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There are 7 different gas cylinders filled with:

  • nitrogen
  • oxygen
  • air
  • methane
  • hydrogen sulfide
  • carbon dioxide
  • sulfur dioxide

How do I determine how much of the respective gases is left?

Nitrogen and oxygen are in a gaseous state, so I reckon a glance at a pressure indicator will do. Air? The same. How about the last four cylinders which are filled with liquids? They have their vapor pressure, but I am not sure that those pressures can be used to accurately calculate the volume of respective liquids. Should the weight of the cylinder be measured?

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  • $\begingroup$ What is their normal pressures? Pressurise nearly any gas enough and it will liquify. Most gas cylinders of those gases don't have liquid contents- it would be most inconvenient for users to have the liquid interfere with valves. $\endgroup$
    – user2617804
    Commented Oct 1, 2016 at 10:12
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    $\begingroup$ @user2617804 Try opening the valve of a CO2 cylinder standing upside down. $\endgroup$
    – Karl
    Commented Oct 1, 2016 at 11:18

2 Answers 2

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Correct.

For the non-liquifying gases (room temperature is far above their critical temperature) knowing the pressure is enough, if you know the volume of your cylinders. (You need a better formula than $pV=nRT$ to calculate the amount of gas at those high pressures. https://en.wikipedia.org/wiki/Van_der_Waals_equation, virial expansion, ...)

Actually the van der Waals equation is not useful, because, although it gives a good idea why the ideal gas law breaks, it gives very wrong numbers esp. for high pressures. BWR or BWRS are supposed to work well, at the cost of using eight or eleven constants.

For those gases which are partially liquified ($\ce{CO2}$ at $57\ \mathrm{bar}$, $\ce{H2S}$ at $12\ \mathrm{bar}$, $\ce{SO2}$ at $3.5\ \mathrm{bar}$), you need to know the weight (and tare weight) of the cylinders, of course.

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  • $\begingroup$ I suggest that if a gas cylinder doesn't have a tare weight on it, be wary of buying from that supplier again. Maybe the discovery of PTFE is connected. $\endgroup$
    – Andrew Morton
    Commented Oct 1, 2016 at 17:12
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To expand on Karl's answer a bit, not only do you need a better equation of state than $PV=nRT$ to accurately measure the contents, depending on the cylinder pressure several of the listed gases may require a correction for effects due to the supercritical phase change.

\begin{array}{c|cc} \mathrm{Species} & T_c \left(^\circ\mathrm C\right) & P_c \left(\mathrm{psi}\right) \\ \hline \ce{N2} & -147 & 493 \\ \ce{O2} & -118.6 & 731 \\ \ce{CH4} & -82.6 & 668.5 \\ \ce{H2S} & 100 & 1300 \\ \ce{CO2} & 31 & 1070 \\ \ce{SO2} & 157.2 & 1143 \\ \hline \end{array}

Data links — $\ce{N2}$ – $\ce{O2}$ – $\ce{CH4}$ – $\ce{H2S}$ – $\ce{CO2}$ – $\ce{SO2}$ $T_c$ $P_c$.

So, even for the gases well above their boiling points at room temperature ($\ce{N2}$, $\ce{O2}$, air, $\ce{CH4}$), if the cylinder pressure exceeds the respective critical pressure for that gas it will still be necessary to weigh the cylinders to get a highly accurate measure of the remaining contents, since supercritical phases generally have liquid-like densities.

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    $\begingroup$ This is a misconception. Unless you are really close to the liquid state, density of a gas (supercritical or not) increases largely linearly with pressure. There also is no phase transition between normal gas and supercritical. $\endgroup$
    – Karl
    Commented Oct 1, 2016 at 17:11
  • $\begingroup$ @Karl Interesting! So, is it that this far above $T_c$ the density will be substantially that of an ideal gas? $\endgroup$
    – hBy2Py
    Commented Oct 1, 2016 at 17:14
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    $\begingroup$ No, but it will not be very far off. The density will always approximately double if you double the pressure... until the gas becomes incompressible. That would be the opposite of what you say. $\endgroup$
    – Karl
    Commented Oct 1, 2016 at 17:22
  • $\begingroup$ @Karl But, regardless, the IGL will not be valid for accurate absolute calculations of the mass of the cylinder contents, correct? $\endgroup$
    – hBy2Py
    Commented Oct 1, 2016 at 17:48
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    $\begingroup$ By way of comparison, at the typical state for a full gas cylinder ($p=200\ \mathrm{bar}$, $T=25\ \mathrm{^\circ C}$), the values for the molar volume are $V_\mathrm m=0.1239\ \mathrm{l\ mol^{-1}}$ (ideal gas), $V_\mathrm m=0.1308\ \mathrm{l\ mol^{-1}}$ (nitrogen), $V_\mathrm m=0.1173\ \mathrm{l\ mol^{-1}}$ (oxygen), $V_\mathrm m=0.1277\ \mathrm{l\ mol^{-1}}$ (air), and $V_\mathrm m=0.1021\ \mathrm{l\ mol^{-1}}$ (methane). $\endgroup$
    – user7951
    Commented Oct 1, 2016 at 20:19

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