CHE 3161 (JUN 10)

CHE 3161: Chemistry & Chemical Thermodynamics

Final Exam - Solution

Date: 06/2010 Total Marks: 100

1) An ideal gas, CP = (7/2)R and CV = (5/2)R, undergoes a cycle consisting of thefollowing mechanically reversible steps:

(a) An adiabatic compression from state A at P1, V1, T1 to state B at P2, V2, T2.

(b) An isobaric expansion from state B at P2, V2, T2 to state C at P3 = P2, V3, T3.

(c) An adiabatic expansion from state C at P3, V3, T3 to state D at P4, V4, T4.

(d) A constant-volume expansion process from state D at P4, V4, T4 to state A at P1, V1 =V4, T1.

(i) Sketch this cycle on a PV diagram

(ii) On the basis of one mole of ideal gas, what is the heat QB→C absorbed in step (b), andthe heat QD→A rejected in step (d)?

(iii) If the thermal efficiency is defined as the ratio of the net work output to the heatabsorbed, show that thermal efficiency is

η = 1 +QD→A

QB→C

(iv) By defining the compression ratio rc ≡ V1/V2, and the expansion ratio re ≡ V4/V3,show that,

η = 1− 1

γ

[(1/re)

γ − (1/rc)γ

1/re − 1/rc

]where, γ = CP/CV .

(v) Determine its thermal efficiency if T1 = 200◦C, T2 = 1000◦C, T3 = 1700◦C, and T4 =601◦C.

(20 marks)

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CHE 3161 (JUN 10)

Figure 1: PV diagram of the cycle undergone by the ideal gas.

Solution:

(i) On a PV diagram, the cycle would appear as shown in the figure above.

(ii) In step (b), the gas undergoes an isobaric expansion. The heat absorbed at constantpressure is equal to the enthalpy change. As a result, since the ideal gas has constantheat capacities,

QB→C = CP (T3 − T2) (1)

In step (d), the gas undergoes an isochoric expansion. The heat rejected at constantvolume is equal to the change in internal energy. As a result,

QD→A = CV (T1 − T4) (2)

(iii) Since the gas undergoes a cycle, the internal energy change ∆U = Q + W = 0, whereQ in the net heat absorbed, and W is the net work input. As a result, the net workoutput is,

−W = QB→C +QD→A

and the thermal efficiency is,

η =QB→C +QD→A

QB→C= 1 +

QD→A

QB→C

(iv) From eqns (1) and (2),

η = 1 +CV (T1 − T4)

CP (T3 − T2)= 1 +

1

γ

(T1 − T4

T3 − T2

)(3)

For the reversible adiabatic expansion, and reversible adiabatic compression of an idealgas, we know that, T V γ−1 = constant. As a result,

T3 Vγ−13 = T4 V

γ−14 and T2 V

γ−12 = T1 V

γ−11

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CHE 3161 (JUN 10)

Since the compression ratio rc ≡ V1/V2, and the expansion ratio re ≡ V4/V3,

T4 = T3

(1

re

)γ−1

and T1 = T2

(1

rc

)γ−1

Substituting these expressions into eqn (3) leads to,

η = 1− 1

γ

T3

(1

re

)γ−1

− T2

(1

rc

)γ−1

T3 − T2

(4)

We know that,

P2 =RT2

V2

and P3 =RT3

V3

Since P2 = P3,T2

T3

=V2

V3

Diving the numerator and denominator of the right hand side by V1 and V4, respectively(since V1 = V4), we have,

T2

T3

=V2/V1

V3/V4

=rerc

(5)

Combining eqns (4) and (5) leads to,

η = 1− 1

γ

(

1

re

)γ−1

− rerc

(1

rc

)γ−1

1− rerc

Which can be rewritten as,

η = 1− 1

γ

(

1

re

)γ−(

1

rc

)γ1

re− 1

rc

(v) Substituting the various temperatures into eqn (3), we have,

η = 0.591

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CHE 3161 (JUN 10)

2) Propylene gas at 127◦C and 38 bar (state 1) is throttled in a steady-state flow processto 1 bar and 90.1◦C (state 2), where it may be assumed to be an ideal gas. Estimate theenthalpy change (∆H) and entropy change (∆S) of the propylene for this change of state.The ideal heat capacity for propylene gas is known to depend on temperature through therelation, CP/R = A + B T + C T 2, where T is in degrees Kelvin. The relevant data forpropylene are given in the Tables below:

A B C1.637 22.706× 10−3 −6.915× 10−6

Tc (K) Pc (bar) ω365.6 46.65 0.140

Table 1: Values of the coefficients A,B and C, and the critical parameters for propylene.

(HR)0/RTc

(HR)1/RTc

Pr = 0.6000 0.8000 1.000 Pr = 0.6000 0.8000 1.000Tr Tr

HHH HHH

1.05 −0.654 −0.955 −1.359 1.05 −0.498 −0.691 −0.8771.10 −0.581 −0.827 −1.120 1.10 −0.381 −0.507 −0.6171.15 −0.523 −0.732 −0.968 1.15 −0.296 −0.385 −0.459

Table 2: Lee/Kesler generalized-correlation values for residual enthalpy

(SR)0/R

(SR)1/R

Pr = 0.6000 0.8000 1.000 Pr = 0.6000 0.8000 1.000Tr Tr

HHH HHH

1.05 −0.439 −0.656 −0.965 1.05 −0.460 −0.642 −0.8201.10 −0.371 −0.537 −0.742 1.10 −0.350 −0.470 −0.5771.15 −0.319 −0.452 −0.607 1.15 −0.275 −0.361 −0.437

Table 3: Lee/Kesler generalized-correlation values for residual entropy

(20 marks)

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CHE 3161 (JUN 10)

Solution: The process undergone by propane can be visualised in the following figure:

Figure 2: The entropy change from the initial to the final state can be calculated by a two-stepprocess. The first is a change from state 1 at 38 bar and 127◦C to an ideal gas at 38 bar and 127◦C.The second is a change from an ideal gas at 38 bar and 127◦C to state 2, which is an ideal gas at90.1◦C and 1 bar.

For the initial conditions,

T1r =T1

Tc=

400.15

365.6= 1.095

P1r =P1

Pc=

38

46.65= 0.815

From the Lee/Kesler generalized-correlation tables given in the problem, at these values ofTr and Pr, using interpolation,(

HR1

)0RTc

= −0.863, and

(HR

1

)1RTc

= −0.534

(SR1)0

R= −0.565, and

(SR1)1

R= −0.496

As a result, since,

HR1 =

(HR

1

)0+ ω

(HR

1

)1and SR1 =

(SR2)0

+ ω(SR1)1

we have,

HR1 = −0.863× 8.314× 365.6 + 0.140× (−0.534× 8.314× 365.6)

= −2.623× 103 + 0.140× (−1.623× 103) = −2.85× 103 J/mol

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CHE 3161 (JUN 10)

SR1 = −0.565×8.314+0.140×(−0.496×8.314) = −4.697+0.140×(−4.124) = −5.275 J/(mol K)

For an ideal gas (with the coefficient D = 0),

∆H ig = R

[A (T2 − T1) +

B

2(T 2

2 − T 21 ) +

C

3(T 3

2 − T 31 )

]It follows that,

∆H ig = 8.314×[1.637 (363.25− 400.15) +

22.706× 10−3

2

[(363.25)2 − (400.15)2

]−6.915× 10−6

3

[(363.25)3 − (400.15)3

]]= −2.85× 103 J/(mol K)

Therefore,∆H = −HR

1 + ∆H ig = 0 J/mol

as is expected from a throttling process at steady-state.

In a similar manner,

∆Sig

R= A ln

(T2

T1

)+B (T2 − T1) +

C

2(T 2

2 − T 21 )− ln

(P2

P1

)As a result, ∆Sig = 22.774 J/(mol K), and,

∆S = −SR1 + ∆Sig = 28.048 J/(mol K)

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CHE 3161 (JUN 10)

3)

(a) The following is a rule of thumb: For a binary system in VLE at low pressure, the equi-librium vapour-phase mole fraction y1 corresponding to an equimolar liquid mixture isapproximately,

y1 =P sat

1

P sat1 + P sat

2

where P sat1 is a pure-species vapour pressure. Clearly, this equation is valid if Raoult’s

law applies. Show that it is also valid for VLE described by the modified Raoult’s law,with,

ln γ1 = Ax22 ; ln γ2 = Ax2

1

(b) Assuming Raoult’s law is an adequate description of the VLE of toluene/ethylbenzenemixtures,

(i) Compute the bubble point pressure and vapor composition for a liquid mixturewith composition xtoluene = 0.6, xethylbenzene = 0.4 at 450 K.

(ii) Compute the dew point pressure and liquid composition for a vapor mixture withcomposition ytoluene = 0.6, yethylbenzene = 0.4 at 450 K.

(iii) Compute the vapor and liquid composition and phase fractions (fraction of thesystem that is in each phase) for an overall composition of ztoluene = 0.6, zethylbenzene =0.4 at 450 K and 3.75 bar.

The vapor pressure of toluene at 450 K is 4.844 bar, and the vapor pressure of ethyl-benzene at 450 K is 2.685 bar.

(20 marks)

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CHE 3161 (JUN 10)

Solution:

(i) From modified Raoult’s law,

yi P = xi γi Psati (i = 1, 2, . . . , N)

and the given equations for ln γ1 and ln γ2,

y1 P = x1 exp(Ax22)P

sat1 and y2 P = x2 exp(Ax2

1)Psat2

These equations sum to give:

P = x1 exp(Ax22)P

sat1 + x2 exp(Ax2

1)Psat2

Dividing the equation for y1 P by the preceding equation yields:

y1 =x1 exp(Ax2

2)Psat1

x1 exp(Ax22)P

sat1 + x2 exp(Ax2

1)Psat2

For x1 = x2 this equation obviously reduces to:

y1 =P sat

1

P sat1 + P sat

2

(b) (i) We will call toluene species 1 and ethylbenzene species 2. Raoult’s law for species 1and 2 is y1 P = x1 P

sat1 and y2 P = x2P

sat2 , from which P = x1 P

sat1 +x2 P

sat2 = 0.6×

4.844 + 0.4× 2.685 = 3.98 bar. Thus, y1 = x1 Psat1 /P = 0.6× 4.844/3.98 = 0.730

and y2 = 1− y1 = 0.270.

(ii) From the Raoults law for each species, we have y1 P/Psat1 +y2 P/P

sat2 = P (y1/P

sat1 +

y2/Psat2 ) = x1+x2 = 1 so P = 1/(y1/P

sat1 +y2/P

sat2 ) = 1/(0.6/4.844+0.4/2.685) =

3.665 bar. Thus, x1 = y1 P/Psat1 = 0.6× 3.665/4.844 = 0.454, and x2 = 1− x1 =

0.546.

(iii) In this case, we know that there will be 2 phases present, since the total pressureis between the bubble point and dew point for this overall composition, calculatedin parts (i) and (ii). So, we will start from the flash calculation equation:

yi =ziKi

1 + V (Ki − 1)and

N∑i=1

yi = 1 =N∑i=1

ziKi

1 + V (Ki − 1)

For a system that obeys Raoults law, Ki = yi/xi = P sat1 /P . So, K1 = 4.844/3.75 =

1.292 and K2 = 2.685/3.75 = 0.716. So, we have

N∑i=1

yi = 1 =0.6× 1.292

1 + 0.292V+

0.4× 0.716

1− 0.284V=

0.7752

1 + 0.292V+

0.2864

1− 0.284V

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CHE 3161 (JUN 10)

Rearranging this equation leads to,

0.08293V 2 − 0.1446V + 0.0616 = 0

Applying the quadratic formula to this last equation gives the fraction of thesystem in the vapor phase as V = 0.740 (or V = 1, which is the trivial solutionthat isn’t physically meaningful). The fraction of the system in the liquid phaseis then L = 1− V = 0.260. The vapor phase mole fractions are then:

y1 =z1K1

1 + V (K1 − 1)=

0.6× 1.292

1 + 0.740× 0.292= 0.637

Thus, y2 = 1− y1 = 0.363, x1 = y1/K1 = 0.493, and x2 = 1− x1 = 0.507.

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CHE 3161 (JUN 10)

4) A single P − x1 data point is available for a binary system at 35◦C:

x1 = 0.389, P = 108.6 kPa

For this system, at 35◦C,

P sat1 = 120.2 kPa and P sat

2 = 73.9 kPa

(a) Verify if Raoult’s law is satisfied

(b) Assume that the excess Gibbs energy obeys the following model:

GE

RT= Ax1x2

Use the known data and the modified Raoult’s law to evaluate the constant A.

(c) Find the corresponding value of y1 at 108.6 kPa and 35◦C.

(d) Find the total pressure at 35◦C for an equimolar liquid mixture.

(c) Is an azeotrope likely at 35◦C?

(20 marks)

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CHE 3161 (JUN 10)

Solution:

(a) According to Raoult’s law,

P = x1Psat1 + (1− x1)P

sat2

From the given data,

P = 0.389× 120.2 + (1− 0.389)× 73.9 = 91.911 kPa

Since this is not equal to the specified pressure, Raoult’s law is not satisfied.

(b) For the given model for the excess Gibb’s energy, we know that the activity coefficientsare given by the expressions,

γ1 = exp(Ax2

2

)and γ2 = exp

(Ax2

1

)From modified Raoult’s law, P = x1 γ1 P

sat1 + (1− x1) γ2 P

sat2 , we have,

P = x1 exp(Ax2

2

)P sat

1 + (1− x1) exp(Ax2

1

)P sat

2

Therefore,

108.6 = 0.389× exp[A× (0.611)2

]× 120.2 + 0.611× exp

[A× (0.389)2

]× 73.9

The only unknown in this equation is A, which can be solved to give,

A = 0.622

(c) We know that from modified Raoult’s law,

y1 = x1γ1P sat

1

P= x1 exp

(Ax2

2

) P sat1

P= 0.389× exp

[0.622× (0.611)2

]× 120.2

108.6= 0.543

(d) Since x1 = x2 and γ1 = γ2, we know from modified Raoult’s law that,

P = x1 γ1 Psat1 + x2 γ2 P

sat2 = 0.5× exp

[0.622× (0.5)2

]× (120.2 + 73.9) = 113.38 kPa

(e) We need to find the relative volatility, α12 =y1/x1

y2/x2

. From the expressions for the

modified Raoults law, it follows that,

α12 =γ1 P1

sat

γ2 P2sat =

120.2× exp [0.622x22]

73.9× exp [0.622x12]

This can be simplified further to,

α12 = 1.6265× exp [0.622(1− 2x1)]

Evaluating this expression at x1 = 0 and x1 = 1 leads to the following values for α12:

x1 α12

0.0 3.0271.0 0.8724

At an azeotrope, since x1 = y1, and x2 = y2, the relative volatility α12 = 1. Since fromthe table of values it is clear that α12 passes through 1.0 for some x1 in the interval0 < x1 < 1, an azeotrope can exist.

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CHE 3161 (JUN 10)

5) Consider the hydrogenation of toluene to give methylcyclohexane:

C7H8(g) + 3 H2(g) C7H14(g)

Starting from a stoichiometric mixture of reactants (3 moles of H2 per mole of toluene):

(a) Compute the equilibrium constant at 350◦C and 1 bar total pressure. You can assumethat the enthalpy change of reaction is independent of temperature and is equal to itsvalue at 298.15 K.

(b) Compute the equilibrium composition at 350◦C and 1 bar total pressure.

The mixture of the reactants and product can be treated as an ideal gas mixture. Enthalpiesand Gibbs energies of formation for toluene and methylcyclohexane are given in the Tablebelow.

C7H8 C7H14

∆Gf298 (J/mol) 122050 27480

∆Hf298 (J/mol) 50170 −154770

(20 marks)

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CHE 3161 (JUN 10)

Solution:

(a) First, we need to compute the enthalpy and Gibbs energy change for the reaction at298 K. For hydrogen, the enthalpy and Gibbs energy of formation are zero because it isan element in its standard state. We are given that, for methylcyclohexane, ∆Gf

298 =27480 J/mol and ∆Hf

298 = −154770 J/mol, and for toluene, ∆Gf298 = 122050 J/mol

and ∆Hf298 = 50170 J/mol. Subtracting the toluene values from the methylcyclohexane

values, we have ∆G◦298 = −94570 J/mol and ∆H◦298 = −204940 J/mol. Neglecting thetemperature dependence of the enthalpy of reaction, we can compute the equilibriumconstant for the reaction as K = K0K1, where K0 = exp(−∆G◦298/(R× 298.15)) is theequilibrium constant at 298 K and K1 = exp(∆H◦298/(R × 298.15) × (1 − 298.15/T )gives the change in the equilibrium constant when the temperature changes from 298K to T . For this case, we have

K0 = exp(94570/(8.314× 298.15)) = 3.71× 1016

and

K1 = exp(−204940/(8.314× 298.15)× (1− 298.15/623.15)) = 1.88× 10−19

From which K = 0.00696.

(b) For a mixture of ideal gases, the composition at equilibrium is related to the equilibriumconstant by ∏

i

(yi)νi =

(P

P ◦

)−νK

which for this problem becomes

yC8H14

yC8H8 y3H2

= K

(P

1 bar

)3

If we initially have 1 mole of toluene and 3 moles of H2, then the number of moles arerelated to the reaction coordinate by

nC8H8 = 1− εe

nH2 = 3− 3εe = 3(1− εe)

nC8H14 = εe

n = 4− 3εe

So, the mole fractions are given by

yC8H8 =1− εe4− 3εe

yH2 =3(1− εe)

4− 3εe

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CHE 3161 (JUN 10)

yC8H14 =εe

4− 3εe

Substituting these into the equilibrium relationship gives:

εe(4− 3εe)3

(1− εe)(3(1− εe))3=εe(4− 3εe)

3

27(1− εe)4= K

(P

1 bar

)3

Using the value of K from part (a) and a pressure of 1 bar, this is

εe(4− 3εe)3

27(1− εe)4= 0.00696

To solve this iteratively, we can rearrange it as

εe =0.1880(1− εe)

4

(4− 3εe)3

Starting from εe = 0 and iterating on this gives εe = 0.00292. Substituting this backinto the expressions for the mole fractions gives yC8H8 = 0.2498, yH2 = 0.7495, andyC8H14 = 0.00073.

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