University of San Carlos
Nasipit, Talamban, Talamban, Cebu City 6000 6000
Department of Chemical Engineering
CHE !!N
Assignment on Low Pressure Vapor-Liquid Equilibria in Non-ideal Mixtures
Submitte" by#
Duterte, $an %ommel T&
Submitte" to#
Engr& 'uis (& Cabatingan )nstructor
September !, *0!+
10.2-1 iven# •
ethyl bromi"e -!. / nheptane -*. mi1ture at 202&!+ ( vap
•
•
P1 = 0.7569 bar ,
vap
P2 = 0.0773 bar
regular solution parameters 3' -cm24mol. + !7
ethyl bromi"e nheptane
5 -cal4cm2.!4* 7&8 &
%e9uire"# a. Compositi Composition on of the vapor vapor in e9uilibrium e9uilibrium :ith :ith a li9ui" li9ui" containing containing &*2 &*2 mol ; ethyl ethyl bromi"e at 202&!+ ( an" 0&+2 bar, assuming the solution is i"eal b. 3apor 3apor composition in part a assuming the solution solution is regular c. 3apor apor composi composition tion in in part part a using using the the UN)<=C UN)<=C mo"el mo"el ". 3apor apor compositio composition n in part a given that a vapor of composit composition ion 7!&+ mol ; ethyl ethyl bromi"e is in e9uilibrium :ith *7&2 mol ; li9ui" ethyl bromi"e solution at a total pressure of 0&2!8 bar at 202&!+ ( Solution# a. Usin Using g %ao %aoul ult> t>ss 'a: 'a:, vap
x1 P1 y 1 = = 0.7879 P vap
x2 P2 y 2 = = 0.0899 P =ssuming the solution is i"eal, ?y i @ 0&77& b. )n or"er to use the regular regular solution mo"el, the volume volume fraction A i must first be "etermine"&
Φ1 =
Φ2 =
x1 V 1 x1 V 1 + x 2 V 2 x2 V 2 x1 V 1 + x 2 V2
= 0.3120
= 0.6880
The activity coefficient B i can no: be calculate"& 2
V1 Φ2 ( δ 1- δ 2)
2
γ1 = e RT
= 0.8997 2
V2 Φ1 (δ 1 -δ 2)
γ2 = e RT
2
= 0.0949
=ssuming the solution follo:s the regular solution mo"el, ?y i @ 0&886& This value is much closer to unity compare" to the value obtaine" from the i"eal solution mo"el, an" this is an in"ication that the mi1ture is not i"eal an" is better "escribe" b y the regular solution mo"el& c. Using the o"ifie" UN)<=C program, am,
γ1 =1.1017
an an"
γ2 =1.0919
& The vap vapor
compositions can then be compute" using the mo"ifie" %aoult>s 'a:& vap
x1 γ1 P1 y 1 = = 0.8681 P vap
x2 γ2 P2 y 2 = = 0.0982 P =ssuming the solution follo:s the UN)<=C mo"el, ?y i @ 0&8662& ". The van 'aar mo"el mo"el is to be use" to calculate calculate the the vapor composit composition& ion& The The activity activity coefficien coefficientt of each component is first calculate" -at 1 ! @ 0&*72, y ! @ 0&7!+ an" @ 0&2!8 bar., then these
2
values are use" to calculate the van 'aar an" F parameters& These parameters are then use" to calculate the activity coefficients at the ne: composition -1 ! @ 0&*2 an" @ 0&+2 bar., an" the mo"ifie" %aoult>s 'a: use" to calculate the ne: vapor compositions&
γ1 =
II
γ2 =
y1 P x1 P
vap 1
y2 P x2 P
vap 2
=
0.815!0.3197" = 1.2108 0.2843!0.7569"
=
0.185!0.3197" = 1.0691 0.7157!0.0773"
Using the van 'aar e9uations for an" F,
( (
II
# = 1+
x2 $% γ2 x1 $% γ
II 1
) )
2 II
$% γ1 = 0.6756
II 2
& = 1+
x 1 $% γ 1 x 2 $% γ
II 2
II
$% γ2 = 0.3052
Calculating the final activity coefficients -for :hen 1 ! @ 0&*2 an" @ 0&+2 bar. # 'I
γ1 = e
(
1+
#x 1 & x2
) = 1.0790
& x2 #x 1
) = 1.1443
2
& 'I
γ2 = e
(
1+
2
The vapor compositions are then 'I
vap
'I
vap
x1 γ1 P1 y 1 = = 0.8502 P
x2 γ2 P2 y 2 = = 0.1029 P =ssuming that the solution follo:s the van 'aar mo"el, ?y i @ 0&8+2!& =ns:er4s# The table belo: summariGes the calculate" values& =ssuming the solution#
is i"eal 0&78 is regular 0&788 follo:s the UN)<=C 0&767! mo"el follo:s the van 'aar 0&7+0* mo"el The regular solution mo"el "escribes the solution best because the closest to
∑ y ( =1
y2
∑y
0&0788 0&088
0&77 0.!"
0&087*
0&8662
0&!0*8
0&8+2!
y1
(
among all the mo"els use", its results are
&
10.2-2 iven# •
:ater -!. / furfural -*. mi1ture at !08&+C an" !&0!2 bar
•
at e9uilibrium, 1 ! @ 0&!0 an" y ! @ 0&7!
•
temperature of the mi1ture is change" to !00&6C, an" some of the vapor con"enses
3
•
vapor pressure "ata Species
vap -bar. !&077 !&02+* 0&!680 0&!!82
Temperature -C. !08&+ !00&6 !08&+ !00&6
Iater
%e9uire"# =ssuming that the vapor phase is i"eal an" the li9ui"phase activity coefficients are in"epen"ent of temperature but "epen"ent on concentration, compute the a. e9uilibrium vapor composition b. e9uilibrium li9ui" composition at the ne: temperature& Solution# The van 'aar mo"el is to be use" to calculate the vapor an" li9ui" compositions& The activity coefficient of each component is first calculate" -at 1 ! @ 0&!0, y ! @ 0&7! an" @ !&0!2 bar., then these values are use" to calculate the van 'aar an" F parameters&
γ1 =
II
γ2 =
y1 P x1 P
vap 1
y2 P x2 P
vap 2
=
0.815!0.3197" = 5.8243 0.2843!0.7569"
=
0.185!0.3197" = 1.2654 0.7157!0.0773"
Using the van 'aar e9uations for an" F,
( (
II
# = 1+
x2 $% γ2 x1 $% γ
II 1
) )
2 II
$% γ1 = 8.5461
II 2
& = 1+
x 1 $% γ 1 x 2 $% γ
II 2
II
$% γ2 = 0.7898
Since the final li9ui" compositions are yet to be foun" an" they are nee"e" to calculate the final activity coefficients -:hich are use" to calculate the final vapor compositions., a trialan"error solution is use"& The follo:ing e9uations must be satisfie" in or"er to fin" the vapor an" li9ui" compositions#
x 1 + x2 = 1) y 1 + y 2 = 1 1.0352x 1 γ1 = 1.013 y 1 ) 0.1193 x 2 γ 2 = 1.013y 2 # 'I
γ1 = e
(
#x 1 1+ & x2
&
))γ 2
'I 2
= e
(
1+
& x2 # x1
)
2
'I
= trial value of 1 ! an" 1* is use" to calculate
γ1
'I
an"
γ2
& The activity coefficients are then use"
to calculate the vapor compositions& The calculation is repeate" until
∑ x ( = ∑ y ( =1
& Using the
icrosoft E1cel Solver function, the vapor an" li9ui" compositions are foun" to be 1 ! @ 0&8!!!, 1 * @ 0&0778, y! @ 0&!*82 an" y * @ 0&766!& =ns:er4s# Ihen the mi1ture is at !00&6C, the :ater content of the li9ui" is 8!&!! mol ; an" that of the vapor is !*&82 mol ;& The calculations tell us that the :ater content of the li9ui" increase" from !0 to 8! mol ; an" in the vapor "ecrease" from 7! to !2 mol ;J this means that the vapor that con"ense" may have primarily been :ater vapor& 10.2-! iven#
4
•
ethanol -!. / benGene -*. mi1ture at +C
•
e1perimental "ata
x1
y1
-bar.
0 0&02 0&08* 0&*!72 0&2!! 0&!+0 0&+!88 0&+*7 0&6!++ 0&07 0&7!0* 0&8!82 0&8+8! !&00
0 0&!86+ 0&*78+ 0&220 0&26*+ 0&27* 0&06+ 0&!0! 0&22 0&+! 0&++6 0&07 0&7*0! !&00
0&*828 0&26!2 0&28+2 0&077 0&!* 0&!*7 0&!00 0&082 0&0*7 0&278! 0&26!+ 0&2026 0&*!! 0&*2*!
%e9uire"# repare an 1y an" 1 "iagram for the system assuming, separately, a. the mi1ture is i"eal b. the mi1ture is regular c. the mi1ture is "escribe" by the UN)<=C mo"el ". the activity coefficients for this system obey the van 'aar e9uation an" the "atum point at
x * =0.6155
is use" to obtain the van 'aar parameters
Compare the results obtaine" in parts -a./-". :ith the e1perimental "ata& =lso compare the compute" van 'aar coefficients :ith those given in Table &+!& Solution# a. The complete 1 / y "iagram at constant temperature can be constructe" by using the vapour pressure "ata to calculate the total pressure an" value of y i for each value of 1 i& s 'a:, the e9uilibrium pressure at each li9ui" propylamine composition 1 ! is given as
P ( x1 ) = x 1 P1 + x 2 P 2 = x 1 P1 + !1-x 1 ¿ P2 vap
vap
vap
vap
:here
P1 =0.2321 bar an"
vap
vap
P2 =0.2939
bar& To calculate the vapourphase mole
fraction, vap
x1 P1 y 1 = P! x 1 "
The follo:ing table summariGes the 9uantities calculate" using the e9uations above& Table !& 3alues obtaine" using the i"eal solution mo"el
x1
x2
P ( x1 )
y1
0&00 0&!0 0&*0 0&20 0&0 0&+0 0&60 0&0 0&70 0&80 !&00
!&00 0&80 0&70 0&0 0&60 0&+0 0&0 0&20 0&*0 0&!0 0&00
0&*828 0&*7 0&*7!+ 0&*+ 0&*68* 0&*620 0&*+67 0&*+06 0&*+ 0&*272 0&*2*!
0 0&070 0&!68 0&*+*8 0&28 0&!2 0&+** 0&67* 0&+86 0&76 !
5
The "iagrams can no: be constructe", as sho:n in
/
,
V =
,
V = +7&28
cm24mol& The solubility parameter can be estimate" from
δ=
√
vap V
,
=
√
vap -RT V
,
The heat of vaporiGation can be estimate" using the appro1imate integrate" ClausiusClapeyron e9uation, :hich is vali" over small temperature ranges#
%$P
vap
P
vap
! T2 "
= -
( T1 )
(
vap 1 1 R T2 T1
)
Using the constants foun" in the 7 th e"ition of erry>s Chemical Engineers> Han"booK to calculate the ethanol vapor pressure at T ! @ +C an" T * @ 6C,
vap
@ !&7+ K$4mol& This is use"
to calculate the solubility parameter for ethanol, :hich is calculate" to be !*&66 -cal4cm 2.!4*& )n summary, 3' -cm24mol. Ethanol +7&28 PenGene 78 The volume fractions can then be calculate" from
Φ( =
5 -cal4cm2.!4* !*&66 8&*
x( V( x 1 V 1 + x2 V 2
The activity coefficients are calculate" from 2
2
2
2
V1 Φ2 ( δ 1- δ 2)
γ1 = e RT V2 Φ1 ( δ 1 -δ 2)
γ2 = e RT The e9uilibrium pressure an" vapor composition is calculate" using the mo"ifie" %aoult>s 'a:# vap
vap
Pe = x 1 γ 1 P1 + x2 γ2 P2 vap
x ( γ ( P ( y ( = Pe
The follo:ing table summariGes the calculate" values& Table *& 3alues obtaine" using the regular solution mo"el
A! 0
A* !
B! 2&02+2
B* !
1! 0&00
e9 0&*828
y! 0
6
0&068 0&82*! *&6*2 !&007 0&!0 0&!08 0&7+8! *&*682 !&02* 0&*0 0&*!8+ 0&70+ !&8668 !&078 0&20 0&202 0&68+ !&!!6 !&!686 0&0 0&286* 0&6027 !&88! !&20* 0&+0 0&860 0&+00 !&2*+7 !&+!6 0&60 0&608 0&28+! !&!782 !&7+ 0&0 0&*! 0&*+8 !&077* *&*76 0&70 0&7++* 0&!7 !&0*26 2&6 0&80 ! 0 ! +&2* !&00 The "iagrams can no: be constructe", as sho:n in
0&2*+ 0&27+ 0&260* 0&26+* 0&26+6 0&26*8 0&2+0 0&27 0&2!+! 0&*2*!
0&!760 0&20*2 0&2702 0&2+* 0&+7 0&+077 0&+!2 0&+760 0&67+ !
c. Using the o"ifie" UN)<=C program, the activity coefficients of benGene an" ethanol :ere "etermine" at various compositions& The follo:ing table summariGes the obtaine" values& Table 2& 3alues obtaine" using the UN)<=C mo"el
B! B* 1* 1! e9 !*&027 ! !&00 0&00 0&*828 &7* !&02! 0&80 0&!0 0&278 *&87+6 !&!2*8 0&70 0&*0 0&0+0 *&!86 !&*+68 0&0 0&20 0&!02 !&2+2 !&*0* 0&60 0&0 0&!!+ !&+78 !&626* 0&+0 0&+0 0&08 !&*+2 !&8*7! 0&0 0&60 0&02 !&!+02 *&22+ 0&20 0&0 0&28*7 !&0667 *&8*8* 0&*0 0&70 0&202 !&0!* 2&708 0&!0 0&80 0&2*+ ! +&20+ 0&00 !&00 0&*2*! The "iagrams can no: be constructe", as sho:n in
y! 0 0&*77 0&2** 0&2687 0&28!+ 0&!2* 0&282 0&+7 0&+2+0 0&6+2! !
". The van 'aar mo"el is to be use" to calculate the vapor composition& The activity coefficient of each component is first calculate" -at 1 ! @ 0&6!++, y ! @ 0&22 an" @ 0&0*7 bar., then these values are use" to calculate the van 'aar an" F parameters& These parameters are then use" to calculate the activity coefficients at the ne: compositions, an" the mo"ifie" %aoult>s 'a: use" to calculate the ne: vapor compositions&
γ1 =
II
γ2 =
y1 P x1 P
=
vap 1
y2 P
0.4343!0.4028" = 1.22455 0.6155!0.2321"
= 2.0164
vap
x 2 P2
Using the van 'aar e9uations for an" F,
( (
# = 1+
& = 1+
II
x2 $% γ2 x1 $% γ
II 1
) )
2 II
$% γ1 = 2.0263
II 2
x 1 $% γ 1 x 2 $% γ
II 2
II
$% γ2 = 1.4998
The final activity coefficients are calculate" using # 'I
γ1 = e
(
1+
#x 1 & x2
)
2
7
& 'I
γ2 = e
(
1+
& x2 #x 1
)
2
=n" the vapor composition is then 'I
vap
x1 γ1 P1 y 1 = P
The follo:ing table summariGes the calculate" values& Table & 3alues obtaine" using the van 'aar mo"el
B! B* 1! 1* e9 &+760 ! 0&00 !&00 0&*828 &6*67 !&0*+8 0&!0 0&80 0&27 2&!0* !&!002 0&*0 0&70 0&0* *&*+0 !&**2 0&20 0&0 0&076 !&+** !&00+ 0&0 0&60 0&086 !&*7 !&6!0 0&+0 0&+0 0&076 !&*6 !&8+80 0&60 0&0 0&00 !&!* *&226 0&0 0&20 0&28*0 !&0+06 *&808+ 0&70 0&*0 0&266! !&0!!7 2&+87+ 0&80 0&!0 0&2!! ! &707 !&00 0&00 0&*2*! The "iagrams can no: be constructe", as sho:n in
y! 0 0&*72+ 0&2+6 0&27! 0&28! 0&087 0&200 0&66! 0&+2*8 0&666+ !
1
0.8
Ideal mixture model
UNIFAC mixture model
0.6
van aar model
'#t%& 0.4
0.2 !e"ular mixture model
0 0.00
0.10
0.20
#x$erimental data
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
x#t%&
8
0.42 0.4 0.38 0.36 Ideal mixture model 0.34
UNIFAC mixture model
van aar model
0.32 () *ar
0.3 0.28 0.26 !e"ular mixture model 0.24
#x$erimental data
0.22 0.2 0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
x#t%&
=ns:er4s# The 1 an" 1y "iagrams are given in
Npentane -!. / acetone -*. system at !&0!2 bar 3aporli9ui" e9uilibrium "ata -bar. 1!
y!
T -C.
0&0*! 0&!07 8&!+ 0&06! 0&20 +&6 0&!2 0&+ 28&+7 0&*!0 0&++0 26&6 0&*8* 0&6! 2&2+ 0&0+ 0&66 2*&7+ 0&+02 0&67 2*&2+ 0&6!! 0&!! 2!&8 0&*7 0&28 2!&82 0&768 0&7!0 2*&* 0&8+2 0&806 22&78 %e9uire"# =re these "ata thermo"ynamically consistentQ Solution# The thermo"ynamic consistency relation is 1
P
vap 1
!&+60 !&28 !&!6 !&026 0&860 0&8!2 0&802 0&77 0&770 0&786 0&8+
vap
P2
0&702 0&02 0&++! 0&82 0&+2 0&*+ 0&*! 0&!2 0&!0 0&!8 0&+
γ
∫ $% γ 2 x 1 =0 0
1
9
γ2 $% γ1
versus mole fraction&
2 1.5 1
))
0.5 ln ,2-,1
0 +0.5
)
+1 +1.5 0
0.2
0.4
0.6
0.8
1
x1
The t:o areas, ) an" )), bet:een the curve an" the
γ2 $% =0 line are e9ual in siGe but opposite in sign γ1
so that the consistency relation can be consi"ere" satisfie", although the value of
γ2 $% γ1
at 1! @ 0&0*!
appears to be an outlier as all the other "ata points are linear& =ns:er# Res, the "ata are thermo"ynamically consistent& 10.2-1! iven# •
*0 mol ; ethanol -!. solution in :ater -*. at 7&!+C
•
=t 7&!+C,
vap
P1 =1.006 bar vap
P2 =0.439 bar
$( γ 1 = γ 1 = 1.6931
x1 0
$( γ 2 = γ 2 = 1.9523
x2 0
%e9uire"# Estimate the a. total pressure an" b. composition of the vapor in e9uilibrium :ith the system& Solution# The van 'aar mo"el uses the follo:ing t:o e9uations to pre"ict activity coefficients#
10
$% γ 1=
$% γ 2=
α
(
α x1 1+ β x 2
)
2
β
(
β x 2 1+ α x1
)
2
x 1 → 0
,
an" γ2
,
x 2 → 0
& These allo: the above e9uations to be simplifie"
to
$% γ1 =
$% γ2 =
#
(
#!0" 1+ & x2
)
2
&
(
&!0" 1+ # x1
)
2
=#
=&
:hich gives @ 0&+*6+6 an" F @ 0&6680!& These values are then use" to calculate the activity coefficients an", subse9uently, the total pressure an" e9uilibrium vapor composition& #
γ1 = e
(
1+
# x1 & x2
) = 1.44433
& x2 # x1
) = 1.01825
2
&
γ2 = e
(
1+
2
The total pressure is vap
vap
P= x 1 γ 1 P1 + x2 γ2 P2 =0.6482 bar
=n" the vapor composition is vap
x1 γ1 P1 y 1 = = 0.4483 P vap
x2 γ2 P2 y 2 = = 0.5517 P =ns:er4s# =t 7&!+C, the vapor in e9uilibrium :ith the system is at 0&6+ bar an" contains &72 mol ; ethanol an" ++&! mol ; :ater& 10.2-1" iven# •
benGene -!. / cyclohe1ane -*. mi1ture
•
base" on vaporli9ui" e9uilibrium "ata, some authors have claime" that the system is "escribe" ex
by
= x 1 x 2 :here
( / :$ ) = 3750-8T for T in (elvin&
%e9uire"# a. Derive e1pressions for the activity coefficients of benGene an" cyclohe1ane
11
b. Determine the enthalpy an" entropy changes on mi1ing :hen ! mole of benGene an" * moles of cyclohe1ane are mi1e" at T @ 200 ( an" constant pressure vap
vap
P1 =¿ 0&2*02 bar an"
c. iven the follo:ing vapor pressure "ata at T @ 2*0 (,
P2 =¿
0&2*!7 bar, "etermine the bubble point pressure of the li9ui" in part -b. at T @ 2*0 (, an" the composition of the vapor in e9uilibrium :ith that li9ui" Solution# ex
a.
ex
ex
= x 1 1 + x 2 2 = x1 RT$%γ 1 + x2 RT$%γ 2 ex
∴
1 = RT $%γ 1 an" ex
= x 1 x 2 =
(
ex
2 = RT $%γ 2
)(
1 1 + 2
ex
= 1 2 ( 1 + 2 )
2 1 + 2
)
= 1 2 ( 1 + 2 )
-2
-1
(
)
ex 2 1 2 ; -2 -1 2 − = = − 1 2 ( 1 + 2 ) + 2 ( 1 + 2 ) = = x 2 2 ; 1 1 + 2 ( 1 + 2 ) ex 1
ex 2
=
∂ N
ex
−2
−1
=− A N 1 N 2 ( N 1 + N 2 ) + A N 1 ( N 1 + N 2 ) = A
∂ N 2
(
N 1 N 1+ N 2
−
N 1 N 2
( N 1 + N 2 )
2
)
2
= A x 1
ex
1 = RT $%γ 1 ,
Since
ex
γ1 =e
1 RT
2
x2 RT
=e
(1- x 1)
2
= eRT
Similarly, ex
γ2 =e b.
x1=
2 RT
2
x1 RT
=e
1 3 an"
(1- x 2)
2
= eRT x2 =
2 3
%ecall, I
ex
<(x = <(x + < =−R
∑ x $% x +< (
ex
(x = (x + =RT ∑ x ( $% x ( + I
ex
(x = (x +T<(x
,
(
ex
, an"
&
]
(x =RT ( x 1 $% x 1 + x 2 $% x 2 "+ x1 x 2 (3750−8T ) =−3862.77
; ex =−< = −8x1 x 2 ;T <(x =− R ( x1 $% x 1 + x2 $% x 2 "+8x 1 x2 ] =21.21
>
(x = (x +T<(x =2500.23 c. Using the e1pressions for the activity coefficients "erive" in part -a.,
12
vap
vap
P= x 1 γ 1 P1 + x2 γ2 P2 =0.3557 bar vap
x1 γ1 P1 y 1 = = 0.3662 P b?bb$e vap
x2 γ2 P2 y 2 = = 0.6338 P b?bb$e =ns:er4s# a. The
activity ( 1-x 1 )
γ1 = e RT
coefficients 2
of
( 1-x 2 )
the
benGene
an"
cyclohe1ane
can
be
"escribe"
by
2
a% γ 2 = e RT
, respectively&
b. Ihen ! mole of benGene an" * moles of cyclohe1ane are mi1e" at T @ 200 ( an" constant
<(x =
pressure,
*!&*! $ ( ! an"
(x =
*+00&*2 $&
c. =t T @ 2*0 (, the bubble point pressure of the li9ui" is 0&2++ bar an" the vapor in e9uilibrium :ith the li9ui" contains 26&6* mol ; benGene an" 62&27; c yclohe1ane& 10.2-21 iven# •
=t *0C, the methyl acetate -!. / methanol -*. system forms an aGeotrope :ith the follo:ing
x 1 = y1 =0.754
properties#
,
x 2 = y2 =0.246
, an" @ !72&+ mm Hg
•
The aGeotrope is to be fe" into a secon" "istillation column operating at +0C
•
urecomponent vapor pressures vap
$:@ 10 P ( @ ) =7.06524 −
1157.63 T ( > ) -53.424
vap
$:@ 10 PeA ( @ ) =8.08097 −
1582.271 T ( > ) -33.424
%e9uire"# a. To "etermine :hether the propose" process is possible, compute the aGeotropic composition of a methyl acetate / methanol mi1ture at +0C& =ssume that the parameters in the activity coefficient you use are in"epen"ent of temperature& Solution# The van 'aar mo"el is to be use" to calculate the final aGeotropic vapor an" li9ui" compositions& The activity coefficient of each component is first calculate" at the initial aGeotropic con"itions -at 1 ! @ y! @ 0&+, 1* @ y* @ 0&*6 an" @ !72&+ mm Hg., then these values are use" to calculate the van 'aar an" F parameters&
γ1
BC(
γ2
=
=
P vap 1
P
P
=
= vap
P2
183.54 = 1.0653 172.2934
183.54 = 1.8829 97.475
Using the van 'aar e9uations for an" F,
(
# = 1+
)
BC( 2
x2 $% γ2 x1 $% γ
BC( 1
BC(
$% γ1
= 1.150046
13
(
)
BC( 2
& = 1+
x 1 $% γ 1 x 2 $% γ
BC( 2
BC(
$% γ 2
= 1.0800
Since the final li9ui" compositions are yet to be foun" an" they are nee"e" to calculate the final activity coefficients -:hich are use" to calculate the final vapor compositions., a trialan"error solution is use"& The follo:ing e9uations must be satisfie" in or"er to fin" the vapor an" li9ui" compositions#
x 1 + x2 = 1) y 1 + y 2 = 1 # BCD
γ1
= e
(
&
#x 1 1+ & x2
)) γ 2
BCD 2
= e
(
1+
& x2 #x 1
)
2
593.425x 1 γ1 = y 1 P) 416.5845 x 2 γ2 = y 2 P vap
vap
P= x 1 γ 1 P1 + x2 γ2 P2
BCD
= trial value of 1 ! an" 1* is use" to calculate γ1
BCD
an"
γ2
& The activity coefficients are use" to
calculate the total e9uilibrium pressure, then the vapor compositions& The calculation is repeate" until
∑ x =∑ y =1 (
(
an"
x ( =y (
& Using the icrosoft E1cel Solver function, the aGeotropic vapor an"
li9ui" compositions are foun" to be 1 ! @ y! @ 0&6+22 an" 1 * @ y* @ 0&26 at +0C an" 62&8 mm Hg& =ns:er4s# The aGeotropic composition of a methyl acetate / methanol mi1ture at +0C is 1 = @ y= @ 0&6+22 an" 1eH @ yeH @ 0&26 at a pressure of 62&8 mm Hg& 10.2-2$ iven# •
PenGene -!. / polyisobutylene -*. mi1ture at !0C
•
Data on physical properties benGene olecular :eight, g4mol
polyisobutylene 0 000 !0 -monomeric. !2!&8 -monomeric.
7
olar volume, cm 24mol
•
77&*6 0&!*66 -at *87&!+ (. 3apour pressure, bar =ssume" to be negligible 0&*28* -at 2!*&+ (. Data for the activity of benGene in )P as a function of the mass ratio of benGene to )P
E / PIE
a E = x E γE
0&722! 0&++2 0&*8!
0&87!! 0&8+8+ 0&7277
%e9uire"# a. Using the "ata in )llustration !0&*, compare the pre"ictions of the
%=
2 2C
= 384.6 ?%(FG
=ssuming that m )P @ !&0, the benGene :eight fractions are given by
HE =
E E + PIE
14
an" are 0&++, 0&2+66 an" 0&**+, respectively& The benGene mole fractions -in terms of the :eight fraction. are then calculate" as
HE xE =
78 H E H PIE + 78 40000
an" are 0&88, 0&886+ an" 0&8822, respectively& The
ϕE =
x E VE x E VE + %x PIE VPIE
The benGene volume fractions are 0&20!, 0&22!2 an" 0&*0+0, respectively& (no:ing this, an" calculating the parameter m as
=
V PIE VE
=574.8
allo:s us to calculate the activity coefficients using the
%$ γE = e
( )
ϕE 1 + 1(1- ϕ E) + (1- ϕE)2 xE
The activities are calculate" by multiplying the activity coefficients :ith the correspon"ing mole fractions& The follo:ing table summariGes the calculate" values& Table +& %esults from the calculation of activities
%E%TED
E / PIE
C='CU'=TED
a E = x E γE
; "ifference
0&722! 0&87!! !&0+!* 6& 0&++2 0&8+8+ !&0!0! +&0 0&*8! 0&7277 0&7+*8 !& b. =s the vapour pressure of benGene at !0C is not given, it :ill be calculate" using the constants foun" in the 7 th e"ition of erry>s Chemical Engineers> Han"booK&
P
vap E
( 10℃) = e
83.107
−6486.2 -9.2194 $%283.15+6.9844J 10-6 ( 283.15 )2 283.15
=0.0605 bar
The benGene partial pressures are given by vap
PE = xE γ E P E
The calculate" partial pressures are sho:n in the follo:ing table& Table 6& Calculation of the benGene partial pressure at "ifferent :eight fractions
:P 1P BP P, bar 0&0 0&0 0&0 0 0&! 0&87*7 0&+*07 0&02!0 0&* 0&88*2 0&7!!* 0&07 0&2 0&88++ 0&86+ 0&0+7! 0& 0&88! !&026 0&06*+ 0&+ 0&887! !&0+7 0&0628 0&6 0&887 !&0++ 0&062 0& 0&888* !&027+ 0&06*7 0&7 0&888+ !&0*0! 0&06! 0&8 0&8887 !&00+ 0&0607 !&0 !&0 ! 0&060+ =ns:er# a. The pre"ictions of the
15
b. The partial pressures of benGene are sho:n in table 6& 10.2-$% iven# •
Component ! an" * mi1ture in vapourli9ui" e9uilibrium
•
=t 80C an" !&7+0+ bar, a vapour of composition y ! @ 0&26 coe1ists :ith a li9ui" of composition 1 ! @ 0&
•
3apour pressures are given by vap
$:@ 10 P( = (−
E( T
for pressure in bar an" T in (, :here = ! @ &!*+, P ! @ !+00, =* @ +&000 an" P * @ !+0 %e9uire"# a. Determine the van 'aar parameters of the system using the given "ata b. Determine :hether the mi1ture has an aGeotrope at 80C an", if so, "etermine its composition an" i"entify :hether it is a ma1imum or minimumpressure aGeotrope c. btain 1y an" 1y "iagrams for the system at 80C ". =n e9uimolar mi1ture of species ! an" * initially at very lo: pressure is compresse" at a constant 80C& =t :hat pressure "oes the first "rop of li9ui" form, an" :hat is its compositionQ =t :hat pressure "oes the last bubble of vapour "isappear, an" :hat :as its compositionQ
Solution# a.
γ1 =
II
γ2 =
y1 P
0.3767!1.8505" = 1.7649 0.4!0.9874"
vap
=
vap
= 1.2670
x 1 P1
y2 P x 2 P2
Using the van 'aar e9uations for an" F,
( (
# = 1+
& = 1+
II
x2 $% γ2 x1 $% γ
II 1
) )
2 II
$% γ1 = 1.4999
II 2
x 1 $% γ 1 x 2 $% γ
II 2
II
$% γ2 = 1.6004
b. The mi1ture has an aGeotrope if 1 i @ yi at any point :ithin the composition range& Table sho:s the calculate" values for species ! using the van 'aar mo"el& Table & Calculating the vapor composition using the van 'aar mo"el
1! 0&00 0&!0 0&*0 0&20 0&2* 0&2 0&26 0&2 0&27 0&0 0&+0 0&60
B! &7!* 2&**2 *&66+ *&!+ *&0+8! !&87* !&80* !&7662 !&72! !&6+0 !&8!2 !&*8+7
B* ! !&0!2 !&0+8 !&!0 !&!6! !&!78 !&*!0! !&**2+ !&*2 !&*60 !&+ !&*!
e9 !&+!* !&*20 !&7! !&767 !&78* !&7+0 !&7+!2 !&7+!2 !&7+!* !&7+0+ !&7286 !&7!+8
y! 0 0&!86! 0&*8!2 0&2* 0&2+!7 0&2+78 0&26+2 0&2672 0&2!* 0&26 0&00* 0&**7
16
0&0 !&!+8* *&!*6 !&672 0&+2! 0&70 !&0677 *&!!! !&6668 0&+06+ 0&80 !&0!0 2&+82+ !&80 0&6*2 !&00 ! &8++0 0&87 ! The highlighte" ro: is the aGeotropic pointJ the mi1ture forms an aGeotrope at appro1imately 0&2 mole fraction of species ! at 80C an" a pressure of !&7+!2 bar& The mi1ture is a ma1imum pressure aGeotrope since the activity coefficients of both species are greater than !&0, an" there are positive "eviations from %aoult>s 'a:& c. Using the values summariGe" in the previous table, the 1y an" 1y "iagrams can be constructe"& 1
0.8
0.6 '1 0.4
0.2
0 0.00
0.20
0.40
0.60
0.80
1.00
x1
1.88 1.78 1.68 1.58 1.48 () *ar
1.38
(+x
(+'
1.28 1.18 1.08 0.98 0.00
0.20
0.40
0.60
0.80
1.00
x1) '1
". The pressures can be rea" from the 1y "iagram of the system& The pressure at :hich the first "rop of li9ui" forms is appro1imately !&6 bar, an" this "rop has a composition 1 ! @ 0&70& The last bubble of vapor "isappears at !&7 bar, an" the bubble ha" a composition of y ! @ 0&0& =ns:er# a. The van 'aar parameters for the mi1ture are @ !&888 an" F @ !&600&
17
b. The mi1ture forms an aGeotrope at appro1imately 0&2 mole fraction of species ! at 80C an" a pressure of !&7+!2 bar& The mi1ture is a ma1imum pressure aGeotrope since the activity coefficients of both species are greater than !&0, an" there are positive "eviations from %aoult>s 'a:& c. See figures an" +& ". The pressure at :hich the first "rop of li9ui" forms is appro1imately !&6 bar, an" this "rop has a composition 1 ! @ 0&70& The last bubble of vapor "isappears at !&7 bar, an" the bubble ha" a composition of y ! @ 0&0& 10.2-$ iven# •
Iater -!. / !,"io1ane -*. mi1ture at 2*2&!+ (
•
3aporli9ui" e9uilibrium "ata -mm Hg. !*0&8 !0&7+ !+!&!6 !+8&! !6&+ !6+&6+ !6&78 !6& !6&8 !6&8+ !66&7 !6+&7 !60&7* !++&! !*&6 !!&6 8*&+!
1! 0&0000 0&0+60 0&080 0&!00 0&*!60 0&*870 0&2660 0&00 0&60 0&70 0&+280 0&6*80 0&80 0&7!!0 0&7800 0&860 !&0000
y! 0&0000 0&!8*0 0&*670 0&2+0 0&2720 0&020 0&*+0 0&20 0&60 0&+!0 0&++0 0&660 0&8+0 0&+20 0&600 0&8+0 !&0000
%e9uire"# a. =ctivity coefficients at each of the reporte" compositions b. =re these "ata thermo"ynamically consistentQ c. lot of e1cess ibbs energy as a function of composition Solution# a. The van 'aar mo"el :ill be use" to calculate the activity coefficients&
P
B
γ1 =
vap 1
P
P
B
γ2 =
vap 2
P
=
167.79 = 1.8138 92.51
= 1.3926
Using the van 'aar e9uations for an" F,
(
# = 1+
(1 −x 1) $% γ2B x1 $% γ
B 1
)
2 B
$% γ1 = 1.7022
!1-x1 B
1+
x 1 $% γ1 2
¿
B
$% γ 2 B
& = ( ¿ ) $% γ 2 = 1.9836 The activity coefficients at every other composition are calculate" using
18
&!1-x1 # x1 1+ ¿ # 'I 1
γ
2
= e ( ¿ )
&!1-x 1 1+¿ & 'I 2
γ
( ¿# x1)
= e ¿
2
The results are summariGe" in the follo:ing table& Table 7& =ctivity coefficients of the :ater"io1ane system
1! B! B* 0&0000 +&76* ! 0&0+60 &60 !&00 0&080 &!662 !&0!* 0&!00 2&*+7 !&0+2 0&*!60 2&08 !&0+* 0&*870 *&8+ !&!+!8 0&2660 *&!08 !&*2* 0&00 !&72+ !&28+ 0&60 !&7!27 !&28*6 0&70 !&6762 !&72 0&+280 !&+*72 !&6 0&6*80 !&2*6 *&00* 0&80 !&!2 *&786 0&7!!0 !&070 2&!02 0&7800 !&0*2 &++** 0&860 !&00*+ 6&*627 !&0000 ! &*680 b.
γ
∫ $% γ 2 x 1 =0 0
1
γ2 $% γ1
versus mole fraction&
19
2.5 2 1.5 1 0.5 ln ,2-,1
0 +0.5 +1
)
+1.5 +2 0.00
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
1.00
x1
The t:o areas, ) an" )), bet:een the curve an" the
γ2 $% =0 line are e9ual in siGe but opposite γ1
in sign, so that the consistency relation can be consi"ere" satisfie"& c. 3alues of ex
ex
ex
are compute" from ex
= x 1 1 + x 2 2 = x1 RT$%γ 1 + !1-x 1 "RT $% γ2
1000
750 ex) /-mol 500
250
0 0.00
0.20
0.40
0.60
0.80
1.00
x1
=ns:er# a. See table 7& b. Pase" on figure +, the "ata are thermo"ynamically consistent& c. See figure 6&
20
