# Equilibrium value for carbonation level in beer

## Homework Statement:

A beer is matured in a vessel at 5 degrees. The vessel has diameter of 5m. Top pressure is 0.1 bar g CO2. Fill level is 100%. Calculate equilibrium value for average carbonation level in beer in g/L.

Data: Atm pressure = 101.3kPa
Henrys constant for CO2 in beer at 5 deg C = 15.75 atm litre mol-1
Acc due to gravity = 9.81 ms-1
Beer density at 5 deg = 1010 kg m-3
Mol mass of CO2 = 44 g mol-1

## Relevant Equations:

Absolute pressure = Gauge press + Atm press + Hydrostatic press
Abs Pressure:
Gauge press: 1 bar G = 100,000 Pa = 0.987 atm
0.1 bar G = 0.0987 atm

Atm press = 101.3 kPa = 1 atm

Hydrostatic press: average CO2 occurs 2.5m up the vessel
(9.81 ms-2)(1010kgm-3)(5/2) = 24,770.25 Pa
1 Pa = 9.869x10^-6 atm
24,770.25 Pa = 0.2445 atm

Abs press= 0.0987 atm + 1 atm + 0.2445 = 1.3432 atm

Henrys Law: P = KhC
1.3432 atm = 15.75 atm litre mol-1 C
C = 0.0853 mol-1 / L

44g CO2 in 1 mole
3.752g CO2 in 0.0853 mol-1 / L

Equilibrium value = 3.752 g/L

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Chestermiller
Mentor
It looks to me like your approach was correct. I haven't checked the arithmetic, however.

haruspex