# Gas and liquid at different temperatures. How do I read the pressure?

I am a homebrewer. My setup holds 6 5-gallon kegs. Previously I had my CO2 inside the chest freezer I've converted into keezer. I've recently acquired a 6th keg and there is not room for both 6 kegs and the CO2 tank inside the keezer.

When I purchase a CO2 refill it is generally around 1000PSI at room temperature and falls to 600PSI a bit in the keezer. Following $PV=nRT$ this makes sense to me. A regulator on the tank exposes 30PSI to the rest of my system.

The keezer is set to keep the beer around 40F/4.4C. Depending on the style of beer I will force carbonate at 8PSI, 12PSI, 16PSI, or 30PSI (for sparkling water). I do this through a series of secondary regulators which can step down the pressure they're fed.

Now my question. If I move the CO2 tank and the regulators to the outside of the keezer unit will I have to adjust the displayed pressure to maintain the same effects? My gas will be at room temperature but it will be dissolved into refrigerated liquid. It's fed into the keezer through vinyl lines fed through holes drilled in the side.

edit: The 600PSI was a number I believe I've remembered incorrectly as was pointed out, it is too far a drop. The reading on the gauge does indeed fall but not that much. I don't believe the exact number there is of direct relevance to the question.

All carbonation pressures would be relative to atmospheric pressure, yes.

• What does seem a bit odd is the you observed such a large drop in pressure for the barrel from 1000 to 600 PSI for a change from 20C to 4.4C. I estimated the pressure drop should have been more like from 1000 to 950 PSI? Any thoughts on that? – JMLCarter Jun 20 '18 at 19:06
• Since atmospheric pressure is about 14.6PSI, I can't see gas at 8 or 12 PSI entering a liquid. Are your pressures relative to atmospheric pressure? – JMLCarter Jun 20 '18 at 19:10
• "When I purchase a CO2 refill it is generally around 1000PSI at room temperature and falls to 600PSI or so inside the keezer. Following PV=nRT this makes sense to me." - Doesn't make sense to me. Did you use ˚C or Fahrenheit in the ideal gas equation? You're supposed to use absolute temperatures in Kelvin in it. In terms of absolute temperatures, the drop in temperature from room temperature to 4.4 ˚C is only about a 7.5% drop, whereas you're reporting that the pressure of the CO2 canister goes down by 40%. – user93237 Jun 20 '18 at 19:22
• Edited for clarity. I've never confirmed the drop against PV=nRT but you're right it's too large. Apologies, 600 was a number I pulled out of thin air and it's likely incorrect. @JMLCarter yes this would all be relative to atmospheric pressure. – brenzo Jun 20 '18 at 19:48
• This probably isn't a good question for the group. No matter what answer is accepted, the OP will want to adjust CO2 settings based on subjective criteria, which is the preferred taste of the beer, and the desired head on the beer. – David White Jun 20 '18 at 19:51