A rigid storage tank of volume 0.0831 ~m^ 3 contains 1.4 ~kg of...

The final absolute pressure of the gas in the tank is the sum of the gauge pressure and the atmospheric pressure.

$P = P_{gauge} + P_{atm} = 5 \mathrm{~atm} + 1 \mathrm{~atm} = 6 \mathrm{~atm} = 6 \times 10^{5} \mathrm{~N} \mathrm{~m}^{-2}$

Using the ideal gas equation, the final number of moles of Nitrogen in the tank is:

$n_{f} = \dfrac{PV}{RT}$

Substituting the given values:

$n_{f} = \dfrac{6 \times 10^{5} \times 0.0831}{8.31 \times 300} = 20 \mathrm{~mol}$

The final mass of Nitrogen remaining in the tank is:

$m_{f} = n_{f} \times M = 20 \mathrm{~mol} \times 28 \mathrm{~g} \mathrm{~mol}^{-1} = 560 \mathrm{~g} = 0.56 \mathrm{~kg}$

The mass of Nitrogen that escaped from the tank is the difference between the initial and final mass:

$\Delta m = m_{i} - m_{f} = 1.4 \mathrm{~kg} - 0.56 \mathrm{~kg} = 0.84 \mathrm{~kg}$

Answer: $0.84 \mathrm{~kg}$