Transcript here

Tutorial 2: Material Properties, Kinetic Theory of
Gasses and Odors.
Question1: [Hardness versus Toughness] In your own words, explain the difference
between hardness and toughness. When a steel knife and a diamond are made to collide at
high speed, which of the two is more likely to break?
Hardness: the resistance of a material to permanent deformation (plastic
deformation). Hardness can be measured in various scales, depending on
needs. For example, in mineralogy, the hardness of a material is measured
against the scale by finding the hardest material that the given material can
scratch, and/or the softest material that can scratch the given material
(Mohs scale).
http://www.allaboutgemstones.com/
mohs_hardness_scale.html
Concept of hardness is opposite to idea of Softness (degree to which the
material is susceptible to deformation).
Toughness: the amount of energy that a material can absorb before rupturing
(the resistance to fracture of a material when stressed). It can be found by
finding the area (i.e. by taking the integral) underneath the stress-strain
curve.
Hooke’s Law
applied
Concept of hardness and toughness are totally two different bulk property
of the materials.
Diamond is the hardest material in the world that can sustain high pressure
without getting deformed. However, steel knife is tougher than diamond, in
the sense that it can sustain high impact (absorb more energy by rearranging
the atom in steel). So when the two collide at high speed, diamond will likely
to shatter first.
Hard, but brittle
Tough and ductile
Question 2: [Strong and Brittleness].
We know that ice is brittle. Does that mean that it behaves like glass and is very
strong when very thin?
No. Brittleness is not the deciding factor. In glass, atoms are
joined by covalent bonds (it is not a lattice of silicon dioxide
molecules but more like one huge molecule with silicon and
oxygen in a 1 to 2 ratio). Recall glass is not crystal, but
amorphous solid (or “liquid”).
In ice, molecules bond due to their polarization (water is a
polar molecule). The water molecules themselves remain
completely intact. The bond due to polarization is relatively
weak and so thin ice isn’t very strong (although strength does
increase when ice is made thinner, but not drastic).
Ice crystal or
snowflake
Question 3: [Grain Size and Hall- Petch equation]
The grain size in steel depends on a number of factors including the speed
with which it is cooled. What kind of grain size (relatively large or relatively
small) should I aim for if I want the steel to be very hard?
During plastic deformation, slip or dislocation must take
place across the boundary between grain A and B in
the picture. Grain boundaries are barriers to
slippage (resistance to slippage) for two reason:
1.
B
in
u
bo
ar
nd
y
The atomic disorder in the grain boundary result in
discontinuity of slip planes from grain into the
other. High-angle boundaries are better in blocking
slip !
grain A
gr
n
ai
a
gr
2.
Since the two grains are of different orientations,
dislocation passing into grain B will have to change
its direction of motion; it become more difficult as
the crystallographic mis-orientation increases
(make steel harder).
slip plane
Adapted from Fig. 7.12, Callister 6e.
(Fig. 7.12 is from A Textbook of Materials Technology, by
Van Vlack, Pearson Education, Inc., Upper Saddle
River, NJ.)
Consequently, smaller grain size: more grain barriers to prevent slip. The
so called Hall- Petch eq. expresses this
 y  0
ky
d
σ0 and ky are material dependent
σy = yield strength
d = average grain diameter
However, if the grains become too small, then the resistance to slippage
decrease and the steel becomes less hard again. Hence, there is a limit for
the grain size if we want to produce a very hard steel.
Question4:[Biceps Muscle and Young’s modulus]
A relaxed biceps muscle required a force of 25.0N for an elongation of 3.0cm; the same muscle under
maximum tension requires a force of 500N for the same elongation. Find Young’s modulus for the muscle
tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with length
0.200m and cross-sectional area 50.0cm2.
Recall:
Given,
Relaxed:
Y
F/A
l / lo
Young’s modulus measures how
difficult a material is to compress
or stretch.
Area, A  50.0cm2  50.0 104 m2
l  0.030 m
Y
25.0 N  (0.200m)
(50.0 104 m2 )  (0.030m)
 3.33104 Pa
Maximum
tension:
Y
500.0 N  (0.200m)
(50.0 104 m2 )  (0.030m)
 6.67105 Pa
The muscle tissue is much more
difficult to stretch when it is under
maximum tension.
Question5: [Elasticity]
A steel cable with cross-sectional area 3.00cm2 has an elastic limit of 2.40x108 Pa. Find the
maximum upward acceleration that can be given a 1200kg elevator supported by the cable if
the stress is not to exceed one-third of the elastic limit.
Given that ,


F 1
 2.40 10 8 Pa  0.80 108 Pa
A 3


 0.8010 Pa3.0010
 F  0.80108 Pa A
8
F=tension
4
m2

 2.40104 N
Free-body diagram
a=acceleration
F  mg  ma
mg
F
2.40104 N
a  g 
 9.81m / s 2
m
1200kg
 10.2m / s 2
The tension in the cable is
about twice the weight of
the elevator.
Question6: [Number Density and Mean free path]
The pressure of residual gases in a vacuum chamber at room temperature is 2.0x10–6 mbar
(note: 1 mbar = 1 millibar = 102 Pa). Assuming that the residual gases are water molecules
(H2O, molecular weight is 18 g mol–1) with an effective collision diameter of 3 Å .
Calculate (a) the number density of H2O molecules (in per cm3) in the vacuum, (b) the mean
free path, and (c) the mean time between collisions of these molecules.
Next consider air at the usual pressure of 1 bar and room temperature. Assume that air is
made of N2, with an effective collision diameter also of 3 Å. Repeat the above calculations
for these molecules. This gives you a feel for how close molecules are in the “normal gas
state”.
1
[Kinetic theory mean free path is given by l  2 n~ d , where n~ is the number density and d
is the effective molecular diameter; Average speed is given by v  8k T ]
2
B
av
m
Assumption: Ideal gases law hold (there is no interaction between the gas
molecules and it’s energy purely due to kinetic)
Given that,
PV  nRT
 Nk BT
PH2O=2x10-4 Pa
Molecular weight of H2O, MH2O=18g/mol,
effective collision diameter, d =3x10-10m,
Temperature, T=298K (in between 293K to 298K)
N
P
n~  
V k BT
2.0010 4 Pa
16
3


4
.
86

10
m
1.3810 23 (298K )
 4.861010 cm3 number density


1
1
l

2
2
~
2 n d
2 4.861016 m3 3 1010 m
 51.46 m Mean free path


 M molar
l
l
m
t

l
l
v
vav
8k BT
8RT
 51.46 m 
 18103
 0.087s

kB 
R
;
NA
M molar  N A m
Mean collision time
8  (8.314)298 Assumption of ideal gas is valid since the gas is
quite dilute and mean free path is considered
long (considered a good vacuum) .
Now, for Nitrogen
PN2=1x105 Pa
Molecular weight of N2, MN2=28 g/mol,
effective collision diameter, d =3x10-10m,
Temperature, T=298K
5
P
1
.
00

10
Pa
n~ 

k BT 1.3810 23 (298K )


 2.431025 m 3  2.431019 cm3
1
1
l

2
~
2 n d
2 2.43 1025 m 3 3 1010 m

 1.03107 m
t l
 M molar
8RT
 2.17 1010 s


2
A good order to remember for mean free path of atom
in gases phase(under room temperature and standard
pressure)

7
 1.0310 m

  28103
8  (8.314)298
Question7: [Diffusion and Random Walk]
We have met some aliens who live in a pure Nitrogen atmosphere.
(i) Base on the notion of diffusion as a random walk, at room temperature and pressure, what
is the (numerical value) if the Diffusion constant D? (show brief workings).
The diffusion coefficient is given by, D 
1
mv
2
3
2
 k BT
2
3k BT
3RT
v 

m
M molar

l v
3
1
1
l

2
~
P 2
4 2n r
4 2
r
k BT
3(8.315)(298)
 515m / s
3
28 10
2.32107 m515m / s 
D
3
 3.98105 m 2 / s
(1.3811023 )(298)

 2.32107 m
5
10 2
4 2 (10 )(10 )
Note: speed here is the root mean square (rms) speed.
(ii) In 10 seconds, approximately how far (on average) from their origin will perfume
molecules have spread out in this atmosphere?
Square root dependent of time is
main characteristic of random walk
motion (compare to ordinary
deterministic linear motion).
R(t )
2
 Dt
 3.98105 m 2 / s  10s 
 0.02m
Question8: [Kinetic Theory of Gasses]
Speed of atoms and molecules at room temperature (293K).
(i) What is the average speed of monoatomic Oxygen (i.e. O not O2) if we know that the
atomic number is 8 and that there are as many protons as neutrons?
v 
3RT
3(8.3145)(293)

M molar
15.9994103
 676m / s
(ii) What would the speeds be at zero Kelvin and 1000K? How many electrons does an
electrically neutral Oxygen atom have in total?
T
1000
T
0
v

v

 676m / s
v 
v R.T 
 676m / s
R .T
TR.T
293
T
293
R .T
 1,249m / s
 0m / s
An electrically neutral Oxygen atom has in total 8 electrons.
(iii) What would be the speed of a water molecule?
v 
3RT
3(8.3145)(293)

M molar
18.01528103
 637m / s
Question8: [Kinetic Theory of Gasses] (Continue)
(iv) How heavy would be a ball be that travels at 1m/s if it had the same kinetic energy as one
mole of monoatomic Oxygen?
m  mmolar  15.9994103 kg
1
1
2
2
m v  (0.0159994)676
2
2
 3655.6 J
KEOxy 
1 2
m v  3655.6 J
2
2(3655.6 J )
m
 7311.2kg
12
Question9: [Olfaction]
Name two key advantages of olfaction with regards to vision?
(i) Permanence
Odors is more permanent than light in the sense that
a scent will remain for some time even after the scent
source has moved away while object can no longer be
seen when it has been moved away.
(ii) Manufacturability
Most organisms produce odor but not light because
many odors are simple organic compounds.
(iii) Detection
The spatial resolution of detected light is more
complicated hence the development of adequate
visual system is more complex than that of olfactory
system.
(iv) Darkness
Odor can be useful in places where light cannot pass
such as muddy water, soil etc
 Odor detection works as well in the dark as it does in
the light
Question9: [Olfaction] (continue)
(ii) Are all odorants organic molecules?
No. Although the vast majority of odorants are organic, some odorants are
mot. A good example of non-organic odorant is ammonia (NH3). It has a
rather pungent smell and is often used in cleaning fluids. Other examples
like Chlorine etc.
Question10: [Odorants]
We have discussed odors that spread through air. Could they spread through water? What does
this mean for fish?
Yes, odors spread through water. The diffusion of odors in water and in
fact in any medium occurs through Brownian motion (random collision
between the odorant and the medium particles. Since diffusion is possible
in water, olfaction sensor is quite highly developed for some fishes for
example shark which can smell as little as one part per million of blood in
water.
http://chaos.nus.edu.sg/teaching/GEM2507/resources/Perfume/JavaApplets/brownian.html
Question11: [Odorants]
How is sensitivity to odorants achieved? In other words, what is so different in the noses of
rabbits when compared to humans? After all, we do know that the design of the noses of all
mammals is nearly identical.
The main method for increasing sensitivity is an increase in the number of
receptor cells. Rabbits have roughly 100times as many receptor cells as
human (1 billion versus 1 million). Another way to increase sensitivity is
by having a larger number of receptor types.