Volume of GAS 1

[hillimonster]

This may be a simple question, but for me I'm not sure what formula(s) to follow.

If I have a 30 Gal tank with a 50% mix of helium and air at 100 psi and I add 100% helium up to 200 psi. What is the volume of helium I have added.

I'm a controls engineer by trade and if I add a feature like this without the Math being right, I open up a hole can of worms with the Mechanical Types.

Thanks

 

[Montemayor]

Your question has little or nothing to do with math but has a lot to do with basic data. You'd better clear up your basic data before any of the guys on this forum start to work their calculators. Otherwise, you're going to get a myriad of answers -any of which could be the right one, or out in left field, somewhere.

1) You say 50% mix; % of what? is it volume or mass? (hint: the easiest would be volume)
2) You say you add 100% Helium. Do you mean you double the amount of original He+air mix - or do you mean that you double the amount of original He? Be specific, please.
3) What is the temperature of the original mix at 100 psi(g?)? Again, BE SPECIFIC! It makes a big difference if the "psi" is gauge or absolute!
4) How is it that you know that the subsequent additional Helium will result in a 200 psi(g?) pressure? Is the final temperature the same as the initial?

As you can see, there are a lot of gaps in the basic data. What specifically is this application for or what does it involve? Is it an academic, laboratory, or industrial application?

I'll await your response.


Art Montemayor
Spring, TX

 

[hillimonster]

Sorry for lack of specifics.

50% mix of air and helium. An evacuated tank with 50psia of air added initially and then 50psia of helium added. Or this could be at 50% mix using a gas analyzer.

The additional 100% helium is added to 200psia.

All pressure indications using a psia pressure transducer.
assuming that the final temp is the same (or constant).

industrial application.

 

[Fzob]

Air is about 7 times heavier than He. So if you are bringing a tank to 50 PSIG with air and then topping it off with He to get to 100 PSIG, you are not getting a 50/50 mix of air and He by volume or mass.

Check out this link

http://wine1.sb.fsu.edu/chm1045/notes/Gases/Mixtures/Gases06.htm

Once you know how much He you had at 100 psig you can do the same calclation to find the volumne of He at 200 psig.

I hope this helped.

Fred

 

[25362]

Although the process by which you attained a 50:50 volume mix of air and helium is not clear, I'll assume your initial (i) 1:1 molar mix at 100 psia, and the final (f) mix at 200 psia. For an isothermal-isochoric (rigid container) process, the gases being chemically inert, premising the ideal gas law applies, and the compressibility factor Z =1:

P_iV=n_iRT, and P_fV=n_fRT, we obtain

P_f/P_i=n_f/n_i​

or:

200/100=2=n_f/n_i​

Meaning you'd have at the end about twice as many moles as you had at the beginning, or a final helium-to-air 3:1 mol ratio.

To obtain the number of moles at the beginning, or at the end, you need to know the temperature using PV=nRT.

 

[katmar]

At these pressures, and assuming that the temperature is 72F, you can safely assume that the ideal gas law applies and it becomes a very simple calculation.

The basic equation is PV = nRT

i.e. the product of absolute pressure and volume equals the product of the number of moles of gas present, the universal gas constant and the absolute temperature.

If the pressure is psia, the volume is US gallon, n is lb mole and temperature is Rankine then R is 80.278 (from Uconeer).

We want to calculate the amount of gas present so we re-arrange the formula to give

n = (PV)/(RT) , and note that V, R and T are constants so we have

n = (P x 30) / (80.278 x 531.67) = P x 0.0007029

When the evacuated tank is pressurized to 50 psia with air this is 50 x 0.0007029 = 0.0351 lb mole of air. With a molecular mass of 28.96 this gives 1.018 lb of air.

Now helium is added until the pressure is 100 psia so the TOTAL number of moles is
100 x 0.0007029 = 0.07029 lb mole. This means 0.0351 lb mole of helium has been added. With a molecular mass of 4 this is 0.1404 lb of helium.

At 200 psia there are 200 x 0.0007029 = 0.1406 lb mole of gas. Or 0.07029 lb mole of helium has been added, with a mass of 0.07029 x 4 = 0.2811 lb of helium.

The important, and non-intuitive, thing to note is that the pressure is proportional to number of moles of gas added, irrespective of the molecular mass of the gas (when the ideal gas law applies). In this example we see that much more air (in mass terms) must be added than helium to achieve the same pressure increase because helium has such a low molecular mass.

If your temperature is not 72 F you can easily rework the above with the right value.

Harvey