If I put container of compressed air into a Coolgardie Safe, and brought the temperature down by say 3°C, how am I supposed to calculate the temperature of the air after it is decompressed back to ambient air pressure?
I'm not asking for calculations, I'm asking for formulae and theories which I can use here.
Before putting the container into the safe, it has a pressure $P_0$ and a volume $V$, and it's at room temperature $T_{room}$. After it's been cooled, its volume is the same, but its pressure and temperature have changed. You know that its temperature is changed to $T_{cool}$, and you need to compute its pressure after cooling $P_{cool}$. Since this is a constant-volume process, you can use the Gay-Lussac law:
$$\frac{P_0}{T_{room}}=\frac{P_{cool}}{T_{cool}}$$
After decompression, its volume increases to some volume and temperature $V_{final}$ and $T_{final}$, and its pressure decreases to $P_{ambient}$. The pressure, volume, and temperature are all changing, which means that where each quantity eventually ends up depends on how the air is decompressed.
If we can assume that the can of compressed air is decompressed quickly enough that there isn't time for it to transfer heat to the surroundings, then it can be approximated as an adiabatic process. In an adiabatic process, we have that
$$P_{cool}V^\gamma = P_{ambient}V_{final}^\gamma$$
where $\gamma$ is a constant that depends on the number of thermodynamic degrees of freedom of the gas. For dry air, we have that $\gamma\approx 7/5$. We also have the combined gas law, which relates the pressure, temperature, and volume of a gas before and after those quantities change:
$$\frac{P_{cool}V}{T_{cool}}=\frac{P_{ambient}V_{final}}{T_{final}}$$
Combining these two equations, we get that:
$$T_{final} = T_{cool}\left(\frac{V_{final}}{V}\right)\left(\frac{P_{ambient}}{P_{cool}}\right)=T_{cool}\left(\frac{P_{ambient}}{P_{cool}}\right)^{2/7}$$
