9
9.3
Generally mixing at constant T and P and mixing at constant T and V are quite
different. However, for the ideal gas we have
PVi = Ni RT (pure fluids) and PV = Ni RT (mixtures)
Thus for the pure fluids (same T and P)
N1RT
N RT
and V2 = 2
P
P
N1RT N 2
9
9.22 (a) The two-constant Redlich-Kister expansion, which leads to the two-constant
equation is
a
k
G ex = x1x2 A + B x1 x2
Margules
fp
Thus
a
f
G ex
= A + B 2 x1 1
x1x2
(1)
which is a linear function of x.
The form of the Wohl Equation which leads to t
9
9.29 We will write the Flory-Huggins expression as
a
G ex
RT
x1x2m
f
= x1 + mx2 1 2 =
x1 + mx2
res
ex
G
RT
1
= N1 ln
x1
comb
ex
G
RT
ex
NG
RT
=
res
+ N 2 ln
=
res
2
x2
x1
mx2
, 2 =
x1 + mx2
x1 + mx2
with 1 =
N1 N 2m
N1 + mN 2
F G I = N m N N m
N G RT J
9
9.28 To check the utility of these models we will use the Gibbs-Duhem equation in the form
x1
FG ln IJ
H x K
+ x2
1
1
FG ln IJ
H x K
1
T,P
=0
2
T,P
For the model
a
2
ln 1 = Ax2 = A 1 x1
f
2
2
ln 2 = Bx1
Gives
x1
FG ln IJ
H x K
+ x2
1
1
T,P
FG ln IJ
H x
9
9.27
a
G ex = ax1x2 x1 x2
N G ex =
f = ALM N N N N OP
aN + N f
N aN + N f Q
L 2 N N N 2c N N N N h OP
= AM
a N + N f PQ
MN a N + N f
a
AN1 N 2 N1 N 2
NG
N1
2
1
2
2
2
1
1
2
2
2
2
1
N 2 ,T , P
2
2
1
2
1
ex
f
2
2
1
2
12
3
2
2
1
2
2
2
2
= A 2 x1x2 x2 2 x1
9
9.25
LM x OP
N
Q
LM x OP
G
NG
=
= N ln M
RT
RT
MN N PPQ
G
FG I
=
GJ
RT N H RT K
LM x OP
LM N dN OP
N
= ln M
N
dN P
NP
N
N M N
PQ
PQ
MN
MN
LM x OP x + x x
G
ln =
= ln
RT
x
N
Q x
L
O x
= 1 lnM x P
N
Q x
C
G ex
= xi ln
RT
i =1
j
ex
ex
ij
j
j
C
ij
j
i
9
9.24 We start with
ln
zc
P
fi
1
=
xi P RT
h
1
RT
Vi Vi IGM dP =
0
Eqn. (7.2 -3a)
z LMMNFGH
P
0
V
Ni
IJ
K
T , P , N j i
OP
PQ
RT
dP
P
Now
dP =
P
RT
P
P
P
1
d ( PV ) dV =
dZ dV = dZ = dV
V
V
V
V
Z
V
Also, by triple product rule
FG V IJ
H N K
i T , P , N
9
9.23
Expression for G ex in this problem is the same as that of Eq. (9.6-6). If we recognize that A and B
a
fc
h
in Eq. (9.6-6) is replaced by ART and BRT here. Also, since 1 2 xAr = xCH 4 xAr , species 1 is
methane and species 2 is argon.
(a) Therefore
9
9.31 Show AEOS ( P ) = C
LM a
Nb
mix
xi
mix
OP
Q
ai
bi
z
A
A V A V = PdV
V T
V
start
P=
( V is convenient since we have ideal gas and ideal gas mixtures)
P-R
RT
a
P=
pure component
V b V (V + b) + b(V b)
z
z
V
A V A V =
V
RT
a
+
dV
V b V (V + b) + b
9
9.32 Starting from
aij
a
bM M = xi x j bij
Q and
RT
RT
ij
G ex = C*
or
LM a
Nb
FG
H
M
xi
M
i
ai
bi
OP
Q
IJ
K
a M G ex
a
a
= * + xi i DRT and bM = M
bM
bi
DRT
C
i
Substituting, we then obtain
aM
a
M = Q so that
DRT RT
and
bM =
aM
D
=Q
1 D
RT
aM
D
Q
+
Solutions to Chemical and Engineering Thermodynamics, 3e
Chapter 9
9.40
File is HCl
i := 0 , 1 . 115
I := i 0.03
i
del := 0.340
1 1.178 I 0.5
( i)
DH := exp
i
1
1 1.178 I 0.5
( i)
EDH := exp
i
0.5
1 + (I )
i
1 1.178 I 0.5
( i) + del I
EEDH := exp
i
9
9.37 a)
Gibbs-Duhem equation.
N G
=0
i
i =1 N
j T, P , N
k j
a i ( T , P, x ) =
G 2
G 2
for a binary mixture x1
+x 2
=0
x1 T,P
x1 T ,P
fi ( T, P, x )
fi ( T, P, x )
=
x i P exp
cfw_ ( G G )
1
RT
i
IGM
i
fi
G G iIGM
a i fi
a i fi
IGM
=
9
9.38
N1
Air
2 bar, 25C
N2
xO2= 0.99, 25C, 1 bar
xO2= 0.05, 25C, 1 bar
No moving parts W = 0
Mass balance (basis 1 mole air)
O2
air
0 = 0.21 + 0.99 N1 + 0.05 N 2
0 = 1 + N1 + N 2
(Note that a mass balance on nitrogen would be redundant with the two mass
9
9.39 Adiabatically and reversibly SIN = SOUT
Will have to do this by trial and error using P-R EOS program
a) At 310 K and 14 bar
H = 187.01 J/mol
S = 21.09 J/mol
Pf = 345 bar T = 594 K S = 21.02 J/mol K H = 12055.05 J/mol
Energy balance
0 = N1 H1 N 2 H
9
9.35
i := 1 , 2 . 1999
x := 0.0005 i
i
VPS := 1000
VT := 1
m :=
VPS
i
phiT :=
(
phiPS := 1 phiT
)
x VT + 1 x VPS
i
i
i
:= 0.6
i
i
VT exp 1
Ti :=
T := 298
VT
x VT
1
phiPS + phiPS 2
( i)
i
m
i
(
(
VPS exp ( 1 m) phiT + phiT
)
PSi :=
x VT + 1 x VPS
i
(
9
9.36 a)
NOUT,1 x1 = 1
x1,y2
NIN
NOUT,2 x2 = 1
x1 N IN + N OUT ,1 = 0
N OUT ,1 = x1 N IN
x 2 N IN + N OUT , 2 = 0
MB on each component
N OUT , 2 = x 2 N IN
E.B. (per mole of feed)
(
)
0 = x1H1 + x 2 H 2 x1 H 2 x 2 H 2 + W + Q
Entropy balance
(
)
0 = x1S1
9
9.33
Equation (9.9-11) is easily derived, is generic, and applies to any mixing rule. This will be used as
the starting point. With the Wong-Sandler mixing rule
Note that derivatives must be taken with respect to mole numbers.
Therefore
Q=
x x
ij
i
j
N
9
9.34
Starting from eqn. (9.2-13)
ln
fk
1
=
x k P RT
z
V = ZRT / P
V =
LM RT N F P I
MN V GH N JK
k
T ,V , N j k
OPdV ln Z
PQ
The Soave-Redlich-Kwong equation of state is
P=
RT
a (T )
NRT
N 2 a (T )
=
V b V (V + b) V Nb V (V + Nb)
with
Nb = Nibi and N 2a
9
9.19 Let M = molality of salt in solution.
i) For KCl: z+ = 1 , z = 1 , M K = M , M Cl = M ;
1
1
I = zi2 Mi = (1 M + 1 M ) = M
2
2
z+ = 3
z = 1
1
ii) For CrCl3 :
I = 32 M + 1 3 M = 6 M
M+ = M M = 3 M
2
z+ = 3
z = 2
1
iii) For Cr2 SO4 3 :
I = 32 2 M + 4
9
9.21
The Gibbs-Duhem equation, written in terms of molalities and using the mean
ionic activity coefficient is as follows:
MSd GS + M Ed G E = 0 where S is solvent and E is electrolyte
but
c
+ RT lna x f
G E = G o + RT ln M + + M
E
o
GS = GS
h
SS
So th
9
9.8
a
fp
NN R
= NG =
S A + BFGH N + N IJK U
V
N +N T
N NW
F G IJ = N R A + BFG N N IJ U
=G
H N K N + N S H N + N KV
T
W
NN
R aN N fU
aN + N f SA + B N + N V
T
W
NN R B
Ba N N f U
+
S
N + N TN + N
aN + N f V
W
= x k A + Ba x x fp x x k A + Ba x x fp + x
9
9.10 i)
One-constant Margules equation
2
RT ln 1 = Ax2
Thus
*
RT ln 1 = RT ln
or
*
1 = exp
a
f
RT ln 1 x1 = 0 = A
af
1 x1
2
2
= Ax2 A = A 1 x2
1 x1 = 0
a
LM Ac1 x h OP
MN RT PQ
2
2
c
f
LM c
MN
2
A 1 x2
x1 1000
exp
ms M1
RT
1 =
and
h
h OP
PQ
ii) Two-
9
9.11 An ideal gas constrained to remain at constant volume and T, is also a system at constant internal
energy and volume, since U is only a function of temperature for the ideal gas. Consequently, at
equilibrium, the entropy should be a maximum.
Suppos
9
9.2
The picture of the process here is as follows
Mixture Ni Moles of gas at T and V !
1 T ,V , P
1
B
2 T ,V , P2
T ,V , Pf
Mix keeping
T and V fixed
C T ,V , PC
(a)
Let
P = initial pressure of species i (pressure in unmixed state)
i
P = final pressure
9
9.1
a
f
PV T , P, N1, N 2 = Ni RT
a
f
Vi (T , P, x) = V i (T , P)
U T , P, N1, N 2 = Ni U i (T , P) Ui (T , P, x) = U i (T , P)
Also Si (T , P, x ) = S i (T , P) R ln xi
x U (T , P, x ) U (T , P ) = x [0] = 0
V = x V (T , P, x ) V (T , P ) = x [ 0] =
9
9.4
We have the following properties for a mixture for mixing at constant T and P:
U (T , P, x ) = Ni U i (T , P)
V (T , P, x) = NiV i (T , P)
S (T , P, x ) = Ni S i (T , P) R Ni ln xi
and S i = S i0 + CV,i ln
Ui
+ R ln
U i0
Vi
V i0
S i0 , U i0 , V i0 a
9
9.6
As a preliminary note that, from Eqns. (6.4-27 and 28)
z
LMT F P I POPdV
GJ
NM H T K QP
LMF P I R OPdV
MNGH T JK V PQ
V = ZRT P
H (T , P) H IG (T , P) = RT ( Z 1) +
V =
and
z
V = ZRT P
S (T , P) S IG (T , P) = R ln Z +
V =
vdw E.O.S. P =
V
V
RT
a
so
9
9.5
(a) Start with eqn. 9.2-13
ln
fi
1
= ln i =
xi P
RT
P=
z
V = ZRT P
V =
LM RT N F dP I OPdV ln Z
GJ
NM V H N K PQ
iV
RT
a
NRT
N 2a
NRT
Ni N j aij
2=
2=
V b V
V Nb V
V Nibi
V2
P
RT
NRT
=
Ni V V Nibi V Nibi
b
=
N
g
2
ab f 2VN a
j ij
i
2
RT
NRTbi 2 N
9
9.7
(a) Start from eqn. (9.2-13)
ln i = ln
fi
1
=
yi P RT
z
V = ZRT P
V =
LM RT N F P I
MN V GH N JK
i T ,V , N j i
OPdV ln Z
PQ
but
PV
B
yi y j Bij
= 1 + mix = 1 +
RT
V
V
P=
RT Bmix RT
+
=
V
V2
P RT
=
+
Ni
V
2
P NRT
=
+
N
Ni
V
Ni RT + RT Ni N j Bij
V