Chapter 4 Vapor Pressure
An important goal of this chapter is to
learn techniques to calculate vapor
pressures
To do this we will need boiling points and
entropies of vaporization
The saturation vapor pressure is defined as the
gas phase pressure in equilibrium with a pure
solid or liquid.
 fi =
 log
i X ipiL* pure liquid
C part
CgasTSP
*
  log piL
 const
 KiH = p*iL/Ciwsat
1
Vapor pressure and Temperature
dGliq = dGgas
from the 1st law H= U+PV
dH = dU + Vdp+pdV
dU= dq - dw
for only pdV work, dw = pdV and from the
definition of entropy, dq = TdS
dU = TdS –pdV
from the general expression of free energy
dG = dU + Vdp+pdV- SdT-TdS
substituting for dU
dG= +VdP - SdT
2
The molar free energy Gi/ni = i
for a gas in equilibrium with a liquid
dliq = dgas
di liq = Vi liqdp - Si liqdT
Vi liqdpi - Si liqdT = Vigasdp - SigasdT
d/dT = (Sigas -Si liq)/Vigas
at equilibrium G = H -S T= zero
so (Sigas -Si liq) = Hi vap/T
substituting
dpi /dT = Hi /( Vigas T)
(Clapeyron eq)
substituting
Vi gas = RT/pi
dpi Hi pi
Hi
1

; d (ln pi )  d (
)
T R
dT RTxT
ln pio
Hi 1

 const
R T
3
Figure 4.3 page 61 Schwartzenbach
4
5
This works over a limited temperature range w/o any
phase change
Over a larger range Antoine’s equation may be used
lnp*i 
B
A
TC
over the limits P1 to P2 and T1 to T2
P
log P2
1

Hvap ( T2  T1)
2.303RT1T2
If the molar heat of vaporization, Hvap of hexane
equals 6896 cal/mol and its boiling point is 69oC, what
is its vapor pressure at 60oC
log
760
6896 cal / mol(342K  333K )

P1
2.303  1.99cal / Kmol  333K  342K
P1= 578 mm Hg
6
Below the melting point a solid vaporizes w/o melting,
that is it sublimes
A subcooled liquid is one that exists below
its melting point.
 We often use pure liquids as the
reference state
 logKp
Log p*i
7
Molecular interaction governing vapor pressure
As intermolecular attractive forces increase
in a liquid, vapor pressures tend to decrease
van der Waals forces
generally enthalpies of vaporization
increase with increasing polarity of the
molecule
Both boiling points and entropies of
vaporization become important parameters in
estimating vapor pressures
8
A constant entropy of vaporization
Troutons rule
Figure 4.5; at 25oC
This suggests that vapS may tend to be
constant
At the boiling point is vapSTb constant?
9
H
const slope = S
T
Tb oC
vapH
kJ mol-1
68.7
174.1
78.3
218
339
80.1
28.9
38.8
38.6
43.7
53.0
30.7
Chlorobenzene 131.7
Hydroxybenzene 181.8
35.2
45.7
vapS
kJ mol-1K-
1
n-hexane
n-decane
ethanol
naphthalene
Phenanthrene
Benzene
10
Predicting vapSTb
Kistiakowsky derived an expression for the entropy of
vaporization which takes into account van der Waal
forces
(eq 4-20)
 Svap= 36.6 +8.31 ln Tb
 for polarity interactions Fistine proposed
Svap= Kf (36.6 +8.31 ln Tb)
Kf= 1.04; esters, ketones
Kf= 1.1; amines
Kf= 1.15; phenols
Kf= 1.3; aliphatic alcohols
Calculating Svap using chain flexibility and
functionality (Mydral et al, 1996)
vapSi (Tb) = 86.0+ 0.04  + 1421 HBN
(eq 4-21)
 = (SP3 +0.5 SP2 +0.5 ring) -1
11
SP3 = non-terminal atoms bonded to 4 other
atoms (unbonded electrons of O, NH, N, S, are
considered a bond)
SP2 = non-terminal atoms singly bonded to two
other atoms and doubly bonded to a 3rd atom
Rings = # independent rings
HBN = is the hydrogen bond number as a
function of the number of OH, COOH, and NH2
groups
HBN 
OH  COOH  .33 NH2
MW
12
13
A more complicated method:
From Zhao, H.; Li, P.; Yalkowski, H.; Predicting the
Entropy of boiling for Organic Compounds, J.
Chem. Inf. Comput. Sci, 39,1112-1116, 1999
Sb= 84.53 – 11 +.35 + 0.052 + i
where:
i = the contribution of group i to the Entropy of
boiling
= the molecular planarity number, or the # of
non-hydrogen atoms of a molecule that are
restricted to a single plane; methane and ethane
have values of 1 and 2; other alkanes, 3; butadiene,
benzenes, styrene, naphthalene, and anthracene
are 4,6,8,10,14
 measures the conformational freedom or
flexibility ability of atoms in a molecule to rotate
about single bonds
= SP3 + 0.5(SP2) +0.5 (ring) –1
14
 = symmetry number; the number of identical
images that can be produced by a rigid rotation of a
hydrogen suppressed molecule; always greater than
one; toluene and o-xylene = 2, chloroform and
methanol =3, p-xylene and naphthalene = 4, etc
15
17
Boiling points can be estimated based on
chemical structure (Joback, 1984)
Tb= 198 + Tb
T (oK)
-CH3
= 23.58 K
-Cl
= 38.13
-NH2 = 73.23
C=O
= 76.75
CbenzH- = 26.73
acetonitrile
acetone
benzene
amino benzene
benzoic acid
toluene
pentane
methyl amine
trichlorethylene
phenanthrene
Joback
(K)
347
322
358
435
532
386
314
295
361
598
obs
(K)
355
329
353
457
522
384
309
267
360
613
18
Stein, S.E., Brown, R.L. Estimating Normal boiling Points
from Group Contributions, J.Chem. Inf Comput. Sci, 34,
581-587, 1994
They start with Tb= 198 + Tb and go to 4426
experimental boiling points in Aldrich
And fit the residuals (Tb obs-Tb calcd)
19
Tb= 198 + Tb
Tb(corr) = Tb- 94.84+ 0.5577Tb0.0007705Tb2 T b< 700 K
Tb(corr) = Tb+282.7-0.5209Tb
Tb>700K
20
Estimating Vapor Pressures
d ln p* vap H

dT
RT 2
To estimate the vapor pressure at a temp lower then
the boiling temp of the liquid we need to estimate
Hvap at lower temperatures.
Assume that Hvap is directly proportional to temp
and that Hvap can be related to a constant, the heat
capacity of vaporization Cp Tb
where vapH/T = Cp Tb
vapHT = vapH Tb + CpTb(T-TTb)
*
lnPiL 
Δ
H
vap Tb
R
(
1
Tb
at the boiling point
ln p iL (
*


 vap STb
R
1
)
ΔC pT
b
T
R
(1 
Tb
T
)
ΔC pTb
(ln
R
Tb
)
T
vapH Tb= Tb vapS Tb

C pTb
R
C pTb
Tb
Tb
)( 1  ) 
(ln
T
R
T
21
)

*
ln piL
(

vap STb
R

CpTb
R
CpTb
Tb
T
)( 1  ) 
(ln b )
T
R
T
for many organic compounds

Cp Tb/vapS Tb ranges from -0.6 to -1
so substituting Cp Tb= -0.8vapS Tb
ln p *iL 

SvapTb
R
[ 1 .8 ( 1 
Tb
T
) 0.8(ln b )]
T
T
if we substitute Svap Tb= 88J mol-1 K-1 and R =8.31
Jmol-1 K-1
ln
*
piL
Tb
Tb
19(1  ) 8.5(ln )] 
T
T
when using Cp Tb/Svap Tb = -O.8, low boiling
compounds (100oC) are estimated to with in 5%, but
high boilers may be a factor of two off
If the influence of van der Waal forces
(Kistiakowky)and polar and hydrogen bonding effects
(Fishtine’s correction factors) are applied

vapS Tb= Kf(36.6 +8.31 ln Tb)
22
*
ln piL
 K f ( 4.4  ln Tb )[1.8(
If we go back to:

*
ln piL
(

vap STb
R

Tb
T
 1) 0.8(ln b )] 
T
T
CpTb
R
CpTb
Tb
T
)( 1  ) 
(ln b )
T
R
T

vapSi (Tb) = 86.0+ 0.04  + 1421 HBN

and Mydral and Yalkowsky suggest that
vapCpi (Tb) = -90 +2.1 in J mol-1K-1

 = (SP3 +0.5 SP2 +0.5 ring) –1
*  (21.2  0.3  177 HBN )(1  Tb )
ln piL
Tb
 (10.8  0.25 )ln
T
HBN 
T 
OH  COOH  .33 NH2
MW
23
24
A vapor pressure calculation for the liquid vapor for
anthracene
ln

*
piL
Tb
Tb
19(1  ) 8.5(ln )] 
T
T
Tb= 198 + Tb ; for anthracene {C14H18}
C14H18
Has 10 =CH- carbons at 26.73oK/carbon
And 4, =C< , carbons 31.01OK/carbon
Tb= 589K; CRC = 613K
At 298K, ln p* = -12.76; p = 2.87 x10-6atm =
and p*iL = 0.0022 torr
What do we get with the real boiling point of 613K ?
ln p*iL = 8.7 x10-7atm
25
Solid vapor pressures (is there a relationship
between solid and liquid vapor pressures??)

For a solid to go directly to the gas phase (sublimation),we
can say that the free energy of sublimation(subGi) has to
be the sum of the free energy need to go from the solid to
the liquid (fusGi) + the free energy of going from a liquid to
a gas (vapGi)
subGi = fusGi + vapGi
also
subHi = fusHi + vapHi and subSi = fusSi + vapSi
back to free energy:fusGi = subGi - vapGi
for both vapGi and subGi
the change in free energy in going from one
temperature to another, like say ambient to the
melting point
phase j Gi = RT ln { pref / p*ij }
so in
fusGi = subGi - vapGi

fusGi = RTln { pismelt / p*is } - RTln { piLmelt / p*iL }

26

fusGi = RTln { p*iL / p*is }
this suggests a relationship between p*iL and p*is
27
Solid Vapor Pressures

subH = fusH + vapH
fusH= Tm fusS
 fusH (s)
fusS
fusH/ Tm = fusS = const?




T





Hsub = Hvap+ Tm Sfus
d ln p*
H

R
d1
T
It can be shown that
*
ln piL
 ln PiS* 
( S fus ) (Tm  Tamb )
R
Tamb
if S= const = 56.4 J mol-1K-1 and R=8.31 J mol1 -1
K , Sfus /R= 6.78
see if you can derive this
28
What is the solid vapor pressure for
anthracene?
Using the correct boiling point we determined the
liquid vapor pressure to be 8.71x10-7 atmospheres
ln
*
piL
 ln
*
piS
( S fus ) (Tm  Tamb )

R
Tamb
if S= const = 56.4 J mol-1K-1 and R=8.31 J mol-1K-1,
Sfus /R= 6.78
ln 8.71x10-7 = ln p*iS+ 6.78 (490.65-298)/298
-13.95 – 4.38 = ln p*iS
7.8x10-9= p*iS
29
Myrdal and Yalkowski also suggest that
a reasonable estimate of fus Si(Tm) is
fus Si(Tm) + 56.5+ 9.2  -19.2 log )
in J mol-1K-1
substitution in to
*
ln piL
ln
 ln PiS*
*
piL
*
piS
( S fus ) (Tm  Tamb )

R
Tamb
gives
 (6.8  1.1  2.3 log  ) (Tm  Tamb )

R
Tamb
30
Measuring solid vapor pressures
31
Using Sonnefeld et al, what is the sold vapor
pressure for anthracene at 289K
log10 p*iS = -A / T + B;
p*iS is in pascals
101,325 pascals = 1atm
A= 4791.87
B= 12.977
log10 p*iS = -4791.87 / T + 12.977
log10 p*iS = -16.0801 + 12.977 = -3.1031
Po = 7.88 x10-4 pascals
p*iS = 7.88 x10-4 /101,325 = 7.8x10-9 atm
32
A chromatographic Method for measuring liquid
vapor pressures
33
A chromatographic Method for measuring liquid
vapor pressures
A
B
Hamilton,J.Chrom.195,
75-83,1980
Hinckley et al., J. Chem.
Eng. Data, 1990;
Yamasaki, 1986
to
t1
t2
retention time A = t1- to=t’1
retention volume (Vr) = column flow x the retention
time and
t’1/t’2= VrA / VrB
and Vr varies inversely with the vapor pressure of a
compound
VrA / VrB = p*BL/p*AL
*
ln p BL
d
 ln p *AL
*
ln pBL

 ln
VrB
eq 2
VrA
vap HB
R
eq1
d(
1
)
T
eq 3
34
d
ln p*AL

vap H A
1 ;
d( )
T
R
eq 4
dividing eq 4 into 3 and integrating
*
ln pBL

ln p*AL
vap HB
vap H A
+ const
eq 5
we said that
*
ln pBL
 ln p *AL
 ln
VrB
eq 2
VrA
combining eqs. 2 and 5
ln
VrB
VrA
 ln
p*AL
(1
 vap H B
 vap H A
)  const
eq 6
1.If we can know or can calculate the vapor
pressure p*AL at different temperatures
2. And run our GC isothermally at these
temperatures and obtain of t’2/t’1 and hence VrB / VrA
3.we can plot ln VrB / VrA vs p*AL , get the slope and
intercept, and plug into eq 5 to obtain p*BL
35
Using vapor pressure and activity
coefficients to estimate organic gas-particle
partitioning (Pankow, Atmos. Environ., 1994)
Gas Atoxic + liquid particle  particle Atoxic +liquid
particle
Aipart
Aigas
Kip = Aipart / (Aigas xTSP);.
TSP has units of ug/m3
Aigas and Aipart have units of ng/m3
log Kip= -log p*iL + const based on solid-gas
partitioning
pi =  p*iL (in atmospheres)
pi = ni/V RTx760 = [Aigas] RTx 760; (mmHg)
[Aigas] = [PAHigas] = ni /V in moles/liter
multiplying by MW (g/mol) and multiplying by 1012=
ng/m3
36
[PAHigas] =  p*iL MWi x 109/(RT x760) in ng/m3
Let’s look at the mole fraction
imoles in the particle phase of i divided by
total moles in the particle phase
Usually we measure the particle phase of
compound i in ng/m3
[Aipart] = [PAHipart]
The number of moles in the particle phase is:
i
Moles = [PAHipart]/ {MWi 109 } = moles/m3
We usually measure TSP as an indicator of total
particle mass and the amt. of liquid in the
particle phase = TSP x fom
The average number of moles in the particle
phase requires that we assume an average
molecular weight for the organic material in the
particle phase, MWavg
total liquid moles of TSP( g/m3) =
37
fomTSP/ {MWavg 106 }
we said before
Moles = [Partipart]/ {MWi 109 } = moles/m3
i

iiMoles/ total moles =
[Partipart] MWavg / {fomTSP MWi 103}
we also said before that
[PAHigas] =
i p*iL MWi x 1012/( 760 RT) in ng/m3
and
Kip = PAHipart / (PAHigas xTSP)
Kip = 760 RT fomx10-6/{p*iLtorrWavg }
p*iL here is in torr
R = 8.2x10-5 m3 atm/(mol K)
38
p* =
log
Kp
BaP
slope = -1
pyrene
phenanthrene
ee
log p*iL-torr
39