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?

  • 2
    $\begingroup$ Our units of volume are $m^3$, not $L$. $\endgroup$ Nov 15 '16 at 7:18

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.


The tank is filled at 23 °C (not sure why that's relevant).

This information is crucial :)

For the initial situation, the number of moles, the volume and the temperature are given.

$$P\,V = n\,\mathrm{R}\,T$$

Plugging these into the ideal gas equation will give you the initial pressure.

Given that the number of moles and the volume don't change, increasing the temperature will raise the pressure until the tank eventually bursts.

Just remember Gay-Lussac's law:

$\Large{\frac{P_1}{T_1} = \frac{P_2}{T_2}}$

  • $\begingroup$ Hmm... Why can't I just plug into the ideal gas law at 100 atm? There would only be 1 unknown. $\endgroup$
    – René
    Nov 15 '16 at 7:37
  • 2
    $\begingroup$ @RenéG You certainly can. I thought that mentioning the underlying, older law might help. $\endgroup$ Nov 15 '16 at 8:07

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