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Equilibrium value for carbonation level in beer

  • Thread starter sci0x
  • Start date
  • #1
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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

Am i correct here please?
 

Answers and Replies

  • #2
20,110
4,191
It looks to me like your approach was correct. I haven't checked the arithmetic, however.
 
  • #3
haruspex
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5,273
mol-1 / L
You mean mol/L, right?
As a matter of style, there are many benefits in working entirely symbolically, only plugging in numbers at the final step.
Also, you cannot justify that many significant digits in the answer. Some values are only given to one sig fig, but then there is the complication of adding the gauge pressure to atmospheric, which makes keeping track of the precision tricky. Looks like the final answer can only be claimed ±5%.
 
Last edited:

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