[rmurray]

I'm trying to figure out how I can use moles of a gas to calculate a pressure to drain a tank down to. Lets say I have:
Initial gas mix of O2: 29.5%, N2: 70.5%, and a Pressure of: 230 bar. This has 2.6374 moles of O2 and 6.3030 moles of N2.
Desired gas mix of O2: 28.0%, N2: 72.0%, and a Pressure of: 230 bar. This has 2.4988 moles of O2 and 6.4256 moles of N2
Topoff gas mix of O2: 20.9%, N2: 79.1% (air).

I know through trial and error that I need to drain the tank to 184.4 bar to be able to add the topoff gas (air) to 230 bar to yield the proper mix. My question is, how can I arrive at this mathematically using moles and not pressure? When the tank is drained to 184.4 bar the tank will have 2.1731 moles of O2 and 5.1935 moles of N2.

Any help would be greatly appreciated.
Thank you,
Rob

[Borek]

Please elaborate on the question, at the moment we have to guess what is what of what.

My bet is that you should construct a set of equations with one or more unknowns and solve for them.

[rmurray]

Hi Borek,
Sorry for the confusion, I'll try to clear it up. I'm basically trying to reverse engineering this problem if possible without iterating to arrive at the answer.
Start mix:
2.6374 mols Oxygen, 29.5% O2
6.3030 mols Nitrogen, 70.5% N2

End mix:
2.4988 mols Oxygen, 28.0% O2
6.4256 mols Nitrogen, 72% N2

Mix used to add to the start mix to achieve the End mix (air by default in this instance, 20.9% Oxygen, 79.1% Nitrogen).

I know that the final answer is to drain the tank down until I have:
2.1731 mols Oxygen
5.1935 mols Nitrogen

I don't need pressures as I can convert this. I just don't know how or where to start in setting up the equation(s) to achieve this. I got close with this:
mols O2 initial-(mols O2 initial-mols O2 End)/(O2% initial-O2% add mix)
2.6374-(2.6374-2.4988)/(.295-.209)
This gives me an answer of about 2.21 mols Oxygen (sorry, at work now so this is off the top of my head). I hope this helps, if not, let me know what else needs clearing up.

Thank you,
Rob

[Borek]

Where did you get moles from? As far as I remember scuba tanks come in volumes around 10 L, at 230 bar and 20 deg C that means around 94 moles - so your numbers are about 10 times too low.

[rmurray]

It is based on 1 L, and all the calcs are figured at 25 deg C. The mols were derived by the GERG2004 equations so they are off a bit from Beattie-Bridgemann or VDWs.

[Borek]

OK, not knowing the exact form of GERG2004 I doubt it will be possible to solve the problem analytically - that is, it will have to be solved numerically, which usually means iterations.

Simplest approach requires that you assume volume (or pressure) is a linear function of number of moles and that volume is additive (both assumptions are correct for ideal gas, but are wrong for non-ideal gas).

Let's assume 1L of mixture. We remove xL, we add xL of other mixture - this way we stay at 1L.

intial amount of oxygen = 1L*initial fraction

removed oxygen = xL*initial fraction

left oxygen = (1L-xL)*initial fraction

added oxygen = xL*fraction in added mixture

final oxygen = (1L-xL)*initial fraction + xL*fraction in added mixture

Solve for xL.

It doesn't have to be 1L, it will work for pressures or moles as well, as these are easily interconvertible.

[rmurray]

I think I follow where you are going with this. It would be too much trouble to iterate through, I just wanted to make sure there wasn't a one pass equation I could implement to arrive at the correct answer.
I'll setup the iteration process but continue to work on this as time permits

Thank you,
Rob