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```					                Chapter 4
Vapor Pressure
pº = Pressure of a substance in equilibrium
with its pure condensed (liquid or solid) phase

Why do we care?
-spills
-pesticide application
-will lead us to Henry’s law constant
KH = PoL/Csatw
Kow = Csato/Csatw                          Air
Koa = Csato/PoL                 A gas is a gas is a gas
T, P

KH                                        Koa

Po L                        Octanol
Water                                                NOM, biological lipids,
other solvents
Fresh, salt, ground, pore                   Kow          T, chemical composition
T, salinity, cosolvents

Pure Phase              Csato
Csat   w        (l) or (s)
Ideal behavior
Ranges of pº (atm)
– PCBs – 10-5 to 10-9
– n-alkanes – 100.2 to 10-16
• n-C10H22 ~ 10-2.5
• n-C20H42 ~ 10-9
–   Benzene ~ 10-0.9
–   toluene ~10-1.42
–   Ethylbenzene ~ 10-1.90
–   propyl benzene ~ 10-2.35
–   carbon tetrachloride ~ 10-0.85
–   methane 102.44

• Even though VP is
“low”, gas phase may
still be important.
Phase diagram

picture of three-phase
diagram
Ideal Gas Law

pV  nRT
p = pressure
V = volume
n = moles of gas
R = gas constant
T = temperature (Kelvin)
Thermodynamic considerations
(deriving the van’t Hoff equation)
consider a gas: if T or P is changed and
equilibrium is re-established:
d1  d 2
the change in chemical potential in the two
systems is equal

d1  S1dT  V1dp     where S = molar entropy
d2  S 2 dT  V2 dp and V = molar volume
dp ( S1  S 2 ) S12
            
dT (V1  V2 ) V12
at equilibrium
G12  1   2  0
G  H  TS

substituting:
dp H 12

dT TV12
for a liquid vaporizing, the volume change can be
assumed to be equal to the volume of gas produced,
since the volume of the solid or liquid is negligible
RT         Q. where did the n go?
V12  Vgas    0         A. this is molar volume
p
dp 0 p 0 (H12 )                    recall (calculus!)
       2                      d ln u 1 du
dT        RT                              
d ln p 0 H12       The van’t        dx     u dx
           Hoff equation
dT      RT 2
where H12 = Hvap (gas) or Hsub (solid)
= energy required to convert one mole of liquid
(or solid) to gas without an increase in T.
Hvap is a function of T.
As T approaches the boiling point, Hvap increases rapidly
At T < boiling point, Hvap increases slowly
from 0-40ºC, Hvap can be assumed to be constant
d ln p 0 H12

dT      RT 2

integrate assuming
Hvap is constant:
Antoine equation
A
ln p    B
0

T
H12
ln p 
0
a
RT
if Hvap is not constant:
b
ln p  
0
a
T c    another Antoine equation
Using Hvap to predict VP at other
temperatures
As we saw in the thermodynamics lecture:

KT 2 H  1 1 
ln           
T T 
KT 1   R  1 2 

Specifically,

pT 2 H vap  1 1 
ln              
pT 1   R  T1 T2 
     
Note the change in slope when the substance is solid
(sublimation)
Hsub = Hmelt (~25%) + Hvap (~75%)
still use liquid phase as reference:
Hypothetical subcooled liquid
= liquid cooled below melting point without crystallizing
compound               pºs    <       pºL
1,4-dichlorobenzene    3.04           2.76
-log P       phenol                 3.59           3.41
22’55’ PCB             7.60           6.64
22’455’ PCB            8.02           7.40

Becomes important later when we talk about solubility
Molecular interactions
affecting vapor pressure
Molecule:molecule interactions in condensed phase
(L or s) have greatest affect on VP
strong interactions lead to large Hvap, low VP
weak interactions lead to small Hvap, high VP
Intermolecular interactions can be classified into
three types:
van der Waals forces (nonpolar)
Polar forces
Hydrogen bonding
van der Waals forces
• nonspecific
• function of size (number of
electrons)
• consist of:
– London dispersive energies
• fleeting areas of charge
– Induced dipoles                         “nonpolar”
• areas of charge arising from
interactions with a polar molecule
Polar interactions:
dipole-dipole interactions
• permanent areas of charge on two molecules
attract

Hydrogen bonds
Specific
donors and acceptors

table 4.3
Part of Table 4.3
compound (class)       (H-donor)  (H-acceptor)
alkanes                         0             0
1-alkenes                       0          0.07
aliphatic ethers                0          0.45
aliphatic aldehydes             0          0.45
aliphatic alcohols           0.37          0.48
carboxylic acids             0.60          0.45
benzene                         0          0.14
phenol                        0.6          0.31
naphthalane                     0           0.2
fluorene                        0           0.2
pyrene                          0          0.29
DCM                           0.1          0.05
Water                        0.82          0.35
Vapor Pressure Estimation Technique

based on regression of lots of VP data, best fit gives:

                n 1  
2

 
2
2/3
ln piL  4.49 V iL          
 n  2    15.1( i )( i )  14.5
*                            Di

                        
2
                 Di

size                             H-bonding
polarizability      ability
pressure in Pa, where:
V iL  molar volume (MW/density)
nDi  refractiveindex
refractive index (response to light) is a function of polarizability.
see table 3.1, also might be available in the CRC
Refractive
index
Difference
between
polarity and
polarizability
Trouton’s rule
At their boiling points, most organic compounds have a
similar entropy of vaporization:          exception: strongly
polar or H-bonding
Svap (Tb) = 85 – 90 J/molK                  compounds

We can be slightly more accurate with Kistiakowsky’s
expression:
Svap (Tb) = KF(36.6 + 8.31ln(Tb))     J/molK Tb in K (eqn 4-20)

KF = 1 for most compounds

At the boiling point:       G  0  H vap  Tb S vap

So if we know Tb, we can estimate Hvap (at the boiling
point) fairly accurately
Table 4.2
Estimating VP at other T (need Hvap)
pT 2 H vap  1 1 
ln              
pT 1    R  T1 T2 
       
Recognize that Hvap is not constant.
Especially if Tb is high (> 100ºC), the
estimate of Hvap from Trouton/Kistiakowsky
may not be valid at the temperature of
interest.
Empirically, Hvap is a function of the VP:

Hvap (T1 )  a log piL (T1 )  b
*
FIG 4.7

From a data set of many compounds, Goss and
Schwarzenbach (1999) get:
Hvap (298K )  8.80 log piL (298K )  70.0
*
Less empirically,
assume Hvap is linearly proportional to T (i.e.
assume that the heat capacity, Cpvap is constant:
H vap (T )  H vap (Tb )  C p vap (Tb )  (Tb  T )   don’t let notation
confuse you.
(Tb) means at
the boiling point.
You do not
substitute this expression into the Clausius-             multiply Hvap by
Clapeyron equation and integrate from Tb to T:            the boiling point

H vap (Tb )  1 1 
ln P 
0
  
T T 
R       b    
C p vap (Tb )  T  C p vap (Tb )  T 
                1  b           ln  b 
R           T         R        T 
Recall:         H vap (Tb )  Tb S vap (Tb )
substitute:
 S vap (Tb ) C p vap (Tb )   Tb 
ln p  
0
                 1  
       R           R           T 
                             
C p vap (Tb )  T 
               ln  b 
R           T 

but we still need to know Cp(Tb)!
generally:
C p (Tb )  0.8  S vap (Tb )   ranges from 1.0 to 0.6

and Svap(Tb)~ 88 J/molK
finally!
  Tb              Tb 
ln p   K F (4.4  ln Tb ) 1.8  1  0.8 ln  
0

 T                T 
KF is the Fishtine factor, usually 1, but         in atm
sometimes as high as 1.3 (see p 113)              Eqn 4-33

the old edition gave (where KF =1):
 Tb            Tb 
ln p  19  1    8.5 ln  
0
in atm
 T            T 
OK for liquids with Tb < 100ºC
High MW compounds, need correction for
intermolecular forces (but we don’t have their boiling
points anyway!) (For refinements see equation 4-33)
Can estimate boiling points, see p. 120
solids?
those previous equations yielded the vapor pressure of
the hypothetical subcooled liquid.
How can we correct this to give the true vapor
pressure of a solid?
Prausnitz (1969):              S fus (Tm )  Tm 
ps0
ln 0                  1
pL          R       T   
Where Sfus(Tm) = entropy of fusion at melting point
unfortunately Sfus is much more variable than Svap
S fus (Tm )  (56 .5  9.2  19 .2 log  )   J/molK
Where  = number of torsional bonds and
 = rotational symmetry number (see p. 125)
the older edition of your book gave this simpler
(but less accurate) equation:

Sfus(Tm) ~ 56.5 + 10.5(n-5)             J/molK

Where n = number of flexing chain atoms.
if n<5, then ignore this term
Estimation of vapor pressures for polychlorinated
biphenyls: a comparison of eleven predictive methods
Lawrence P. Burkhard, Anders W. Andren, and David E.
Armstrong
Environmental Science and Technology 1985, 19, 500 -
507
conclusions:
• non-correlative methods have poor predictive ability (error
increases as VP decreases)
• correlative methods requiring a set of compounds with
known P are much better
• best method: determine VP as function of GC retention
times
Determination of vapor pressures for
nonpolar and semipolar organic compounds
from GC retention data (Hinckley et al, 1990)
• Chromatographed 2 reference compounds (eicosane
and p,p’DDT) having known VP and Hvap versus a host
of unknowns (PAHs, organochlorines, etc)
• Isothermal runs allow determination of RRT at several T
• Comparison of RRT with reference compounds allows
determination of VP at given T
• Comparison of changes in RRT with T and knowledge
of Hvap for reference compound allows calculation of
Hvap for all unknowns
Problem 4.2
• In a dump site, you find an old 3-liter pressure bottle
containing FREON 12 with a pressure gauge that
reads 2.7 bar. (First, you realize that this gauge
was not manufactured in the US.) The temperature
is 10ºC. What mass of FREON 12 is in the bottle?
• Also estimate the free energy, enthalpy, and entropy
of condensation of FREON 12.
• You find the following info for FREON 12 in the
CRC:
T deg C p/kPa
-25         123
0         308
25         651
50       1216
75       2076
Problem 4.6
• estimate VP at 0C based on VP at 25ºC
or based solely on Tb and Tm

log(p) @25C Tm (degC) Tb (degC)
dimethyl phthalate          0.38     5.5      283.7
2,3,7,8-TCDD                 -6.7    305      446.5
(hint  = 4)
Homework
• Do problems 4.3 and 4.4
• Due 2/2/10

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