If I had a container, full with air, and I suddenly decreased the volume of the container, forcing the air into a smaller volume, will it be considered as compression, will it result in an increase in temperature, and why?
P.S: the container is strong enough to bear high pressure and not to explode.
Yes, it is compression and yes, it will heat up the gas.
If there's no heat exchange between the gas and the container (or the environment), we call it an adiabatic process. For an adiabatic process involving an ideal gas (which is a very good approximation for most common gases), $pV^\gamma$ is constant where $\gamma$ is an exponent such as $5/3$. Because the temperature is equal to $T=pV/nR$ and $pV/pV^\gamma=V^{1-\gamma} $ is a decreasing function of $V$, the temperature will increase when the volume decreases.
Macroscopically, the heating is inevitable because one needs to perform work $p\,|dV|$ to do the compression, the energy has to be preserved, and the only place where it can go is the interior of the gas given by a formula similar to $(3/2)nRT$.
Microscopically, the gas molecules are normally reflected from the walls of the container with the same kinetic energy. However, the molecules that hit the wall moving "against them" during compression will recoil with a greater velocity. If one calculates the average energy gain for the molecules, he gets the same temperature increase as one that follows from the macroscopic calculation.
When you increase the pressure of steam you increase its temperature. If to start with, the steam has a pressure of 105 kPa. What does it need to increase to, to raise the temperature to 120oC (Hot enough to kill all microbes).
