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Volume of Compressed Gas in a Cylinder

To find the volume of gas available from a compressed gas cylinder, we apply the Ideal Gas Law (PV =nT!" In a high#pressure cylinder, the volume will be affected by the content$s compressibility factor %(PV = %nT!" &or e'ample, an L cylinder of pure helium may contain )*+ cu" ft" of gas while the samecylinder of pure air may contain )++ cu" ft" under the same conditions" &or these practical calculations,

however, we assume ideal gas behavior for simplicity"The Ideal Gas Law PV = nThere-P is pressureV is volumen is the number of moles is the gas constantT is the absolute temperature

hen the temperature is .ept constant, we can derive the e/uation-P()! ' V()! = P(0! ' V(0!

here-P()! is the pressure of the compressed gas in the cylinder (psi!V()! is the internal volume of the cylinder, often referred to as water volume (liter!1

P(0! is the atmospheric pressure () atm # )+"2 psi!V(0! is the volume of gas at pressure P (0! (liter!

&or e'ample, an L si3ed cylinder is filled with nitrogen at 0444 psi" hat is the gas volume of nitrogenfrom the cylinder5

P()! is 0444 psiV()! is the internal volume of L cylinder 06"7 liter1P(0! is )+"2 psiV(0! is the un.nown volume of gas

8olving the e/uation above for V(0! gives-V(0! = 9p()! ' V()!:;P(0! = (0444 psi ' 06"7 liters!;)+"2 psi = +4)* liters(appro'imately )+4 cu" ft"!

1The water volume of the high#pressure cylinders can be found on this

Conversion of PPM (mole basis) to Grams per Liter

for Gas Mixtures

ompound in nitrogen, howmuch of >ompound will be contained in a ) liter volume5

?ey ssumptions-)" Temperature of sample is assumed to be at 0)")@> (24@&!"0" Pressure of sample is assumed to be ) atm Aust before inAection" t around atmospheric

pressure, gases behave in close to ideal manner"

>alculation-Bsing the Ideal Gas Law (PV = nT! for a temperature of 06+")@? (0)")@>!, a pressure of ) atm, andthe gas constant of 4"4C0) liter ' atm;mole ' degree ?, we find that ) mole of ideal gas occupies0+")7 liters"

Dne liter of gas will then contain ();0+")7 ! moles" 8ince the concentration of >ompound is )44 ppm,the total number of moles of >ompound in ) liter is-

(total E moles per liter! ' (concentration of >ompound !"

The concentration of )44 ppm (parts per million! is unit#less, and e/uals )44 mole#parts per ),444,444total moles = 4"444)44 in decimal formF thus the amount of moles of >ompound in one liter of mi'tureis-

();0+")7! moles per liter ' 4"444)44 =4"44444+ moles of >ompound per liter

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In order to find the weight of >ompound , we need to .now its molecular weight" &or e'ample, if>ompound is hydrogen sulfide, with a molecular weight of *+"4C gram;mole, we obtain the followingconcentration-

4"44444+ moles per liter ' *+"4C gram per mole =4"444)+4 gram per liter or 4")+4 milligram per liter

General &ormula for >onversion of ppm (mole! to grams per liter for Gas i'tures(for temperature 0)")@> (24@&! and pressure ) atm!

>onversion of grams per liter to ppm (mole basis! for Gas i'ture-(for temperature 0)")@> (24@&! and pressure ) atm!

The concentration e'pressed in decimal form is unit#less" To find the concentration of >ompound inppm, multiply the answer from e/uation above with ),444,444" &or e'ample, a concentration of>ompound of 4"444)44 (decimal form from e/uation above! would be )44 ppm, while a concentrationof 4"4)4444 would be )4,444 ppm or )H"