Page 898 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

Appendix H: Dissolved Gases 853

Example H.4 Determine Partial Pressure H.1.4 ATMOSPHERIC PRESSURE VERSUS ELEVATION
of Oxygen in Air
Figure H.2 gives atmospheric pressure as a function of eleva-
The mole fraction of oxygen in air is 0.2095 (Table H.1). tion, with plotting data obtained from Lide (1996, pp. 14–17).
Equation H.6, a best fit polynomial that accurately depicts the
a. Determine the partial pressure of oxygen at sea plot of Figure H.2, was from the Kladiographt software,
level. which was used to develop the plot from the data provided.
1. Apply Dalton’s law for p(air, sea level), The actual pressure at any elevation may vary depending on
the local conditions. For example, a mercury barometer
n(O 2 ) located at the Engineering Research Center, Colorado State
p(air, sea level)
p(O 2 ) ¼
n(air) University, Fort Collins, Colorado, at a ground elevation of
0:2095 mol O 2 1585 m (5200 ft) reads 634 mm Hg (which varies a few mm
1:0 atm air
mol gas mixture(air) Hg from day-to-day with weather conditions); when com-
¼
pared, for this same elevation, Equation H.6 calculated 83.7
¼ 0:2095 atm O 2
kPa, or 628 mm Hg (a 0.9% discrepancy). For any given
b. Determine the partial pressure of oxygen at eleva- elevation, the pressure will vary about a mean as weather
tion, 1800 m, conditions change.
1. Apply Dalton’s law for p(air, 1800 m) ¼ 81.49
kPa (Equation H.24 or Figure H.2), which is P(atm) ¼ M0 þ M1 Z þ M2 Z þ M3 Z 3
2
0.804 atm,
5
4
þ M4 Z þ M5 Z þ M6 Z 6 (H:6)
n(O 2 )
p(air, 1800 m)
p(O 2 ) ¼
n(air)
where
0:2095 mol O 2 P(atm) is the atmospheric pressure (kPa)
0:804 atm
mol gas mixture (air) Z is the elevation (m)
¼
M0, M1, M2, M3, M4, M5, M6 are polynomial
¼ 0:168 atm O 2 (17:0 kPa O 2 )
coefficients
Comments M0 ¼ 101.325
In applying Henry’s law, the partial pressure of the gas M1 ¼ 0.011944
species of interest must be determined first. M2 ¼ 5.3142 10 07
TABLE H.1
Composition of Air and Calculation of Molecular Weight
Gas Law Data P(atm) ¼ 101,300 Pa R ¼ 8.31451 N m=K mol T ¼ 208C ¼ 293.15 K
e
a
3
d
3
c
Gas a MW (g=mol) X(gas) b r(molar) (mol=m ) r (kg=m ) MW Fraction (g gas=mol air)
N 2 28.0134 0.78084 32.45226 0.9091 21.8740
31.9988 0.209476 8.70597 0.2786 6.7030
O 2
Ar 39.948 0.00934 0.38818 0.0155 0.3731
44.0098 0.000314 0.01305 0.00057 0.0138
CO 2
Ne 20.1797 0.00001818 0.00076 1.5247E 05 0.00037
He 4.0026 0.00000524 0.00022 8.7168E 07 2.0974E 05
Kr 83.80 0.00000114 0.00005 3.9703E 06 0.000095
Xe 131.29 0.000000087 0.00000 4.7472E 07 1.1422E 05
16.0428 0.000002 0.00008 1.3335E 06 3.2086E 05
CH 4
H 2 2.01588 0.0000005 0.00002 4.1891E 08 1.0079E 06
b 8
O 3 47.999 1.0 10
Rn b 222 6.0 10 20
Sum 0.999997147 41.5606 1.2038 28.964 f
a
Lide, D. R., Handbook of Chemistry and Physics, 77th edn., CRC Press, Inc., Boca Raton, FL, 1996, pp. 4–37:98.
b
Weast, R. C. (Ed.), Handbook of Chemistry and Physics, 59th edn., CRC Press, Inc., Boca Raton, FL, 1978, p. F205.
c
r(molar) ¼ n=V ¼ P(gas)=RT ¼ X(gas) P(atm)=RT.
d
r ¼ (molar) MW(gas)=1000.
e
MW fraction ¼ MW(gas) X(gas).
f
MW(air) ¼ sum[MW(gas) i X(gas) i ].