Environmental Engineering Reference
In-Depth Information
4.585Torr. The boiling point of the liquid
T
b
at a given pressure is the temperature
at which
P
P
l
while the normal boiling point is that at which
P
l
1 atm. BA
is the solid-vapor equilibrium line. Similar to the normal boiling point for liquids,
the normal melting point is the temperature at which
P
s
=
=
1 atm. Beyond point C,
liquid and vapor phases cannot co-exist in equilibrium; this is called the
critical
point.
The critical temperature,
T
c
, critical pressure,
P
c
, and critical volume,
V
c
, are
unique to a compound. For water
T
c
is 647 K and
P
c
is 218 atm. The phase above
the critical point of a compound is called
super
-
critical state
. Extending the line
BA beyond A to B
A, one obtains the hypothetical
sub-cooled liquid state
, which
is important in estimating the solubility of a solid in a liquid. The ratio of the
sub-cooled liquid vapor pressure to the solid vapor pressure is the fugacity of the
solid.
The equilibrium at any point along the equilibrium
P
-
T
line in Figure 3.6 can
be described in terms of chemical potentials. If I and II represent two phases in
equilibrium, then we have the following criteria:
=
S
I
d
T
V
I
d
P
S
II
d
T
V
II
d
P
.
−
+
=−
+
(3.32)
Hence
d
T
=
Δ
V
=
Δ
d
P
S
H
V
,
(3.33)
Δ
T
Δ
where
V
are the entropy and volume changes for the phase transitions I
and II. We also used the definition of entropy change,
Δ
S
and
Δ
Δ
S
= Δ
H/T
.
For molar changes we then have the following equation:
d
P
d
T
=
Δ
H
m
T
Δ
V
m
.
(3.34)
This is the
Clausius-Clapeyron equation
. For both vapor-liquid and vapor-solid
equilibrium it gives the change in vapor pressure with temperature (Table 3.8). Since
Δ
vapor transitions, d
P
/d
T
is always positive, that is, both liquid and solid vapor pressure increase with temper-
ature. For both liquid
H
m
and
Δ
V
m
are positive for liquid
→
vapor and solid
→
vapor transitions, the molar volume of
the gas is much larger than that of the liquid or solid. Hence,
→
vapor and solid
→
V
m
=
Δ
V
m
≈
RT/P
.
Thus
dln
P
d
T
=
Δ
H
m
RT
2
.
(3.35)
Δ
The quantity
H
m
is not independent of
T
. However, over small ranges of
temperature, it can be assumed to be constant. Then we can integrate the above
equation to obtain
ln
P
2
P
1
1
T
2
−
.
=−
Δ
H
m
R
1
T
1
(3.36)
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