Determining how much gas to purchase

[JWF239]

I am ordering some calibration gases to develop a method for my gas cell equipped FTIR. The gas cell is roughly 2 liters in volume and I vacuum it down to -28 in Hg. I want to run the samples at 20 psig. I have another cylinder from the company I am going to order from that says it is 34 L when it is clearly not. It says 34 L @ 70 F and 500 psig. I assume this means it was 34 L of gas at atmospheric pressure condensed to 500 psig to fit into the 1 L-ish cylinder?

Assuming that is what it means then:

I can take the -28 in Hg and convert to psi using the factor 0.49 psi = 1 in Hg. Then I can add this number to the 20 psi that I want in my gas cell to get 33.72 psig total.

So lets call it 40 psi and 2.5 L that I need to be safe. So this makes the left side of my ideal gas law 9.3046 atm*L and I will use the same temperature that they have on theirs so 70 F = 294.26 K. So using the R of 0.082057 L*atm*mol^-1*K^-1 I only need 0.3853 moles of my gas.

Again, assuming the 34 L @ 70 F and 500 psig means they pressurized 35 L of atmospheric pressure gas to 500 psig, this means their cylinder had 1.4 moles of gas in it.

Is there anything I am missing or over simplifying here? I really don't want to get the cylinders and not have enough gas.

 

[TD2K]

You can ignore any temperature effects for a first approximation. I'm assuming you vacuum the cell down to -28 in Hg and then pressurize the cell with your calibration gas (the 70L) to 20 psig.

You are right, the 34 L would be the volume of the gas at atmospheric pressure and 70F that has been compressed (not condensed) into the cylinder.

You're overcomplicating the volume sample calculation of the gas cell.

I'd assume (for simplicity) that there is nothing in the cell at -28" Hg vacuum. So you "need" 2 L to bring it up to atmospheric pressure. 20 psig = (20 + 14.7)/14.7 = 2.4 "volumes" or another 2.4 * 2 L = 4.7 L. Add the 2 L to bring the cell up to atmospheric pressure gives you 5.7 L needed per run. If atmospheric pressure is significantly different from 14.7 psia, you'll need to adjust the numbers.

1 cylinder should give you about 12 runs. This doesn't include the residual gas in the compressed gas cylinder. Since the lowest pressure you can get down to is 20 psig (since that is what you need in the cell) but for 500 psig to 20 psig, that's not really significant.

 

[mbt22]

(34 l @ 0 psig) = 34 l / (500 psig + 14.7 psia) * (14.7 psia) = 1.0 l @ 500 psig. Gas quoted as (normal/standard/arbitrary) volume, as a historical substitute for being sensible and using mass units.

You can run your cylinder down to 20 psig, so you can't use 1.0 l (20 psig + 14.7 psia) / (14.7 psia) = 2.4 l @ 0 psig

At your final cell conditions, you have 2 l @ 20 psig, or 2 l * (20 psig + 14.7 psia) / (14.7 psia) = 4.7 l @ 0 psig

- 28 in Hg ~= 1 psia, so there is already 2 l * (1 psia) / (14.7 psia) = 0.14 l @ 0 psig

So, you can run your experiment (34 l - 2.4 l) / ( 4.7 l - 0.14 l) = 6.9 ~= 6 times

All assuming constant temperature, 14.7 psia local pressure and ideality.

Or, a generous eye to say its a 1 litre 35 bara cylinder, I want to take a 2 litre cylinder to about 2.5 bara, so there's plenty. If I'm feeling particularly rigorous, 35 / (2*2.5) ~= 7. Close enough for me.

Matt