If this cannot be said free expansion then also this paradox arises.

Does this answer your question? Work done by free expansion of gas and the energy of the piston

Without a piston, the pressure becomes non -uniform and makes the gas accelerate. The work done by the pressure gradients provides the kinetic energy of the gas. The gas temperature drops so that the internal energy loss equals the gain in kinetic energy of the rapidly moving gas.

It looks like there are a few misunderstandings. (1) It seems like you think the internal energy of a gas is constant when it adiabatically expands. This is just wrong. When a gas adiabatically expands internal energy of the gas turns into work. (2) It seems like you think "free expansion" can also be adiabatic. It's not entirely clear to me what free expansion means to you, but usually it is essentially the opposite of adiabatic. Gas is suddenly exposed to a region of lower pressure and allowed to expand into it, and the adiabatic equations do not apply.

@AXensen Free exapansion is usually treated as adiabatic in fluid dynmics as the gas is almost in local equilibrium at each point in space. See rocket engine nozzles

@mikestone I wasn't entirely sure there... but after some googling it seems the definition of free expansion in thermodynamics, especially WRT ideal gasses, conforms to my understanding. "The free expansion of a gas is an irreversible process" from: chem.libretexts.org/Bookshelves/…

@AXensen In the experiment where gas expands from one cylinder into another it becomes irreversible at the point when the rapidly moving gas loses energy to friction and comes to rest. This is what is usually meant by "free expansion" in thermo books. They ignore the reversible conversion of internal energy to kinetic energy and focus only on the friction part.. Most of the "pardoxes" relating to free expansion come from not thinking about the kinetic energy of the bulk motion of the gas. A gas flow without friction or heat inflow is isentropic.

@mikestone Say a gas in a container of volume V is suddenly exposed to a second chamber of volume V. The gas gains kinetic energy, which may initially be isentropic. But there's no way it remains isentropic in the long term. The only way I could think of is if it somehow is retained as waves ripping through the gas forever. But even in the absence of friction, over time from the complex shape of the container the waves will turn into just heat. I think this is why my link just says outright that free expansion of gasses is not reversible. They're thinking of the equilibria before vs after.

@AXensen I dont think there will be waves until the gas hits the far wall. Then there will sound waves and complicated back and forth viscous sloshing motion. Free expansion into the vacuum of actual outer space , as with a rocket engine, will be almost isentropic. Bernouili will hold: $1/2 v^2 +h$ ($h$=enthapy per unit mass) is constant along a streamline,

The case which i am referring to is a gas is placed in a chamber with perfectly non conducting walls and the piston is also non conducting and it has a weight. It is allowed to expand against vacuum . Isn't it adiabetic expansion. Actually it is my fault to relate this paradox with free expansion. You can call my case as free or non-free adiabetic expansion but it really have no relation with the paradox. The only thing I can see is that internal energy of the gas is not constant . It should be constant because dq and dw ( p(vaccume)×dv ) are both 0.

Your theory of "bulk motion " couldnot solve my paradox

And I never said that adiabetic process is one where internal energy is constant

It is not like other paradoxes which arised due to not thinking about bulk velocity or velocity gradient. The velocity gradient during non equilibrium will lead to different pressure at different parts of the chamber. But external pressure is always constant. The velocity gradient has nothing to do with external pressure

Do you think that the piston continues moving forever, or, even with a frictionless piston, does it eventually come to rest with zero KE?

I have assumed nothing which can or will stop the piston. So 0 KE is only there when the expansion has not started. And I will ignore Friction in any case till I understand the basics of thermodynamics

@mikestone.. BTW I think I have got the answer . Here p(ext) is not p(vacuum). It p(ext)= F( contact force between gas and piston)/area of the piston because the piston is the immediate external environment for the gas. According to newton's law of equal and opposite force, p(ex)=(p(in) on the piston). Am I true?

When the gas expands rapidly the pressure in the gas is largest away from the (massless?) piston and essentially zero at the piston face. It is the pressure gradient that accelerates the gas and is responsible for converting internal energy (adiabtically) to kinetic energy.. My point is that when gas expands into a vacuum, as with a rocket in space. the expansion is adiabatic (isentropic), but there is no one $P$ and no one $T$ in the fluid. That is why we have fluid dynamics equations, and why the thermo books fudge the issue.

Eventualy, viscous stresses within the gas will cause the kinetic energy of the piston to dissipate, and the piston will stop moving. When this happens, the change in internal energy U from the initial state to the final state in the process you described will be zero: $$\Delta U=0$$

The piston is not massless. If you say that pressure is 0 at the pistons face then the net pressure and force on the piston is 0 . Newton said that force is required accelerate. This means that pressure gradient cannot accelerate the piston if net force on it is 0. If the piston is massless then only pressure at the face of the piston is 0. Reason 1) from the equation of elastic colision, we can conclude that nothing can exchange momentum with massless object. So d(momentum)/dt = force =0. 2) when the gas starts expanding , many molecules hit the piston. When the fastest molecule hits it.....

the piston being massless gains a velocity double that of fastest molecule. After that no molecule can collide with the piston. This pressure near it is 0

But if the piston is not massless then there will always be some molecules that hits the piston. So pressure will be exerted. Pressure is different at different parts of the fluid, that is why I exclusively said "(p(in) on the piston)".

@ chet Miller I think the piston will never stop because , the vacuume can have no viscous stress. Only the gas can have viscous stress. That viscous stress can oppose the internal pressure , even nullify it but can never exceede the internal pressure (suggested by molecular mechanism of viscous stress) . Thus there will never be force in opposite direction of pistons motion. So it will never stop.

I know there is viscous stress in gas but just imagine ,at any state can the gas start pulling the piston back. It cannot because gas has molecules colliding with the piston. Those colisions can only push the piston forward not pull it backward. And I don't think the ideal gas law is true during non equilibrium . P is different at different points.

Sorry when I said that the piston will experience pressure if it is not weightless I was wrong. If it is lighter than the gas molecules then also it will not experience colisions.

And if it is not much heavier than the gas molecule then also a position will come when colisions with it will stop but it will take some time.