CHE3161 - Semester1 - 2011 - Solutions
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CHE 3161 (JUN 11)
Solution for CHE3161_S1_2011 Question 1.
(20 Marks)
One mole of ideal gas with C p = (5/2)R (5/2)R and C v = (3/2)R (3/2)R undergoes the following two sequential steps: (i) Heating from 200 K to 600 K at constant pressure of 3 bar, followed by (ii) Cooling at constant volume. To achieve the same amount of Work produced produced by this two-step process, a single isothermal expansion of the same gas from 200 K and 3 bar to some final pressure, P can be performed.
(1) Draw all the processes on a P-V a P-V diagram. diagram.
[4 marks]
(2) What is the final pressure, P of the isothermal expansion process assuming mechanical reversibility for both the processes?
[14 marks]
(3) Comment on the value of P of the isothermal expansion process assuming mechanical reversibility for the two-steps process while mechanical irreversibility for the isothermal expansion process.
[2 marks]
Solution:
(1)
P 1
Isobaric
2
Isochoric
Isothermal
3
4
V Page 1 of 12
CHE 3161 (JUN 11)
(2) For the two-steps process: !
!!"
!
! !"
! !
!!!! !!! !!! ! !! ! !!! !! ! !! !!
! ! !
Since P 1 = P 2, hence W 12
! P 2V 2 + P 1V 1
=
Applying Ideal Gas Law: W 12
! RT 2
=
= R
W 23
1
(T 1 ! T 2 )
" ! 32 PdV
=
=
+ RT
0
Therefore, Total W 13
= W 12 + W 23
= R
(T 1 ! T 2 )
For the isothermal expansion process: W 14
=
RT 1 ln
P 4 P 1
If the two works have to be the same: RT 1 ln
ln P 4
P 4 P 1 =
P 4 P 1
=
=
R(T 1
' T 2 )
(1)
(T 1 ' T 2 ) T 1
& (T 1 ' T 2 ) # ! % T 1 " & (200 ' 600) # 3 exp $ ! 200 % "
P 1 exp $
=
=
0.406bar
(3) P increases since the left hand term of Eqn (1) will be multiplied by a factor of less that 1 (1 =100% efficiency).
Page 2 of 12
CHE 3161 (JUN 11)
Question 2.
(20 Marks)
Calculate the compressibility (Z), residual enthalpy (HR ), residual entropy (S R ), and residual Gibbs energy (GR ) of propane at 80 oC and 15 bar using the Soave/Redlich/Kwong equation of state. The critical properties of propane are T c = 369.8 K, P c = 42.48 bar, and ! = 0.152.
[20 marks]
Solution:
For the given conditions: T r
=
80 + 273.15 369.8
=
15
0.9550 Pr
=
=
0.3531
42.48
The dimensionless EOS parameters for the R/K EOS are: !
=
!
Pr
=
0.08664
T r
Pr
=
0.0320
T r
(T r ;" ) = "#1 + (0.480 + 1.574" ! 0.176" ) 1 ! T r 2
!
q
=
!! (T r )
=
0.4278! (T r )
=
0.08664T r
"T r
(
1/2
2 $ )% = 1.0328
5.3403
We can now solve iteratively for Z using the equation: ( Z ! ! ) ( Z ! 0.0320) Z = 1 + ! ! q! = 1 + 0.0320 ! 0.1709 Z ( Z + ! ) Z ( Z + 0.0320)
Starting with an initial guess of Z = 1, and iterating gives, Z = 0.8442 Then the integral I is: I =
1
! ! "
ln
Z + #$ Z + !$
=
0.0372
The derivative is:
Page 3 of 12
CHE 3161 (JUN 11) d ln ! (T r ) d ln T r
0.5 " % T 2 = !(0.480 + 1.574" ! 0.176" ) $ ' = !0.6877 # ! (T ) & r
r
Next, we can use these values to calculate the residual enthalpy and entropy from: H R RT
=
" d ln ! (T r )
% ! 1' qI = !0.4915 # d ln T r &
Z ! 1 + $
R
S
ln( Z ! ! ) +
=
R
d ln " (T r ) d ln T r
qI = !0.3448
Therefore, R
H
=
R
S
=
!1443.031 J.mol-1;
!2.867 J.mol-1.K -1
Knowing these, G
R
R
=
H
! TS R
=
!430.572 J.mol-1
Page 4 of 12
CHE 3161 (JUN 11)
Question 3.
(20 Marks)
(1) Prove: An equilibrium liquid/vapour system described by Raoult’s law cannot exhibit an azeotrope. (2) A liquid mixture of cyclohexanone(1)/phenol(2) for which x1 = 0.6 is in equilibrium with its o
vapour at 144 C. Determine the equilibrium pressure P and vapour composition y1 from the following information: ln ! 1
!
2
=
A x2 ; o
ln ! 2 Sat
2
=
A x1
Sat
!
At 144 C,
!
The system forms an azeotrope at 144 oC for which
P1
=
75.20 and P2
=
31.66 KPa az
x1
=
az
y1
=
0.294
Solution:
(1) For a binary system obeying Raoult’s law, sat
(1)
y2 P x 2 P2
(2)
y1P x1 P1 =
sat
=
equations (1) + (2) give, sat
y1P + y2 P = x1 P1
+
sat
x2 P2
As y1 + y2 =1 and x1 + x2 = 1, therefore sat
P = P2
+ x 1
sat
(P1
sat
! P2 )
(3)
Equation 3 predicts that P is linear in x1. Thus no maximum or minimum can exist in this relation. Since such an extremum is required for the existence of an azeotrope, no azeotrope is possible. (2) Based on the known information, we can first determine the value for A, and then calculate equilibrium pressure and vapour composition. From modified Raoult’s law, yi P x i ! i Pi sat =
Page 5 of 12
CHE 3161 (JUN 11)
At the azeotrope, yi= xi, then, P ! i
=
Pi
sat
Therefore, sat
! 1
=
P2
sat
P1
! 2
Given the conditions, 2
ln ! 1
=
A x2 ;
ln ! 2
2
=
A x1
Then,
ln
! 1
2
A ( x2
=
! 2
! x12 )
Therefore, ln A
! 2
=
2
x2
sat
! 1
! x12
ln =
P2
sat
P1
2
! x12
x2
Putting in the known numbers for satuation pressures and compositions at the azeotrope: A = -2.0998
Next, at x1 = 0.6, x2 = 1-x1 = 0.4, ! 1 ! 2
=
=
2 exp( A x2 ) 0.7146 =
2
exp( A x1 ) sat
P = x1 ! 1P1
=
+ x
0.4696 sat
! 2 P2
2
=
38.1898 kPa
The vapour composition y1 is: sat
y1
=
x1! 1P1 P
=
0.8443
Page 6 of 12
CHE 3161 (JUN 11)
Question 4.
(20 Marks)
The molar volume (cm 3 mol-1) of a binary liquid system of species 1 and 2 at fixed T and P is given by the equation V = 120 x1 + 70 x2 + (15 x1 + 8 x2) x1 x2. (a)
(b)
Determine an expression as a function of x1 for (i)
the partial molar volume of species 1,
(ii)
the partial molar volume of species 2, V 2 .
[8 marks]
V 1 .
Using the expressions obtained in (4a), calculate the values for (i)
the pure-species volumes V 1 and V 2 .
(ii)
the partial molar volumes at infinite dilution
!
V 1
and
!
V 2
[12 marks]
.
Solutions:
(a)
V = 120 x1 + 70 x2 + (15 x1 + 8 x2) x1 x2 But x1 + x2 = 1 x2 = 1 – x1
!
V = 120 x1 + 70(1 – x1) + [15 x1 + 8(1 – x1)] x1(1 – x1) Reagreement and simplification of equation will lead to: V = -7 x13 – x12 + 58 x1 + 70 dV dx1
(i)
=
!21 x12 ! 2 x1 + 58
Using Eq. (11.15),
V 1
=
V 1
= V + x
dV 2
dx1
!7 x13 ! x12 + 58 x1 + 70 + (1 ! x1 )(!21 x12 ! 2 x1 + 58)
Reagreement and simplification of equation will lead to:
Page 7 of 12
CHE 3161 (JUN 11) V 1
(ii)
= 14 x
3
1
! 20 x12 ! 2 x1 + 128
Using Eq. (11.16),
V 2
=
V 2
=
V ! x1
dV dx1
!7 x13 ! x12 + 58 x1 + 70 ! x1 (!21 x12 ! 2 x1 + 58)
Reagreement and simplification of equation will lead to: V 2
= 14 x
3
+ x
1
2
1
+
70
(b) (i)
For pure species volume, V 1 x1 = 1 3
2
Thus, V 1 = 14(1) – 20(1) – 2(1) + 128 V 1 = 120 cm 3 mol-1
For pure species volume, V 2 x2 = 1 or x1 = 0 2
Thus, V 2 = 14(0) + 0 + 70 V 2 = 70 cm 3 mol-1
(i)
For partial volume at infinite dilution,
!
V 1
x1 = 0 Thus,
!
V 1
!
V 1
= 14(0)3 – 20(0) 2 – 2(0) + 128 = 128 cm 3 mol-1
For partial volume at infinite dilution,
!
V 2
x2 = 0 or x1 = 1 Page 8 of 12
CHE 3161 (JUN 11)
Thus,
!
V 2
!
V 2
= 14(1) + 1 2 + 70 = 85 cm3 mol-1
Page 9 of 12
CHE 3161 (JUN 11)
Question 5.
(20 Marks)
Equilibrium at 425 K and 15 bar is established for the gas-phase isomerisation reaction:
n-C4H10( g ) "iso-C4H10( g )
If there is initially 1 mol of reactant and K = 1.974, calculate the compostions of the equilibrium mixture ( yn-C4H10 and yiso-C4H10) by two procedures:
(a)
Assume an ideal-gas mixture.
(b)
Assume an ideal solution.
[6 marks] [14 marks]
For n-C4H10: #1 = 0.200; T c,1= 425.1 K; P c,1= 37.96 bar For iso-C4H10: #2 = 0.181; T c,2 = 408.1 K; P c,2= 36.48 bar
Solutions:
Given T = 425 K, P = 15 bar, K = 1.974, no = 1, v
=
!v
i
=
1"1
=
0
Assume species 1 $n-C4H10, species 2 $iso-C4H10. y1
=
y 2
=
1 " !
=
1 + 0(! )
1 " !
!
(a)
1 + 0(! )
= !
For an ideal-gas mixture: !v
)
( yi )
vi
=
i
v
v
y1 1 " y22
=
K
(1 ! * ) !1 " * K
* =
1 ! *
( P % && ## K ' P o $
=
=
K
1.974
Page 10 of 12
CHE 3161 (JUN 11)
Thus,
= 0.664
%
y1= 1 - % = 0.336 y2= % = 0.664
(b)
For an ideal solution: "v
! ( y ) )
vi
i i
( P % & o # K ' P $
=
i
For species 1 $n-C4H10: #1 = 0.200; T c,1= 425.1 K; P c,1= 37.96 bar
P r ,1 ! P c ,1
=
P
T r ,1 ! T c ,1
15
P r ,1
=
=
37.96
0.395
T r ,1
Using Equation (3.65) to determine B o
B1
=
0.083 !
0.422 1.6
=
1
1 =
0.139 !
0.172 4.2
=
1
T
425 =
=
425.1
1
o
!0.339
Using Equation (3.66) to determine B B1
=
1
!0.033
Using Equation (11.68) to determine " 1. ( 1
=
& 0.395 {' 0.339 + 0.2('0.033)}#! = 0.872 % 1 "
exp$
For species 2 $iso-C4H10: #2 = 0.181; T c,2 = 408.1 K; P c,2= 36.48 bar
P r , 2 P r , 2
! P
c, 2
=
P
15 =
=
36.48
T r , 2
0.411
Using Equation (3.65) to determine B
T r , 2
! T
c,2
=
T
425 =
=
408.1
1.041
o
Page 11 of 12
CHE 3161 (JUN 11) o
B2
0.083 !
=
0.422 1.6
=
1.041
!0.313
Using Equation (3.66) to determine B1 1
B2
=
0.139 !
0.172 4.2
=
1.041
!6.29 " 10
!3
Using Equation (11.68) to determine " 2. ) 2
=
& 0.411 # { ( 0.313 + 0.181((6.29 ' 10(3 )}! = 0.883 % 1.041 "
exp$
!v
) ( y + ) i i
i
vi
=
( P % & o # K ' P $
v v ( y1+ 1 ) 1 " ( y2+ 2 ) 2
=
1.974
[(1 ! * )(0.872)]!1[* (0.883)]1
Thus,
=
1.974
= 0.661
%
y1= 1 - % = 0.339 y2= % = 0.661
Page 12 of 12
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