=== Fermentation ===
1 mol glucose -> 2 mols ethanol + 2 mols CO2
=== Molar Mass ===
1 mol glucose = 0.180 kg glucose
1 mol CO2 = 0.044 kg CO2
1 mol ethanol = 0.046 kg ethanol
=== Thermodynamics ====
pV = nRT, where
p is pressure (Pascals = newtons per square meter)
V is volume (cubic meters)
n = moles of gas
R = gas constant = 8.314 N*m/( mol * K )
T = temperature (Kelvin)
How much CO2 does your priming sugar make (assuming it all gets converted)? If we add 4 ounces glucose which is 113 g, then we have added 0.113/0.180 mols = 0.628 mols of glucose. So, 2*0.628 mols = 1.256 mols of CO2 will be produced. So, now assume that the temperature is 20 Celsius (293 Kelvin) and we want to know the volume of CO2 at atmospheric pressure (1 atm = 101,325 Pascals). then V = nRT/p = (1.256*8.314*293)/(101,325) = 0.0302 m^3 = 30.2 liters of CO2.
So, 30.2 liters of CO2 are dissolved in your 5 gallons = 19 liters of beer! Well, in reality only 75% of the glucose gets consumed, so it's more like 0.75*30.2 = 22.7 liters of CO2, but still, that's a lot.
How much CO2 goes out of the airlock during fermentation? You could use the above procedure if you knew the equivalent amount of glucose that was originally in your wort. Since 1 kg of glucose dissolved to make 1 gallon of wort has a gravity of 1.097, divide your gravity in points by 97 points to get the equivalent kilograms of glucose per gallon of wort. Then multiply by the number of gallons to get the total kilograms of equivalent glucose. So if your gravity is 1.055 and you have 5 gallons of wort, then you have 55/97*5 = 2.84 kg of equivalent glucose that's available to ferment. If you assume that 75% of it ferments, then you get 0.569 m^3 = 569 liters of CO2 at room temperature and atmospheric pressure. No wonder my fermenter exploded last time I made that German ale...
1 mol glucose -> 2 mols ethanol + 2 mols CO2
=== Molar Mass ===
1 mol glucose = 0.180 kg glucose
1 mol CO2 = 0.044 kg CO2
1 mol ethanol = 0.046 kg ethanol
=== Thermodynamics ====
pV = nRT, where
p is pressure (Pascals = newtons per square meter)
V is volume (cubic meters)
n = moles of gas
R = gas constant = 8.314 N*m/( mol * K )
T = temperature (Kelvin)
How much CO2 does your priming sugar make (assuming it all gets converted)? If we add 4 ounces glucose which is 113 g, then we have added 0.113/0.180 mols = 0.628 mols of glucose. So, 2*0.628 mols = 1.256 mols of CO2 will be produced. So, now assume that the temperature is 20 Celsius (293 Kelvin) and we want to know the volume of CO2 at atmospheric pressure (1 atm = 101,325 Pascals). then V = nRT/p = (1.256*8.314*293)/(101,325) = 0.0302 m^3 = 30.2 liters of CO2.
So, 30.2 liters of CO2 are dissolved in your 5 gallons = 19 liters of beer! Well, in reality only 75% of the glucose gets consumed, so it's more like 0.75*30.2 = 22.7 liters of CO2, but still, that's a lot.
How much CO2 goes out of the airlock during fermentation? You could use the above procedure if you knew the equivalent amount of glucose that was originally in your wort. Since 1 kg of glucose dissolved to make 1 gallon of wort has a gravity of 1.097, divide your gravity in points by 97 points to get the equivalent kilograms of glucose per gallon of wort. Then multiply by the number of gallons to get the total kilograms of equivalent glucose. So if your gravity is 1.055 and you have 5 gallons of wort, then you have 55/97*5 = 2.84 kg of equivalent glucose that's available to ferment. If you assume that 75% of it ferments, then you get 0.569 m^3 = 569 liters of CO2 at room temperature and atmospheric pressure. No wonder my fermenter exploded last time I made that German ale...
