GEK2507: Physical Questions from everyday Life

1. GEK2507: Physical Questions from everyday Life Tutorial 2: Material Properties, Kinetic Theory of Gasses and Odors.

2. 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).

3. 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.

4. 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

5. 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

6. 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: • 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). • 2. 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 ! 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.)

7. Consequently, smaller grain size: more grain barriers to prevent slip. The so called Hall- Petch eq. expresses this 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. σ0 and ky are material dependent σy = yield strength d = average grain diameter

8. 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. Young’s modulus measures how difficult a material is to compress or stretch. Recall: Given, Relaxed: Maximum tension: The muscle tissue is much more difficult to stretch when it is under maximum tension.

9. 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. F=tension Free-body diagram a=acceleration The tension in the cable is about twice the weight of the elevator. mg

10. 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”.  [Kinetic theory mean free path is given by , where is the number density and d is the effective molecular diameter; Average speed is given by ] Assumption: Ideal gases law hold (there is no interaction between the gas molecules and it’s energy purely due to kinetic) Given that, 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)

11. number density Mean free path Mean collision time Assumption of ideal gas is valid since the gas is quite dilute and mean free path is considered long (considered a good vacuum) .

12. Now, for Nitrogen PN2=1x105 Pa Molecular weight of N2, MN2=28 g/mol, effective collision diameter, d =3x10-10m, Temperature, T=298K A good order to remember for mean free path of atom in gases phase(under room temperature and standard pressure)

13. 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, 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).

14. 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? (ii) What would the speeds be at zero Kelvin and 1000K? How many electrons does an electrically neutral Oxygen atom have in total? An electrically neutral Oxygen atom has in total 8 electrons. (iii) What would be the speed of a water molecule?

15. 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?

16. 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

17. 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.

18. 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

19. 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.