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Met 5011 Thermodynamics and Kinetics of Materials II
Sample Questions and Answers for Final Exam, 2008
1.Calculate the equilibrium constant for the reaction <NiO> + (H2) = <Ni> + (H2O)
At 750˚C (1023˚K) from the following data:
<Ni> +½ (O2) = <NiO> ; ∆G˚ = -58450 + 23.55T cal
(H2) + ½ (O2) = (HO) ; ∆G˚ = -58900 + 13.1T cal
could pure nickel sheet be annealed at 750˚C in an atmosphere containing 95% H2O and 5% H2
by volume without oxidation?
Ans; Chapter 5 Example 5.B (Page 96)
2.Calculate the vacuum needed for obtaining molybdenum metal according to reaction
MoS2 = Mo + S2 at 800˚C and 1000˚C.
Given: MoS2 + 2H2 = Mo + 2H2S ; K1
2H2S = 2H2 + S2 ; K2
Temperature C : 800˚
1000˚
K1 : 7.311 x 10-6
1.439 x10-3
K2 : 2.2 x 10-4
5.6 x 10-3
Ans; MoS2 + 2H2 = Mo + 2H2S
K1
2H2S =2H + S2
K2
------------------------------------------------MoS2 = Mo + S2
K1 +K2
At 800˚C(1073K)------- ∆G˚ = -RT Ln K1K2
= -1.987 x 1073 x Ln (7.311 x 10-6 x 2.2 x 10-4) = 43169.79 cal
∆G˚ = -RT Ln K1K2
Ln K = (43169.79)/ (-1.987 x 1073) = -20.25
K = 1.61 x 10-9
K = 2Mo PS2/2MoS2 = 1.61 x 10-9
At 1000˚C(1273K)------- ∆G˚ = -RT Ln K1K2
= -1.987 x 1273 x Ln (1.439 x 10-3 x 5.6x 10-3) = 29667.41 cal
∆G˚ = -RT Ln K1K2
Ln K = (29667.41)/ (-1.987 x 1273) = -11.73
K = 8.16 x 10-6
K = aMo PS2/aMoS2 = 8.06 x 10-6
3 .Calculate the CO /CO2 ratio in equilibrium with carbon at 727˚C assuming that the total
pressure to be 2 atm.
Given : 2C + O2 = 2CO ; ∆G˚1000 = -95.3 Kcal
C + O2 = CO2 ; ∆G˚1000 = -94.0 Kcal
Ans; 2C + O2 = 2CO ; ∆G˚1000 = -95.3 Kcal
C + O2 = CO2 ; ∆G˚1000 = -94.0 Kcal
-------------------------------------------C + CO2 = 2CO ; ∆G˚1000 = -1.30 Kcal = -1300cal
∆G˚ = -RT Ln K1K2
Ln K = (-1300)/ (-1.987 x 1000) = 0.654
K = 1.924
K = PCO2/aC PCO2 = PCO2/PCO2 = 1.924
Assuming the total pressure to be 2atm:
PCO + PCO2 =2
PCO = 2-PCO2
K = (2-PCO2)2/PCO2 = 10.24
2
4-4PCO2 + PCO2 = 1.924 PCO2
PCO2 2 – 5.924 PCO2 + 4 = 0
Compare with ax2 + bx + c = 0
A= 1, b = -5.924 ,c = 4
2
− b ± b 2 − 42c − (−5.924) ± (−5.3924) − 4 x1x 4
=
2a
2 x1
= 0.777
PCO2 =0.777, PCO = 2-0.777 = 1.223
PCO/PCO2 = 1.223/0.777 = 1.574
PCO2 =
4.(a) The activities of aluminium at different concentrate concentrations in Al-Cu solutions at
1100C are given below:
XAL : 0.90 0.75 0.63 0.53 0.45 0.38 0.30 0.20 0.10
aAl : 0.89 0.69 0.50 0.31 0.20 0.10 0.30 0.005 0.008
Calculated the activity coefficient of aluminium in a solution containing 51 atm % copper.
Solution
Chapter 6 Example 6.E (Page 129)
(b) Will a blast furnace gas anslyzing 28%CO, 13% CO2 and 59% N2 reduce wustite (FeO) at
727˚C (1000K)?
Given: <Fe> + ½ (O2) = <FeO> ; ∆G˚ = -62050 + 14.95T cal
(CO) + ½ (O2) = (CO2) ; ∆G˚ = -67500 + 20.75T cal
Ans;
<Fe> + ½ (O2) = <FeO> ; ∆G˚ = -62050 + 14.95T cal
(CO) + ½ (O2) = (CO2) ; ∆G˚ = -67500 + 20.75T cal
-------------------------------------------------------------<Fe> + (CO2) = <FeO> + (CO) ; ∆G˚ = 5450-5.8T cal
∆G˚ = 5450-5.8T cal at 727 ˚C(1000K)
=5450-5.8 x1000 = -350cal
Ln K = -350/-1.987 x1000 = 0.176
K = 1.193
K = aFeO PCO/aFe PCO2 = 1.193
PCO/PCO2 = 1.193
At 28% CO, 13% CO2 and 59% N2
PCO/PCO2 = 2.154
Equilibrium partial pressure is greater than equilibrium constant. Therefore a blast furnace gas
analyzing at 28% CO, 13% CO2 and 59% N2will reduce wusite.
5.(a) Vanadium melts at 1720˚C (1993K). The Raoultain activity coefficient of vanadium of
infinite dilution in liquid dilution in liquid iron at 1620˚C (1893K) is 0.068. Calculate the free
energy change accompanying the transfer of the standard state from pure solid vanadium to the
infinitely dilute, weight percent solution of vanadium in pure iron at 1620˚C.
Given : (a) Heat of fusion of vanadium = 4500 cal/mole
(b) Atomic weight of vanadium and iron are 50.95 and 55.85 respectively.
Ans; Similar Chapter 6 Example 6.G (Page 136)
(b) The activity coefficient of copper in iron at infinite dilution, relative to pure copper as the
standard state, is 8.5 at 1600˚C (1873K).
{Cu} = [Cu]Fe, wt% at 1600˚C. Assume the atomic weights of copper and iron to be 63.5
and 55.85 respectively.
Ans; {Cu} = [Cu]Fe, wt% at 1600˚C.
M Fe
∆G˚ = RT Ln r0Cr + RT Ln
100 M CU
55.85
= -9652.01 cal
= 1.987 x 1873 x Ln 8.5 + 1.987 x 1873 x Ln
100 x63.5
The free energy change for reaction is -9652.01 cal
6(a) The peritectic decomposition of the compound ZrZn 14 at 545˚C (818K) may be represented
as
ZrZn14 =ZrZn6 + 8[Zn]Zn-Zr alloy
The alloy of the peritectic temperature contains 99.2 atom% Zinc. Calculate the activity
coefficient of zinc in the alloy.
Given: At 545˚C, (∆G˚ZrZn14 – ∆G˚ZrZn6) = -678 cal
Ans; ZrZn14 =ZrZn6 + 8[Zn]Zn-Zr alloy
At T 545˚C (818K), (∆G˚ZrZn14 – ∆G˚ZrZn6) = -678 cal
Atom % Zn = XZn = 99.2
∆G˚ = RT Ln r0Zn . XZn / wt% Zn
-678 = 1.987 x 818 x Ln r0Zn 99.2/0.8 , r0Zn = 0.9966
(b) Calculate the free energy charge when the standard state of manganese is transferred from
the pure liquid state to the infinitely dilute, weight percent solution of manganese in iron at
1627˚C(1900K).
Given: (a) Melting point of manganese = 1245˚C(1518K)
(b) Atomic weights of manganese and iron are 54.94 and 55.85 respectively
(c) Fe-Mn system can be regarded as an ideal solution.
Ans; {Mn}pure sub .ss = [Mn]Fe, wt%
M Fe
∆G˚ = RT Ln r0Mn + RT Ln
100M MN
55.85
= -17323.88 cal/mole (non= 1.987 x 1900 x Ln 1 + 1.987 x 1900 x Ln
100 x54.94
ideal)
K = wt% Fe/wt% Mn = (1/55.85) / (1/54.94)
x 100 = 98.37
∆G˚ = RT Ln K = -1.987 x1900 x Ln 98.37 = -17323.88 cal/mole (ideally)
7(a) The solubility of oxygen in liquid silver at 1075˚C (1348K) is given below for a range of
partial pressure of oxygen:
Partial Pressure Solubility
of
of oxygen
oxygen
mmHg
c.c/100g Ag
128
81.5
488
156.9
760
193.6
1203
254.8
Show that the solubility of oxygen in silver follows Sievert’s Law,
and calculate the amount of oxygen dissolved in 100g of liquid silver at 1075˚C, from
atmosphere air, assuming that the air contains 21% oxygen by volume.
Ans; Chapter 7 Example 7.A (Page 144)
(b) The excess partial molar free energy of zinc in liquid Cu-Zn alloys at 1027C (1300K) can be
represented as G Zn -xs (cal/mol) = -5150 (1-XZn)2
Calculate the activity of copper at 1027˚C (1300K) in an equiatomic solution.
Ans; In an equiatomic solution, XZn = X Cu = 0.5
G -XSZn = -5151(1- 0.5)2
= -1287.5 cal/mole
G –XSCu = -5151XCU2 (XZn + XCu = 1)
= -1287.5 cal/mole
G -XSZn = RT Ln rCU
-1287.5 = 1.987 x 1300 Ln rcu
Ln rcu = -0.498
rcu = 0.607
rcu = acu / Xcu
acu = 0.607 x 0.5 = 0.304
Thus the activity of copper at 1027C (1300K) in an equiatomic solution is 0.304.
8.From the e.m.f. measurements at 527C(800K), the following values of the activity coefficient
of cadmium in zinc-cadmium solutions have obtained:
XCd: 0.2
0.3
0.4
0.5
rCd : 2.153 1.817 1.544 1.352
Determine whether the Zn-Cd solution exhibits regular behavior
(i)
(ii)
If the Zn-Cd solution is regular, calculate the valves of the excess partial molar free
energy of zinc and cadmium, excess integral molar free energy of solution, and
excess integral molar entropy of solution at 527C (800K) for the equiatomic Zn-Cd
solution.
Ans; Chapter 7 Example 7.B(i) (Page 115) & Example 7.C
9. The reaction for the electrochemical cell,oxygen dissolved in liquid
l
l
Cu (ZnO2)0.85(CaO)0.15 Ni,NiO may be represented as NiO = Ni + [O]Cu,liquid.
The e.m.f. is 138mV at 1200˚C (1473K). The standard free energy of formation of NiO at
1200˚C w.r.t. a reference state of oxygen as pure gas at 1 atm.
Solution
Chapter 8 Example 8.E (Page 198)
10 (a) The e.m.f. of the reversible cell
l
l
Zn(l,pure) KCl-NaCl – LiCl – ZnCl2 Zn (in Cd-10.3 atom% Zn alloy) has been found to be
85.80x10-3 V at 800˚C (1073K). Calculate (i) the activity of zinc in the alloy; (ii) the value of
the partial molar free energy , entropy and enthalpy of mixing of zinc in the alloy, assuming that
the temperature coefficient of e.m.f. is 0.111x10-3 V/deg.
Solution
Chapter 8 Example 8.B(i) (Page 192)
(b) The reversible e.m.f. between pure magnesium and magnesium-zinc alloy containing 63.5
atom% magnesium in a fused KCl-LiCl-MgCl2 electrolyte may be represented by
E = 16.08x10-3 +1.02x 10-5T , where E and T are in V and K respectively. Calculate the
activity coefficient and excess partial molar free energy of mixing of magnesium in the above
alloy at 727C(1000K).
Solution
Chapter 8 Example 8.C (Page 193)
11(a)Chapter 8 Exercise No. 6
Solution
Similar Chapter 8 Example 8.C (Page 193)
(b) Chapter 8 Exercise No. 7
Solution
Similar Chapter 8 Example 8.C (Page 193)
12. Calculate the equilibrium constant at 727˚C (1000K)for the reaction
<FeO> + (CO) = <Fe> + (CO2) from the following data:
<FeO> + (H2) = <Fe> + (H2O); ∆G˚ = 3150 – 1.85T cal
(CO2) + (H2) = (CO) + (H2O); ∆G˚ = 8600 – 7.8T cal will FeO form if an iron
sheet is annealed at 727˚C in an atmosphere containing 10%CO,2%CO2 and 88%N2?
Ans; <FeO> + (H2) = <Fe> + (H2O); ∆G˚ = 3150 – 1.85T cal
(CO2) + (H2) = (CO) + (H2O); ∆G˚ = 8600 – 7.8T cal
----------------------------------------------------------------<FeO> + (CO) = <Fe> + (CO2); ∆G˚ = -5450 + 5.95T cal
∆G˚ = -5450 + 5.95T cal at T = 727C (1000K)
= -5450 + 5.95 x 1000 = 500cal
LnK = 500/(-1.987 x 1000) = 0.252
K = 0.778
K = aFe .PCO2 /aFeO.PCO = 0.778
PCO2 / PCO = 0.778 (equilibrium)
At the gas atmosphere contains 10% CO,2% CO2 and 88% N2.
PCO2 / PCO = 0.02/0.1 = 0.2 which is less than equilibrium value of 0.778 at 727 C(1000K).
Thus Iron sheet is not annealed.
13.Chromium plates are bright annealed at 727˚C in a wet hydrogen atmosphere. The pressure
of wet hydrogen is 1atm.
(i) Calculate the permissible water content in the hydrogen if there is to be no oxidation at
727˚C.
(ii) Will annealed chromium plates be oxidized when cooled to 727˚C in the furnace
atmosphere, as calculated in (i)? Neglect the possibility of dissolved of hydrogen in chromium.
Given: 2<Cr> + 3 (H2O) = <Cr2O3)> + 3(H2) ; ∆G˚ = -91050 + 22.80T cal
Ans;PH2 + PH2O =1
2<Cr> + 3 (H2O) = <Cr2O3)> + 3(H2) ;
∆G˚ = -91050 + 22.80T cal at 727C (1000K)
= -91050 + 22.80 x1000 = -68250 cal
Ln K = -68250 / (-1.987 x 1000) = 34.35
K = 8.265 x 1014
K = a Cr2O3 . PH23 / aCr2. PH2O 3 = 8.265 x1014
PH23 / PH2O 3 = 8.265 x 1014
PH2/PH2O = 93847.02
PH2O = PH2/ 93847.02 = 1-PH2O/ (93847.02)
93847.02 PH2O = 1-PH2O
PH2O = 1.07 x 10-5 atm
Volume percent, H2O = 1.07 x 10-3 %
∆G˚ = -91050 + 22.80T cal at 227C (500K)
= -91050 + 22.80 x500 = -79650 cal
Ln K = -79650 / (-1.987 x 500) = 80.17
K = 6.57 x 1034
K = a Cr2O3 . PH23 / aCr2. PH2O 3 = 6.57 x 1034
PH23 / PH2O 3 = 6.57 x 1034
PH2/PH2O = 4.04 x1011
1-PH2O =
4.04 x1011 PH2O
PH2O = 2.48 x 10-12 atm
14. Al-Zn alloys exhibit the following reaction at 477˚C (750K): RT Ln rZn = 1750(1-XZn)2,
where R and T are expressed in cal/deg/mole and K, respectively. Calculate the activity of
aluminium at 477C in an Al-Zn alloy containing 40 atm % Zinc.
Solution
Chapter 7 Example 7.G (Page 169)
15.Calculate the CO /CO2 ratio in equilibrium with carbon at 727˚C assuming that the total
pressure to be 2.5 atm.
Given : 2C + O2 = 2CO ; ∆G˚1000 = -95.3 Kcal
C + O2 = CO2 ; ∆G˚1000 = -94.0 Kcal
Ans; Similar with no.3
16(a)Liquid iron contains 0.2% aluminium and 0.05 % nitrogen by weight at 1620C (1893K).
Determine the possibility of precipitating aluminium nitride, AlN, by thermodynamic
calculation.
Given: AlN = [Al] Fe,wt% + [N] Fe,wt% , K1893 = 6 x 10-3
∆G˚ = RT Ln K = 1.987 x 1893 x Ln 6 x10-3 = 19243.26 cal /mol
∆G˚> 0; The reaction is right to left that will precipitate.
(b)The melting point of cobalt is 1480C (1753K). Calculate the free energy change for the
transfer of one g-atm of cobalt from pure liquid to a 1wt% solution in liquid iron at 1500C
(1773K). Assume that cobalt behaves ideally in iron at that temperature. Atomic weight of Co
and Fe are 58.9 and 55.85 respectively.
Ans; [Co] pure sub ss = [Co] Fe wt%
M Fe
∆G˚ = RT Ln r0Co + RT Ln
100 M C 0
55.85
= -16411.11 cal/mole
= 1.987 x 1773 x Ln 1 + 1.987 x 1773 x Ln
100 x58.9
The free energy change for transfer of one g-atm of cobalt from pure liquid to a 1wt% solution
in liquid iron at 1500C(1773K)is -16411.11 cal/mole.
17(a)Chapter 6 Example (1-F) Page 14
(b)Chapter 6 Example (1-G) Page 15
18. Chapter 6 Example (1-D) Page 9
19.Chapter 7 Example (2-L) Page 43
20.Chapter 8 Example (3-A) Page 53
21. Chapter 6 Example (1-C) Page 7
22.Que; Chapter 8 Exercise 5
Ans; Cd (1,pure) = Cd (in cd-Pb alloy)
Ln aCd = -ZCd FE /RT = (- 2 x 23061 x 14.35 x 10-3)/(1.987 x 773) = -0.431
aCd = 0.65
rCd = aCd/KCd = 0.65/0.4 = 1.625
∆G˚ = -ZFE = - 2 x 23061 x 14.35 x 10-3 = -661.85 cal
23. Que; Chapter 8 Exercise 6
Ans; Cd (pure) = Cd (in Cd –Sb alloy)
∂ E/∂ T = 33.63µV/deg = 33.63 x 10-6 V/deg , E = 28.48mV = 28.48 x 10-3 V
Ln aCd = -ZCd FE /RT = (-2 x 23061 x 28.48 x 10-3) / (1.987 x 773) = 0.8552
aCd = 0.4252
G-MCd = RT Ln aCd = 1.987 x 773 x Ln 0.4252 = -1313.55 cal/mole
S-MCd = ZCd F (∂ E/∂ T) = 2 x 23061 x 33.63 x 10-6 = 1.551 cal/deg
24.Chapter 8 Example (3-D) Page 58
25. Calculate the CO /CO2 ratio in equilibrium with carbon at 727˚C assuming that the total
pressure to be 2.25 atm.
Given : 2C + O2 = 2CO ; ∆G˚1000 = -95.3 Kcal
C + O2 = CO2 ; ∆G˚1000 = -94.0 Kcal
Ans; Similar with no.3
26.Liquid iron contains 0.22% aluminium and 0.04 % nitrogen by weight at 1620C (1893K).
Determine the possibility of precipitating aluminium nitride, AlN, by thermodynamic
calculation.
Ans; Similar with 14(a)
27.The melting point of cobalt is 1480C (1753K). Calculate the free energy change for the
transfer of one g-atm of cobalt from pure liquid to a 1wt% solution in liquid iron at 1500C
(1773K). Assume that cobalt behaves ideally in iron at that temperature. Atomic weight of Co
and Fe are 58.9 and 55.85 respectively.
Ans; Similar with 14(b)
28. Chapter 7 Example (2-K) Page 43
29.Calculate the equilibrium constant for the reaction <NiO> + (H2) = <Ni> + (H2O)
At 750˚C(1023˚K)from the following data: <Ni> +½ (O2) = <NiO> ; ∆G˚ = -58450 + 23.55T cal
(H2) + ½ (O2) = (HO) ; ∆G˚ = -58900 + 13.1T cal
could pure nickel sheet be annealed at 750˚C in an atmosphere containing 95% H2O and 5% H2
by volume without oxidation?
Ans; Chapter 5 Example 5.B (Page 96)
30.Calculate the vacuum needed for obtaining molybdenum metal according to reaction
MoS2 = Mo + S2 at 800˚C and 1000˚C.
Given: MoS2 + 2H2 = Mo + 2H2S ; K1
2H2S = 2H2 + S2 ; K2
Temperature C : 800˚
1000˚
1.439 x10-3
K1 : 7.311 x 10-6
K2 : 2.2 x 10-4
5.6 x 10-3
Ans; MoS2 + 2H2 = Mo + 2H2S
K1
2H2S =2H + S2
K2
------------------------------------------------MoS2 = Mo + S2
K1 +K2
At 800˚C(1073K)------- ∆G˚ = -RT Ln K1K2
= -1.987 x 1073 x Ln (7.311 x 10-6 x 2.2 x 10-4) = 43169.79 cal
∆G˚ = -RT Ln K1K2
Ln K = (43169.79)/ (-1.987 x 1073) = -20.25
K = 1.61 x 10-9
K = 2Mo PS2/2MoS2 = 1.61 x 10-9
At 1000˚C(1273K)------- ∆G˚ = -RT Ln K1K2
= -1.987 x 1273 x Ln (1.439 x 10-3 x 5.6x 10-3) = 29667.41 cal
∆G˚ = -RT Ln K1K2
Ln K = (29667.41)/ (-1.987 x 1273) = -11.73
K = 8.16 x 10-6
K = aMo PS2/aMoS2 = 8.06 x 10-6
Met 5013 Characterization of Materials
Sample Questions and Answers for Final Exam, 2008
1.(a) Define or briefly explain the following .
(1) Moles
(4) Miliequivalents
(2) Molarity (5) Normality
(3) Molality
(b) How many milliliters of the following solutions need to be taken to give 1.00 gram of each
substance?
(1) 0.00126 M NH3 ( N = 14, H = 1 )
(2) 0.2468 N Na2CO3 (Na=23, C=12, O=16)
(3) 0.1106 N Ba(OH)2 (Ba=137.36. O = 16, H= 1)
2.(a) Calculate the normality of the following solutions and how many milliequivalents of each
substances are there in each part?
(i) 10.10 grams of KHP in 136 ml of solution (Mol Wt of KHP= 204)
(ii) o.3062 grams of NaH2P4 in 150ml of solution ( Na= 23, H = 1, P=31)
(b)Dilute solution of HCl and NaOH are to be standardized. To rate 0.9862 gram of pure
potassium acid phthalate (KHP,mol.wt = 204 ), 46.24 ml of the NaOH is needed. In another
titration 25.42 ml of the NaOH solution exactly neutralizes 32.46 ml of HCl solution. What are the
normalities of the NaOH and HCl solutions?
Ans; Example 1-7 (Page 6)
3.(a) (i) What is a gravimetric factor?
(ii) A mineral containing magnesium is dissolved and the magnesium precipitates as
MgNH4PO4, magnesium ammonium phosphate . This precipitate upon ignition is converted to
Mg2P2O7, magnesium pyrophosphate. If a 0.5000 gram sample of the mineral yielded 0.7400 gram
of Mg2P2O7, What is the percentage of magnesium oxide in the mineral?
Ans; Example 2-1 (Page 12)
(b) In a certain gravimetric procedure for fluoride a 0.6822 gm sample of impure ferric fluoride
was weighed out. The ferric fluoride was then converted to ferric sulfate by evaporation with
sulfuric acid.
Fe2 (SO4)3 + 6HF
2FeF3 + 3H2SO4
The iron from the ferric sulfate was subsequently precipitated the hydrous ferric oxide and
ignited and weighed as Fe2O3. The final weight of Fe2O3 was 0.18449gm. What is the percentage of
F in the sample? ( F=19.Fe=55.85, O=16)
Ans; Example 2-3 (Page 13)
4.(a) Define (i) Lambert's Law (ii)Beer's Law and (iii) Beer-Lambert Law .
(b) The % transmittancy of an aqueous solution of sodium fumarate at 250 m µ, and 25˚ is 19.2%
for 5 x 10-4 M solution in a 1 cm cell. Calculate the absorbency and molar absorbancy index. What
will be the percent age transmittancy of a 1.75 x 10-4 M solution?
5.(a) What are the components of a spectrophotometer? And illustrate with schematic diagram .
Ans; Page 27
(b) For acetone vapor at 50˚ the value of K in the expression
It = Io e-ktc is 6.63 for light of 3130 A, where "c" is expressed in moles per liter and 't' is in
centimeters. What percent of light will be absorbed by acetone vapor at 100mm pressure at 50˚
when a beam of light of 3130 A is passed through a cell 15 cm long?
6. Describe the basic principles of thermo gravimetric analysis (TGA) and Differential thermal
analysis (DTA).
Ans; Example 2 , Page 12
7.(a) Example 11 (Page 8)
(b) Example 8 (Page 6)
8.(a) Example 4 (Page 13)
(b) Which conditions must fulfill a reaction in volumetric analysis? Describe the requirements of
primary standard substance.
9(a) Define the sampling and suggest a layout of equipment for sample preparation.
Ans; Volumetric analysis (the preparation of mine sample,equipment and layout , pulverizers and
sample splitters)
(b) Briefly explain calorimetric and spectrophotometer analysis.
10.(a)How many method do you know to characterize the various element present in the unknown
mixture from bulk sample?
Ans; Gravimetric analysis
(b)When a spectrometer is used it is unnecessary to make comparison. How do you use it?
Ans; Typical spectrometer with diagram
11.(a)Briefly explain the use of photoelectric colorimeter with neat sketch.
(b) In many analytical determinations using light absorption, absorbance measurements are made
using radiation of only one wavelength. What instrument can be employed?
12.From our course of study, how do you determine the desired component(substance) from the
bulk of unknown sample?
Ans; Write definitions for each analysis and Compare
13.How many steps are involved in fire assaying method?
Ans; From definition of fire assaying to advantages of fire assaying.
14.(a)Define fire assaying. What are the factors upon which the fire assaying of silver and gold
mainly depend?
Ans; Fire assaying and 4 factors
(b) How do you analyze by using atomic absorption spectrometry.
Ans; Page 21 9-20 Analysis by atomic absorption spectrometry.
15.How do you characterize analysis the component from bulk sample by X-ray fluorescence
spectrometry?
Ans; X-ray fluorescence spectrometry
16.What kind of DTA and TGA traces would you expect to obtain an heating samples of following
until they became liquid.(a) beach sand (b)window glass (c) salt (d) ferroelectric BaTiO3
Ans;DTA , reason from Page 109,110
17.(a)Define thermal analysis and briefly explain the two main thermal analysis techniques?
(b) Write down the specific application of DTA and TGA.
18.Which of the following would you expect to give a reversible DTA effect with as without
hystersis (a) melting of salt (b) decomposition of CaCl3 (c)melting of beach sand (d) oxidation of
metallic Mg (e) decomposition of Ca(OH)2. Why?
Ans; (a) No effect DTA (b)Page 104 (c)Page 108(d)Page 110(e)Page 112
19.How many milliliters(ml) of reagent grade porchloric acid, HClO4, whose density is 1.668 g/ml
and which contains 70% HClO4 by weight, will be needed to prepare 1500ml of an exactly 0.5019
N solution?
Ans; Example 4 Page 5
20.How many milliliters of a 0.7692 N solution of NaOH must be diluted to exactly 500ml to
produce a 0.18007 N solution?
Ans; Example 5 Page 5
21.Dilute solution of HCl and NaOH are to be standardized. To titrate 0.9862 g of pure potassium
acid phothalate 46.24 ml of the NaOH is needed. In another titration 25.42 ml of the NaOH solution
exactly neutralizes 32.46ml of the HCl solution. What are the normalities of the NaOH and HCl
solution?
Ans; Example 7 Page 6
22.To a 0.2422 g sample of pure Na2CO3 was added 42.68 g of an acid solution. This quanity of
acid was enough to react with both equivalents of carbonate and still provide excess hydrogen ion in
solution. This excess hydrogen ion was then titrated with 10.08ml of a solution of base, 1.00ml of
which is equivalent to 26.89 mg of KHP. calculate the normality of the acid and base solutions.
Ans; Example 8 Page 6
23.A sample of impure soda ash (Na2CO3) whivh weighs 0.6211g is dissolved in 75.00 ml of 0.101
NHCl. The excess acid is titrated with 11.25ml of a NaOH solution, 1.000ml of which is equivalent
to 0.4628ml of the HCl solution. What is the % of sodium oxide,Na2O in the sample?
Ans; Example 10 Page 7
24.Calculate the NH3 liter of HCl solution whoch has a Na2O liter of 6.50 mg/ml.
Ans; Example 12 Page 8
25.A mineral containing Mg is dissolved and the Mg precipitate as MgNH4 PO4. This precipitate
upon igniton is converted into Mg2P2O7 . If a 0.5000g sample of the mineral yield 0.7400 g of
Mg2P2O7 . What is the % of MgO in the mineral?
Ans; Example 1 Page 12
26.A 0.4264 g sample of cadmium carbonate was heated to constant weight at 800C and when
cooled, was found to weight 0.3869 g. Assuming that the only gaseous was CO2, what is the % of
metallic Cd in the sample?
Ans; Example 2 Page 12
27. Example 3 Page 13
28.How many milliliters of a 0.4628 M solution of NH3 are necessary to prepare 1.00liter of a
0.1500M solution?
Ans; For prepare NH3 solution,Molarity (M) = m mole of NH3 /ML of NH3
M mole of NH3 = 0.1500M x 1000 ml = 150m- mole
Molarity (M) = m mole of NH3 /ML of NH3
ml of NH3 = 150m-mole/ 0.4628 M = 324.114ml
29.How many milliliters of a 1.5 M solution of H2SO4 are required to make up 2.00liter of an
exactly 0.200M solution of H2SO4?
Ans;Since 1 moles of H2SO4 can react with either 1 or 2 moles of OH ions. Assuming that 2 moles
of OH ions.Equivalent wt of H2SO4 = ½ x molecular wt of H2SO4 = ½ x 98 = 49
Normality (N) = m Equivalent of H2SO4/ ml of H2SO4
C of H2SO4 = 0.200N x 2000ml = 400 m Equivalent
m Equivalent of H2SO4 = w of in mg/ Equivalent wt of H2SO4
w of in mg = 400 m Equivalent x 49 mg/ m Equivalent = 19600-Mg
m moles = w of in Mg/ Molecular wt = 19600mg/98 mg/m –mole = 200 m-mole
Morality(M) = m mole/ml of solution
ml of solution = 200m.mole/1.5 M = 133.33 ml
30.The % tramsmittancy of an aqueous solution of sodium fumarate a 250mµ,and 25C is 19.2% for
5 x 10-4 M solution in a 1cm cell. Calculate the absorbancy and molar absorbancy index. What will
be the % I at 1.75 x 10-4M.
Ans; % I(It/Io) = 19.2%, It/Io = 0.192 , Io/ It = 4.
absorbancy(A) log (Io/ It) = 0.72
log Io/ It = ε.t.C
log 5.21 = ε x 1x 5 x 10-4
ε = 1433.4
Molar absorbancy index(ε) = 1433.4
log Io/ It = ε.t.C = 1433.4 x 1x 1.75 x 10-4 = 0.251
Io/ It = 1.783
It/Io = 0.5612
% It/Io = %I = 56.12%
Met-5021 Failure analysis of Engineering Materials
Sample Questions and Answers for Final Exam, 2008
1. How many information you can get from investigation on the failure of a wire rope on a crane?
Ans; The failure of a wire rope on a crane (summary, conclusion & recommendation)
2. Briefly explain the performing tests in studying of failure of a wire rope crane?
Ans; The failure of a wire rope on a crane( tensile testing)
3. What are the performing tests studying in failure of a wire rope crane?
Ans; The failure of a wire rope on a crane( tensile testing)
4. What information do you get from studying in failure of a main under carriage leg on a crop
duster air craft ?
Ans; A failure of a under carriage leg on a crop duster air craft (summary & conclusion)
5. Write short notes on a failure of a crop duster air craft?
Ans; A failure of a under carriage leg on a crop duster air craft (Theory background ,tensile & shear
load, summary & conclusion)
6. How do you studying the tensile and shear load at failure of a main under carriages on a crop
duster air craft?
Ans; A failure of a tensile duster air craft (tensile & shear load at failure)
7. How do you investigate the failure of an aircraft during landing altitude?
Ans;Crop duster air craft (Theory background and 3 Figure)
8. What information can you get from investigation on the failure of an air craft tow bar ?
Ans; The failure of an air craft tow bar (Background, second section third section )
9. Write short notes of an aircraft tow bar.
Ans; An aircraft tow bar (Theory background ,tensile & shear load, summary & conclusion)
10.How many observation do you get from case study of failure of an aircraft tow bar?
Ans; 6 factors from conclusion
11. Why do a crop duster aircraft suffer and accident during landing?
Ans; A failure of a under carriage leg on a crop duster air craft (First section and second section)
12. Why do you investigate the failure of a wire rope on a crane?
Ans; Failure of a wire rope on a crane (Theory background , summary & conclusion)
13.Briefly discuss the Fractography of wire in failure of a wire rope on a crane?
Ans; Failure of a wire rope on a crane (Fractography of wire)
14.How do you think about the failure of a wire rope on a crane?
Ans; The failure of a wire rope on a crane (Theory background ,tensile test, summary)
15. How many steps involved in observation of case study at failure of main under carriages leg a
crop duster aircraft?
Ans; A crop duster aircraft ( 1st ,2nd and 3rd Section)
16. Briefly explain the three sections investigative at failure of an aircraft tow bars?
Ans; The failure of an aircraft tow bars(three sections)
17.What information can you get from investigation at failure of a main under carriages leg a crop
duster aircraft?
Ans; A crop duster aircraft( 6 factors from 2nd Section)
18. How many sections are investigative at failure of an aircraft tow bar?
Ans; The failure of an aircraft tow bar( 3 sections)
19.How do you summarize the finding of the fracture and mechanics information?
Ans; A crop duster aircraft ( 3 factors)
20.How do you understand the fracture modes at the associated fracture surface?
Ans; The fracture modes
21. As a sample is compression and tension, how do you think about this type of failure?
Ans; From bending
22. Write down the differences between torsion, bending and tension.
Ans; Definitions of torsion and bending and tension
23.Define the following;
i) Failure
ii)Fracture
24.Short notes on i) Creep
Separation
iii)Fatigue
ii)Slip iii)Twining
Iv)Failure analysis
iv)Quasicleavage
v)Intergranular
25.Why failure analysis and prevention are important function to all of the engineering discipline?
26.How do you examine the failure of saddle components used in crop duster air craft?
Ans; A crop duster air craft (2nd Section)
27.Describe the applications of SEM fractography at the failure of a main under carriages by a crop
duster aircraft.
28.From your course of failure analysis, discuss the important function of failure analysis and
prevention in all of the engineering discipline.
Ans; Overview of failure analysis
29.Describe the factors regarding the failure at failure of an aircraft tow bar.
Ans: The failure of an aircraft tow bar (6 factors)
30.Crop duster aircraft suffered an accident during landing, in which the landing gear collapsed.
Which material used in landing gear lug? Why the landing gear had collapsed during landing?
Ans; A crop duster aircraft ( 1st ,2nd and 3rd Section)
Met-5022 Industrial Management and Control
Sample Questions and Answers for Final Exam, 2008
1.What is the objective of inventory control? What types of costs have to be considered in
controlling inventory?
Ans: Chapter 8 Page 38,eight factors
2.If C = 17,000 kyats, S = 2,000 kyats, n = 5 years and depreciation rate 0.4 .Find out the
depreciation ratc for an asset wing declining balance method .
Depreciation Rate = 2(1/n)
Ans;
Year Depreciation Beginning
Depreciation Accumulated Ending book
Rate
Book Value Charge
Depreciation Value
1
0.4
17000
6800
6800
10200
2
0.4
10200
4080
10880
6120
3
0.4
6120
2448
13320
3672
4
0.4
3672
1469
14797
2203
5
0.092
2203
203
150000
2000
Ans; Chapter 7 Example 7.3.3
3. (a) What ıs a process chart and why ıs it used?
(b) What advantage is it to write down the steps in a job?
Ans; Chapter 6 Page 32, ten factors
4. Determine the economic order quantity for item 335,
Given :
Annual requirement- 2000 units
Cost per unit
-K 6.4
Inventory holding cost - 30%
Ordering cost pert order- K 14
Assumıng 250 working days per year, how after must item 335 be ordered ?
Ans; R= 2000 units, C = K 6.4
I = 30 % , A = K 14
2 RA
2 x 2000 x14
E.O.Q =
=
= 171
CI
6 .4 x 0 .3
R
2000
= 11.8 orders/year
=
E.O.Q
171
11.8 order ----250 day
1 order
----250 x
1
= 21.2 days
11.8
Therefore, order must be placed every 20th and 21st days alternately.
5.What are PERT’s advantages to management. ?
Ans; Chapter 9, Page 44 ,9.3
6.Defıne the following
i. Distribution overhead
Ans; Chapter 7 Page 34,35
ii. Direct Material
iii. Salvage value
7. What is the objective ot motion study? Time study?
Ans; Chapter 6 Page 30
8.Defıne the following
i.Trade credit
ii.Short-term loan iii.Commercial paper .
Ans; Chapter 10 Page 49
9. A project consists of 13 activities for which optimistic,most likely and pessimistic duration is
given in the dependency table below. (i)Draw a network for the project (ii) Find the expected
time of each activity and determine the critical path (iii) Condense the network without
distorting the dependency relationship. T (B + 4m+P)/6
Duration (weeks)
Activity
A
B
C
D
E
F
G
H
I
J
K
L
M
Depends on
None
None
A
A
B
B
B
C
D,E
F
H
H,S,L
J
Optimistic (B)
1
2
2
1
2
2
5
2
2
1
1
3
1
Ans; with figure
Activity Depends on
A
None
B
None
C
A
D
A
E
B
F
B
G
B
H
C
I
D,E
J
F
K
H
L
H,S,L
M
J
Expected time
2
3
3
2
4
3
10
3
4
2
1
6
1
Most Likely(M)
2
3
3
2
4
3
9
3
4
2
1
6
1
Pessimistic(P)
3
4
4
3
6
4
19
4
6
3
1
9
1
10. John Son’s emporium sells 9000 Yo Yo’s per year. Each Yo Yo’s cost 2.5 kyats, the
holdıng cost is 8% and the reordering cost is 9 kyats per order. What is the E.O.Q and how
after should John Son order Yo Yo per year? ( 5-marks)
Soln: R=9000
C = 2.5 kyats
A = 9 kyats
I = 0.08
E.O.Q = 900
R/E.O.Q = 9000/900 = 10times/year
11.The M/C has an estimate life ot one million stamps and cast, kyats 100,000. The depreciation
charge per stamp is them 100,000/1000000 = 0.1 kyats.
Assuming the zero salvage value, the depreciation charges using depreciation by use method ore
as shown below.
Given;
Year
1
2
3
4
5
6
Total Yearly Cost per Yearly depreciation Accumalated
Stamps
Stamp
chorge
depreciation
150,000
300,000
200,000
200,000
100,000
50,000
Ending
book value
Total Yearly
Stamps
150,000
300,000
200,000
200,000
100,000
50,000
Ending
book value
85,000
55,000
35,000
15,000
5,000
0
Ans;
Year
1
2
3
4
5
6
Cost per
Stamp
0.1
0.1
0.1
0.1
0.1
0.1
Yearly depreciation Accumalated
chorge
depreciation
15,000
15.000
30,000
45,000
20,000
65,000
20,000
85,000
10,000
95,000
5,000
100,000
12.Briefly explain the “optimum operating capacity " with a neat sketch at break-even curve for
a manutacturing plant.
Ans; Chapter 10 Page 51 ,10-6
13. Determine the economic order quantity for item 555.Annual requirement is 4000 units.Item
555 of cost per unit is K 12. Inventory holding cost is 60% and ordering cost pert order is K 28.
Assumıng 250 working days per year, how after must item 555 be ordered ?
Ans;R= 400 wits, C =K 12.8
I = 60 % , A = K 28
2 x 400 x 28
= 171
E.O.Q =
12.8 x0.6
R
4000
=
= 23.4 orders/year
E.O.Q
171
23.4 order ----250 day
1
= 10.68 days
23.4
Therefore ,order must be placed every 10th and 11st days alternately.
1 order
----250 x
14. Define an activity, an event and slack time.
Ans; An activity
An activity is any portion at a project, which consume time or resources and has a
definable beginning and ending. Activity may involve labor, paper work, contractual
negotiations, machinery operations. etc. (or)
An activity does involve time and labor and precedes an event. An activity is
represented in a PERT network by an arrow and connects two events.
An event
The beginning and ending points of activities are called events. (or)
An event is frequently represented by a circle. An event takes no time and represents no work, it
does signify that something has occurred.
Slack Time
The difference between the expected time for an event and longest allowable time is the
slack time.
15. Find out the depreciatıon rate for an asset using ‘declining balance method’. If the cost of
asset is 17,000 kyats estimated salvage value is 2000 kyats and the asset’s estimated life is 5
years, construct the table and fınd the ending book value for each year.
Ans;
Year
1
2
3
4
5
Depreciation
Rate
0.4
0.4
0.4
0.4
0.092
Beginning
Book Volve
17000
10200
6120
3672
2203
Depreciation
Charge
6800
4080
2448
1469
203
Accumulated
Depreciation
6800
10880
13320
14797
150,000
Ending book
value
10200
6120
3672
2203
2000
16.Explain the basic prerequisites for an inventory control system.
Ans; Chapter8, Page 36,8.2
17. Define the Following;
i. Slack time
ii.An activity
iii. Motion study
18. How can flow diagram aid in the analysis of a work process?
Ans; Chapter 6 Page 31, four factors, page 32, ten factors.
19.(a)A certain product is made to stock. Annual demand rate is 12,000 unit. Invoice price of
one unit of the item is 10 kyats. The cost of plocing one order is 12 kyats and the holding cost
rate 24% mper year. Determine the E.O.Q
Ans;
R = 12,000 units
C= 10 kyats
I= 0.24
A= 12 kyats
2 x12000 x12
= 346.41
E.O.Q=
10 x0.24
(b) George Door knob supply have sell 75 modal door-knob year at 35 kyats each .Each order
cast 3kyats and the inventory cost is 6 % Calculate the E.O.Q
Ans; R = 75, C = 35, A = 3, I = 0.06
2 x75 x3
= 14.64
E.O.Q =
35 x0.06
20. Clearly define PERT and explain the advantages of PERT.
Ans; Chapter 9, Page 44,9.3
PERT Program
Evalution and Review Technique (PERT) is a sophisticated new tool
used to
help management
21.A die-casting machine has an estimated life of 1 million casting and cost 100,000kyats. The
depreciation charge per casting is then 100000/1000000 =0.1 kyats. Assuming the zero salvage
value, complete the followıng table using depreciation by use method .
Ans;
Year
Castıng year Cost castıng Yearly
Accumulated Endıng
deprecıatıon deprecıatıon Book Value
1
2
3
4
5
6
150,000
300,000
200,000
200,000
100,000
50,000
0.1
0.1
0.1
0.1
0.1
0.1
15,000
30.000
20,000
20,000
10,000
5,000
15.000
45,000
65,000
85,000
95,000
100,000
85,000
55,000
35,000
15,000
5,000
0
22. Define the followings:
(i) Inventory Theory
E.O.Q and R.O.P
(i)
Ans; Chapter 8,Page 41
23. Define the followings:
(i) Capital cast
(ii) Long term debt
(iii) Mortgage bonds
Ans;(i)Capital cost
The Capital cost at a project is the total amount of money invested .It consists of two
main components: the fixed capital, that used to build the plant and the working capital, that tied
up in day-to-day operating expense.
(ii) Long term debt Ans; Chapter 10,Page50
(iii) Mortgage bonds Ans; Chapter 10 ,Page51
24. Find out the depreciation rate for magnetic separation machine using" declining balance
method".
It magnetic separation machine is perchase for K 40,000 the estimated life is 5 years and
esrtimated salrage value is K 5,000 , construct the table and find the "Ending book value" for
each year.
Depreciation Rate = 2(1/n)
Soln:
Year
Depreciation
Rate
0.4
0.4
0.4
0.4
0.0355
1
2
3
4
5
BeginningBook
Value
40000
24000
144000
8640
5184
Depreciation
Charge
16000
9600
5760
3456
184
Accumulated
Depreciation
16000
25600
31360
24816
35000
Ending book
vallue
24000
14400
8640
5184
5000
25.(a)If an investmnt of 20,000 kyats has an annual cash inflow of 6000 kyats per year. What
would be required time of payback period?
20,000kyats
= 3.3 years
Ans; Pay bock period =
6000kyats
Projects are ranked with quicker payback being preferred .The longer the payback period is the
greater the risk.
(b) Jone's air conditioner sells 1000. Deplux air conditioner per years at 250000, kyats each.
Each order cast 50,000 kyats and the inventory cast is 5% Calculate the E.O.Q
Ans;
R = 1000, I = 5%, A = 50000, C = 250000
2 x1000 x50000
= 89.44
E.O.Q =
250000 x0.05
26.Illystrate the fıve components of process chart symbol under which all work can be
classifıed.
Ans; Chapter 6 Page 31 To write Process chart symbol and steps in a process (6.3)
27. The machine has an estimate life of 2,000,000 stamps and cost kyats 200,000. The
depreciation charge per stamp is 0.1 kyats. Assuming the zero salvage value, complete the
following table using depreciation by use method:
Given;
Year Total Yearly Cost per Yearly depreciation Accumulated Ending
Stamp
stamp
charge
depreciation
book Value
1
300,000
2
600,000
3
400,000
4
400.000
5
200,000
6
100,000
Ans;
Year
1
2
3
4
5
6
Total Yearly
Stamp
300,000
600,000
400,000
400.000
200,000
100,000
Cost per
stamp
0.1
0.1
0.1
0.1
0.1
0.1
Yearly depreciation Accumulated
charge
depreciation
30,000
30000
60000
90000
40000
130000
40000
170000
20000
190000
10000
200000
Ending
bopk Value
170000
110000
70000
30000
10000
0
28.(a)Jefferson's quality tonb stones sells 55 tonb stones per years at 200 kyats each. Each order
cast of 10 kyats and the inventory cost is 1% .Calculate the E.O.Q.
Ans;
A= 10, I = 0.01. C = 200, R = 55
2 x55 x10
E.O.Q =
= 23. 45
200 x0.01
(b) Write done the formula of Economic Order Quantity
2 RA
E.O.Q ==
CI
Where, E.O.Q = Economic Order Quantity
R = Annual requirement for the item in unsts or demand per period
A = Acquisition cost per order or the cast of plocing one order
C = Invoice price of one unit of the item or cost per item
I = Holding cost of stocks, express as a percentage of the average stock or inventory cost
29. Find out the depreciation rate for paper cutting machine using declining balance method. It
paper cutting machine is perchase for 85000 kyats, the estimated life is 5 years and estimated
salvage value is 10000kyats, construct the table and find the ending book value for each year .
Depreciation rate = 2(1/n)
Ans;
Year Depreciation
Rate
1
0.4
2
0.4
3
0.4
4
0.4
5
0.0922
Beginning
Book Volve
85,000
51,000
30,600
18,360
11016
Depreciation
Charge
34,000
20,400
12,240
7344
1016
Accumulated
Depreciation
34,000
54,400
66,640
73984
7500
Ending book
value
51,000
30,600
18,360
11016
10000
30.Determine the economic order quantity for item 445.
Given- Annual requirement 1000 unit
Cost per unit
K 3.2
Inventory holding cast 15 %
Ordering cost per order K 7.0
Assuming 250 working days per year,how often must item 445 be ordered?
Ans;R = 1000 unit, C = K 3.2
I =15%, A=K 7.0
2 RA
E.O.Q=
CI
2 x1000 x7
= 171
3.2 x0.15
1000
units = 5.8 orders/year
R/E.O.Q =
171
5.8 order ------250day
1
1 order------250 x
=43.103 days
5.8
Therefore order must be placed every 43rd and 44th day alternately .
=
Met 5023 Nanomaterials and Materials Technology
Sample Questions and Answers for Final Exam, 2008
1. In some applications, if the material units are so small, how can they be very useful?
Ans; Chapter 1 page 1
2. Express the uses of nanotechnology (at least any five examples)
Ans; Chapter 1 page 2
3.What are two main techniques used in nanotechnology? Briefly explain them.
Ans; Chapter 2 page 3
4. Short notes the procedure of solution –GEL ( SOL-GEL) synthesis method?
Ans; Chapter 3 page 5
5. Define carbon nanocones. What methods are used to produce nanocones?
Ans; Chapter 3 page 6
6.Briefly explain (i) about the carbon nanotubes (ii) how to produce nanotubes (iii) types of
nanotubes.
Ans; Chapter 3 page 6
7. What are the properties of carbon nanotubes ?
Ans; Chapter page 7
8. What are the industrial applications for carbon nanotubes ?
Ans; Chapter 3 page 7
9. Briefly explain (i) nanowiress (ii) types of nanowires (iii) applications of nanowires.
Ans; Chapter 3 page 8
10. How can nanomatrials impact on the health and environment?
Ans; Chapter 4 page 9
11. What safety guidelines should we know handling nanomaterials in laboratory?
Ans; Chapter 4 page 9
12 In solar energy conversion systems, what are the research directions?
Ans; Chapter 5 page 10
13. Express any three applications of nanocomposites.
Ans; Chapter 6 page 11
14. Briefly explain the synthetic processing of clay-based nanocomposites with a net sketch.
Ans; Chapter 6 page 12
15 In the materials industry, the development of ceramic and polymer nanocomposites is a rapidly
expanding multi- disciplinary research activity. Discuss about the polymer nanocomposites.
Ans; Chapter 6 page 12
16. Why nanocomposite characterization techniques are developed?
Ans; Chapter 6 page 12
17. (a) How many steps are required to achieve nanotechnology- produced goods?
Ans; Chapter 2 page 3
(b) What is nanometer? Give any five examples used to measure things i8n nanometer.
Ans; Chapter 1 page 1
18.(a) What are the advantages of AFM over SEM?
Ans; Chapter 2 page 4
(b) What is nanotechnology?
Ans; Chapter 1 page 1
(c) Why nanotechnology is referred to as a general-purpose technology.
Ans; Chapter 1 page 1
19.(a) Name the different types of electron microscope.
Ans; Chapter 2 page 4
(b)Define atomic force microscope with a neat sketch.
Ans; Chapter 2 page 4
20. (a) What are nanomaterials?
Ans; Chapter 3 page 5
(b) What methods are used to produce nanomaterials?
Ans; Chapter 5 page 5
21. (a) What are nanomaterials?
Ans; Chapter 3 page 5
(b) What methods are used to produce nanomaterials?
Ans; Chapter 3 page 5
22.(a) Write down the basic forms of carbon nanomatrials.
Ans; Chapter 3 page 6
(b) Define carbon nanocones. What methods are used to produce nanocones.
Ans; Chapter 3 page 6
23. (a) Describe three novel nanocomposites design.
Ans; Chapter 6 page 11
(b) Describe the advantages of nano-sized additions in nanocomposites.
Ans; Chapter 6 page 11
24.(a) What are two main techniques used in nanotechnology? Briefly explain them.
Ans; Chapter 2 page 3
(b)Describe the disadvantages of nano-sized additions.
Ans;Chapter 6 page 11
25.(a) Express any three applications of nanocomposites.
Ans; Chapter 6 page 11
(b) What is nanometer? Give any five examples used to measure thigs i8n nanometer.
Ans; Chapter 1 page 1
26.(a) Briefly explain (1) about the carbon nanotubes (ii) how to produce nanotubes (iii) types of
nanotubes.
Ans; Chapter 3 page 6
(b) Write down the basic forms of carbon nanomatrials..
Ans; Chapter 3 page 6
(c) What is solar energy?
Ans; Chapter 5 page 10
27.(a) In some applications, if the material units are so small, how can they be ver useful?
Ans; Chapter 1 page 1
(b) How many steps are required to achieve nanotechnology-produced goods?
Ans; Chapter 2 page 3
28.(a) What are the advantages of AFM over SEM?
Ans; Chapter 2 page 4
(b) Name the different types of electron microscope.
Ans; Chapter 2 page 4
29.(a)What are the properties of nanomaterials?
Ans; Chapter 3 page 5
(b) Short nots the procedure of solution- GEL (SOL-GEL) synthesis method?
Ans; Chapter 3 page 5
30.(a) What is solar energy?
Ans; Chapter 5 page 10
(b) Express any three applications of nanocomposites.
Ans; Chapter 6 page 11
Met-5024 Materials Aspects in Design
Sample Questions and Answers for Final Exam, 2008
1.How can the fracture toughness of ceramic depend on the structure of zirconia ?
Ans; Chapter 14 Page 628,14-2.5 Section
2.Briefly Explain the structures of silicates in glasses?
Ans; Chapter 14 Page 634 ,14-3.2
3.Short notes on the formation and secondary processing glasses with neat sketch?
Ans; Chapter 14 Page 644,14-4.1 a,b
4. What are the design techniques in selection of ceramics and glasses?
Ans; Chapter 14 Page 650, 14-5
5.Chapter 14 Example 14-7 Page 652
6.Briefly explain the crystal structure of ceramic.
Ans; Chapter 14 Page 605, 14-2.1
7.How do you select the glass reinforced plastics?
Ans; Chapter 14 Page 797, 16.6
8. Short notes on the selection of ceramics and glasses.
Ans; Chapter 14 Page 650, 14-5.1
9. Classify the types of refractories ?
Ans; Chapter 15 Page 657,14-5.3
10.How do you select the refractory used in metallurgical engineering?
Ans; Page 658, 14-5.3
11.What service condition are required for the refractory?
Ans; Page 658, 14-5.3
12.Short notes on the structure of zirconia.
Ans; Page 628, 14-2.5(similar withQ1)
13.How do zirconia depend the fracture toughness of ceramics?
Ans; Page 658, 14-2.5
14.Classify the types of plastics.
Ans; Page 678, 15-4
15. What are the types of thermoplastics?
Ans; Chapter 15 Page 681,15-4.1
16. Short notes on the following.
i) Thermosets
ii) Elastomers
Ans; Page 678,683
iii) Styrenic Plastics
i)Vinyl Plastics
17.(a)Define the composites and reinforced plastics
Ans; Page 701, 15-6
(b) Classify the types of composites.
Ans; Page 720, 15-8.1
18.What differences between thermoplastics and Thermosetting plastics?
Ans; Page 678, 15-4
Ans; Chapter 15 Page 681, 15-4.1
19.What are the properties of structured steels?
Ans; Chapter 16 Page 755,16-2.1
20.How do you investigate the case of stainless steel for reinforcement of concrete in highway
bridge?
Ans; Chapter 16 Page 766,16-2.2(a)
21. How would you use aluminium in skins of aircraft?
Ans; Chapter 16 Page 767,16-3.1
22.What material do you select to use skins of aircraft?
Ans; Chapter 16 Page 767,16-3.1
23.Describe some of the property we would consider in selecting the material for skins of aircraft.
Ans; Chapter 16 Page 681,15-4.1
24.How do you investigate the case study of aluminium alloy selection from the concord?
Ans; Chapter 16 Page 774,16-3.1(a)
25.What materials would you select for concord?
Ans; Chapter 16 Page 774, 16-3.1(a)
26.How do you select an aluminium alloy in concord?
Ans; Chapter 16 Page 774,16-3.1
27.What materials are selected for automotive industry? Why?
Ans; Chapter 16 Page 778 ,16-4
28.What materials are selected for automotive fenders?
Ans; Chapter 16 Page 795 ,16-5.2
29.What materials are selected for beverage containers with flow diagram?
Ans; Chapter 16 Page 795 ,16-5.1
30.(a) What materials would you select for Nasa space shuttle?
Ans; Chapter 16 Page 774,775
(b)Design the analytic hierarchy process (AHP) in life-cycle analysis.
Ans; Chapter 16 Figure 16-33