Solutions Exercises Lecture 4/II – AES1030
1) The vapour pressure of n-hexane (C6H14) is described by the Antoine equation given
below:
•, vap
ln ( phexane
[bar ]) = 16.6734 −
5000.758
T [K ]
The molar enthalpy of evaporation of hexane at 180 K and its specific vapour pressure
•
•
∆ H hexane
, vap (180 K , phexane ) = 37.79
kJ
mol
Further, sublimation pressures of hexane at two different temperatures are given:
is given:
• , sub
phexane
= 7.49 ⋅10−6 bar at 176 K
• , sub
phexane
= 9.11⋅10−6 bar at 177 K
a. Compute the molar sublimation enthalpy starting with the Clausius-Clapeyron
equation. Give all the assumptions and derivations.
Solution:
We start with the Clausius-Clapeyron equation:
•
dpi• ∆ Si ,trs
=
. The Clausius-Clapyeron equation describes the slope of all the
dT ∆Vi ,•trs
equilibrium curves. In the Clausius-Calpeyron equation, the transition pressure ptrs the
specific equilibrium pressure and the transition temperature Ttrs the equilibrium
temperature. For the given exercise it is the sublimation pressure ptrs = p = pi•
(sublimatiedruk) en the sublimation temperature. The sublimation curve describes an
equilibrium curve so that ∆Gi•, sub = 0 . During the phase transition the temperature and
the pressure are constant so that (based on the total differential and the definition of
•
i , sub
the Gibbs energy): ∆G
= 0 = ∆H
•
i , sub
− Tsub ⋅ ∆ S
•
i , sub
⇒ ∆S
•
i , sub
=
∆ H i•, sub
Tsub
=
∆ H i•, sub
T
Incorporating this in the Clausius-Clapeyron equation gives:
•
∆ H i•, sub
dpi• ∆ Si , sub
=
=
dT ∆Vi •, sub T ⋅ ∆Vi ,•sub
The molar enthalpy of sublimation can be computed using the equation. However, first
the equation needs to be integrated. Before this can be done, some assumptions need to
be made:
1) ∆ H i•, sub =constant
2) ∆Vi ,•sub = Vi ,•V − Vi •, S ≈ Vi ,•V because Vi ,•V ≫ Vi •, S
3) For not too high pressure (valid around the triple point) the molar volume of
the gas phase can be described by the ideal gas law (note: here it is already
accounted for that: Vi ,•V =
R ⋅T
pi•
Incorporating this into the equation (with the red frame) and integrating gives
the equation with which the molar enthapy of sublimation can be
computed(note: instead of Tsub we write here T):
∆ H i•, sub
∆ H i•, sub ∆ H i•, sub ∆ H i•, sub ⋅ pi•
dpi•
=
≈
=
=
dT T ⋅ ∆Vi ,•sub T ⋅Vi ,•V T ⋅ R ⋅ T
R ⋅T 2
•
pi
•
dp • ∆ H i , sub dT
⇒ •i =
⋅ 2 ⇒
pi
R
T
pi• (T2 )
T
•
dpi• ∆ H i , sub 2 dT
=
⋅
∫ pi•
∫ T2
R
T1
p • (T )
 pi• (T2 ) 
∆ H i•, sub
⇒ ln  •
=−
 p (T ) 
R
 i 1 
i
1
1 1
⋅ − 
 T2 T1 
 p • (T ) 
1
⇒ ∆ H i•, sub = − R ⋅ ln  i• 2  ⋅


 pi (T1 )   1 − 1 


 T2 T1 
With the given data (sublimation pressures at 176 K and 177 K) we can compute the
molar enthalpy of sublimation:
T1= 176 K and pi•(176 K) = 7.49·10-6 bar
T2 = 177 K and pi•(177 K) = 9.11·10-6 bar
•
∆ H hexane
, sub = −8.314
 9.11 ⋅10 −6 bar 
J
1
J
kJ
⋅ ln 
= 50712.8
= 50.71
⋅
−6
mol ⋅ K
mol
mol
 7.49 ⋅10 bar   1 − 1  1


 177 K 176  K
b. By using the given data compute for hexane the triple temperature (in K) and the
pressure of the triple point (in bar).
Solution:
The triple point is the point in the pressure, temperature phase diagram at which the
three equilibrium curves, namely the vapour pressure curve, the melting curve and the
sublimation curve, cross each other. Thus for the computation of the triple point
temperature and pressure at least two of the equations for the description of the
equilibrium curves need to be equalize with each other. Here the given equation for
the vapour pressure curve and the in exercise a derived equation describing the
sublimation curve are equalized with each other. Note that now in the equation for the
sublimation curve the temperature T2 is the triple point temperature and the belonging
pressure the triple point pressure; phexane• (Ttr). For T1 an arbitrary temperature and its
belonging sublimation pressure can be used. However, the difference between this
chosen temperature and the expected value for the triple point temperature should not
be too large.
•
 1

∆ H hexane
1
, sub
•
•
ln phexane
ln
176
T
=
p
T
=
K
−
⋅ −
( tr )
)
hexane ( 1
 T T ( = 176 K ) 
R
1
 tr

J
50713
•
mol ⋅  1 − 1 
⇒ ln phexane
(Ttr ) = ln 7.49 ⋅10−6 bar −


J
 Ttr 176 K 
8.314
mol ⋅ K
1  1
1 
•
⇒ ln phexane
(Ttr ) = −11.802 − 6099.71 ⋅  −

K  Ttr 176 K 
(
)
(
(
)
(
)
)
(
)
 1
1 
⋅ −

 Ttr 176 K   ⇒

5000.758
•, vap

ln ( phexane (Ttr ) ) = 16.6734 −

Ttr

•, sub
ln ( phexane
(Ttr ) ) = −11.802 − 6099.71
1
K
 1
1 
5000.758
⋅ −
 = 16.6734 −
Ttr
 Ttr 176 K 
6099.71 6099.71 5000.758 1
= ⋅ ( 6099.71 − 5000.758 )
⇒ −11.802 − 16.6734 +
=
−
176
Ttr
Ttr
Ttr
−11.802 − 6099.71
⇒ 6.182 =
1
K
1098.95
1098.95
⇒ Ttr =
= 177.8 K
6.182
Ttr
•, vap
ln ( phexane
,tr ) = 16.6734 −
5000.758
= −11.45 ⇒ phexane ,tr = exp ( −11.54 ) = 10.65 ⋅10−6 bar
177.8
c. Compute the molar enthalpy of fusion at the triple point. Give all the assumptions and
derivation you make for the computation.
Solution:
With the available data the molar enthalpy of fusion cannot be determined as the
molar enthalpy of sublimation or molar enthalpy of evaporation. Firstly, there are no
temperatures and belonging pressures on the melting curve given. Secondly, there is
no equation of state given for the description of the molar volumes of the solid or the
liquid phase. This the molar enthalpy of fusion needs to be computed from the molar
enthalpy of evaporation and the molar enthalpy of sublimation. We remember, that the
enthalpy is a state function. Thus, for the computation of the change of enthalpy it
does not matter which path is chosen as long as the initial and end state are the same. .
With this in mind, the molar enthalpy of fusion which describes the phase transition
from solid to liquid can be computed at the triple point by the following process; first
phase transition from solid to gas (sublimation) followed by the phase transition from
gas to liquid (condensation = reverse evaporation. If we further assume that the molar
enthalpy of sublimation, the molar enthalpy of evaporation and the molar heat fusion
are constant, the molar enthalpy of fusion can be computed by:
∆ H i•,vap = −∆ H i•,cond
∆ H i•,vap = H i•,V − H i•, L 

•
•
• 
∆ H i , sub = H i ,V − H i , S  ⇒ ∆ H i•, fus = H i•, L − H i•, S = H i•,V − H i•, S − H i•,V − H i•, L = ∆ H i•, sub − ∆ H i•,vap

∆ H i•, fus = H i•, L − H i•, S 

(
)
Giving
•
•
•
∆ H hexane
, fus = ∆ H hexane , sub − ∆ H hexane ,vap = ( 50.71 − 37.79 )
kJ
kJ
= 12.92
mol
mol
d. Which assumptions do you need to make to allow the computation of the molar
enthalpy of evaporation from the Clausius-Clapeyron equation and two different
temperatures?
Solution:
1) ∆ H i•,vap =constant (this is realistic for T < 0.7 Tc)
2) ∆Vi ,•vap = Vi ,•V − Vi ,•L ≈ Vi ,•V realistic because for not too high temperatures
Vi ,•V ≫ Vi •, L
3) For not too high pressures, e.g. at the triple point, the molar volume of the
gas can be described by the ideal gas law: Vi ,•V =
R ⋅T
pi•
Incorporating these assumptions and equations into the Clausius-Clapeyron
equation with T = Tvap gives:
∆ H i•,vap
∆ H i•,vap ∆ H i•,vap ∆ H i•,vap ⋅ pi•
dpi•
=
≈
=
=
dT T ⋅ ∆Vi ,•vap T ⋅ Vi ,•V T ⋅ R ⋅ T
R ⋅T 2
pi•
•
dp • ∆ H i ,vap dT
⇒ •i =
⋅ 2 ⇒
pi
R
T
pi• (T2 )
T
•
dpi• ∆ H i ,vap 2 dT
=
⋅
∫ pi•
∫ T2
R
T1
p• (T )
i
1
 pi• (T2 ) 
∆ H i•,vap  1 1 
⇒ ln  •
=−
⋅ − 
 p (T ) 
R
 T2 T1 
 i 1 
(
)
(
)
⇒ ln pi•,vap (T2 ) = ln pi•,vap (T1 ) −
∆ H i•,vap  1 1 
⋅ − 
R
 T2 T1 
2)
a) Determine the slope of the melting curve of water, dT/dP, at the melting point (T =
273 K and p = pi• = 1atm). The molar enthalpy of fusion of water at T=273 K and
p = pi• = 1 atm is 6010 J mol-1. At the same temperature and pressure the molar
•
volume of fusion is ∆Vi , fus = -1.63 cm3 mol-1.
b) Next, compute the melting temperature of water at 1000 atm.
c) Which assumptions need to be made to allow the computation of the molar enthalpy
of fusion with the given data?
The following data are given:
Tfus (p = 1atm) = 273 K
∆Vi •, fus (273K, 1atm) = -1.63 cm3 mol-1.
∆ H i•, fus (273K, 1atm) = 6010 J mol-1
Wanted:
a) dT /dp at Tfus (p = 1atm) = 273 K
b) Tfus (p = 1000 atm)
Solution:
Ad a) Before the slope of the melting curve can be determined, first the equation for the
description of the melting curve needs be derived. The melting curve describes the
equilibrium between the liquid and the solid phase. The equilibrium is described by the
equaltiy of the molar Gibbs energy of the solid and of the liquid phase:
•
•
Gwater
, L ( T , p ) = Gwater , S ( T , p )
For small changes of the temperature and the pressure with the system always in equilibrium,
we can also write:
•
•
d Gwater
, L ( T , p ) = dGwater , S ( T , p )
In general, the change of the molar Gibbs energy can be described by:
G
d Gi• ( T , p ) = d   = Vi • ⋅ dp − Si• ⋅ dT ⇒ Vi ,•L ⋅ dp L − Si•, L ⋅ dT L = Vi ,•S ⋅ dp S − Si•, S ⋅ dT S
n
If the system changes its state in such a way that it always displays an equilibrium between
the liquid and the solid state, the changes in the temperature and the pressure in the coexisting
phases are the same:
dpL = dpS = dp en dTL = dTS = dT
and:
(
)
(
)
Vi ,•L ⋅ dp − Si•, L ⋅ dT = Vi ,•S ⋅ dp − Si•, S ⋅ dT ⇒ Vi ,•L − Vi •, S ⋅ dp − Si•, L − Si•, S ⋅ dT = 0
(
)
(
)
)
)
⇒ Vi ,•L − Vi ,•S ⋅ dp = Si•, L − Si•, S ⋅ dT
(
(
Vi ,•L − Vi ,•S
dT
⇒
=
dp
Si•, L − Si•, S
The phase transition from solid (S) to liquid (L) is described so that the last equation can be
rewritten:
dT fus
•
dpwater
(V
=
(S
•
water , L
•
− Vwater
,S
•
water , L
•
− S water
,S
) = ∆V
) ∆S
•
water , fus
•
water , fus
.
Note that in this equation it is already incorporated that the phase transition from solid to
liquid is described; the transition pressure p is replaced by pwater• and the transition
temperature T by Tfus.
•
•
In the exercise ∆Vwater , fus is given but not ∆ S water , fus . The change of the molar entropy of
fusion can be replaced by properties which can be determined experimentally:
•
On the melting curve the liquid and the solid phase is in equilbrium so that: ∆ Gwater , fus = 0 .
The molar Gibbs energy of fusion describes an isothermal and isobaric process and can be
described by:
•
•
•
∆ Gwater
, fus = ∆ H water , fus − T fus ⋅ ∆ S water , fus = 0
•
•
•
⇒ ∆ H water
, fus = T fus ⋅ ∆ S water , fus ⇒ ∆ S water , fus =
•
∆ H water
, fus
T fus
Thus the molar entropy of fusion is replaced by the molar enthalpy of fusion (=molar heat of
fusion) and the melting temperature:
•
•
•
dT fus
∆Vwater
∆Vwater
T fus ⋅ ∆Vwater
, fus
, fus
, fus
=
=
=
.
•
•
•
•
dpwater ∆ S water , fus ∆ H water , fus
∆ H water , fus
T fus
The equation given above describes the slope of the melting curve. The slope at the melting
temperature Tfus(1atm)= 273 K can be computed with the given data:
dT fus
•
dpwater


cm3 
m3 
273K ⋅  −1.63
273K ⋅  −1.63 ⋅10−6


T fus ⋅ ∆V
mol 
mol 


=
=
=
•
J
Nm
∆ H water
, fus
6010
6010
mol
mol
K
K
K
= −7.4 ⋅10−8
= −7.4 ⋅10−8
= −0.0075
N
1
atm
atm
2
5
m
1.013 ⋅10
•
water , fus
Ad b) Wanted: Tfus (p = 1000 atm)
In exercise a the equation for the description of the slope of the melting curve has been
derived. Integration of this equation gives an expression with which the melting temperature
•
at 1000 atm can be computed. However, in order to do this we need to assume that ∆Vi , fus =
•
constant and ∆ H i , fus = constant…this is quite a rough assumption seeing that the pressure
changes from 1 atm to 1000 atm. However, if this assumption is not made the computation of
the melting temperature is not possible. So, we use these values but keep the assumption in
mind:
dT fus
•
dpwater
⇒
=
•
T fus ⋅ ∆Vwater
, fus
•
∆ H water
, fus
•
T fus ( pwater
,2 )
•
pwater
,2
∫
∫
•
T fus ( pwater
,1
dT
=
T
fus
)
•
pwater
,1
=
•
∆Vwater
, fus
•
∆ H water
, fus
•
∆Vwater
, fus
∆H
•
water , fus
⋅ T fus
⋅ dp =
•
∆Vwater
dT
, fus
⇒
=
⋅ dp
•
T fus ∆ H water
, fus
•
∆Vwate
r , fus
∆H
•
water , fus
•
pwater
,2
⋅
∫
•
dpwater
•
pwater
,1
•
•
 T fus ( pwater
 ∆Vwater
,2 )
, fus
•
•


⇒ ln
=
⋅ ( pwater
,2 − pwater ,1 )
•
•
 T fus ( pwater ,1 )  ∆ H
water , fus


⇒
•
T fus ( pwater
,2 )
•
T fus ( pwater
,1 )
•
 ∆Vwater

, fus
•
•

= exp 
⋅ ( pwater
−
p
)
,2
water ,1
 ∆H •

water , fus


•
 ∆Vwater

, fus
•
•
•
•

⇒ T fus ( pwater
=
T
p
⋅
exp
⋅
p
−
p
( water ,2 water ,1 ) 
,2 )
fus ( water ,1 )
 ∆H •
water , fus




cm3
−
1.63


mol ⋅ (1000atm − 1atm ) 
⇒ T fus ( p = 1000atm ) = 273K ⋅ exp 
 6010 J



mol


3

−6 m
−
⋅
1.63
10

mol
= 273K ⋅ exp 
Nm
 6010

mol


N
N 

⋅ 1000 ⋅1.013 ⋅105 2 − 1 ⋅1.013 ⋅105 2  
m
m 



T fus ( p = 1000atm ) = 265.61K
(Note: Tfus(pi•) is the melting temperature at the pressure pi•)
The melting temperature can also be determine in a different manner and is given below. For
this approach it is assumed that the slope of the melting curve is constant over the complete
pressure range.
dT fus
K
= −0.0075
= co ns tan t
•
dpwater
atm
(
•
T fus pwater
⇒
∫
)
•
T fus ( pwater
,0 )
dT = T fus ( p
•
water
) −T ( p
fus
•
•
T fus ( pwater
) = T fus ( pwater
,0 ) − 0.0075
p•
•
water ,0
water
K
K
•
•
=
−
⋅
⋅ ( pwater
− pwater
0.0075
dp = −0.0075
)
,0 )
∫
atm p•
atm
water ,0
K
•
•
⋅ ( pwater
− pwa
ter ,0 )
atm
•
•
⇒ T fus ( pwater
= 1000atm ) = T fus ( pwater
,0 = 1atm ) − 0.0075
•
T fus ( pwater
= 1000atm ) = 273K − 0.0075
K
⋅
atm
(( p
K
⋅ 999atm = 265.5 K
atm
•
water
•
= 1000atm ) − ( pwater
,0 = 1atm )
)
3) The following (transition) pressures are given:
ice
water
T [K]
p = pi•
[mmHg]
transition
269.15
3.280
Sublimation
271.15
3.880
Sublimation
275.15
5.294
Evaporation
277.15
6.101
Evaporation
First derive the necessary equations from the Clausius-Clapeyron equation and then
Compute the following properties:
a) Molar heat of sublimation
b) Molar heat of evaporation
c) Molar heat of fusion
d) Triple point temperature and pressure
Solution:
Ad a) and b) Note: In the following text the derivations and computations are given
for component i (indicated by the index i). For the given exercise this component I is water, i
= H2O.
For the derivation of the equations we start with the Clausius-Clapeyron equation:
•
dpi• ∆ Si ,trs
=
. The Clausius-Clapyeron equation describes the slope of all equilibrium
dT ∆Vi ,•trs
curveswith the transition pressure ptrs= p = pi• and the transition temperature Ttrs= T.
On the equilibrium curves we know that ∆Gi•,trs = 0 . Further, we know that on the
equilibrium curves we can state that T and p are constant so that:
•
i ,trs
∆G
= 0 = ∆H
Leading to:
•
i ,trs
− Ttrs ⋅ ∆ S
•
i ,trs
⇒ ∆S
•
i ,trs
=
∆ H i•,trs
Ttrs
=
∆ H i•,trs
T
•
∆ H i•,trs
dpi• ∆ Si ,trs
=
=
dT ∆Vi ,•trs T ⋅ ∆Vi ,•trs
This equation is used for the computation of the molar enthalpy of sublimation and
evaporation.
Ad a) Before the molar enthalpy of sublimation can be computed using the above
dervied equation, the equation needs to be rewritten. Therefore, a few assumptions
need to be made:
1) ∆ H i•, sub =constant
2) ∆Vi ,•sub = Vi ,•V − Vi ,•S ≈ Vi ,•V because Vi ,•V ≫ Vi •, S
3) For not too high pressure, the molar volume of the gas phase is described by
the ideal gas law: Vi ,•V =
R ⋅T
pi•
With these assumptions we get (with T = Tsub):
∆ H i•, sub
∆ H i•, sub ∆ H i•, sub ∆ H i•, sub ⋅ pi•
dpi•
=
≈
=
=
dT T ⋅ ∆Vi ,•sub T ⋅Vi ,•V T ⋅ R ⋅ T
R ⋅T 2
pi•
•
dpi• ∆ H i , sub dT
⇒ • =
⋅ 2 ⇒
pi
R
T
pi• (T2 )
T
•
dpi• ∆ H i , sub 2 dT
∫ pi• = R ⋅ T∫ T 2
p • (T )
1
 p • (T ) 
∆ H i•, sub
⇒ ln  i• 2  = −
 p (T ) 
R
 i 1 
⇒ ∆H
•
i , sub
i
1
1 1
⋅ − 
 T2 T1 
 pi• (T2 ) 
1
= − R ⋅ ln  •
⋅
 p (T )   1 1 
 i 1 
 − 
 T2 T1 
Using the given data:
T1= 269.15 K and pi•(269.15 K) = 3.28 mmHg = 3.28 · 750 · 105 Pa
T2 = 271.15 K and pi•(271.15 K) = 3.88 mmHg= 3.88 · 750 · 105 Pa
We get:
∆ H i•, sub = −8.314
 3.88mmHg 
J
1
J
⋅ ln 
= 50964.99
⋅
1
1 1
mol ⋅ K
mol
 3.28mmHg  
−


 271.15 K 269.15  K
Ad b) For the computation of the molar enthlpay of evaporation we make the
following assumptions:
1) ∆ H i•,vap =constant (realistic for T < 0.7 Tc)
2) ∆Vi ,•vap = Vi ,•V − Vi ,•L ≈ Vi ,•V because Vi ,•V ≫ Vi ,•L
3) For not too high pressures the molar volume of the gas phase can be
described by the ideal gas law: Vi ,•V =
R ⋅T
pi•
Incorporating these assumptions in the derived equation gives (with T = Tvap):
∆ H i•,vap
∆ H i•,vap ∆ H i•,vap ∆ H i•,vap ⋅ pi•
dpi•
=
≈
=
=
dT T ⋅ ∆Vi ,•vap T ⋅ Vi ,•V T ⋅ R ⋅ T
R ⋅T 2
pi•
•
dp • ∆ H i ,vap dT
⇒ •i =
⋅ 2 ⇒
pi
R
T
pi• (T2 )
T
•
dpi• ∆ H i ,vap 2 dT
=
⋅
∫ pi•
∫ T2
R
T1
p• (T )
i
1
 pi• (T2 ) 
∆ H i•,vap  1 1 
⇒ ln  •
=−
⋅ − 
 p (T ) 
R
 T2 T1 
 i 1 
 pi• (T2 ) 
1
•
⋅
⇒ ∆ H i ,vap = − R ⋅ ln  •

 p (T )   1 1 
 i 1 
 − 
 T2 T1 
With the given data, the molar enthalpy of evaporation can be computed:
T1= 275.15 K en pi•(275.15 K) = 5.294 mmHg= 5.294 · 750 ·105 Pa
T2 = 277.15 K en pi•(277.15 K) = 6.101 mmHg = 6.101 · 750 ·105 Pa
∆ H i•,vap = −8.314
 6.101mmHg 
J
1
J
⋅ ln 
= 44976.05
⋅
1
1 1
mol ⋅ K
mol
 5.294mmHg  
−


 277.15 K 275.15  K
Ad c) With the available data the computation of the molar enthalpy of fusion cannot
be done as for the molar enthalpy of sublimation and of evaporation. The molar
enthalpy of fusion is therefore determined from the molar enthalpy of evaporation and
of sublimation. This is only possible because the molar enthalpy is a state function and
thus the change of the molar enthalpy depends only on its initial and its end state. The
phase transition from solid to liquid (fusion) is now split into two processes, namely
first the phase transition from solid to gas followed by the phase transition from the
gas to the liquid state. The latter process, the condensation, is the reverse process of
evaporation. Again because the enthalpy is a sate function, we know that the enthalpy
of evaporation is the equal to the negative enthalpy of condensation. If further it is
assumed that all mola renthalpies of transition are constant, we get:
∆ H i•,vap = H i•,V − H i•, L 

•
•
• 
∆ H i , sub = H i ,V − H i , S  ⇒ ∆ H i•, fus = ∆ H i•, sub − ∆ H i•,vap

∆ H i•, fus = H i•, L − H i•, S 

J
J
J
∆ H i•, fus = 50964.99
− 44976.05
= 5988.94
mol
mol
mol
Ad d) In this exercise it is asled to compute the triple point temperature and
pressure. The triple point describes the phase equilibriuim between a liquid, a
solid and a gas phase. Further it is the point at which the melting curve, the
vapour pressure curve and the sublimation curve cross. By equalizing two of
the equations describing the equilibrium curves with each other, the triple point
temperature and pressure can be determined.
In exercise a and b the equations for the description of the vapour pressure
curve and of the sublimation curve have been derived and are now used to
determine the triple point temperature and pressure. The vapour pressure at the
triple point temperature is described by:
∆ H i•,vap  1 1 
ln p (Ttr ) = ln p (T1 ) −
⋅ − 
R
 Ttr T1 
For T1 an arbitrary temperature slightly larger than the expected triple point
temperature is chosen. Thereby, the vapour pressure at this temperature should
be known. For example we choose T1 = 275.15 K with pi•(275.15 K) = 5.294
mmHg (computed with the expression for the description of the vapour
pressure curve). With the molar enthalpy of evaporation
J
∆ H i•,vap = 44976.05
we have all data to allow the use of the expression
mol
given above.
(
)
•
i
(
•
i
)
The same approach is used for the sublimation curve. The sublimation pressure
at the triple point temperature is given by:
∆ H i•, sub  1 1 
ln p (Ttr ) = ln p (T2 ) −
⋅ − 
R
 Ttr T2 
With T2 a temperature slightly smaller than the expected triple point
temperature, e.g. T2 = 269.15 K, and the belonging sublimation pressure
pi•(269.15 K) = 3.28 mmHg and the molar enthalpy of sublimation
J
∆ H i•, sub = 50964.99
we have all data to describe the sublimation curve
mol
completely.
Equalizing the vapour pressure with the sublimation pressure curve at the triple
point temperature gives:
∆ H i•,vap  1 1  
•
•
ln pi (Ttr ) = ln pi (T1 ) −
⋅ −  
R
 Ttr T1  

•
∆
H



1
1
i , sub
ln pi• (Ttr ) = ln pi• (T2 ) −
⋅  − 
R
 Ttr T2  
(
)
(
(
)
(
)
(
)
(
)
•
i
(
⇒ ln p (T1 )
•
i
)
•
i
)
∆ H i•,vap  1 1 
∆ H i•, sub  1
1
•
−
⋅  −  = ln pi (T2 ) −
⋅ − 
R
R
 Ttr T1 
 Ttr T2 
(
)
 pi• (T1 )  ∆ H i•,vap ∆ H i•, sub
1
⇒ ln  •
+
−
=
 p (T )  R ⋅ T
R ⋅ T2
Ttr
1
 i 2 
∆ H i•,vap
⇒ Ttr =

ln 


 ∆H •
∆ H i•, sub
i ,vap

⋅
−
 R
R





∆ H i•, sub
−
R
R
•
•
pi (T1 )  ∆ H i ,vap ∆ H i•, sub
−
+
R ⋅ T1
R ⋅ T2
pi• (T2 ) 
Filling the given data into the equation allows the computatio of the triple point
temperature:
J
J
50964.99
mol ⋅ K −
mol ⋅ K
J
J
8.314
8.314
mol ⋅ K
mol ⋅ K
Ttr =
J
J
44976.05
50964.99
 5.294mmHg 
mol ⋅ K
mol ⋅ K
ln 
+
−

J
J
 3.28mmHg  8.314
⋅ 275.15 K 8.314
⋅ 269.15 K
mol ⋅ K
mol ⋅ K
Ttr = 273.28 K
44976.05
The triple point pressure can be computed by filling the determined triple point
temperature into the equation either describing the sublimation pressure or the
vapour pressure:
(
)
(
)
ln pi• (Ttr ) = ln pi• (T2 ) −
(
)
(
)
ln ptr = pi• ( 273.28 K )
ln ptr = pi• ( 273.28 K )
∆ H i•, sub  1 1 
⋅ − 
R
 Ttr T2 
J
mol ⋅ K
= ln ( 3.28mmHg ) −
J
8.314
mol ⋅ K
= 1.532
⇒ ptr = pi• ( 273.28 K ) = 4.627 mmHg
50964.99
1
1


⋅
−

 273.28 K 269.15 K 