Fugacidad Mezclas Equilibrio L-V L-L V-V

1. Consider a solution of two species S1/S2 at 25
Example 6.1
o
C such that x1 1V= 0.4. If = 40 x 10-6
m3
2V
/mol,
find . The solution specific gravity is = 0.90 and the molecular weights of the species are
32 and 18 respectively.
Molar mass of solution = x1M1 + x2M2
= 0.4 x 32 + 0.6 x 18
(M = MW)
= 23.433 gm/mol = 23.6 x 10-3
Solution molar volume =
kg/mol
Density
massMolar
⇒
=
3
3
23.6 10 /
900 /
x kg mol
kg m
−
= 26.2 x 10-6
m3
Now V = x
/mol
1 221 VxV +
⇒
( ) 6
2 1 1 2
26.2 - 0.4 40 10
( ) /
0.6
x x
V V xV x
−
=− =
= 17.0 x 10-6
m3
/mol.
The molar enthalpy of a binary solution is given by:
Example 6.2
V = 500 x1 + 1000 x2 + x1x2 (50 x1+40x2) cm3
1 1, and .V V ∞
/mol. Find the expressions for
Putting x1 = 1- x2
V = 500 + 550 x
; it follows:
2 – 60 x2
2
+ 10 x2
2
2 2
2
550 120 30
dV
x x
dx
= − +
3
1 2
2
dV
V V x
dx
= −
= 500 x1 + 1000 x2 + x1x2 (500x1 + 40 x2) -x2 [550 – 120 x2 + 30 x2
2
Putting x
]
2 = 1-x1; and simplifying:
3
1 1 1540 60 x 20 xV = − +
1
1 1
0
lim 540 J / mol
x
V V∞
→
= =

2. Alternately one may also use the generic definitions:
2
1
1 , ,n T P
nV
V
n
∂
=
∂
V = 500 + 550 x2 – 60 x2
2
+ 10 x2
Putting x
3
2 = 1-x1
V = 1000 – 460 x
and simplifying:
1 – 30 x1
2
– 10 x1
V = 1000 – 460
3
3
3
1
2
2
11
n
n
10
n
n
30
n
n
−−
Where,n = n1 + n2
2
1
1 n
nV
V
n
∂
=
∂
(moles of mixture)
; note that:
2
1 n
nV
n
∂
∂
= 1
2
2 3
1 1 1
1 2 3
1
10
1000 460 30
n
n n n
V n
n n n n
  ∂
= − − −  
∂   
=
2
2 3
1 1
1 2
1
10
1000 460 30
n
n n
n n
n n n
 ∂
− − − ∂  
= 640 – 60
2 3
1 1
2 3
20
n n
n n
+
= 640 – 60 x1
2
+ 20 x1
3
The same exercises may be carried for obtaining
… [same as the earlier expression]
2.V
What is the change in entropy when 0.6 m
Example 6.3
3
of CO2 and 0.4 m3
of N2, each at 1 bar and 25o
For an ideal gas, mole fraction = volume fraction
C
blend to form a gas mixture at the same conditions? Assume ideal gases.
CO2 (1) / N2(2); y1 = 0.6, y2=0.4
( ) ln 5.5 /mix i iS R y y J molK∆ =− ∑ =
Estimate the fugacity of ethane at 122.2 K and 5 bar using the truncated virial EOS.
Example 6.4
For ethane Tc = 305.4K, Pc = 48.84 bar, ω = 0.099
Now,

3. Estimate the fugacity of ammonia vapor at 4.0 MPa and 321K assuming that it obeys the RK
equation of state.
Example 6.5
Now,
In the above equation T = 321.55K, V is not known.
So solve RK-EOS for Vvapour
V
at the given temperature and pressure, i.e., at P =1.95 MPa, T =
321.55 K (see example 2.3)
vapour turns out to be ≈ 1.1987 * 10-3
m3
Thus on substitution in eq. (1) ln ϕ = -0.1189 → ϕ = 0.888
/mol
Estimate the fugacity of methane at 32C and 9.28 bar. Use the generalized correlation
approach.
Example 6.6
For methane Tc = 190.7 K, Pc
For given T
= 46.41 bar, ω = 0.011
r and Pr, read off ϕ0
and ϕ1
Then
from figures of fugacity coefficients.
Example 6.7

4. Estimate the fugacity of cyclopentane at 110 C and 275 bar. At 110 C the vapor pressure of
cyclopentane is 5.267 bar.
For cyclopentane Tc = 511.8 K, Pc = 45.02 bar, ω = 0.196, Zc = 0.273, Vc = 258 cm/mol,
Tn
T = 383K, P = 275 bar, P
= 322.4 K
3
T
(383K) = 5.267 bar
r 0.117sat sat
r CP P P= == 0.7486,
Calculate fsat
Virial EOS
at the given Vapour pressure by
ln sat sat sat sat sat
BP RT f Pφ φ⇒ = ⇒ =
Here T = 383K, B is obtained as in problem 18, by B0
, B1
Final ϕ
.
S
Now by Rackett equation
= 0.9
0.2857
(1- )nbrTsat
C CV V Z=
( )
3
322.4 / 54.8 0.63 97.092
.exp 10.79
sat
nbr
sat sat sat sat sat
cm
T V
mol
f P V P P RT barφ
= = ⇒ =
 ∴= −= 
For the following system compute the species fugacity coefficients for an equimolar mixture
at 20 bar and 500K.
Example 6.8
Tc P(K) c V(
bar)
c X 103
(m3
Z/mol) ωc yi
Propane (1) 369.9 42.57 0.2 0.271 0.153 0.4
Pentane (2) 469.8 33.75 0.311 0.269 0.269 0.6
P = 20 bar, T = 500 K, y1 = y2 = 0.5

5. Similarly, for pure components
Following the same procedure above (K11 = K22= 0), [Kij
B
= 0 if i = j] it may be shown that:
11 = - 1.1833×10-4
m3
/mol, B22 = - 3.4407×10-4
m3
Thus:
/ mol
Similarly,
Calculate the fugacities of ethylene and propylene in a mixture of 70 mole percent ethylene
and 30 mole percent propylene at 600 K and 60 bar. Assume that the mixture follows the
Redlich-Kwong equation of state.
Example 6.9

6. Tc P(K) c ( bar)
Ethylene (1) 283.1 51.17
Propylene (2) 365.1 46.0
R-K parameters for pure species and mixture are obtained first
Similarly,
Now, solve for Z from cubic EOS,
It follows,
Since one real root only exists,

7. Methanol (1)-acetone (2) system is described by the Van Laar activity coefficient model. At
60
Example 6.10
0
12 210.47; 0.78A A= =C, the model parameters are . Estimate the activity coefficients for a
solution containing 10mole% of methanol.
12 21
1 22 2
12 1 21 2
21 2 12 1
ln ;ln
1 1
A A
A x A x
A x A x
γ γ= =
   
+ +   
   
For x1 = 0.1, γ1 = 1.5219, γ2 = 1.0032
Use of Regular Solution Model to estimate activity coefficients for an equimolar benzene (1)
/ cyclohexane (2) solution 350
Example 6.11
o
K. The solubility parameters are: δ1 = 9.2 (cal/cm3
)1/2
; δ2 =
8.2 (cal/cm3
)1/2
. The molar volumes: V1
L
= 88 cm3
/mol; V2
L
= 107 cm3
Volume fraction =
/mol
( )1 1 1 1 1 2 2/L L L
xV xV x VΦ= + = 88 / (88 + 107) = 0.45
2 11 0.55Φ = − Φ =
[ ]
22 2 2
1 1 2 1 2ln ( ) 88 0.55 9.2 – 8.2L
RT V xγ δ δ= Φ − =
R = 1.987 cal/mol, T = 350o
Hence ln γ
K
1 = 0.038 ⇒ γ1
Similarly RT ln
= 1.04
2γ = 2 2
2 1 1 2( )L
V δ δΦ −
Hence γ2 = 1.03

8. Use UNIFAC model to estimate activity coefficients for an equimolar n-pentane (1) /acetone
(2) solution 350
Example 6.12
o
( )
( )
3
11 22 12
12 12 11 22
2
11 1 2 12
1
2
22 2 1 12
2
963, 1523, 52 /
2
exp
exp
S
S
B B B cm mol
B B B
B P P Py
RT
B P P Py
RT
δ
δ
δ
=− =− =
∴ = − − =
 − +
 Φ =
  
 − +
 Φ =
  
K.
Follow the algorithm provided in the text for bubble pressure calculation.
Final Answer: 1 285.14 , 0.812, 0.188bP KPa y y= = =