Now according to whatever knowledge I have, when change in internal energy is zero, the kinetic energy of particles should remain the same.
Internal energy can consist of both molecular kinetic energy and molecular potential energy. Thus a zero change in internal energy means the sum of the changes in kinetic and potential energy is zero. Only in the case of ideal gas behavior, where all the internal energy is considered kinetic energy, can we say that the change in average kinetic energy of the particles will be the zero.
Does this mean that particles aren't moving from one partition to another?
There is collective movement of the particles from the pressurized side to the vacuum side when the plug is first removed. Molecules away from the opening do work pushing molecules in front through the opening giving them kinetic energy while giving up kinetic energy of their own for a net change in kinetic energy of zero.
Once equilibrium conditions are eventually established the continued random motion of the particles on both sides means some particles will continue to move between the two sides, but the net movement between sides will be zero.
No change in internal energy, kinetic energy of molecules means Δ𝑇=0
Therefore Δ𝑆=0 & Δ𝐺=0 and process maintains equilibrium. Because of
all the above process is reversible as well. Reversible processes are
slow.
The process is not reversible. The fact that $Q=0$ for the adiabatic free expansion of an ideal gas in a vacuum does not mean $\Delta S=0$. Entropy can increase without any heat transfer during an irreversible process when it only involves the generation of entropy with no transfer of entropy. That is what is happening here in the free expansion.
You use the definition for entropy change by assuming any convenient reversible path connecting the same two equilibrium states that the irreversible path took. The convenient path need not bear any resemblance to the actual path. You can do this because entropy is a state function. The difference in entropy is then calculated by applying the following entropy definition for entropy change to the alternate reversible path between initial and final states.
$$\Delta S=\int_i^f\frac{\delta Q_{rev}}{T}$$
Since $\Delta T=0$, that means, for an ideal gas, we can assume a reversible isothermal expansion instead of the irreversible expansion. Thus
$$P_{f}V_{f}=P_{i}V_{i}$$
The heat transfer is
$$Q=RT\ln\frac{V_{f}}{V_{i}}=RT\ln\frac{P_{i}}{P_{f}}$$
and the entropy generated is
$$\Delta S=\frac{Q}{T}=R\ln\frac{V_{f}}{V_{i}}=R\ln\frac{P_{i}}{P_{f}}$$
Hope this helps.
