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This is the calculation for the number of moles of water vapor it would take to produce 5.9psi at 70f in a chamber with a volume of 339cmm. It translates to .1015g.
Step-by-step solution:
Converting pressure to SI units.
P = 40679.0663 Pa
Converting volume to SI units.
V = 0.000339 m3
Converting temperature to SI units.
T = 294.26111111111 K
R = 8.3144621 J/(mol*K)
Computing number of moles using the selected equation.
n = 0.0056364238741574
Don’t be that impressed. There is a calculator for everything.
This page has lots with options if anyone wants to play with some numbers.
This is the calculated co2 pressures for run 2. They are based 3% converted CBD and 7% THC.
[TABLE="width: 128"]
[TR]
[TD]Temp[/TD]
[TD]Pressure [/TD]
[/TR]
[TR]
[TD]70[/TD]
[TD]2.855[/TD]
[/TR]
[TR]
[TD]100[/TD]
[TD]3.017[/TD]
[/TR]
[TR]
[TD]150[/TD]
[TD]3.286[/TD]
[/TR]
[TR]
[TD]200[/TD]
[TD]3.555[/TD]
[/TR]
[TR]
[TD]230[/TD]
[TD]3.717[/TD]
[/TR]
[TR]
[TD]250[/TD]
[TD]3.825[/TD]
[/TR]
[/TABLE]
Back to the H2O pressures, at 250f and 37psi there would be 0.48g of vapor. These are some numbers that were present in test 1. They are not posted yet.
Both runs lost 0.4 grams on a scale that weighs in 0.1 increments.
The decarb accounts for 0.12g of the 0.4g of weight loss. Another 0.1 of water vapor is possible at the final 5.9 psi.
That leaves 0.27g plus or minus 0.1 as loss. That can be accounted for from condensation on the chamber walls and what was in it. Other possibilities are errors, terpene and aromatic loss.
I never know
Now I have to ask, didn't posting all those calculations make the geeky part of your scientific brain jump for joy? Canyon, I'm in awe at how you guys fell into the deciphering of the decarb process and infused oils. A while ago I googled my study hall and went looking at what other sites thought of the work we do over here at 420 Magazine. What I found made me chuckle in appreciation. On one site someone said it was too technical. Made me proud to be part of this team.