And the opposite follow-up question: why does opening the air cylinder makes the air cooler?
What I know is that I can't find these answers using the ideal gas law, because that is an equation of state. I cannot use Charles' law (it requires pressure to remain constant) nor Boyle's law (it requires the temperature to remain constant).
Similarly, I cannot use Gay-Lussac's law ($P \propto T$), because that law requires both mass and volume to be constant (when filling an air cylinder I'm adding mass and/or I'm reducing the gas volume).
So, where can I find a physical justification for this effect?
Best Answer
Because what you are doing is a flow process, with mass inflow and no mass outflow, you need to use the thermodynamic equation:
$dU_{cv}={m_{in}d}{H}_{in}-{m_{out}d}H_{out}+\delta Q-\delta W_{shaft}$
If you insulate your air cylinder well enough, $\delta Q = 0$.
Assuming that your air cylinder does not deform, $\delta W = 0$.
Since you are filling your cylinder with air and assuming no air escapes, ${m_{out}d}H_{out} = 0$
Therefore, the enthalpy of the gas which you are filling adds to the internal energy of the gas in the cylinder, and because the internal energy is positively correlated to temperature, the temperature in of the gas in the cylinder rises.
$\Delta{U_{cv}}={H_{in}}>0$, so $\Delta{T} >0$
You may apply the reverse for the release of air from the cylinder. In this case:
$\Delta{U_{cv}}={-H_{out}}<0$, so $\Delta{T} <0$
http://en.wikipedia.org/wiki/Thermodynamic_system#Flow_process