$V\propto T$ if pressure,$P$ and amount of gas,$n$ are constant. This arises from the equation $PV=nRT$, the ideal gas law. The pressure, outside the piston is irrelevant. What matters is the pressure inside the container. In 'normal experiments' the outside pressure (atmospheric pressure) is always constant. You could always increase the pressure inside the container by having heavier pistons (if the pistons have mass, the pressure inside the container would be larger than the pressure outside the container, assuming the arrangement of the pistons are like in this animation: the important thing is for the pressure inside the container to remain constant, while varying the temperature, to get the change in volume.
1) These sorts of things are actually being studied nowadays with trapped atomic gases. These gases are released from optical traps and expand into what is essentially a perfect vacuum. If you want, they can be recaptured, although a typical trap does not have sharp ``walls'', but harmonic confinement.
2) Yes, if you expand into a vacuum then you just convert internal energy into (collective) kinetic energy $\frac{1}{2}\rho u^2$, where $\rho$ is the mass density and $u$ is the fluid velocity. If you expand forever then temperature will go to zero, and all energy is kinetic.
3) If you recapture then the result will depend somewhat on the nature of the gas and the walls.
i) Ideal gas, soft walls (harmonic confinement). The systems rebounds from the wall and starts to undergo periodic oscillations. The energy remains kinetic. Since the gas is ideal, the motion is undamped.
ii) Real gas, soft walls. Same thing, but collective oscillations are damped by viscosity, Systems settles down, energy goes back to internal (that means for ideal gas equation of state $T$ goes back to initial value), entropy has increased.
iii) Ideal gas, sharp walls. Rebound creates shock waves. This is a complicated process, since you will get a messy systems of interacting shocks. Even ideal shocks create entropy, so this systems may settle down as in ii).
iv) Real gas, sharp wall. As in iii), but this definitely settles down as in ii).
Best Answer
The temperature of the gas will eventually reach equilibrium with the walls of the container, and since a perfect insulator is not possible, the gas, walls and outside environment will, given enough time, be at the same temperature.