وزارة التعليم العالي و البحث العلمي

Ministry of high education& scientific Research

University of Technology

Materials engineering department لوجيتوالجامعت التكن

قسم هندست المواد

Date: 27/05/2012

Final Examination

(2011-2012)

Class: Second Year

Subject: Chemical Metallurgy

Allowed time: 3 hrs Examiner: Dr. Mohammed Abdul Hussein

Note: Attempt only Five Questions.

Q1/ (A) Determine two parameters used to predict spontaneity of the process, then determine when each

one from them is useful in dealing with different conditions of the process. State their

relationship with enthalpy change. (B) Is the reaction spontaneous under standard conditions?

4KClO3 (s) 3KClO4 (s) + KCl (s)

i ∆Hf (kJ/mol) S o

KClO3 -397.7 143.1

KClO4 -432.8 151.0

KCl -436.7 82.6

(20 Marks)

Q2/ (A) Explain in details the Reversible and Irreversible processes, give an example.

(B) Water at 33 C is pumped from storage tank at the rate of 0.1 m3 /sec. The motor for the pump

supplied work at the rate of 60 KJ/Kg. The water passes through heat exchanger of 1000 KJ/Kg, and is delivered to elevation of 10 m above the first tank at the velocity 5 m/sec. What is the

temperature of the water delivered to this elevation? (0.94 Btu = 103J). Density of water=1000

Kg/m3. Sketch Flow Diagram.

(20 Marks)

Q3/ (A) For closed system adiabatic processes for an ideal gas, by beginning from the differential first

law of Thermodynamic prove that:

1

1

2

1

2

P

P

T

T

(B) One Kilogram of air is heated reversibly at constant pressure from an initial state of 300 K and 1 bar until its volume triple. Calculate Q, W, ∆U and ∆H for the process. Assume that air

obeys the relation PV/T=83.14 bar cm3 mol 1 K 1 , Cp=29 J mol 1 K 1 , and Molecular Weight of air=29 Kg/Kmol.

(20 Marks)

Enthalpy, Btu/kg Temperature

C Enthalpy, Btu/kg Temperature

C

160 50 201 90

179 70 226 110

Q4/ For the following Figure, calculate for each process ∆U, ∆H, Q & W.

Given that: T1=T2=1600 K

P1= 10 bar, P2= 3 bar, P3= 2 bar Cp=3.5R, Cv=2.5R R=8.418 J/mol. K

(20 Marks)

Q5/ Sulfur dioxide gas is oxidized in 90 percent excess air with 80 % conversion to Sulfur

trioxide. The gases enter the reactor at 410 oC and leave at 460

oC. How much heat must be

transfer from the reactor on the basis 1 mole of entering gas?

SO2 + O2 SO3

Given data; Standard heat of formation at 25 C, and Constants.

i ∆H298 J/mole A 10 3 B 106 C 10-5 D

SO2 -296830 5.699 0.801 --- -1.015

O2 ---- 3.639 0.506 ---- -0.227

SO3 -395720 8.060 1.056 ---- -2.028

N2 ----- 3.280 0.593 ---- 0.040

(20 Marks)

Q6/ (A) One mole of an ideal gas, Cp= (7/2)R, Cv=(5/2)R, is compressed adiabatically in a

piston/cylinder device from 1 bar and 40 C to 5 bar. The process is irreversible and requires

35 percent more work than a reversible adiabatic compression from the initial state to the same final pressure. What is the final temperature and entropy change of the gas?

(B) For an ideal gas with constant heat capacities show that: For a pressure change from P1 to P2, the sign of ∆S of an isothermal change is opposite that for a constant volume change.

(20 Marks)

Good Luck

Pre

ssure

Volume

1

2

4 3

3

Question Solutions of the Final Examination (2011-2012)-Chemical Metallurgy

University of Technology,Materials engineering department

Dr. Mohammed Abdul Hussein

Q1/ (A)

The major parameters to predict spontaneity of the process are Entropy

change (∆S) and Free Energy (∆G), the first one is useful in dealing with different

conditions of the process, but usually the second (∆G) be more useful for certain

conditions of temperature and pressure.

Note that ΔG is composite of both ΔH and ΔS: ΔG = ΔH -TΔS If ΔH < 0 and ΔS > 0….spontaneous at all T •A reaction is spontaneous if ΔG < 0. Such that: If ΔH > 0 and ΔS < 0….not spontaneous at any T If ΔH < 0 and ΔS < 0….spontaneous at low T If ΔH > 0 and ΔS > 0….spontaneous at high T

Q1/ (B)

ΔH°rxn = ∑ᵧi ΔHi

= 3(−432.8kJ) + (−436.7kJ) − 4(−397.7kJ)

= −144kJ

ΔS°rxn = ∑ᵧi Soi

= 3(151.0J/K) + (82.6 J/K) − 4(143.1J/K)

= −36.8J/K

ΔG°rxn = ΔH°rxn − TΔS°rxn

= −144kJ − (298K)(−36.8J/K)(1kJ/1000J)= −133kJ

ΔG°rxn < 0; therefore, reaction is spontaneous under standard conditions.

Q2/ (A)

A process is reversible when its direction can be reversed at any point by an infinite change in external conditions. Once the process is initiated, no infinite change

in external conditions can reverse its direction; the process is irreversible. The apparatus is shown in figure (a gas in piston/cylinder). Expansion processes result

when mass is removed from the piston (we assume the piston without friction), then the piston will rise and new balance of the piston level will be achieved. The oscillation of the peiton level due to that mass m is suddenly removed to a shelf. If we

change the mass by powder the piston gradually rise and the other hand will return to initial level in the same path.

m

Figure : Gas Expasion ∆l

Question Solutions of the Final Examination (2011-2012)-Chemical Metallurgy

University of Technology,Materials engineering department

Dr. Mohammed Abdul Hussein

Q2/ (B)

∆H + ∆Ep + ∆Ek = Q – Ws

Mass flow rate= Density x Volumetric flow rate= 0.1 m3 /sec X 1000 Kg/m3=100

Kg/sec.

∆Ep= m ∆Z g= 100 Kg/sec x 10 m x 9.81 m/s2= 9810 kJ/sec.

∆Ek =m x ∆u2/2= 100 Kg/sec (52-0)/2= 1250 kJ/sec. (u1 could be neglected; tank).

Q = 1000 kJ/Kg x 100 Kg/sec = 100000 kJ/sec. (positive sign assume heat absorbed

by the system).

Ws = 60 kJ/Kg x 100 Kg/sec = 6000 kJ/sec.

Then ∆H= Q – Ws – ∆Ep – ∆Ek = = 100000 kJ/sec – 6000 kJ/sec – 9810 kJ/sec – 1250

kJ/sec = 82940 kJ/sec. = H2 – H1

By extrapolation method H1 (at 33 C) = 144 Btu/Kg x (1 kJ/0.94 Btu) x 100 Kg/sec=

15300 kJ/sec.

Then H2= ∆H + H1 = 82940 kJ/sec. + 15300 kJ/sec. = 98240 kJ/sec.

H2 = 98240 kJ/sec. x (0.94 Btu/ 1 kJ) / 100 Kg/sec = 171.5 Btu/kg.

By extrapolation method T2 (have H=923.5 Btu/kg ) = 66.8 C.

Q3/ (A)

First law of thermodynamic, ∆U = Q – W (Adaibatic Q =0)

Cv dT = – P dV, P=RT/V

Then Cv dT = – RT(dV/V)

dT/T= – R/Cv (dV/V)

R/Cv = (Cp – Cv)/ Cv = Cp/ Cv – 1 =ᵞ – 1

dT/T = – (ᵞ – 1) dV/V (Integration)

ln(T2/T1)=ln(V2/V1)-(γ -1)

)1(

1

2

1

2

V

V

T

T

P1V1/T1 = P2V2/T2, then V2/V1 = (T2/T1 )(P1/P2) substitute in the last equation

Then:

1

1

2

1

2

P

P

T

T

Question Solutions of the Final Examination (2011-2012)-Chemical Metallurgy

University of Technology,Materials engineering department

Dr. Mohammed Abdul Hussein

Q3/ (B)

No. of moles = 1/29 Kg/Kmol =0.0 34 mole

P1 V1/T1= P2 V2/T2 = 83.14 bar cm3 mol 1 K 1 , constant pressure P1= P2

V1/T1= V2/T2 , and V1/T1= 3V1/T2, T2=3T1=3 x 300= 900

V1= 300 X 83.14 bar cm3 mol 1 K 1 = 24942 cm3 mol 1 and V1= 848 cm3

V2 = 3V1= 2544 cm3

At constant pressure Q = ∆H=nCp(T2 – T1)= 0.034 x 29 (900-300) = 600 J

W=P∆V=1 X 105 (2544 cm3 - 848 cm3 ) = 1696 x 105 N/m2 . cm3=169.6 J ∆U = Q – W = 600 J – 169.6 J = 430.4 J.

Q4/

1 2 T1 = T2 (Isothermal process), ∆H=0, ∆U=0

Q=W= RT ln

2

1

P

P=8.314 x 1600 ln

3

10= 16015 J.

2 3 Constant Volume (W=0)

3

3

2

2

T

P

T

P , constant volume. T3= 1600 x 2/3= 1066.6 K

∆H= Cp (T3 – T2)= 3.5x8.314(1066.6 – 1600)= – 15519.4 J

∆U= Cv (T3 – T2)= 2.5x8.314(1066.6 – 1600)= – 11085.3 J Q = ∆U = –11085.3 J

4 1

Adiabatic Process (Q=0), and P4=P3= 3 bar

1

4

1

4

1

P

P

T

T,

1

4 3

101600

T, and =1.4

Then

4T = 1134.3 K

∆H= Cp (T1 – T4)= 3.5x8.314(1600 – 1134.3)= 13551.4 J ∆U= Cv (T1 – T4)= 2.5x8.314(1600 – 1134.3)= 9679.6 J W= - ∆U= – 9679.6 J

3 4

Constant pressure (Isoparic), Q= ∆H ∆H= Cp (T4 – T3)= 3.5x8.314(1134.3 – 1066.6)= 1970.0 J ∆U= Cv (T4 – T3)= 2.5x8.314(1134.3 – 1066.6)= 1407.1 J

W = Q – ∆U = 1970.0 – 1407.1= 562.8 J

Question Solutions of the Final Examination (2011-2012)-Chemical Metallurgy

University of Technology,Materials engineering department

Dr. Mohammed Abdul Hussein

Q5/ Basis: 1 mole of SO2 entering., T1=410+273.15= 683.15 K, T2= 460+273.15= 733.15 K. O2 required (computed on base completely conversion)

According reaction equation ( 1 mole SO2 Vs 0.5 mole O2 required) Excess O2= 0.5 x 0.9= 0.45 mole

Total O2 entering = 0.5 + 0.45= 0.95 mole.

N2 entering= 0.95 x

21

79= 3.57 mole.

Leaving the reactor: SO2= 1 x (1 – 80% conversion) = 0.2 mole

O2 reacted = 1 x 80% x 0.5= 0.4 mole. O2 output = input-reacted = 0.95 – 0.4 = 0.55 mole. SO3 produced = No. of mole SO2 reacted = 0.8 mole.

∆H=∆HR + ∆H298 + ∆HP

∆H298 = ∑ᵧi ΔHi= - 395720 + (- 1) (- 296830)=- 98890 J

Tam=(T1+T2)/2=708.15 K.

∆HR= ∑niCpm,hi(298.15 – Tin)

∆HP= ∑niCpm,hi(Tout–- 298.15)

∑niCpm,hi=R(∑niAi + (∑niBi)Tam + 21TT

Dn ii)

∑niAi= 1 x 5.699 + 0.95 x 3.639 + 3.57 x 3.280= 20.86

∑niBi= 3.4, ∑niDi= - 1.088

∑niCpm,hi= 8.314(20.86+3.4 x 708.15 +733683

088.1

x)= 173.5

The:

∆HR= ∑niCpm,hi(298.15 – 683)= – 66771.5 J

And the same way ∆HP= 78665.3 J

Finally

∆H=– 66771.5 J + (– 98890 J) + 78665.3 J

= – 86996.2 J

Tin=410 oC

SO2 1 mol

O2 0.95 mol

N2 3.57 mol

Tout=460 oC

SO2 0.2 mol

SO3 0.8 mol

O2 0.55 mol

N2 3.57 mol

∆H298

∆HP ∆HR

Q=∆H

Question Solutions of the Final Examination (2011-2012)-Chemical Metallurgy

University of Technology,Materials engineering department

Dr. Mohammed Abdul Hussein

Q6/ (A) Compressed adiabatically, Q=0

∆U = – W

The first step we find reversible and adiabatic work (i.e ∆S=0)

∆S=0=Cp ln

1

2

T

T – R ln

1

2

P

P

7/2Rln

15.313

2T=Rln

1

4, T2= 465.3 K

W=– ∆U= Cv(T2 – T1)= 5/2R(465.3-313.15)=

Wrev.= 3163.2 J

Wirrev=1.3 x Wrev (More 30 %).= 4112.2 J

= Cv(T2irrv. –T1)=5/2R(T2

irrv– 313.15), T2irrv.= 511 K

∆S=7/2R ln

15.313

511 – R ln

1

4

= 2.7 J/K > 0 spontaneous process.

Q6/ (B)

∆S=Cp ln

1

2

T

T – R ln

1

2

P

P,

1. Isothermal process T2=T1

Then (∆S)iso= – R ln

1

2

P

P

2. Constant volume

2

2

1

1

T

P

T

P , and

1

2

1

2

P

P

T

T

And ∆S=Cp ln

1

2

P

P – R ln

1

2

P

P

= (Cp-R)ln

1

2

P

P= +Cvln

1

2

P

P

Therefore (∆S)T opposite sign of (∆S)V