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CHE 3161 (JUN 11)
Page 1 of 12
Solution for CHE3161_S1_2011 Question 1. (20 Marks)
One mole of ideal gas with Cp = (5/2)R and Cv = (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 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 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
V
1 2
3
4
Isobaric
Isothermal
Isochoric
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CHE 3161 (JUN 11)
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(2) For the two-steps process:
!!" = !!! !" = !! = !!(!! !!) = !!!! + !!!!
Since P1 = P2, hence
112212 VPVPW +=
Applying Ideal Gas Law:
)( 211212
TTRRTRTW
=
+=
0
3223
=
= PdVW
Therefore,
)( 21231213
TTRWWWTotal
=
+=
For the isothermal expansion process:
1
4114 ln PPRTW =
If the two works have to be the same:
bar
TTTPP
TTT
PP
TTRPPRT
406.0200
)600200(exp3
)(exp
)(ln
)1()(ln
1
2114
1
21
1
4
211
41
=
=
=
=
=
(3) P increases since the left hand term of Eqn (1) will be
multiplied by a factor of less that 1 (1
=100% efficiency).
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CHE 3161 (JUN 11)
Page 3 of 12
Question 2. (20 Marks)
Calculate the compressibility (Z), residual enthalpy (HR),
residual entropy (SR), and residual Gibbs
energy (GR) of propane at 80oC and 15 bar using the
Soave/Redlich/Kwong equation of state. The
critical properties of propane are Tc = 369.8 K, Pc = 42.48 bar,
and = 0.152. [20 marks]
Solution:
For the given conditions:
Tr =80+ 273.15369.8 = 0.9550 Pr =
1542.48 = 0.3531
The dimensionless EOS parameters for the R/K EOS are:
! =!PrTr= 0.08664 PrTr
= 0.0320
!(Tr;") = 1+ (0.480+1.574" ! 0.176" 2 ) 1!Tr1/2( )"#
$%2=1.0328
q = !!(Tr )"Tr
=0.4278!(Tr )0.08664Tr
= 5.3403
We can now solve iteratively for Z using the equation:
Z =1+! ! q! (Z !!)Z(Z +!) =1+ 0.0320! 0.1709(Z ! 0.0320)Z(Z +
0.0320)
Starting with an initial guess of Z = 1, and iterating
gives,
Z = 0.8442
Then the integral I is:
I = 1! !"
ln Z +#$Z +!$ =0.0372
The derivative is:
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CHE 3161 (JUN 11)
Page 4 of 12
d ln!(Tr )d lnTr
= !(0.480+1.574" ! 0.176" 2 ) Tr!(Tr )
"
#$
%
&'
0.5
= !0.6877
Next, we can use these values to calculate the residual enthalpy
and entropy from:
HRRT = Z !1+
d ln!(Tr )d lnTr
!1"#$
%
&'qI = !0.4915
SRR = ln(Z !!)+
d ln"(Tr )d lnTr
qI = !0.3448
Therefore,
HR = !1443.031J.mol-1;
SR = !2.867 J.mol-1.K-1
Knowing these,
GR = HR !TSR = !430.572 J.mol-1
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CHE 3161 (JUN 11)
Page 5 of 12
Question 3. (20 Marks)
(1) Prove: An equilibrium liquid/vapour system described by
Raoults law cannot exhibit an
azeotrope.
(2) A liquid mixture of cyclohexanone(1)/phenol(2) for which x1
= 0.6 is in equilibrium with its
vapour at 144oC. Determine the equilibrium pressure P and vapour
composition y1 from the
following information:
ln!1 = A x22 ; ln!2 = A x12
At 144oC, P1Sat = 75.20 and P2Sat = 31.66KPa
The system forms an azeotrope at 144oC for which x1az = y1az =
0.294
Solution:
(1) For a binary system obeying Raoults law,
y1P = x1P1sat (1)
y2P = x2P2sat (2)
equations (1) + (2) give,
y1P + y2P = x1P1sat + x2P2sat
As y1 + y2 =1 and x1 + x2 = 1, therefore
P = P2sat + x1 (P1sat !P2sat ) (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 Raoults law,
yiP = xi ! iPisat
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CHE 3161 (JUN 11)
Page 6 of 12
At the azeotrope, yi= xi, then,
! i =PPisat
Therefore,
!1!2=P2satP1sat
Given the conditions,
ln!1 = A x22 ; ln!2 = A x12 Then,
ln !1!2= A(x22 ! x12 )
Therefore,
A =ln !1
!2x22 ! x12
=ln P2
sat
P1satx22 ! x12
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 = exp(A x22 ) =0.7146 !2 = exp(A x12 ) = 0.4696 P =
x1!1P1sat + x2!2P2sat = 38.1898 kPa
The vapour composition y1 is:
y1 =x1!1P1satP = 0.8443
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CHE 3161 (JUN 11)
Page 7 of 12
Question 4. (20 Marks) The molar volume (cm3 mol-1) of a binary
liquid system of species 1 and 2 at fixed T and P is given
by the equation V = 120x1 + 70x2 + (15x1 + 8x2) x1x2.
(a) Determine an expression as a function of x1 for [8
marks]
(i) the partial molar volume of species 1, V1 .
(ii) the partial molar volume of species 2,V2 .
(b) Using the expressions obtained in (4a), calculate the values
for [12 marks]
(i) the pure-species volumes V1 and V2 .
(ii) the partial molar volumes at infinite dilution V1! and V2!
.
Solutions:
(a) V = 120x1 + 70x2 + (15x1 + 8x2) x1x2
But x1 + x2 = 1
x2 = 1 x1
V = 120 x1 + 70(1 x1) + [15x1 + 8(1 x1)] x1(1 x1)
Reagreement and simplification of equation will lead to:
V = -7x13 x12 + 58x1 + 70
58221 121
1
+= xxdxdV
(i) Using Eq. (11.15), V1 =V + x2dVdx1
V1 = !7x13 ! x12 + 58x1 + 70+ (1! x1)(!21x12 ! 2x1 + 58)
Reagreement and simplification of equation will lead to:
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CHE 3161 (JUN 11)
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V1 =14x13 ! 20x12 ! 2x1 +128
(ii) Using Eq. (11.16), V2 =V ! x1dVdx1
V2 = !7x13 ! x12 + 58x1 + 70! x1(!21x12 ! 2x1 + 58)
Reagreement and simplification of equation will lead to:
V2 =14x13 + x12 + 70
(b)
(i) For pure species volume, V1
x1 = 1
Thus, V1 = 14(1)3 20(1)2 2(1) + 128
V1 = 120 cm3 mol-1
For pure species volume, V2
x2 = 1 or x1 = 0
Thus, V2 = 14(0) + 02 + 70
V2 = 70 cm3 mol-1
(i) For partial volume at infinite dilution, V1!
x1 = 0
Thus, V1!= 14(0)3 20(0)2 2(0) + 128
V1! = 128 cm3 mol-1
For partial volume at infinite dilution, V2!
x2 = 0 or x1 = 1
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CHE 3161 (JUN 11)
Page 9 of 12
Thus, V2! = 14(1) + 12 + 70
V2!= 85 cm3 mol-1
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CHE 3161 (JUN 11)
Page 10 of 12
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. [6 marks]
(b) Assume an ideal solution. [14 marks]
For n-C4H10: 1 = 0.200; Tc,1= 425.1 K; Pc,1= 37.96 bar
For iso-C4H10: 2 = 0.181; Tc,2 = 408.1 K; Pc,2= 36.48 bar
Solutions:
Given T = 425 K, P = 15 bar, K = 1.974, no = 1,
=== 011ivv
Assume species 1 n-C4H10, species 2 iso-C4H10.
=+
=
=+
=
)(01
1)(01
1
2
1
y
y
(a) For an ideal-gas mixture:
974.11
)1(
)(
12121
=
=
=
=
=
K
KKyy
KPPy
vv
v
o
vi
i
i
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CHE 3161 (JUN 11)
Page 11 of 12
Thus, = 0.664
y1= 1 - = 0.336
y2= = 0.664
(b) For an ideal solution:
=i
v
ov
ii KPPy i)(
For species 1 n-C4H10: 1 = 0.200; Tc,1= 425.1 K; Pc,1= 37.96
bar
395.096.3715
1,
1,1,
==
=
r
cr
P
PPP
11.425
4251,
1,1,
==
=
r
cr
T
TTT
Using Equation (3.65) to determine Bo
339.01422.0083.0 6.11 ==
oB
Using Equation (3.66) to determine B1
033.01172.0139.0 2.4
11 ==B
Using Equation (11.68) to determine 1.
{ } 872.0)033.0(2.0339.01395.0exp1 =
+=
For species 2 iso-C4H10: 2 = 0.181; Tc,2 = 408.1 K; Pc,2= 36.48
bar
411.048.3615
2,
2,2,
==
=
r
cr
P
PPP
041.11.408
4252,
2,2,
==
=
r
cr
T
TTT
Using Equation (3.65) to determine Bo
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CHE 3161 (JUN 11)
Page 12 of 12
313.0041.1422.0083.0 6.12 ==
oB
Using Equation (3.66) to determine B1
32.4
12 1029.6041.1
172.0139.0 ==B
Using Equation (11.68) to determine 2.
{ } 883.0)1029.6(181.0313.0041.1411.0exp 32 =
+=
974.1)]883.0([)]872.0)(1[(
974.1)()(
)(
112211
21
=
=
=
vvi
v
ov
ii
yy
KPPy i
Thus, = 0.661
y1= 1 - = 0.339
y2= = 0.661
