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8/18/2019 Solution Thermodynamics Theory-Ch 11
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8/18/2019 Solution Thermodynamics Theory-Ch 11
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topics
• Fundamental equations for mixtures
• Chemical potential
• Properties of individual species insolution (partial properties)
• Mixtures of real gases
• Mixtures of real liquids
8/18/2019 Solution Thermodynamics Theory-Ch 11
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A few equations
dT nS dP nV nGd
PV U H
from
dT nS nS Td nH d nGd TS H G
)()()(
d(nH)obtain
)()()()(
−=
+≡
−−=−≡ For a closed system
Total differential form, what are (nV) and (nS)
Which are the main variables for G??
What are the main variables for G in an open system of
k components?
8/18/2019 Solution Thermodynamics Theory-Ch 11
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G in a mixture (opensystem)
+
∂
∂+
∂
∂= dT
T
nGdP
P
nGnGd
n P nT ,,
)()()(
8/18/2019 Solution Thermodynamics Theory-Ch 11
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G in a mixture of components at !and P
i
k
iidndT nS dP nV nGd
∑=+−= 1)()()( µ
ow is this e!"ation red"ced if n #$
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" phases (each at ! and P) in a closeds#stem
i
k
i
idndT nS dP nV nGd ∑=+−= 1)()()( µ
β α
)()()( nM nM nM +=
dT nS dP nV nGd )()()( −=
Apply this equation to each phase
Sum the equations for each phase, take into account that
In a closed system:
8/18/2019 Solution Thermodynamics Theory-Ch 11
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$e end up with
0=+ ∑∑ β β α α µ µ ii
ii
i
i dndn
How are dni and dni
β related at constant n?
8/18/2019 Solution Thermodynamics Theory-Ch 11
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For " phases% components atequili&rium
β α
β α
β α
µ µ ii
P P
T T
=
=
=
%or all i # $, &,'k
Thermal e!"ilibri"m
echanical e!"ilibri"m
hemical e!"ilibri"m
8/18/2019 Solution Thermodynamics Theory-Ch 11
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'n order to solve the Ppro&lem
• eed models for µi in each phase
• xamples of models of µi in the vaporphase
• xamples of models of µi in the liquidphase
8/18/2019 Solution Thermodynamics Theory-Ch 11
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ow we are going to learna&out
• Partial molar properties
• *ecause the chemical potential is a partial
molar propert#
• At the end of this section thin a&out this * $hat is the chemical potential in ph#sical terms
* $hat are the units of the chemical potential
* +ow do we use the chemical potential to solve aP (phase equili&rium) pro&lem
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Partial molar propert#
i jnT P i
i n
nM M
≠
∂
∂=
,,
)(
Solution property
Partial property
Pure-species property
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example
i jnT P i
innV V
≠
∂∂= ,,)(
0lim
~
)(
)(
→∆
∆=∆
∆=∆
wn
ww
ww
nV nV
nV nV
pen !eaker: ethanol " water, equimolar
#otal $olume n%
# and P
Add a drop of pure water, nw
&i', allow for heat e'chan(e, until temp #
)han(e in $olume ?
8/18/2019 Solution Thermodynamics Theory-Ch 11
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!otal vs, partial properties
i
i
i
i
i
i
M nnM
M x M
∑
∑=
=
See deri$ation pa(e *+
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-erivation of Gi&&s.-uhemequation
ii
i
i
i
i
x P xT
M x M
dx M dT T
M dP
P
M dM
∑
∑
=
+
∂∂
+
∂∂
=,,
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Gi&&s.-uhem at constant !/P
P&Tconstant0=∑ ii
i M d x
seful for thermodynamic consistency tests
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*inar# solutions
1
12
1
21
dxdM x M M
dx
dM
x M M
−=
+=
See deri$ation pa(e *+.
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!tain d&/d'0 from 1a2
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xample 11,0
• $e need "% cm0 of antifree2esolution3 0 mol4 methanol in water,
• $hat volumes of methanol and water
(at "5oC) need to &e mixed to o&tain"% cm0 of antifree2e solution at "5oC
• -ata3
water /07.18 /77.17
methanol /7.!0 /".8
1
2
1
1
mol cmV mol cmV
mol cmV mol cmV
==
==
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solution
• Calculate total molar volume
• $e now the total volume% calculate thenum&er of moles required% n
• Calculate n1 and n"
• Calculate the volume of each pure species
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3ote cur$es for partial molar $olumes
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From Gi&&s.-uhem3
P&Tconstant0=∑ ii
i M d x
02211 =+ V d xV d x
4i$ide !y d'0, what do you conclude respect to the slopes?
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6ead and wor example11,7
• Given +87x19:x"9x1x"(7x19"x")determine partial molar enthalpies asfunctions of x1% numerical values for pure.
species enthalpies% and numerical valuesfor partial enthalpies at in;nite dilution
• Also show that the expressions for the
partial molar enthalpies satisf# Gi&&s.-uhem equation% and the# result in thesame expression given for total +,
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ow we are going to start looingat models for the chemical
potential of a given component in amixture• !he ;rst model is the ideal gas mixture
• !he second model is the ideal solution
• As #ou stud# this% thin a&out thedi
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The ideal-gas mixturemodel
• => for an ideal gas
• Calculate the partial molar volumefor an ideal gas in an ideal gas
mixture
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or an# par a mo ar proper # o er
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or an# par a mo ar proper # o erthan volume% in an ideal gas
mixture3
P to p fromintegrate
T constant at P Rd dS
pT H P T H
pT M P T M
i
ig
i
i
ig
i
ig
i
i
ig
i
ig
i
ln
),(),(
),(),(
−=
==
=
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Partial molar entrop# (igm)
i
ig
ii
ig
i
ig
i
i
ig
ii
ig
i
y R P T S pT S P T S
y R P T S pT S
ln),(),(),(
ln),(),(
−==
−=
8/18/2019 Solution Thermodynamics Theory-Ch 11
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Partial molar Gi&&s energ#
P y RT T y RT P RT T
P RT T G P and 1atm P etween g integratin
T constant at P RTd dP P
RT dP V dG
y RT GG
y RT TS H S T H G
iiii
ig
i
i
ig
i
ig
i
ig
i
i
ig
i
ig
i
ig
i
i
ig
i
ig
i
ig
i
ig
i
ig
i
ln)(lnln)(
ln)(
ln
ln
ln
+Γ =++Γ =
+Γ = =
===
+==
+−=−=
µ
µ )hemical potential of
component i in an
ideal (as mi'ture
#his is µ for a pure component 555
+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++
8/18/2019 Solution Thermodynamics Theory-Ch 11
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Pro&lem
• $hat is the change in entrop# when ,?m0 of C=" and ,0 m0 of "% each at 1 &arand "5oC &lend to form a gas mixture atthe same conditions@ Assume ideal gases,
i
i
i
i
ig
ii
ig
i
i i
i
ig
ii
ig
i
i
i
ig
i
ig
i
ig
i
y y RS yS
y y RS yS yS
y RS S
ln
ln
ln
∑∑∑ ∑∑−=−
−==
−=
We showed that
8/18/2019 Solution Thermodynamics Theory-Ch 11
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solution
∑−=∆i
ii y ynRS ln
n # -V./T# $ bar $ m0. 1/ 2 &34 56
∆S # &7894: ;.5
8/18/2019 Solution Thermodynamics Theory-Ch 11
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Pro&lem• $hat is the ideal wor for the
separation of an equimolar mixtureof methane and ethane at 1?5oC and0 &ar in a stead#.ow process into
product streams of the pure gases at05oC and 1 &ar if the surroundingstemperature !σ 8 0B@02 6ead section 78+ 1calculation of ideal work292 #hink a!out the process: separation of (ases and chan(e of state
First calculate H and S for methane and for ethane chan(in( their state from
P0, #0, to P9#9
Second, calculate H for de-mixing and S for de-mixing from a mi'ture of ideal(ases
8/18/2019 Solution Thermodynamics Theory-Ch 11
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solution
ig
mixdei
i
i
ig
mixdei
i
i
i
i
i
ig
mixde
ig
mixde
ig T
T
ig
i
ig T
T
ig
i
S S yS
H H y H
y y RS H
P
P
T
dT T !pS
dT T !p H
−
−
−
−
∆+∆=∆
∆+∆=∆
=∆=∆
−=∆=∆
∑
∑
∑
∫ ∫
ln0
ln)(
)(
1
22
1
2
1
#
8/18/2019 Solution Thermodynamics Theory-Ch 11
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ow we introduce a new concept3fugacit#
• $hen we tr# to model realDs#stems% the expression for thechemical potential that we used for
ideal s#stems is no longer valid
• $e introduce the concept of fugacit#that for a pure component is the
analogous (&ut is not equal) to thepressure
8/18/2019 Solution Thermodynamics Theory-Ch 11
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$e showed that3
)ln()( P y RT T iiig
i +Γ = µ
P RT T G iig
i ln)( +Γ =
iii f RT T G ln)( +Γ =
Pure component i , ideal (as
)omponent i in a mixture
of ideal (ases
et;s define:
For a real fluid, we define
Fu(acity of pure species i
8/18/2019 Solution Thermodynamics Theory-Ch 11
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6esidual Gi&&s free energ#
i
R
i
iig
ii
R
i
RT G
P
f RT GGG
φ ln
ln
=
=−=
%alid for species iin any phase and
any condition
8/18/2019 Solution Thermodynamics Theory-Ch 11
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>ince we now how to calculateresidual propertiesE
P
dP " RT
G
RT G
P
ii
Ri
i
R
i
)1(ln
ln
0−==
=
∫ φ φ
aals, etc
8/18/2019 Solution Thermodynamics Theory-Ch 11
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examples
• From irial =>
• From van der $aals=>
RT
P #iii =φ ln
i
iiiii
" T R
P a
RT
P " "
22ln1ln −
−−−=φ
8/18/2019 Solution Thermodynamics Theory-Ch 11
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Fugacities of a ".phases#stem
l
ii
l
i
$ii
$i
f RT T G
f RT T G
ln)(
ln)(
+Γ =
+Γ =
ne component, two phases:
saturated liquid and saturated $apor at Pisat and #i
sat
>hat are the equili!rium conditions for a pure component?
8/18/2019 Solution Thermodynamics Theory-Ch 11
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Fugacit# of a pure liquid at Pand !
sat
i sat
i
l
i
l
i
sat
i
$
i
sat
i
l
i
sat
i
sat
i
$
il
i P
P f
P f
P f
P f
P
P f P f
)(
)(
)(
)()()( =
8/18/2019 Solution Thermodynamics Theory-Ch 11
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Fugacit# of a pure liquid at Pand !
dP V
RT
P P f P
P
l
i
sat
i
sat
i
l
i sat i
∫ =1
e#$)( φ
8/18/2019 Solution Thermodynamics Theory-Ch 11
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example• For water at 0oC and for P up to 1% Pa
(1 &ar) calculate values of f i and φ i fromdata in the steam ta&les and plot them vs, P
−−
−=−= )(1
)(1
ln%
%%
% iiii
ii
i
i
S S T
H H
RGG RT f
f
et Hi@ and Si
@ from the steam ta!les at *o) and the lowest P, 0 kPa
At low P, steam is an ideal (as BC f i@ BP@
#hen (et $alues of Hi and Si at *o) and at other pressure P and calculate f i 1P2
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8/18/2019 Solution Thermodynamics Theory-Ch 11
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Pro&lem
• For >=" at : B and 0 &ar% determine goodestimates of the fugacit# and of G66!,
>=" is a gas% what equations can we use to
calculate f 8 φP
Find !c% Pc% and acentric factor% ω, !a&le *1% p, :H
Calculate reduced properties3 !r% Pr
!r81,0I0 and Pr80,H5
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+igh P% high !% gas3 use Jee.Bessler correlation
• From ta&les 15 and 1: ;nd φ and φ1
• φ 8 ,:?"K φ1 8 1,057
• φ 8 φ φ1ω = 0.724
• f 8 φ P 8 ,?"7 x 0 &ar 8 "1?,17 &ar
• G66! 8 lnφ = −0.323
8/18/2019 Solution Thermodynamics Theory-Ch 11
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Pro&lem
• stimate the fugacit# of c#clopentane at11oC and "?5 &ar, At 11 oC the vaporpressure of c#clopentane is 5,":? &ar,
• At those conditions% c#clopentane is a high P liquid
dP V RT
P P f P
P
l
i
sat
i
sat
i
l
i sat i∫ =
1e#$)( φ
%ind Tc, -c, >c, Vc and acentric factor, ω, Table $, p9
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%ind Tc, -c, >c,, Vc and acentric factor, ω, Table $, p9@47
alc"late red"ced properties Tr , -r sat
Tr # 79384@ and -r sat # 79$$3
At - B -r sat we can "se the virial CDS to calc"late φisat
2.!
1
".1
0
10
172.01&.0'
!22.008.0
)(e#$
r r
r
r sat
i
T #
T #
# #
T
P
−=−=
+= ω φ
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φisat # 79:
-
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coeLcient
ω
φ φ φ
φ ω φ φ
ω φ
φ
φ
φ
)(
lnlnln
)1(ln
)1(ln
)1(ln
ln
10
10
0
1
0
0
0
0
=
+=+−=
−=
=
−==
=
∫ ∫
∫
∫
i
r
r P
r
r P
i
r
r P
ii
r c
P ii
R
i
i
R
i
P
dP
" P
dP
"
P dP "
P P P
P dP "
RT G
RT G
r r
r
#a!les =0* to =0.
ee-Eessler
+$ 0 -ue Monda#
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+$ 0% -ue Monda#%>eptem&er 1?
• Problems 11.! 11."! 11.#! 11.1!11.1$
W E 8, F"e onday, September &8
• Pro!lems 0080+, 0080 1!2, 00890, 008991a2, 00891a2, 00897