Adiabatic filling of a container Suppose a thermally insulated container is filled with atmospheric air until the pressure reaches 5000 psi.  This could represent the filling of a diving cylinder, before thermal dissipation becomes significant.
Initially, then, some mass of air has atmospheric temperature $T_1$ and $p_1$.  When forced in to the tank adiabatically, we expect its final temperature to be
\begin{align}
T_2 &= T_1\left(\frac{p_2}{p_1}\right)^{1-1/\gamma} \\
&= (300\,\mathrm{K})\left(\frac{5000\,\mathrm{psi}}{14.5\,\mathrm{psi}}\right)^{1-1/1.4} \\
&\approx 1590\,\mathrm{K}
\end{align}
This final temperature seems far too hot.  A diving cylinder would likely melt at this temperature.
More likely, though, something is wrong with my reasoning.  Is this not the correct use of the temperature-pressure relations, assuming an ideal gas?  If not, what other information is needed to predict the final temperature of the container?
 A: You're trying to ask a question about real life (would the tank melt) with a model that approximates away the very thing that you need (adiabatic). With the assumptions you have made, you are in fact using the correct equations.
The adiabatic assumption is only valid if it is thermally insulated which is not the case in real life. There will be losses of heat in the system while filling the tank unless the filling occurs virtually instantaneously. 
A quick search indicates that tanks should be filled at around 500 psi/minute, so it would take about 10 minutes to fill up a tank from empty to 5000 psi. Over the course of those 10 minutes, the heat that is added to the air as it is compressed is absorbed into the steel and conducted away to the atmosphere. So the process is far from adiabatic, which is why the tank does not actually melt. 
Lastly, I would point out that steels melt at much higher temperatures, over 2000K so even if adiabatic, a steel tank would not melt. Aluminum on the other hand melts around 1200K, so that one wouldn't work so well. 
A: Your calculation looks correct to me (although 5000 psi is a little high for a typical diving cylinder). But this calculation gives the final temperature of the air inside the cylinder. If -- after rapid compression -- we wrap insulation around the cylinder and allow the cylinder and the gas to reach equilibrium, the final temperature will be...
...tricky to calculate without tables of specific heats at high temperatures and pressures...
A: I think you forget that when you fill the tank, you add mass to it. Your relations assumes that the mass in the system is constant.
A: The practical thing that you are missing here is that the air compressor is cooled constantly so that it doesn't melt during operation (can't be adiabatic). So the upper limit on the temperature is driven by the operating temperature of the compressor. Then the other posters comments about heat transfer out of the tank apply.
