Can you make 2.5 gallons of soda in 5 g. keg?

If you want to make soda in a 5 gallon keg but only want to make 2.5 gallons, does this turn out OK?

Does it carb up and serve OK?

Does it take faster or slower to carb up 2.5 vs. 5 gallon in a 5 gallon keg?

Thanks, Doug

That should be just fine. You will use more CO2 because you have to pressurize more empty space. It should carb quicker because of the smaller volume.

be sure to also purge all the O2 out of the keg. CO2 is heaver so once you put force carb pressure on it just pull the relief valve for some time!!! to get all the O2 out. otherwise CO2 may fall out of suspension!

duganderson said:

If you want to make soda in a 5 gallon keg but only want to make 2.5 gallons, does this turn out OK?

Does it carb up and serve OK?

Does it take faster or slower to carb up 2.5 vs. 5 gallon in a 5 gallon keg?

Thanks, Doug

Click to expand...

jeepinjeepin said:

That should be just fine. You will use more CO2 because you have to pressurize more empty space. It should carb quicker because of the smaller volume.

Click to expand...

+1. As stated you will use more CO2. Basically the way it works is you have two volumes, the liquid and the headspace at some pressure (say 30 PSI). The physics here can basically be done using the Ideal Gas Law which says PV=nRT (P = Pressure, V = Volume, n = # of Moles of Gas, R = Constant, T = Temperature)... so if you think about it you have a Volume of the headspace at a Pressure, and we can assume that R and T are constant, so you will use more Moles of Gas (e.g. more gas) to fill it. Does that make sense?

As for the liquid... when you first start out you will have a liquid with no CO2 in it and a chamber of gas with some partial pressure of CO2 (if you purge it should be almost 100% CO2).. Again assuming ideal gases we can say that

P = P(O2) + P(CO2) + P(N2) ... etc (whatever else is in the gas).... assuming you purge them the pressure on the liquid will be 30 PSI of CO2.... The liquid doesn't like having a different pressure of CO2 than the gas so the CO2 will diffuse into the liquid (carbonating it)... the rate of diffusion is dictated by several things, but suffice it to say one of them is surface area. So over time the CO2 will dissolve into the liquid... and eventually carbonate... the math for this part is a little more complex (Volumes of CO2, Rate of Diffusion, Equilibrium, etc), so I will just leave it at that...

Does that all make sense?

The short answer is Yes. It will work fine.

xokeman said:

be sure to also purge all the O2 out of the keg. CO2 is heaver so once you put force carb pressure on it just pull the relief valve for some time!!! to get all the O2 out. otherwise CO2 may fall out of suspension!

Click to expand...

See above.... the CO2 doesn't "fall" out of suspension, it is in equilibrium with the gasses in the headspace... the O2 will change the partial pressure of CO2, but... the major issue with the O2 in the headspace is that a lot of bacteria require O2 to survive and thus can't survive in CO2, so to help kill them and prevent their growth we remove the O2... that and O2 oxidizes things in the beer/soda usually leading to flavors that are... well... to quote Alton Brown, "Not Good Eats".

Would it really require a lot more CO2 because once the head space is full, wouldn't all the extra CO2 that gets absorbed once the head space is pressurized be the same?

Maybe, however, I'm underestimating how much gas it would take to pressure 1/2 of an empty 5 gallon keg.

duganderson said:

Would it really require a lot more CO2 because once the head space is full, wouldn't all the extra CO2 that gets absorbed once the head space is pressurized be the same?

Maybe, however, I'm underestimating how much gas it would take to pressure 1/2 of an empty 5 gallon keg.

Click to expand...

Well... we can actually figure it out exactly... but it is almost pointless.... since ultimately you will use exactly the same volume once the liquid is dispensed. Let me explain...

Assuming the seltzer is already at equilibrium... you have a keg "A" with 5 gallons of seltzer so about 0 gallons of headspace.... and a keg "B" with 2.5 gallons of 2.5 gallons of headspace.... as you dispense the liquid the liquid will be displaced by CO2 at whatever pressure you are carbonated at (I recommend dispensing at the same pressure you carbonate at)... so keg A will go from 0 gallons of headspace to 5 gallons of headspace.... for a difference of 5 gallons of CO2 @ 30 PSI.... keg B went from 2.5 gallons of headspace to 0 gallons of headspace thus a total of 5 gallons of headspace... for again 5 gallons of CO2 at 30 PSI. If we assume the same temperature, and use PV = nRT ... we see that P, V, T, and R are all the same, THUS n (number of moles of gas) must also be the same. So ultimately it uses exactly the same amount of gas to dispense it. As far as carbonation... if we assume that the liquids are at the same temperature both under 30 PSI then they will have the same number of volumes of CO2 dissolved in them PER gallon... so the 5 gallons will absorb exactly twice as much as the 2.5 gallons.

What does this all mean.... assuming exactly ideal conditions the 2.5 gallons will actually use 1/2 as much gas to carbonate and the same amount to dispense as the 5 gallons. The reason we say that you will use more gas is because to dispense the 2.5 gallons you used as much gas as if you had dispensed 5 gallons. And when you go to refill the tank you "vent" all that gas off... thus it is lost. So mathematically we can say the volume of CO2 to dispense was D (5 gallons) and the volume to carbonate was C (5 gallons).... so for the 5 gallons of seltzer we get

C + D = Total....

And to dispense 5 gallons in 2.5 gallon increments we get

[(.5)C + D] + [(.5)C + D] => C + 2D = Total... So the difference to dispense 2 tanks of 2.5 gallons versus 5 gallons is exactly D (theoretically of course).

Does that make sense? You can figure out exactly how much gas that is if you want... just plug in P, V, R, and T and you will get n... ... Lets do that real quick.

P = 30 PSI = 2.04137892 atm
V = 5 gallons = 18.9270589 liters
T = 34 degrees Fahrenheit = 274.261111 kelvin
R = 0.08205746 (L*atm)/(mol *K) (see wikipedia for gas constant)...

(sorry for the odd units on P, V, and T.... they have to match the ones in R so everything cancels out)...

So we get n = PV/RT => n = 1.716818793 mol of CO2... since C is 12 g/mol and O is 16 g/mol we get CO2 at 44 g/mol... so that comes out to 75.54002691 grams of CO2... which is 0.166537252 lbs...

So assuming everything above you will use an extra .16 lbs of CO2.

Sorry for the very long winded answer, but if you are interested in the math there it is... if not then just go with .16 lbs of CO2 extra. (obviously not including any extra CO2 used during the purging etc... when you fill up the second batch of 2.5 gallons... just dispensing).

Any questions?