processes.
A: 
My question is: whether this thermodynamic expansion is a reversible
process or not? Why is it reversible or irreversible?

It is irreversible for two reasons:

*

*A necessary condition for a process to be reversible is it must be carried out extremely slowly, i.e., quasi-statically. The free expansion of a gas occurs rapidly and therefore the process is irreversible.


*You can't return the gas to its original state (volume, pressure and temperature) after the free expansion unless you compress it. That requires heat transfer to the surroundings while the gas is being compressed. But that can't happen because the container is thermally insulated.

suppose, somehow, we get back to the initial state of the system.
Then, instead of releasing the piston, we just hold the piston and
allow the force that holds the piston to slowly decrease so that the
piston moves slowly...As much as I know, this slow process is mainly an isothermal process.

It would not be an isothermal process. By applying a resisting force to cause the gas to expand slowly, the gas now does work against that resisting force, whereas the gas in the first example did not have to do work expanding against a vacuum.
Since the gas is doing expansion work, and the container is insulated so that heat cannot transfer to the gas while it is doing work, the internal energy and thus the temperature of the gas will decrease.
Hope this helps.
