ENVE 731 - PROCESS FUNDAMENTALS OF ENVIRONMENTAL SYSTEM
EXERCISE QUESTIONS - 2
2.1)
KH = 722 × 10-3 atm·m3/mole
Since almost 21% of air in atmosphere consists of O2, mole fraction of O2 (XO2) in
atmosphere is 0.21.
Partial pressure of O2 = PO2 = XO2 × PT = 0.21 × 1 atm = 0.21 atm
Solubility of O2 can be found by using Henry’s Law; PO2 = KH × CS,O2
CS,O2 = PO2 / KH =
0.21 atm
= 0.291 mole/m3
-3
3
722  10 atm·m /mole
CS,O2 = 0.291 mole/m3 × 32 × 103 mg/mole × 1m3/1000liter = 9.33 mg/l
2.2)
(a) K H,N2 = 1.447 atm·m3/mole (from the book, page 43)
or you can find from N2 solubility K H,N2 =
PN2
1 atm

 0.0526 atm  L/mg
C N2 19 mg/L
Dry air contains roughly (by volume) 78.08% nitrogen, 20.95% oxygen, 0.93% argon,
0.038% carbon dioxide, and trace amounts of other gases. This shows that molar fraction
of nitrogen to air is about 0.781 and the partial pressure of N2:
Partial pressure of N2 = PN2 = XN2 × PT = 0.781 × 1 atm = 0.781 atm
PN2 = K H,N2 × CN2
CN2 = PN2 / K H,N2 =
0.781 atm
= 0.54 mole/m3
1.447 atm·m3 /mole
= 0.54 × 10-3 mole/L × 28 × 103 mg/mole
= 15.1 mg/L
(b) The size and the depth of the ocean make equilibrium difficult, although at the surface
waters equilibrium is likely. At lower levels, there is little mixing, and various chemicals
and biological reactions taking place in these zones either produce or consume N2.
Therefore, it is unlikely to have equilibrium with atmosphere.
2.4)
(a) Most soluable to least soluable in water (with respect to KOW values):
nitrobenzene > benzene > chlorobenzene > orthoxylene
1
ENVE 731 - PROCESS FUNDAMENTALS OF ENVIRONMENTAL SYSTEM
EXERCISE QUESTIONS - 2
(b) Most soluable to least soluable in organic liquids (with respect to KOW values):
orthoxylene > chlorobenzene > benzene > nitrobenzene
(c) Most volatile to least volatile (with respect to KH values):
benzene > orthoxylene > chlorobenzene > nitrobenzene
2.5)
Because Cs = 1.1 mg/L, the 500 ml of water can dissolve no more than 0.55 mg before it
reaches saturation conditions. The solid phenanthrene remaining will be 9.45 mg at
equilibrium with water. No phenanthrene will escape to air, because solid form is denser
than water and it will settle to the bottom of the flask. The KH is too low for volatilization
from water phase.
Thus C = Cs = 1.1 mg/L
2.6)
Initial n-decane =
7.4 g
= 0.052 moles
142 g/mole
Equilibrium will be reached in 10 weeks, thus;
Pdec = PV,dec = 1.74×10-5  0.052
Cdec, in air =
PV,dec
RT
=
1.74  10 5 atm
= 7.12×10-5 mole/L
0.082 atm  L/mole K  298K
Liquid n-decane left = 0.052 moles - 7.12×10-5 mole  0.052 moles
12.1)
log KOW, CCl4 = 2.73 → KOW, CCl4 = 102.73 = 537
log KOW, C6H4Cl2 = 3.38 → KOW, C6H4Cl2 = 103.38 = 2399
ρCCl4 = 1.594 g/mL
Total amount of added CCl4 = 0.05 mL × 1.594 g/mL = 0.0797 g = 79.7 mg (A+B)
Total amount of added C6H4Cl2 = 0.02 g = 20 mg (C+D)
2
ENVE 731 - PROCESS FUNDAMENTALS OF ENVIRONMENTAL SYSTEM
EXERCISE QUESTIONS - 2
KOW, CCl4 = 102.73
A mg
= 0.2 L
B mg
0.3 L
KOW, C6H4Cl2 = 103.38
A
 358
B
C mg
= 0.2 L
D mg
0.3 L
A = 79.48 mg and B = 0.22 mg
A
 1599
B
C = 19.9875 mg and B = 0.0125 mg
In octanol:
C CCl4 = 79.48 mg / 0.2 L = 397.4 mg/L (  99.7%)
C C6H4Cl2 = 19.9875 mg / 0.2 L = 99.9 mg/L (  100%)
In water:
C CCl4 = 0.22 mg / 0.3 L = 0.73 mg/L (  0.28%)
C C6H4Cl2 = 0.0125 mg / 0.3 L = 0.041 mg/L
12.7)
Octane, C8H18, MW = 114 g/mole, Pv = 0.015 atm
(a)
At equilibriums condition the gas pressure of empty space of the tank reaches to vapor
pressure of octane. Then:
PvVempty = noctaneRT
noctane =
Pv Vempty
RT

0.015 atm  70L  0.75
 0.033 mole
0.082 atm  L/mol K  294K
Moctane = noctane × MWoctane = 0.033 mole × 114 g/mole = 3.76 grams
(b)
Poct = Pv × Xoct = 0.015 atm × 0.002 = 3.0 × 10-5 atm
PvVempty = noctaneRT
noctane =
Pv Vempty
RT
3.0  105 atm  70L  0.75

 6.5 ×10-5 mole
0.082 atm  L/mol K  294K
Moctane = noctane × MWoctane = 6.5 ×10-5 mole × 114,000 mg/mole = 7.45 mg
3