Question

Oxygen , at a pressure of 200 bars, will be contained in a steel container at a temperature of 20 degrees celcius.
The volume of the container is 0.04 m3. Assuming that Oxygen is an ideal gas, calculate the mass of Oxygen which can
be contained in the container.
The container is protected against excessive pressure by a fusible plug which will melt if the temperature
goes too high.
Calculate at what temperature the fuse plug will melt if the pressure is limited to 240 bar.
(The molar mass of oxygen = 32 Kg / Kmol and Ru = 8314 J / Kmol K)

Answer 1

According to the ideal gas law (see https://en.wikipedia.org/wiki/Ideal_gas_law)

$PV=\nu R T.$ We know that $P=200\,\mathrm{bars} = 200\cdot 10^5\,\mathrm{Pa} = 2\cdot10^7\,\mathrm{Pa}, \; T = 20+273.15 = 293.15\,\mathrm{K}, \; V = 0.04\,\mathrm{m}^3.$ We can calculate the amount of oxygen $\nu$ :

$\nu = \dfrac{PV}{RT} = \dfrac{2\cdot10^7\,\mathrm{Pa}\cdot0.04\,\mathrm{m^3}}{8.31\,\mathrm{J/mol/K}\cdot293.15\,\mathrm{K}} = 328.4\,\mathrm{mol}.$

The molar mass of molecular oxygen is 32 g/mol, so the mass of oxygen is $328.4\,\mathrm{mol}\cdot32\,\mathrm{g/mol} = 10.5\,\mathrm{kg}.$

The volume is constant, so the process of heating will be isochoric.

$P_1V = \nu R T_1, \;\; P_2V = \nu RT_2 \;\Rightarrow \; \dfrac{T_2}{T_1} = \dfrac{P_2}{P_1}, \;T_2 = T_1 \cdot\dfrac{P_2}{P_1} = 293.15\cdot\dfrac{240}{200} = 351.8\,\mathrm{K}.$