Page 213 - Materials Chemistry, Second Edition

196 Practical Design Calculations for Groundwater and Soil Remediation

Example 5.24: Determine the Oxygen Concentration in Air
Determine the mass concentration of oxygen in ambient air at 20°C. Express
the answer in the following units: mg/L, g/L, and lb/ft .
3
Solution:
The oxygen concentration in the ambient air is approximately 21% by
volume, which is equal to 210,000 ppmV. Equations (2.1) or (2.2) can
be used to convert it to a mass concentration,

MW
3
°
1 ppmV= [mg/m] at T = 20 C
24.05
(2.1)
32
= = 1.33 mg/m = 0.00133 mg/L
3
24.05
or
MW
3
1 ppmV = x10 − 6 [lb/ft] at T =68F°
385
(2.2)
32
−
−
6
6
= x10 = 0.083 x10lb/ft 3
385
Therefore,
210,000 ppmV = (210,000)(0.00133 mg/L) = 279 mg/L = 0.28 g/L
= (210,000)(0.083 × 10 ) = 0.0175 lb/ft 3
−6
Discussion:
The oxygen concentration in the ambient air, 279 mg/L, is much higher
than the saturated dissolved oxygen (DO) concentration in water, 9
mg/L at 20°C.

Example 5.25: Determine the Necessity of Oxygen Addition
for In Situ Soil Bioremediation
A subsurface contains 5,000 mg/L of gasoline. The air in the subsurface is
relatively stagnant. The total bulk density of the soil is 1.8 g/cm ; the degree
3
of water saturation in the soil is 30%; and the porosity is 40%.
Demonstrate that the oxygen in the soil void is not sufficient to support the
complete biodegradation of the intruding gasoline.