Relation of pressure and temperature for an ideal gas at constant volume I have a tank of ideal gas with volume $3.1\ \mathrm L$; I have $R = 8.314$, and 11 moles. The tank is filled at $23\ \mathrm{^\circ C}$ (not sure why that's relevant) and the tank will rupture at $P = 100\ \mathrm{atm} = 1.013\cdot 10^7\ \mathrm{Pa}$. At what temperature does the tank rupture?

I divided $PV$ by $nR$. However, plugging those numbers in gives 

$$\frac{3.1 \times 1.013\times 10^7}{8.314\times 11} = 342374$$
in kelvin. However, my book gives $70.2\ \mathrm{^\circ C}$ as the correct answer. I assume my units are somehow profoundly messed up. Or maybe that mysterious 23 has something to do with it? What am I doing wrong?
When using the value of $R=8.314$, the units are $\mathrm{\frac{J}{K\cdot mol}}$. One joule is equal to:

$$\mathrm{J=\dfrac{kg\cdot m^2}{s^2}=N\cdot m = Pa\cdot m^3 = W\cdot s = C\cdot V}$$

If you want to use volume in liters, you need to use a different value for R.