Systems with aqueous-electrolyte-solution liquid phases
The treatment presented in the previous section can be extended directly to elec-
trolyteH2O systems. For example, eq 1 for a NaClH2O system can be written
*g
*g
*g
dpg + d H
0 = Sm,H
(14)
dT Vm,H
2O
2O
2O
0 = SmdT Vmdp + xH2OdH2O + xNaCld l aCl
l
l
l
l
l
(15)
N
0 = Sm,H2OdT Vm,H2Odps + d*s 2O .
*s
*s
(16)
H
l
l
For a simple NaCl aqueous solution, xNaCl = 1 xH2O , the following relation is de-
rived by steps similar to those that led to eq 9:
S*s
Sm xH2OSm,H2O
*g
l
l
*s
m,H2O Sm,H2O
dT
l
Vm,H O V *g
Vm xH2OVm,H2O
l
*s
*s
m,H2O
2
*g
ls
sg
Vm,H
l
sg dAs,m
ls dAs,m
Vm
2O
=
d γ
d γ
l
g
l
*g
l
*s
*s
dVm Vm xH 2OVm,H 2O
dVm
Vm,H
Vm,H 2O
2O
l
xNaCl
d l aCl
+
N
l
l
*s
.
(17)
Vm
xH2OVm,H2O
dAs,g
s
*g
m
l
>> Vm,H2O
sg
= 0,
and by assuming that γ
As before, by noting that
Vm,H O
g
2
dVm
the following relation is obtained:
ls
(
) dT
ls dAs,m
l
l
*s
l
xNaCld l aCl
l
γ
= Vmd
l
Sm
xH2OSm,H2O
(18)
N
dVm
which yields on rearrangement
ls
d l aCl
ls dAs,m
l
l
l
*s
l
= S xH2OSm,H2O + xNaCl
γ
N
dT .
(19)
l m
Vmd
dT
dVm
The ice-solution capillary pressure can be calculated as a function of temperature
with
d l aCl
l
l
*s
l
+ xNaCl
N
T Sm
xH2OSm,H2O
dT
pc = ∫
dT .
(20)
l
Vm
T0
To calculate ice-solution capillary pressure, the quantities in the integrand must be
calculated. These quantities are
1. The mole fractions of the solute and solvent
2. The melting point of the solution phase in bulk
3. The molar entropy of ice
4. The molar entropy of the liquid solution
5. The molar volume of the liquid solution
6. The temperature derivative of the solute chemical potential.
4
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