Lei de Henry e Soluções

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122 5 SIMPLEMIXTURES
�e Henry’s law constant for CH4 in benzene is 44.4 × 103 kPa kg mol−1 and
the density of benzene is 0.879 g cm−3.
cN2 =
ρpN2
KN2
= (0.879 × 103 kgm−3) × (100 kPa)
44.4 × 103 kPa kg mol−1
= 1.97... mol m−3
�e molar concentration is therefore 2.0 × 10−3 moldm−3 .
E5A.10(b) In Section 5A.3(b) on page 152 it is explained that for practical applications
Henry’s law is o�en expressed as pB = KBbB, where bB is the molality of the
solute, usually expressed in mol kg−1.�e molality is therefore calculated from
the partial pressure as bB = pB/KB, and the partial pressure is expressed as
pB = xBp, where p is the total pressure�eHenry’s law constant for N2 in water
is 1.56 × 105 kPa kg mol−1, and for O2 the constant is 7.92 × 104 kPa kg mol−1
�e total pressure is assumed to be 1 atm (101.325 kPa)
bN2 =
xN2 p
KN2
= 0.78 × (101.325 kPa)
1.56 × 105 kPa kg mol−1
= 5.1 × 10−4 mol kg−1
bO2 =
xO2 p
KO2
= 0.21 × (101.325 kPa)
7.92 × 104 kPa kg mol−1
= 2.7 × 10−4 mol kg−1
E5A.11(b) As explained in Exercise E5A.9(b) the concentration of a solute is estimated
as cB = ρpB/KB where ρ is the mass density of the solvent. �e Henry’s law
constant for CO2 in water is 3.01 × 103 kPa kg mol−1 and the density of water
is 0.997 g cm−3 or 997 kgm−3.
cCO2 =
ρpCO2
KCO2
= (997 kgm−3) × (2.0 atm) × (101.325 kPa/1 atm)
3.01 × 103 kPa kg mol−1
= 67.1... mol m−3
�e molar concentration is therefore 0.067 mol dm−3 .
Solutions to problems
P5A.2 �e Gibbs–Duhem equation [5A.12b–146], expressed in terms of partial molar
volumes is nAdVA + nBdVB = 0.�is rearranges to
dVB = −
nA
nB
dVA
Division of both the top of bottom of the fraction on the right by (nA + nB)
allows this equation to be rewritten in terms of the mole fractions
dVB = −
nA/(nA + nB)
nB/(nA + nB)
dVA = −
xA
xB
dVA = −
xA
(1 − xA)
dVA