I am trying to calculate a cost per hour use of a gas we buy in a cylinder. The details I have been working with are:

Gas used CP-grade N2 material number: 110628-L can

  • Gas: N, density = 1.251 g/L @ 0.101 MPa M (molar mas) = 14.007
  • Gas cylinder: Pressure = 200 bar or 20 MPa Cylinder volume 9.45 m3 or 9450 L
  • Operating pressure of gas from cylinder = 4.5 bar or 0.45 MPa and flow rate = 60 cL/min

I have been applying the ideal gas law to PV=nRT to figure out how many total litres of N are in the can at 4.5 bar pressure to then get a rough estimate of how many hours of flowing gas I get. However I seem to be getting unreasonable numbers, potentially from me using the equation or reasoning incorrectly:

When I apply the gas law under isothermal V2=V1/(P2/P1) conditions to see how much the internal gas volume expands from the 20 MPa pressure, as a sanity check I wanted to see how many L at 1 atm came out. The result was 1.89 ML which makes little sense as the conversion of 1.89 ML N2 to mass = 2.364 Mg.

Similarly it equates to 420 kL @ 0.42 MPa.

Where am I going wrong?

Maybe the specifications mean that its 9.45m3 or 9450 L of compressed gas to 200 bar?

  • 1
    $\begingroup$ 9450 L is about 10 cubic meters, so unless your cylinder is really big -- say like 1 meter in diameter and 12 meters high, it isn't 9450 L big. So you are probably right that what you are receiving is 9450 L of gas at STP. A further note: nitrogen gas is $\textrm{N}_2$, a diatomic molecule, so its molecular weight is 28 g/mol, not 14. $\endgroup$
    – Curt F.
    Apr 15 '15 at 14:23
  • $\begingroup$ I worked it out in the end. The specifications given of 9450 L are indeed the volume of the uncompressed gas, which compresses to 20 MPa results in about 47 L which fits the dimensions of the L gas cylinder. I simply understood the information backwards. Noted for the 28 g/mol. Thank you. $\endgroup$
    – Fiztban
    Apr 15 '15 at 14:27
  • $\begingroup$ Have a look at this video. This Should you help you out. youtube.com/watch?v=Yo9IZ2RJCCU $\endgroup$
    – Anup
    Dec 31 '16 at 17:38

Turns out my assumptions were wrong about the stated $9\,450\ \mathrm L$ of $\ce{N2}$ gas. The $9\,450\ \mathrm L$ is the total volume of standard pressure gas compressed into the cylinder to $20\ \mathrm{MPa}$.

Using $V_2=V_1/(p_2/p_1)$ having $p_1=0.101\ \mathrm{MPa}$, $p_2=20\ \mathrm{MPa}$, $V_1=9\,450\ \mathrm L$. The total volume of gas within the cylinder is about $47\ \mathrm L$ a reasonably fitting size for the cylinder dimensions.

Hence to answer my own question, at the pressure of usage ($0.45\ \mathrm{MPa}$) the volume of gas (using same equation as above) is $2\,121\ \mathrm L$ which at $60\ \mathrm{cL/min}$ equates to $59\ \mathrm h$ of flow roughly.


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