There's actually not one simple answer to your question, which is why you are a bit confused. To specify your problem fully, you must specify exactly how and whether the gas swaps heat with its surroundings and how or even whether it is compressed. You should always refer to the full gas law $P\,V=n\,R\,T$ when reasoning. Common situations that are considered are:
Charles's Law: The pressure on the volume gas is constant. No work is done by the gas on its surroundings, nor does the gas do any work on its surroundings or piston or whatever during any change. The gas's temperature is that of its surroundings. If the ambient temperature rises / falls, heat is transferred into / out from the gas and its volume accordingly increases / shrinks so that the gas's pressure can stay constant: $V = n\,R\,T/P$; with $P$ constant, you can retrieve Charles's Law;
Isothermal: the gas is compressed / expanded by doing work on / allowing its container to do work on its surroundings. You think of it inside a cylinder with a piston. As it does so, heat leaves / comes into the gas to keep the temperature constant. As the gas is compressed, the work done on it shows up as increased internal energy, which must be transferred to the surroundings to keep the temperature constant. At constant temperature, the gas law becomes $P\propto V^{-1}$;
Adiabatic: No heat is transferred between the gas and its surroundings as it is compressed / does work. AGain, you think of the gas in a cylinder with a plunger. This is prototypical situation Feynman talks about. As you push on the piston and change the volume $V\mapsto V-{\rm d}V$, you do work $-P\,{\rm d}V$. This energy stays with the gas, so it must show up as increased internal energy, so the temperature must rise. Get a bicycle tyre pump, hold your finger over the outlet and squeeze it hard and fast with your other hand: you'll find you can warm the air up inside it quite considerably (put you lips gently on the cylinder wall to sense the rising temperature). This situation is described by $P\,{\rm d}V = -n\,\tilde{R}\,{\rm d} T$. The internal energy is proportional to the temperature and the number of gas molecules, and it is negative if the volume increases (in which case the gas does work on its surroundings). But the constant $\tilde{R}$ is not the same as $R$: it depends on the internal degrees of freedom. For instance, diatomic molecules can store vibrational as well as kinetic energy as their bond length oscillates (you can think of them as being held together by elastic, energy storing springs). So, when we use the gas law to eliminate $P = n\,R\,T/V$ from the equation $P\,{\rm d}V = -n\,\tilde{R}\,{\rm d} T$ we get the differential equation:
$$\frac{{\rm d} V}{V} = - \frac{\tilde{R}}{R}\frac{{\rm d} T}{T}$$
which integrates to yield $(\gamma-1)\,\log V = -\log T + \text{const}$ or $T\,V^{\gamma-1} = \text{const}$, where $\gamma=\frac{R}{\tilde{R}}+1$ is called the adiabatic index and is the ratio of the gas's specific heat at constant pressure to the specific heat at constant volume.
1) If we take the heat out of the system (i.e. with a cooling circuit, like happens in car engines) would gas temperature go down to the initial state, without decompressing it?
Is the piston fixed? Then the volume will remain constant during the cooling. Gas temperature would go down and pressure will fall.
How far down the temperature will go depends on the cooling. If the cooling circuit is bound to a lower temperature, then it will continue cooling until the gas has reached that. Independent of any initial temperature of the gas.
2) How long could you repeat that compression-heating-cooling process? Would it result in liquefying the gas?
If I understand it correctly, you remove the heat caused by compression (and return to initial temperature), then compress it again, then remove the heat and repeats like that.
This is exactly what happens in a usual pressure cylinder - like and air cylinder for diving. That will have a large pressure inside because a lot of air is compressed, but the temperature is equal to the surroundings.
Yes, it will liquify at some point. You compress it, pressure rises and the boiling point also rises. It will stay as gas, if the temperature is allowed to rise also. But by removing heat and cooling it down, you will at some point have a temperature below the new boiling point. In a phase diagram of a substance you will be able to see where the transition points are from liquid to gas - that point is tightly bound to volume, pressure and temperature.
3) How can I calculate the relation between the piston speed, pressure, and the heat generated?
Why piston speed? If you move the piston fast, you add kinetic energy to the molecules not caused by the volume decrease. You will have a hard time figuring out how much is due to piston movement.
Usually in models of this kind of volume decrease in a pressure champer the piston is assumed to move so slowly that kinetic energy is negligible. Then piston speed will have no relation to the other parameters.
Best Answer
Macroscopic temperature is a measure of the average kinetic energy of a group of molecules. Those molecules which are colliding with the face of the piston have their velocity increased due to their elastic collision with a moving boundary - the piston face. Therefore, as a group their rebound velocity and their temperature is increased. As a group, these molecules do have a higher temperature than those at the other end of the cylinder and there is a temperature gradient in the cylinder. This gradient is reduced by heat that is transferred through the working gas.
If you speed up your observations, you will realize that the increase in molecular or atomic velocity immediately after rebound from the piston face is not random. The increase is entirely along the axis of piston face movement. As this is not purely random motion, one can ask if it is really thermal energy at all or if it is not more correctly viewed as flow energy and you can reasonably ask what portion of it is reflected as temperature at any particular moment in time. The answer depends a bit on the speed of the gas molecules in relation to the speed of the piston face and on what time increment and size scale you want to observe the process.
The increased kinetic energy of these molecules can not propagate through the gas and away from the piston face at a speed any faster than the speed at which the molecules possessing it are themselves moving. Generally speaking, it is limited by the speed of sound in the gas. When the speed of the piston face approaches the speed of the gas molecules, very large gradients can build up creating supersonic shock waves.
Furthermore, the initial increase in kinetic energy is all oriented along the direction of the piston face movement. As these faster molecules collide at various and random angles with slower gas molecules, the slower molecules are sped up in random directions and the increase in kinetic energy is randomized. This energy is therefore re-partitioned among the three translational axes until the average energy in each axis is equal and relative uniformity is reached. Likewise, energy is also internally re-partitioned between molecular translation, rotation, and vibration energy storage modes within the molecule. This internal molecular re-partition is extremely fast - nano or pico seconds if I recall correctly. Now the increased molecular energy is clearly all thermal.
Bottom line, temperature and pressure gradients do occur on compression. For subsonic processes they can generally be ignored. These gradients are important in some cases. As one example, they form the basis for thermo-acoustic phenomenon. As another, they can form hot spots on objects traveling at supersonic speeds.