ALWAYS include units!!!
let's start with pressure
P @10m = P CO
2 guage + P hydrostatic
= Patm + PCO
2 + ρgh
Patm = 101325 Pa
PCO
2 = 0.1 bar * (10
5 Pa / bar) = 10000 Pa
Phydro = (1008 kg/m
3) * (9.80m/s
2) * (10m) * (1Pa / 1kg/ms
2) = 98784 Pa
Ptotal = 101325Pa + 10000 Pa + 98784 Pa = 210109 Pa = 2.10x10
5 Pa
then let's find kH...
I'm going to use a reference value first to verify the 6.1g/L # kH* for CO
2 in H
2O @ 25°C = 29 atm/M * (101325 Pa / atm) = 2.94x10
6 Pa/M @25°C
* note: that is from the wikipedia henry's law page.
then at 2°C,
kH @2°C = kH@25°C * exp((-dHsol/R)(1/T
2 - 1/T
1))
= 2.94x10
6 Pa/M * exp(-2400K)*(1/275K - 1/278K))
= 1.50x10
6 Pa/M
and finally
[CO
2] = 2.10x10
5 Pa * (1M / 1.50x10
6 Pa) * (44.01g/mol) = 6.2 g/L
meaning.. we
verified the answer of 6.1g/L using an alternate kH
*********
now let's go back to your solution
(1) ALWAYS INCLUDE UNITS !!!
(2) your atmospheric pressure is 10
5 Pa not 105 Pa. 101325 Pa
(3) take a look at your kH CO
2 @ T = 2° value.. that's NOT Pa / M
that value is Pa per MOLE FRACTION of CO
2 in the liquid phase <
sigh>
**********
ok.. so let's see if we can solve this with YOUR kH value
We're going to simplify this by making 2 assumptions
(1) that beer is essentially water
(2) The mole fraction of CO
2 in H
2O is going to be very small
(I can easily show - and you should be able to also - that 6g/L is about 0.14 mol CO
2 per 55.5 mol H
2O. So that is a reasonable assumption).
making this equation approximately true.
mol CO
2 mol CO
2 χ CO
2 = ----------------------- ≈ ------------
mol CO
2 + mol H
2O mol H
2O
and now I can rewrite that kH value as
84.1x10
6 Pa 84.1x10
6Pa * (mol H
2O)
kH = -------------- ≈ ----------------------------
χ CO
2 (mol CO
2)
and now we can finish up.
2.10x10
5 Pa mol CO
2 44.01g CO
2 55.5 mol H
2O
[CO
2] = -------------- * ------------------------------ * -------------- * --------------- = 6.1 g / L
1 (84.1x10
6 Pa) * (mol H2O) mol CO
2 1 L H
2O