Material Science Assignment with Questions and Answers

This assignment includes questions and answers related to Material Science, covering topics like crystal structures, ideal gases, freezing point depression, and phase diagrams.

Question Answer
Material science

Question Answer
Contents
Q1..........................................................................................................................................................3
Q2..........................................................................................................................................................4
Q3..........................................................................................................................................................6
Q4..........................................................................................................................................................7
Q6........................................................................................................................................................11
Q7........................................................................................................................................................13
Q8........................................................................................................................................................14

Question Answer
This assignment (the first of two) is due 5.00 pm Monday , September 10 . You may hand it in at
the lecture or hand it in directly to my office before September 10. This assignment is worth 17%
of your final mark. You must sign and submit the attached declaration sheet. Please print this
question sheet and write your answers directly onto it. When appropriate, give details about how
you obtained your answer.
Q1. The metal titanium, which is useful in making alloys to resist high temperatures and
extremely corrosive environments, adopts two crystal structures – close-packed hexagonal (α-phase)
at room temperature and body-centred cubic (β-phase) above 882 °C. Under favourable
circumstances, the α phase can dissolve up to 32 atomic % (nearly 15 wt %) of oxygen whereas the β
phase can dissolve very little oxygen at the transition temperature (882 °C) and a little more at
higher temperatures. Oxygen atoms have a radius of approximately 0.7 x 10-10 m. a) Given that the
“distance of closest approach” between two titanium atoms is approximately 2.92 x 10-10 m,
calculate the radii of the octahedral and tetrahedral interstices in both the hexagonal and cubic
forms of titanium. (5 marks).
Sol.
As given in question,
The distance of closest approach = 2.92 x 10-10 m
For BCC structure, edge length= 2 x 2.92 x 10−10
√ 3 =3.372 x 10−10 m
Since, for BCC structure = radius r = √3 a
4 = 1.732 x 3.372 x 10−10
4 =1.46x 10−10 m Ans
For HCP
radius r = √2 a
4 = 1.414 x 3.372 x x 10−10
4 m= 1.1921 x 10-10 m Ans.
a) to explain why the solubility of oxygen is different in the two structures of titanium? (3 marks).
Answer
The solubility of oxygen in with titanium in α phase is explained by large number of unfilled shells
and the donor-acceptor properties oxygen atoms and titanium, this case explanation is also viable
for other analogue of titanium such as zirconium and hafnium of α -phase situation. Due to this
reason, there is wide range of interstitial solid solutions is being made. It also helps in transformation
of ordering of structure and some lower oxide-suboxide is also created due to solubility of oxygen in
α-phase and
b) Use your answers to part Answer: When heating of titanium goes above 882oC, the
intermolecular spaces are getting wider as compared to α -titanium, in this condition, there is little
more spaces for oxygen occurs and results in little more solubility. But it also decreases the ductility
of titanium because of slip is converted from waviness to plane slip. This causes less ductile material
as compared to pure titanium.

Question Answer
c) OPTIONAL BONUS question – Can you explain how a metal that is capable of dissolving so much
oxygen below 882 °C can simultaneously have excellent oxidation resistance? (2 marks).
Answer – This is due to reason that, titanium has passive oxide films on their surface. This property is
like stainless steel, which completely depends upon oxide films for its corrosion resistance property.
Q2. a) In a 10 litre container, 1 mole of CO2 gas and 3 moles of N2 gas are mixed at 25 °C.
Assuming they are ideal gasses, what is the partial pressure of CO2 in the container? (3 marks).
Answer- As given in question,
CO2 = 1 mole
N2 = 3 moles,
V = 10 lt
T = 273oC + 25oC = 298oK
R = 8.206 x 10-2 L atm /(K.mol)
For ideal gas law we know that
PV = RT (For one mole of mixture)
P = 8.206 x 10-2 X 298/ 10 = 2.445388
Now, P(CO2) = 1 x 2.445388 = 2.445388
P(N2) = 3 * 2.445388 = 7.336164
Total pressure = P(CO2)+ P(N2) = 2.445388 + 7.336164 = 9.781552 atm Ans
b) An engineer (trained at a different university) wants to use a brass (Copper-Zinc alloy) component
at a temperature of 400 °C. She remembers something about zinc having low melting and boiling
points from her studies and asks you if there will be any problems. Calculate the equilibrium vapour
pressure of Zn over brass at 400 °C given that the boiling point of zinc at 1 atm. pressure is 910 °C
and the latent heat of sublimation ∆Hf = 13 kJ/mol. What advice would you give the engineer? (3
marks).
Answer- As given in question,
T1 = 400+273 = 673oK, T2 = 910+273 = 1183o K, ∆Hf = 130000, P = 1 mol K =8.314 J/mol
From Clausius Clapeyron equation,

Question Answer
ln ( P1
P2 )=( Δ H f
R )( 1
T2
− 1
T1
) Putting the value as per given above
ln ( 1
P2 ) =−( 13000
8.314 ) ( 1
1183 − 1
673 )
ln ( 1
P2 )= ( 1563.628 ) (0.000845−0.00064)
ln ( 1
P2 ) =1.00162
1
P2
=e1.00162
P = 0.3673 atm Ans
In this condition, we can suggest that, brass component can be made at closed environment.
c) Ethylene glycol (C2H4(OH)2) dissolved in water provides the standard ‘anti-freeze’ coolant for
water-cooled engines. How many grams of ethylene glycol would need to be dissolved in 5 kg of
pure water in order to depress the freezing point of water by 18 °C? (The molal freezing point
depression constant for water Kf = 1.86 K mol-1 kg and the relevant atomic masses are: C = 12g, H =
1g and O = 16g.) Note: ethylene glycol is an organic compound and does not break up or dissociate
when it dissolves in water i.e. i = 1. Hint: first calculate the molality (m) of the ethylene glycol
solution. (4 marks).
Answer: The molar mass of glycol (C2H4(OH)2) = {(12 x 2) + (1x6) +(16x2)} = 62 g/mol
Now Solvent (b) = 5 kg = 5000 gm
Δ T f =¿
Now molality concentration (m) = ax 1000
bxM = ax 1000
5000 x 62 …………….(i)
Freezing point depression = Δ T f =K f .Cm … … … … (ii)
Δ T f =0− (−18 )=18o C…
Putting the value from equation (i) and (ii)
18 = 1.86x a
5 x 62

Question Answer
Now a= 18 x 5 x 62
1.86 =3000 g
Therefore, 3000 g of ethylene glycol needed to dissolve 5 kg of pure water
Q3. The fuel economy of vehicles is an important consideration as we approach the end of our
fossil fuel reserves and experience growing environmental concerns about emissions. a) The
molecule: iso-octane C8H18 is reasonable “representative molecule” for petrol. Liquid iso-octane
can burn to produce carbon monoxide CO and water according to the following unbalanced
reaction:
C8H18 (l) + O2(g) → CO (g) + H2O (l)
First, balance this equation. Next, calculate how much heat in kJ is evolved in the burning of one
mole of C8H18 to CO? The enthalpies of formation are: ∆Hf° (C8H18 (l)) = -208.5 kJ mol-1, ∆Hf° (CO
(g)) = 110.6 kJ mol-1, ∆Hf° (H2O (l)) = -284.9 kJ mol-1. (3 marks).
Answer-
The balanced equation can be given as
2 C8 H18 ( l ) +17 O2 ( g ) ⇢16 CO ( g ) + 18 H2 O(l)
As given in question
(i) 8 Cs +9 H2 ( g ) ⇢C8 H 18 ( l ) Δ Hf
o=−208.5 KJ /mol
(ii) C ( s ) + 1
2 O2 ( g ) ⇢CO ( g ) Δ Hf
o=−110.6 KJ /mol
(ii) H2 +1
2 O2 ( g ) ⇢ H2 O ( l ) Δ Hf
o =−284.9 KJ /mol
From equation adding the above equation,
Δ Hr
o=18 ( −284.9 ) +16 ( −110.6 ) −2 ( −208.5 )
Δ Hr
o=−6480.8 KJ for 2 mole The for one mole Δ Hr
o=3240.4 KJ /mol Ans
b) CO can oxidize further to carbon dioxide CO2 according to the reaction:
CO(g) + ½ O2 (g) → CO2 (g) ∆H = -285 kJ mol-1.
Sum the above two chemical equations in an appropriate way in order to determine the amount of
heat evolved when one mole of C8H18 burns completely to form CO2. (3 marks).
Answer – As per previous equation,
C8 H18 ( l )+ 17
2 O2 ( g ) ⇢ 8 CO ( g ) +9 H2 O ( l ) Δ Hr
o =−3240.4 KJ /mol

Question Answer
CO(g) + ½ O2 (g) → CO2 (g) ∆H = -285 kJ mol-1.
Adding both the reaction
Δ Hr
o=−3240.4−2280=5520.4 KJ /mol Ans
c) A small-medium car in Australia currently typically uses about 9 litres of fuel per 100 km in a city
environment. Given that the density of iso-octane C8H18 is 705.5 kg m-3 at 25°C and the atomic
masses of carbon and hydrogen are C = 12g and H = 1g, (i) how many moles of C8H18 are there in
these 9 litres of C8H18? (ii) what is the maximum amount of energy released by burning them? (iii) If
the work done in moving the car around the city is on average 300 kJ/km, how efficiently is the fuel
used (in % of the chemical energy liberated)? (4 marks).
Answer- As given in question,
Density of C8H18 = 705 kg-m3
(i) I mole of C8H18 = 12x8 + 1 x 18 = 114 gm.
We know that mass = volume x density = 0.702 x 1000 = 702 gm
Since 1 litre of C8H18 contains 702 gm
Then 9 litres, = 9 *x 702 = 6318 gm
Number of moles in 6318 gm of C8H18 = 6318/114 = 55.42 moles
(ii) Energy evolved by burning 55.42 moles of C8H18 = -3240.4 x 55.42 = 179582.968 KJ Ans.
(iii) Movement of car = 100/9 km/litre
Distance travelled by consuming chemical energy = 179582.968/300 = 598.61 km
Efficiency of car w.r.t chemical energy = 100 * 100/598.61 = 16.705% Ans.
Q4. (10 marks, 2 per section). The zinc-tin (Zn-Sn) phase diagram is important in the canned goods
industry as alloys from within it are used to coat the inside of food containers and food processing
equipment. Answer the following questions about the Zn-Sn phase diagram shown overleaf. Express
all compositions as Zn-x wt.% Sn, e.g. the eutectic composition (which is explicitly shown in this
particular diagram) is written as Zn-91.2 wt% Sn.
Begin by labelling all parts of the phase diagram with Greek characters α, β etc as done in Chapter 2
of the Text.

Question Answer
Answer:
The phase diagram of given Zn-SN is labelled above, since there is not significant α and β, phase,
therefore only α+L and β+L and α + β phase is available. To identify the composition at different
point we have assigned points from B1 to B9, their composition is as given below.
Points Composition
B1 Zn 10 wt.% Sn
B2 Zn 20 wt.% Sn
B3 Zn 30 wt.% Sn
B4 Zn 40 wt.% Sn
B5 Zn 50 wt.% Sn
B6 Zn 60 wt.% Sn
B7 Zn 70 wt.% Sn
B8 Zn 80 wt.% Sn
B9 Zn 91.2 wt.% Sn (Given)
a) An industrially important composition is Zn-48.5 wt% Sn. What is the chemical composition (in wt
%) of the very first solid to appear in a liquid alloy of this overall composition that is cooled slowly
from 400°C?
Answer: AT temp TL 397oC and at Zn 12.4 wt.% Sn
b) At what temperature does this solid begin to form in the liquid?
Answer: .AT temperature TE = 198.5 oC
c) For the same composition alloy:
(ii) What are the compositions (in wt %) of the two phases that are present at 250°C?
Answer: AT 250o C, the phases are (α+L) with Zn 85 wt.% Sn and (liquid) from Zn 85 wt.% Sn to Zn
99.9 wt.% Sn.
(iii) What are the relative fractions of the two phases that are present at 250°C?

Question Answer
Answer: At 250o C, the phases are (α+L) with f(α+L) is 0.85 and fl = 0.15
d) For the same alloy:
(iii) What are the compositions (in wt %) of the two phases present at 180°C?
Answer: Only one phase present i.e. i.e. Solid (α + β) phase, starting from Zn 0 wt.% Sn to Zn 100 wt.
% Sn
(iv) What are the relative fractions/amounts of the two phases present at 180°C?
Answer: Since only one phases is present at 180oC and its fraction is 0 to 1.
e) What is the lowest temperature a liquid Zn-Sn alloy can exist at and what is the composition of
that liquid?
Answer: The lowest temperature is eutectic temperature which is TE = 198.5 oC. The composition is
Zn-91.2 wt.% Sn
f) OPTIONAL BONUS question – for this alloy, draw the microstructure just above and just below the
eutectic temperature. Calculate the fraction of free α phase, beta phase and eutectic alpha phase in
this alloy at room temperature. (5 marks).
Figure 1 Microstructure of Zn-Sn above eutectic temperature
Figure 2-Microstructure of Zn-Sn below eutectic temperature
Fraction of Free alpha phase if given by W α =91.2−0.8
100 =0.904

Question Answer
And free beta phase = W β =100−91.2
100 = 8.8
100 =0.088
Alpha in eutectic mixture = Total α – primary α = 0.904-0.088 = 0.90312
Q5. The concentration of a reactant labelled ‘A’ in a new polymerization reaction (i.e. to form a new
polymer material) is observed every 14 s and the results are given in the table below.
a) Plot the data appropriately on the two graph templates given and use the graphs to decide
whether this is a first order or a second order reaction. (5 marks).
Answer: After plotting the graph with the given data, which is as follows,
T A lnA 1/lnA
0 4
1.38629
4
0.72134
8
14 3.3333
1.20396
3 0.83059
28
2.85714
3
1.04982
2
0.95254
2
42 2.5
0.91629
1
1.09135
7
56 2.2222
0.79849
8
1.25235
2
70 2
0.69314
7
1.44269
5
84
1.81818
2
0.59783
7
1.67269
6
98
1.66666
7
0.51082
6
1.95761
4
112
1.53846
2
0.43078
3
2.32135
3
126
1.42857
1
0.35667
5
2.80367
6
140 1.3333
0.28765
7
3.47636
2
154 1.25
0.22314
4 4.48142
168
1.17647
1
0.16251
9
6.15311
6

Question Answer
182
1.11111
1
0.10536
1
9.49122
2
196
1.05263
2
0.05129
4
19.4955
7
210 1 0 #DIV/0!
0 50 100 150 200 250
0
0.5
1
1.5
f(x) = − 0.00628262550506181 x + 1.20805754278434
T vs Ln[A]
0 14 28 42 56 70 84 98 112 126 140 154 168 182 196
0
5
10
15
20
25
T Vs 1/ln[A]
The second graph which is showing time Vs 1/ln[A], is not showing any significant result, but from
first graph, the order of the reaction is zero order reaction.
b) What is the rate constant? (5 marks).
The rate constant provides the relation between rate of the reaction and concentration of the
reactant. The rate constant is proportionality factor in the mathematical formulation of rate law.
rate constant (k )= Rate
[ A ]x [ B ] y
From the above equation, y = -0.0063x + 1.2081, the term -0.0063 is the rate constant for given data
Q6. (Difficult, 10 marks) A (low carbon steel) camshaft needs to be case-hardened by the process
of ‘carburization’ of the region near the surface. In order to do this, the component is embedded in
carbon at a high temperature. This gives a constant surface concentration Cs of carbon of 4.6 x 10-29
atoms m-3. The carbon is allowed to diffuse into the steel for a certain time t (120 minutes) at a
temperature of 975°C. The parameters that describe the diffusion coefficient (D) of carbon in Fe are

Question Answer
the following: Do = 2.7 x 10-7 m2 sec-1 and Q = 75 kJ mol-1. Assume there is no carbon already in
the component. First, using the appropriate equation in Chapter 3, calculate the diffusion coefficient
of carbon in Fe at 975°C. The ideal gas constant R is: 8.314 Jmol-1K-1 and °K = °C +273. Make sure
that you properly convert any units.
Answer: As given in question,
Do = 2.7 x 10-7 m2/sec, Q = 75 kJ/mol, t = 975+273 = 1248 KR = 8.314 J/mol/K
The diffusion constant can be calculated by given formula
DC=D0 e
−Qd
RT Putting the appropriate value
DC=2.7∗10−7 e
−−75
8.314 x1248
DC=2.68055 x 10−7 m2/sec Ans
Next, calculate the average depth of penetration d (the average distance an average atom moves in
the diffusion time t of 120 minutes), see the equation in Chapter 3 of the Text.
Answer The average depth of penetration is given as
d= √ 2 Dt= √ 2 x 2.68055 x 10−7 x 7200=0.06213 m Ans
Finally, calculate the detailed concentration depth profile C(x) of the diffusion of carbon into Fe after
the diffusion time of 120 minutes. To calculate C(x), use the relevant solution to Fick’s Second Law
(also called the Diffusion Equation) for the case of diffusion from a constant surface concentration
Cs. This solution is:
C(x,t) = Cs [1-erf (x/(2(Dt)1/2))]
where erf is known as the error function, x is the distance into the material from the surface and t is
the diffusion time. Note that the time t is fixed in this case. It’s best to present this depth profile by
calculating the ratio C(x)/Cs and plotting this ratio as a function of distance x. Unless you are a bit
lucky and your calculator happens to have the erf function built in, it’s probably best to use MATLAB
or EXCEL (which have the function erf immediately available) for this calculation. For a suitable range
for the depth x, you should choose 0 < x < 4d or thereabouts.
Answer: As per given equation
C(x,t) = Cs [1-erf (x/(2(Dt)1/2))]
Or, Cx−C0
Cs−C0
=1−erf ( x
2 √ Dt )
Cs = 4.6 x 10-29 atoms/m3, C0 = 0, x = 0 to 4*0.06213, t = 5000, 6000, 7000, 8000 sec
Now putting value in equation

Question Answer
Cx=4.6 x 10−29
(1−erf ( x
2 √ Dt ) )
Now I must make a table for different value of x and Cx which is as follows,
x X/2*(dt)^0.5 Cx
0 0 4.6E-29
0.04 0.455251925 2.39E-29
0.08 0.910503851 9.1E-30
0.12 1.365755776 2.46E-30
0.16 1.821007701 4.61E-31
0.2 2.276259627 5.91E-32
0.24 2.731511552 5.15E-33
0.28 3.186763478 3.03E-34
0.32 3.642015403 1.19E-35
0.36 4.097267328 3.15E-37
The graph plotted against Cx Vs t is as follows.
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4
0
5E-30
1E-29
1.5E-29
2E-29
2.5E-29
3E-29
3.5E-29
4E-29
4.5E-29
5E-29
Cx Vs X
x
Cx
As per Cs profile for depth x the suitable Cs value is 1.4 x 10-29 atoms/m3
Q7. a) Label the following schematic stress-strain diagram with the yield strength (σy), ultimate
tensile strength (UTS), strain to fracture (εf) and Young’s modulus (E). (2 marks).

Question Answer
Answer:
b) From
the graph, determine the numerical values of the quantities in part (a). In calculating Young’s
modulus, you need to know the strain at the yield point which is 0.0012 (difficult to read from the
graph). (4 marks).
Answer:- Young’s Modules = Stress
Strain = 240
0.0012 =200000 MPa Ans
c) Force is applied to a cylinder of this material until it reaches its yield stress and held at that level.
The cylinder was initially 100 mm long and 5 mm in diameter. What is its new length with the force
applied and what happens when the force is relaxed? (2 marks).
Answer : E = 200000 MPa, L = 100 mm,
Area of cross section = π /4 d2= 3.14 x 5 x 5
100 x 100 x 4 =0.001963 m2
Now, Stress = 0.0012 x 200000 = 240 MPa
We know that F = Stress x area = 240 x 105
0.001963 =¿ 1.223 x 1010 Newton.
The change in length = ∆ l=0.0012 x 100=0.12 ,
Then new length will be = 100+0.12 = 100.012 mm Ans
d) A strip of this material is stretched by 25% in tension and the stress is then relaxed. What is the
new yield strength of the material? (2 marks).
Answer: Generally yield strength is derived from graphical solution; In general 0.2% offset is taken
for elastic limit, i.e. As per above said, if the material is stretched by 25%, then its yield strength.

Question Answer
Then their new lengths will 24.8% bigger that previous length. Its yield point is also shifted to some
higher end.
e) OPTIONAL BONUS question – if Poisson’s ratio for the material is 0.33, what is the diameter of the
stressed rod in part (c)? (2 marks).
Answer: It is clear that when there is longitudinal strain on length of cylinder, its diameter will
reduce i.e. it will be under lateral strain,
From the formulae, Poisson ratio μ = Lateral strain/longitudinal strain
Putting the value of longitudinal and poison ration,
Lateral strain = 0.0012 x 0.33 = 0.00396
The new diameter will be = 5 – 0.00396*5 = 5-0.0198 = 4.9802 mm Ans
Q8. a) Write a brief summary of each of the strengthening mechanisms that can operate in
metals. Use dot points and try to stay within the space provided. (5 marks).
Answer: There are number of strengthening mechanism is applied, but the main operation is listed
below.
 Grain Size reduction- It is well known that grain boundaries are the barrier for slip in any
material. If barrier is more disoriented, the strength of the barrier is increased. In the same
way if grain side is reduced the barrier will also get strength.
 Solid solution- A solid solution with little impurity can distorts the atom and stress is
generated, On the other hand stress can produce heavy barrier to dislocation movement
 Strain Hardening (Cold work)-If any material passes through yield strength then it comes
under the cold work. This process is generally happened at room temperature and
commonly the cross sectional area is being deformed. In this process, the material entangles
with each other internally, and makes difficult to dislocate. This process makes overall
material stronger.
 Annealing- Annealing is well known process in metal working, in which material is heated to
above the recrystallization temperature and then cooled down. The main purpose behind
this process is to improve cold work property by increasing ductility and retaining most of
the hardness. This also help to reduce the dislocation significantly
 Precipitation hardening.-When supersaturated solid solution is processed in several steps,
the fine particle is precipitated from coarse particle. In this process strength and hardness
can be increased by precipitating fine particle
b) Consider the grain size strengthening effect in high-strength steel with a = 350 MPa and b=20
MPa/mm-1/2 what yield strength will an alloy with a mean grain diameter of 20x10-6 m have? (3
marks).
Answer-From Hall petch relation we know that,

Question Answer
σ y=σ i+ k
√d (Putting the value in given equation)
σ y=350 MPa+ 20
√20 x 10−6 = 350MPa + 4472MPa = 4822.136 Ans.
c) What will the yield strength be if the manufacturing process is changed so that the mean grain
diameter of the steel produced is now 5x10-6 m have? (2 marks).
Answer- Again putting the Value in above equation
σ y=σ i+ k
√d
σ y=350 MPa+ 20
√5 x 10−6 = 350 + 8944.272 = 9294.272 Ans.
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Ghosh, A, Textbook of materials and metallurgical thermodynamics. in , New
Delhi, Prentice-Hall of India, 2003.
Hartog, J, Strength of Materials. in , Dover Publications, 2012.

Question Answer
Hopkins, D, Problems in Metallurgical Thermodynamics and Kinetics. in, Elsevier
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Peters, A, & J Warn, Concise chemical thermodynamics. in , Boca Raton, Taylor &
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