Civil Engineering Reference
In-Depth Information
Values for Henry's constant at 68
F (20
C) and the associated temperature correc-
tion factors are provided in Table 9-1 for various gases. Examples 9-1 and 9-2 illus-
trate the use of Henry's law to calculate the equilibrium concentration of gases in
water and an enclosed reservoir. The lower the value of Henry's constant, the more
soluble the gas. The polarity and molecular weight of a gas strongly affect its solubility,
with more polar and higher-molecular-weight gases being more soluble.
Example 9-1: Solubility of Oxygen in Water
The partial pressure of oxygen in the
atmosphere is approximately 0.21 atm. Determine the concentration of oxygen in water
at 68
F
(
20
C
)
and at 41
F
(
5
C
).
Table 9-1 shows that, at 20
C, H
4.3
10
4
atm for oxygen. Therefore,
C
(0.21 atm) / (4.3
10
4
atm)
4.9
10
6
mol O / mol H O
2
2
To convert to milligrams O
2
per liter,
C
(4.9
10
6
mol O / mol H O)(55.56 mol H O / L)(1,000 mg / g)(32 g / mol)
2
2
2
8.7 mg O / L
2
By using the van't Hoff equation, determine the Henry's constant at 5
C. From Table
9-1,
H
1.45
10
3
kcal / kmol and
b
7.11.
1.45
103 kcal / kmol
log
H
7.11
(1.987 kcal / K
kmol)(278 K)
4.49
H
3.06
10
atm
4
C
(0.21 atm) / (3.06
10
4
atm)(55.56 mol H O / L)(1,000 mg / g)(32 g / mol)
2
12.2 mg O / L
2
Example 9-2: Hydrogen Sulfide in an Enclosed Reservoir
Hydrogen sulfide ex-
ists in groundwater at concentrations as great as 2 mg / L. Determine the concentration
of H
2
S in the headspace of an enclosed reservoir at 68
F
(
20
C
).
Table 9-1 shows that at 20
C,
H
5.15
10
2
atm for H
2
S. First, convert the
concentration in milligrams per liter to a mole fraction:
C
(2 mg / L) / [(55.56 mol H O / L)(1,000 mg / g)(34 g / mol H S)]
2
2
1.1
10
mol H S / mol H O
2
6
2
P
(1.1
10
6
mol H S / mol H O)(5.15
10
2
atm)
2
2
5.5
10
4
atm
550 ppm
(parts per million on a volume basis)
v
Note:
The Occupational Safety and Health Administration's peak 10-min exposure
limit is 50 ppm
v
.
