Pre-U STPM Physics Term 1 CC039242a

Physics Term 1 STPM Chapter 8 Deformation of Solids Solution: (a) Young’s modulus, E = F A e l = F e ( l A) Since wires P and Q have the same length l, and cross-sectional area A, EP EQ = ( F e )P ( F e )Q F e = gradient of graph = ( 200 N 2.0 mm) ( 250 N 4.0 mm) = 1.60 (b) (i) E silver E tin = 8 5 = 1.60 E copper E silver = 13 8 = 1.625 E steel E copper = 21 13 = 1.615 Hence, the possible pairs are: P Silver Copper Steel Q Tin Silver Copper (ii) If P: silver and Q: tin Tensile strength = Maximum force Area Maximum force = Tensile strength × Area Maximum force of P (silver) = (29 × 107) (2 × 10–6) = 580 N Maximum force of Q (tin) = (2 × 107) (2 × 10–6) = 40 N Hence, Q is not tin, since the graph show F up to 250 N. If P: copper and Q: silver Maximum force of P (copper) = (12 × 107 ) (2 × 10–6 ) = 240 N > 200 N Maximum force of Q (silver) = 580 N > 250 N Hence, P is copper wire and Q is silver wire. If P: steel and Q: copper Maximum force of P (steel) = (40 × 107 ) (2 ×10–6) = 800 N > 200 N Maximum force of Q (copper) = 240 N < 250 N Hence, Q is not copper. This confirms that P: copper wire Q: silver wire 194 8

Physics Term 1 STPM Chapter 8 Deformation of Solids Quick Check 2 1. The stress-strain graphs of two cords P and Q, which are of different materials, up to the breaking points are shown below. The cords are of the same length and diameter. 0 Stress Strain P Q Which statement is true? A Cord P is stiffer than cord Q. B Cord P is ductile and cord Q is brittle. C It is more difficult to break cord P than cord Q D For the same force, cord P extends more than cord Q. 2. The figure below is the stress-strain graph for three specimens X, Y and Z which have the same dimensions. Z X Y Stress 0 Strain Which of the following combinations could be the materials from which the specimens are made? X Y Z A Glass Copper Rubber B Glass Rubber Copper C Rubber Copper Glass D Rubber Glass Copper 3. The figure shows a mass m hanging from the mid-point of a wire of natural length l and cross-sectional area A. m   l What is the stress in the wire? A mg A sin α B mg A cos α C mg 2A sin α D mg 2A cos α 4. Among the three materials, steel, glass and rubber, which two best illustrate the properties described below. A B C D Obey Hooke’s Law up to the breaking point Steel Steel Glass Rubber Still exhibits elastic behaviour when the strain is large Glass Rubber Rubber Steel 5. A uniform rubber band, marked with equal divisions is stretch from both ends as shown in the figure. F F Which one of the following correctly shows the separation of the divisions when the rubber band is extended to twice its original length. A B C D 195 8

Physics Term 1 STPM Chapter 8 Deformation of Solids 6. The graphs below show how the force F varies with extension for samples of four materials. The breaking point is indicated as X. Which graph represents the behaviour of the typical brittle material? A F X 0 e C F 0 e X B F 0 e X D F 0 e X 7. Which of the following is the correct dimension for Young’s modulus? A M L–2 C M L–1 T–2 B M L T–2 D M L–2 T–2 8. The tension in a stretched wire is given as T = ke l where e is the extension, l the original length and k is a constant. If the cross-sectional area of the wire is A, which of the following represents the Young’s modulus of the material of the wire? A kl A C kA l B k e D ke lA 9. A wire has natural length l, cross-sectional area A and E the Young’s modulus of its material. Which of the following relates correctly the Young’s modulus E to the force constant k of the wire. A E = Al k C E = kA l B E = l kA D E = kl A 10. Two steel wires, P and Q joined in series to support a weight, W at the lower end. The length of P is l and the length of Q is 2l. The diameter of Q is twice than that of P. What is the ratio of the extension of P to that of Q? A 1 4 C 2 B 1 2 D 4 11. A copper wire of diameter d is stretched by a load hanging vertically from its lower end. The strain of the wire is 8.0 × 10–4. What is the strain in another copper wire of the same length but of diameter 2d? A 2.0 × 10–4 C 1.6 × 10–3 B 4.0 × 10–4 D 3.2 × 10–3 12. In an experiment to measure the Young’s modulus of steel, a student uses a steel wire. Length of wire = (3.00 ± 0.01) m Diameter of wire = (0.55 ± 0.01) mm Mass of load = (8.00 ± 0.01) kg Extension = (3.00 ± 0.01) mm Which of the following contributes the greatest uncertainty in the calculated value of the Young’s modulus? A Measurement of length B Measurement of diameter C Measurement of load D Measurement of extension 13. A light rod of length L hangs from the lower ends of two vertical wires M and N which are of the same natural length and diameter but have different Young’s moduli E1 and E2 respectively. A load is placed on the rod at a distance x from the wire M, so that the rod remains horizontal. What is the value of x in terms of L, E1 and E2? A E1L (E1 + E2) C (E1+ E2)L E1 B E2L (E1 + E2) D (E1+ E2)L E2 14. A steel wire and a copper wire of equal length and diameter are joined at one end. A longitudinal force F is applied to the free ends, as shown. F F steel Steel copper Copper F F If the Young’s modulus of steel is twice that of copper, which statement is true? A The extension of the copper wire is twice of the steel wire. 196 8

Physics Term 1 STPM Chapter 8 Deformation of Solids B The stress of the steel wire is twice of the copper wire. C The strain of the steel wire is twice of the copper wire. D The final length of the copper wire is twice of the steel wire. 15. The stress-strain graph for a material under increasing, and then decreasing stress is as shown below. Which statement is correct? Stress Increasing Decreasing stress stress Strain 0 A The elastic limit had been exceeded. B Heat is dissipated in the process. C Permanent deformation is produced in the process. D The material undergoing plastic deformation. 16. Which material does not undergo plastic deformation? A Glass C Plastic B Steel D Rubber 17. The force constant of a spring P is 6 N cm–1, and that of another spring Q is 3 N cm–1. 6 N P Q Figure (a) (a) If the springs are connected in series as shown in Figure (a) and a force of 6 N is applied, calculate (i) the extension in each spring, (ii) the force constant of the combination. 6 N Q P Figure (b) (b) The springs P and Q are then joined in parallel as shown in the Figure (b), and a force of 6 N is applied. If the extensions of the springs are equal, calculate (i) the extension in the spring system (ii) the force constant of the combination 18. The supporting rod of a hydraulic jack is made of steel with length 0.50 m and crosssectional area 1.0 × 10–3 m2 . The rod is vertical and supports a load of 2.4 × 104 N at the top. Calculate (a) the stress on the rod, (b) the compression of the rod. (Young’s modulus of steel = 2.0 × 1011 Pa) 19. A wire of radius 0.40 mm extends by 0.1% of its natural length when a load of 1 kg hangs from its lower end. Calculate the Young’s modulus of the material of the wire. 20. (a) On the same axes, sketch the forceextension graph for (i) a copper wire, (ii) a rubber band which have the same dimensions. (b) Explain the differences in the behaviour of the two materials by using the molecular theory of matter. 21. (a) In an experiment to determine the Young’s modulus of the material of a wire, the wire is fixed at one end, and masses are hung from the lower end. State three sources of error and suggests a way to reduce or eliminate each error. (b) The table shows the results obtained from the above experiment. The natural length of the wire is 2.40 m and its diameter is 0.520 mm. Load/kg Scale reading/mm 0 3.15 0.500 3.70 1.000 4.20 1.500 4.65 2.000 5.10 2.500 5.55 3.000 6.00 Draw a suitable graph to determine the Young’s modulus of the material of the wire. (c) Explain why metals with high melting points have high values for their Young’s moduli. 197 8

Physics Term 1 STPM Chapter 8 Deformation of Solids 8.3 Strain Energy Students should be able to: • derive and use the formula for strain energy • calculate strain energy from force-extension graphs or stress-strain graphs Learning Outcomes 1. The energy stored in a stretched wire or spring is known as strain energy. 2. The work done in stretching a wire is converted into strain energy of the stretched wire. 3. The strain energy of a stretched wire can be deduced from the extension-force graph (Figure 8.6). 4. Strain energy = ∫ 0 e F dx = Area between the extensionforce graph and the extensionaxis = Shaded area in Figure 8.6 5. Within the elastic limit, the graph is a straight line, Strain = Shaded area = 1 2 Fe 6. From stress = F A Force, F = Stress × A and Strain = e l extension, e = l × Strain where A = cross-sectional area l = original length of wire From strain energy = 1 2 Fe = 1 2 (stress × A) × (l × strain) Strain energy Al = 1 2 (stress × strain) Al = volume of wire Strain energy per unit volume = Area below stress-strain graph (Figure 8.7) 7. If the extensions are e1 and e2 when the forces are F1 and F2 respectively (F2 > F1), Increase in strain energy = Shaded area in Figure 8.8 = 1 2 (F1 + F2 )(e2 – e1) Extension, x e F 0 Force Strain energy Figure 8.6 Stress Strain Strain energy per unit volume 0 Figure 8.7 Extension e1 F1 F2 A B 0 Force e2 Figure 8.8 2010/P1/Q17, 2011/P1/Q17, 2017/P1/Q9 198 8

Physics Term 1 STPM Chapter 8 Deformation of Solids 8. If the wire is stretched beyond the elastic limit, work done to produce an extension e = ∫ 0 e F dx = Shaded area in Figure 8.9 Example 8 The table below shows how the extension e of a nylon string changes with the tension F in the string. e/m 0 1.9 2.8 3.4 3.8 4.1 4.3 F/kN 0 2.0 4.0 6.0 8.0 10.0 12.0 (a) Draw the graph of force F against extension e. (b) Deduce from the graph, the extension when the force is 7.0 kN. (c) Use your graph to calculate the strain energy of the string when it is stretched by a force of 7.0 kN. (d) A mountain climber can withstand a maximum force of 7.0 kN without experiencing any injury. The upper end of a nylon string is fixed and the lower end is attached to the mountain climber. Calculate the maximum vertical distance fallen by the climber (before the string becomes taut) without experiencing any injury caused by the force of the string. Solution: (a) (b) From the graph, when F = 7.0 kN, extension, e = 3.6 m F / kN 0 e / m 12.0 10.0 8.0 6.0 4.0 2.0 1.0 2.0 3.0 4.0 5.0 A1 A2 A3 A4 A5 A6 A7 A8 Extension, x e 0 Force,F Figure 8.9 (c) Strain energy = Area under the graph from e = 0 to e = 3.6 m = A1 + A2 + A3 + A4 + A5 + A6 + A7 + A8 = 1 2 (0.5 × 0.4) + 1 2 (0.5)(0.4 + 0.8) + 1 2 (0.5)(0.8 + 1.4) + 1 2 (0.5)(1.4 + 2.2) + 1 2 (0.5)(2.2 + 3.2) + 1 2 (0.5) (3.2 + 4.6) + 1 2 (0.5)(4.6 + 6.5) + 1 2 (0.1)(6.5 + 7.0) = 8.6 kJ (d) If h = height fallen by climber under free fall, using the principle of conservation of energy. Loss in gravitational potential = Gain in strain energy mg (h + 3.6) = 8 600 1 000 (h + 3.6) = 8 600 h = 8.6 – 3.6 = 5.0 m 199 8

Physics Term 1 STPM Chapter 8 Deformation of Solids 1. A force F stretching a wire produces an extension l. The energy stored in the stretched wire is 1 2 Fl only if A the elastic limit is not exceeded B the extension is directly proportional to the force C the cross-sectional area of the wire remains unchanged D the weight of the wire is negligible 2. The extension-force graph of a wire is as shown below. 5 20 40 60 80 10 15 20 Extension / mm 0 Force / N What is the work done to produce an extension of 20 mm? A 0.150 J B 0.400 J C 1.200 J D 1.600 J 3. The graph below shows the variation of the extension x of a length of rubber for different force F. 54321 80 60 40 20 0 x / m F / N The strain in the rubber for an extension of 5 m is A less than 200 J B 200 J C between 200 J and 400 J D 400 J 4. The graph below shows how the length l of a wire varies with the force stretching it. l /m l1 A1 A3 A4 A2 F1 F2 0 Force / N l2 Which of the area(s) shown represents the increase in strain energy when the length of the wire increases from l1 to l2? A A2 C A3 B A1 + A2 D (A3 + A4) 5. A wire is gradually stretched to just beyond its elastic limit, and it is then gradually released. Which of the following graphs best illustrates the variation of stress with the longitudinal strain ? A C B D 6. When a spring fixed at one end is pulled by a force of 8 N, the extension is 40 mm. Two such springs are joined in series and is pulled to produce a total extension of 40 mm. What is the strain energy in the springs? A 0.04 J C 0.16 J B 0.08 J D 0.32 J Quick Check 3 200 8

Physics Term 1 STPM Chapter 8 Deformation of Solids 7. A thread suspended vertically is loaded with increasing load to a maximum. The load is then reduced gradually to zero. The graph below shows the results obtained. P Q O R Force Extension Which of the areas shown in the graph represent Work done in Energy dissipated from stretching the wire during loading and unloading A R P B Q + R P C P + Q R D P + Q Q 8. The area under which of the following graphs represents the strain energy per unit volume? A C 0 Extension Force 0 Force Extension B D 0 Strain Stress 0 Stress Strain 9. One end of an elastic string is fixed and increasing loads are hung from the lower end. The extension x for each load F is shown below. F/N 0.5 1.0 1.5 2.0 2.5 3.0 3.5 x/cm 1.3 2.5 3.7 5.0 6.8 9.5 13.5 (a) Plot a graph of extension x against load F. (b) (i) What is the range of extension when Hooke’s Law is obeyed? (ii) Explain how you would measure the extension. (iii) State one precaution in the measurement of the extension. (iv) Use the graph to calculate the work done to produce an extension of 10.0 cm. 10. A support cable on a bridge has a crosssectional area of 0.0067 m2 and a length of 24 m. It is made of high tensile steel of Young’s modulus 2.8 × 1011 Pa. The tension in the cable is 720 kN. Calculate (a) the extension of the cable, (b) the strain energy stored in the cable. 11. (a) Define stress and strain. (b) Sketch the stress-strain graph for an elastic cord that obeys Hooke’s Law. (c) Deduce the relation between the strain energy in the stretched elastic cord and the area under the stress-strain graph. (d) An elastic cord of cross-sectional area 2.0 mm2 and unstretched length 40.0 cm is fixed across a horizontal runway of width 40.0 cm as shown below. The Young’s modulus for the elastic cord is 6.0 × 108 N m–2. A trolley of mass 1.50 kg, moving along the central line of the runway with a speed of 0.60 m s–1, hits the cord and is brought momentarily to rest. Calculate the maximum extension of the cord. 12. A catapult consists of two rubber bands with unstretched length 0.20 m. Each rubber band is stretched by 0.10 m by a tension of 50 N. A stone of mass 0.060 kg is projected vertically upwards from the catapult after each rubber band has been stretched to a total length of 0.35 m. (a) Calculate the energy stored in the stretched catapult. (b) Find the maximum height reached by the stone. Neglect air resistance (g = 10 m s–2). 201 8

Physics Term 1 STPM Chapter 8 Deformation of Solids Important Formulae 1. Stress = Longitudinal force, F Cross-sectional area, A Strain = Extension, e Original length, l 2. Young’s modulus, E = Stress Strain = F / A e / l 1. Three different wires X, Y and Z have the following properties: Wire X: undergoes plastic deformation before breaking Wire Y: a large stress produces a small strain Wire Z: breaks immediately after elastic limit The material of which wire is ductile, rigid and brittle? Ductile Rigid Brittle A Wire X Wire Y Wire Z B Wire X Wire Z Wire Y C Wire Y Wire Z Wire X D Wire Z Wire Y Wire X 2. A rod of length 100.0 cm and negligible weight is suspended horizontally by two wire P and Q of the same length attached at the ends. The cross-sectional area of wire P is twice that of wire Q, and ratio of the Young’s modulus of P to that of Q is 0.75. W P x Q At what distance x from wire P should a weight W be hung so that the rod remains horizontal? A 33.3 cm B 40.0 cm C 50.0 cm D 75.0 cm 3. A steel wire of length 2.00 m has a diameter of 2.0 mm. The Young’s modulus of steel is 2.0 × 1011 N m–2. A load of 5.00 kg hangs from one end of the wire. What is the strain energy in the wire? A 9.5 × 10–4 J C 3.8 × 10–3 J B 1.9 × 10–3 J D 4.8 × 10–3 J 4. The total maximum mass of a lift and its load is 2500 kg. The maximum upwards acceleration of the lift is 22.0 m s–2. What is the minimum tensile strength of the lift cable? (g = 10 m s–2) A 2.5 × 107 N m–2 C 5.5 × 107 N m–2 B 3.0 × 107 N m–2 D 8.0 × 107 N m–2 5. A ductile material is exerted by a stress until it reachs its elastic limit. If the stress is increased, which statement is true about the deformation of the material? A The material will undergo further deformation which obeys Hooke’s law. B The material has undergone an irreversible deformation. C The material fractures without undergoing deformation. D The material will not undergo further deformation. 6. The length of a spring is l1 when it is stretched by a force F1, and its length is l2 when it is stretched by a force F2. Which shaded area in the graph represents the work done to stretch the spring from l1 to l2? 3. Hooke’s law, F = ke 4. Within the limit of proportionality, Strain energy = 1 2 Fe Strain energy per unit volume = 1 2 (stress)(strain) STPM PRACTICE 8 202 8

Physics Term 1 STPM Chapter 8 Deformation of Solids A Length Force F1 F2 l1 l2 0 C Length Force F1 F2 l1 l2 0 B Length Force F1 F2 l1 l2 0 D Length Force F1 F2 l1 l2 0 7. A wire of natural length l and cross-sectional area A is made of material of Young’s modulus E. What is the strain energy per unit volume when the extension of the wire is e? A EeA l C Ee2 2l 2 B EeA l 2 D Ee2 A 2l 8. An elastic string of natural length l and force constant k is fixed at one end O. A mass m is attached to the other end. The mass is whirl in a horizontal circle about O on a smooth horizontal surface. If the angular velocity of the mass is w, what is the radius of the circle? A kl mω2 B k + mω2 kl C kl k + mω2 D kl k – mω2 9. The relation between the force F and the extension x of a spring is shown below. F 0 x Which graph best shows how the work done W depends on the extension x? A W 0 x C W 0 x B W 0 x D W 0 x 10. A light uniform rod of length 100.0 cm is supported horizontally by two wires A and B of the same natural length and cross-sectional area. When a mass is hung from the rod at the point P, the rod stays horizontal. 35 cm 10.0 cm Mass A P B 5.0 cm If the Young’s moduli of the wires A and B are E1 and E2 respectively. Calculate the value of E1 E2 . 11. A wire of cross-sectional area 0.52 mm2 and natural length 60.0 cm is stretched at both ends by a force of 50 N as shown in the figure. Wire F = 50 N F = 50 N (a) What is the stress on the wire? (b) If the extension is 0.12 cm, what is the strain? (c) Calculate the Young’s modulus. (d) What is the strain energy of the stretched wire? 12. The table below shows the Young’s modulus and tensile strength of steel and copper. Material Young’s modulus (N m–2) Tensile Strength (N m–2) Steel 2.0 × 1011 4.0 × 108 Copper 1.2 × 1011 2.5 × 108 203 8

Physics Term 1 STPM Chapter 8 Deformation of Solids (a) Discuss which of the materials is (i) more rigid, (ii) more ductile. (b) A copper wire has a length of 2.00 m and diameter of 1.2 mm. What is the extension of the wire when it is subjected to a tensile stress of 150 N? 13. (a) On the same axes, sketch the stress-strain graphs of (i) a copper wire, and (ii) a glass fibre until they break. From the graphs, deduce three difference properties of copper wire and fibre glass. (b) A steel wire of length 3.50 m and diameter 0.60 mm is stretched until the strain is 1.20 × 10–3. Calculate (i) the force stretching the wire, (ii) the strain energy, and (iii) the fraction change in the crosssectional area of the wire. [Young’s modulus of steel = 2.00 × 1011 N m–2] 14. (a) Explain the following features of Searle’s method for the determination of Young’s modulus of the material of a wire. (i) The test wire used is long and thin. (ii) Another identical wire hangs along the side of the test wire. (b) Sketch the extension-force graph for a copper wire up to the breaking point. (c) Explain how you use the graph to determine (i) the Young’s modulus of copper, (ii) the work done to stretch the wire. 15. (a) A wire fixed at one end is stretched by an increasing longitudinal force. With the aid of a suitable sketch graph, explain the stages in the stretching process when the work done (i) is completely recoverable, (ii) partially recoverable, (iii) not recoverable at all. (b) A load of 10 kg hangs from the lower end of a vertical steel wire of natural length 1.00 m. If the Young’s modulus of steel is 2.0 × 1011 N m–2 and the diameter of the wire is 0.030 cm, calculate the strain energy in the wire. 16. (a) Explain why metals with high values of Young’s moduli have high melting points. (b) A tungsten wire P is of natural length 1.00 m and diameter 0.20 mm. A copper wire Q of length 1.50 m is of diameter 0.50 mm. One end of wire P is fixed and a load of 0.10 kg hangs from its lower end. Attached to the bottom of the 0.10 kg load is the wire Q and a load of 0.25 kg hangs at the lower end of wire Q. Assuming that the elastic limits of the wires are not exceeded, calculate the total strain energy in both wires. (Young’s modulus of tungsten = 3.6 × 1011 Pa, Copper = 1.2 × 1011 Pa) 17. (a) Explain what is meant by elastic limit and yield point. (b) The stress on a wire is increased gradually from zero until the wire shows plastic behaviour. The stress is then slowly reduced to zero. Sketch the stress-strain graph for the wire. Mark on your graph, the limit of proportionality P, and the yield point Y. (c) `During plastic deformation of a crystalline solid, the stress would cause the bonds between atoms to break simultaneously and atomic planes would slide over each other.’ Explain why the above statement is not true for most crystalline solids. For what crystalline solid is the statement true? (d) A uniform rod of cross-sectional area 1.5 cm2 and coefficient of linear expansion α = 2.4 × 10–5 K–1 is heated from 20°C to 80°C. The rod is prevented from expanding. The force F exerted on the rod at different temperatures θ are shown below. F/kN 0.0 7.0 14.5 22.0 29.5 37.5 45.5 θ/°C 20.0 30.0 40.0 50.0 60.0 70.0 80.0 (i) Derive an equation for F in terms of θ, α, A, the cross-sectional area, and E, Young’s modulus of the rod. (ii) Use a graph to determine the Young’s modulus of the rod. 204 8

Physics Term 1 STPM Chapter 8 Deformation of Solids 18. (a) Explain, in terms of molecular behaviour of matter, (i) elastic deformation (ii) plastic deformation (b) The variation of applied force F with extension e of a steel wire is shown by the graph below. F/N e/mm 160 120 80 40 2.5 0 5.0 The cross-sectional area of the steel wire is 7.85 × 10–7 m2 and its length is 150 cm. (i) Calculate the stress when the load is 120 N. (ii) Determine the Young’s modulus of the material of the wire. (iii) Calculate the elastic strain energy of the wire. (iv) Estimate the plastic strain energy of the wire. 19. (a) Distinguish between elastic and plastic deformation. A cylinder of length 1.00 m, area of cross-section 8.06 × 10–5 m2 , and mass 0.650 kg falls from rest so that its base drops from a height of 2.50 m above the floor as shown in the figure. (b) By ignoring air resistance, calculate the kinetic energy of the cylinder, just before it hits the ground. (c) On hitting the ground, the cylinder behaves perfectly elastically and obeys Hooke’s Law. It undergoes a maximum contraction of 1.60 × 10–3 m. Calculate (i) the maximum strain in the cylinder, (ii) the maximum elastic potential energy stored in the cylinder, (iii)the maximum force on the rod, (iv) the Young’s modulus of the material of the cylinder. (d) If the collision with the ground is not perfectly elastic, explain how your answer in (c)(iii) and (iv) might be affected. 2.50 m 1.00 m 1 1. (a) Stress = F A = mg A = 1.64 × 107 N m–2 (b) Strain = e l = 1.24 × 10–3 2. Stress = F A , A = πd2 4 Stress ∝ 1 d2 (stress)1 = d2 (2d)2 (5.0 × 106 ) N m–2 = 1.25 × 106 N m–2 3. Stress, s = F A , strain, n = e l (i) In series: for each wire, force = F, extension = e Hence stress = s and strain = n (ii) In parallel: for each wire, force = F/2, extension = e/2 Stress = F/2 A = s/2 Strain = e/2 l = n/2 2 1. A 2. A 3. D 4. C 5. C ANSWERS 205 8

Physics Term 1 STPM Chapter 8 Deformation of Solids 6. A 7. C 8. A 9. D 10. C 11. A 12. B 13. B 14. A 15. B 16. A 17. (a) (i) P:1.0 cm Q: 2.0 cm (ii) 2 N cm–2 (b) (i) 0.67 cm (ii) 9 N cm–2 18. (a) 2.0 × 107 N m–2 (b) 6.0 × 10–3 m 19. 1.95 × 1010 N m–2 20. (a) Force 0 Extension Rubber Copper (b) Copper: strong molecular attraction Rubber: Chains of molecules are straightened when being pulled. 21. (a) – Thermal expansion, support yielding. Use an identical wire parallel to test wire. – Measurement of extension. use spherometer or micrometer. – Measure of diameter of wire at various points. (b) 1.20 × 1011 N m-2 3 1. B 2. C 3. A 4. B 5. B 6. B 7. D 8. B 9. (a) 0 0.5 2.0 4.0 6.0 8.0 10.0 12.0 14.0 1.0 1.5 2.0 2.5 3.0 3.5 F / N A3 A4 A2 A1 x / cm (b) (i) 0 to 5.0 cm (ii) Extension, x = l = l0 (iii) Lower load slowly (iv) Work done = A1 + A2 + A3 + A4 = 0.179 J 10. (a) 9.21 × 10–3 m F = EA l e (b) 3.32 × 103 J W = 1 2 Fe 11. (a) Refer to page 185 (b) Refer to pafe 188 (c) Strain energy = (Area under stress-strain graph) × Volume of wire (d) 1.34 cm 1 2 mv2 = 1 2 Fe 12. (a) 11.25 J (b) 18.75 m mgh = 2 ( 1 2 Fe) STPM Practice 8 1. A 2. B : Tension in wire Q, TQ = (YA l )e Tension in wire P, TP= ((0.75Y)(2A) l )e = 1.5TQ Take moments about the weight, (TP)x = (TQ)(100.0 - x) 1.5x = 100.0 – x, hence x = 40.0 cm 3. C : F = mg = (YA l )e, e = mgl YA Strain energy E = 1 2 Fe = 1 2 (mg) (mgl YA ) = 1 2 (5.0 × 9.81)2 ( (2.00) (2.00 × 1011)(π)(1.00 × 10–3) 2)J = 3.8 × 10-3 J 4. D : Tension in cable F = m(g + a) Minimum tensile strength = F A = (2500)(10 + 22) (10.0 × 10–4) = 8.0 × 107 N m-2 5. B : Plastic deformation after elastic limit. 6. C : Work done = average force × extension = 1 2 (F1 + F2)(l2 – l1) 7. C : Strain energy, U = 1 2 Fe U per unit volume = Fe 2Al = (EAe l )( e 2Al) = Ee2 2l 2 8. D : mrω2 = ke = k(r – l), r = kl k – mω2 9. B : W = 1 2 Fe = 1 2 (ke)e = 1 2 ke2 10. 2.4 T1 × 25 = T2 × 60, T = ( EA l ) e 11. (a) 9.62 × 107 N m–2 (b) 2.00 × 10–3 (c) 4.81 × 1010 N m–2 (d) 3.00 × 10–2 J Strain energy = 1 2 × stress × strain × volume 12. (a) (i) Steel is more rigid. Reason: Young’s modulus greater. 206 8

Physics Term 1 STPM Chapter 8 Deformation of Solids (ii) Steel is more ductile: Reason: Higher tensile strength. Can withstand greater stress before breaking. (b) Young’s modulus, Y = F/A e/l Extension, e = Fl AY= (150)(2.00) π(0.6 × 10–3)2 (1.2 × 1011) m = 2.21 mm 13. (a) Copper wire Fibre glass Ductile material Brittle material Lower Young’s modulus Gradient higher, higher Young’s modulus Less stiff Greater stiffness, greater stress for the same strain (b) (i) F = (EA l )e = (2.00 × 1011)π(0.30 × 10–3) 2 (1.20 × 10–3) N = 67.9 N (ii) Strain energy = 1 2 Fe = 1 2 F(l × strain) = 1 2 (67.9)(3.50)( 1.20 × 10–3) J = 0.143 J (iii) Volume = A0l0 = A1l1 A1 A0 = l0 l1 = 3.50 3.50 + (0.00120 × 3.50) = 0.9988 Cross sectional decreases by (1 – 0.9988) = 0.0012 which is 1.2% 14. (a) (i) Larger extension smaller % error in measurement of extension (ii) Eliminate error due to thermal expansion, or support yielding (b) Force 0 Extension (c) (i) Young’s modulus = ( F e ) ( l A) = 1 gradient ( l A) A = cross-sectional area l = natural length (ii) Work done = Area between curve and extension-axis. 15. (a) (i) Below elastic limit (ii) Beyond elastic limit, but before wire breaks, (iii) Wire breaks (b) 0.340 J 16. (b) 7.13 × 10–4 J 17. (b) 0 P Y Strain Stress (c) Impurity atoms cause atomic planes to slide instead of bonds breaking. Pure material. (d) (i) F = EA α (θ – 20) (ii) E = 2.12 × 1011 N m–2 Gradient of F against θ graph = EAα 18. (a) (i) When stretched, separation between atoms increases. When force is removed, force of attraction between atoms causes the atoms to return to the initial positions. No permanent increase in separation. (ii) During plastic deformation, atomic planes slide over each other. Sliding of atomic planes is due to dislocations produced by impurity atoms. Atoms do not return completely to their initial positions when the force is released. (b) (i) Stress = F A = 1.53 × 108 N m–2 (ii) Young’s modulus, E = F/A e/l = 1.53 × 108 (2.5 × 10–3 / 1.50) N m–2 = 9.18 × 1010 N m-2 (iii) Strain energy = 1 2 Fe = 0.15 J (iv) Plastic strain energy = 0.15 J + 1 2 (120 + 160)(5.0 – 2.5)10-3 J = 0.50 J 19. (a) Elastic deformation : Object returns completely to original dimensions. Plastic deformation : Permanent deformation is produced. (b) 15.9 J Kinetic energy = loss in potential energy (c) (i) 1.6 × 10–3 (ii) 15.9 J (iii) 2.00 × 104 N 1 2 Fe = 15.9 J (iv) 1.55 × 1011 N m–2 (d) Maximum force smaller Value of Young’s modulus smaller because part of the kinetic energy converted to heat. 207 8

CHAPTER Concept Map KINETIC THEORY OF 9 GASES Bilingual Keywords Absolute temperature: Suhu mutlak Average: Purata Degree of freedom: Darjah kebebasan Equation: Persamaan Equipartition: Pembahagi sama Ideal gas: Gas unggul Mean: Min Molar gas constant: Pemalar gas molar Pressure: Tekanan Root-mean-square (r.m.s.): Punca-min-kuasa-dua (pmkd) Translational: Translasi Ideal Gas Equation pVm = RT pV = nRT pV = ( N NA )RT Molecular Kinetic Energy 1 2 mc2 = 3 2 kT R.m.s Speed Molecular speed distribution 0 n(v ) T1 T2 >T1 v Pressure of a Gas Assumptions of the kinetic theory p = 1 3 ρ < c2 > p = 1 3 (nm) < c2 > Degree of Freedom Law of Equipartition of Energy Kinetic energy of molecule = f 2 kT Internal Energy Internal energy of n moles, U = f 2 nRT Kinetic Theory of Gases 208

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 209 INTRODUCTION 1. In this chapter, you will learn about the macroscopic and microscopic properties of gas. 2. The macroscopic properties of a gas include volume, pressure and temperature which can be measured. 3. The microscopic properties are based on the behaviour of gas molecules. 4. The macroscopic properties can be explained using the microscopic properties. 9.1 Ideal Gas Equation Students should be able to: • use the ideal gas equation, pV = nRT Learning Outcome 1. Boyle’s law (a) Boyle’s law states that for a fixed mass of gas at constant temperature, its pressure is inversely proportional to its volume. Pressure, p ∝ 1 V if temperature remains constant or pV = constant The value of the constant depends on (i) the mass of the gas (ii) the type of the gas (iii)the temperature of the gas (b) The variation of pressure p with volume V according to Boyle’s law can be illustrated by the graphs shown in Figure 9.1. 0 V p At temperature T1 At temperature T2 >T1 0 At temperature T2 > T1 At temperature T1 p 1 V –– 0 At temperature T2 > T1 At temperature T1 pV p Figure 9.1 Boyle’s Law 2009/P1/Q19, 2013/P1/Q10, 2015/P1/Q10, 2017/P1/Q8

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 210 2. Charles’ law (a) Charles’ law states that for a fixed mass of gas at constant pressure, its volume increases by 1 273 of its volume V0 at 0°C for each degree centigrade rise in temperature. V = V0 (1 + 1 273 θ) V0 = volume of gas at 0°C V = volume at θ°C V V0 0 –273°C θ / °C 0 V T / K (a) (b) Figure 9.2 (b) The graph of volume V against temperature θ°C is as shown in Figure 9.2(a). When the graph is extrapolated, the volume is zero when θ = –273°C which is known as the absolute zero of temperature or zero kelvin (0 K). (c) Figure 9.2(b) is the graph of volume V against temperature T in K, where T/K = 273 + θ/°C. Hence, the volume V ∝ T. (d) Alternatively, Charles’ law may be stated as: The volume of a fixed mass of gas at constant pressure is directly proportional to its absolute temperature T. V T = constant 3. Ideal gas (a) An ideal gas is a gas which obeys Boyle’s law and Charles’ law at all pressures. An ideal gas only exists in theory. Real gases such as hydrogen, helium, oxygen and nitrogen deviate from the behaviour of an ideal gas as shown in Figure 9.3. N2 pV 0 P Ideal gas He H2 O2 Figure 9.3 Behaviour of real gases (b) For an ideal gas, it obeys Boyle’s law: pV = constant at constant temperature for all pressures and temperatures. The real gas that deviates least from the behaviour of an ideal gas is hydrogen. For very low pressures (p → 0), all real gases behave almost like the ideal gas.

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 211 Ideal Gas Equation Boyle's law p1V1 T1 Charles' law p2V T1 p2V2 T2 (a) (b) (c) Figure 9.4 1. Charles’ law and Boyle’s law can be combined in an equation known as the ideal gas equation. 2. Figure 9.4(a) shows a fixed mass of ideal gas at pressure p1, temperature T1 and volume V1. Suppose that the volume of the gas changes to V under constant temperature T1 and pressure p2. By Boyle’s law p2V = p1V1.............................. a Suppose that the temperature of the ideal gas is changed to T2 at constant pressure p2, then using Charles’ law, the final volume V2 is given by V2 T2 = V T1 .............................. b Equation a × b: (p2V) ( V2 T2 ) = (p1V1) ( V T1 ) p2V2 T2 = p1V1 T1 or pV T = constant The value of the constant depends on the mass of the gas and the type of the gas. 3. For a mole of gas, the value of the constant is the same for all gases, and is known as the molar gas constant, R. 4. For one mole of gas at standard temperature and pressure (s.t.p.), 273 K and 101.3 kPa, the volume is 22.4 dm3 (22.4 × 10–3 m3 ). Using pV T = R Molar gas constant, R = pV T = (101.3 × 103 )(22.4 × 10–3) 273 = 8.31 J K–1 mol–1 5. Hence, the ideal gas equation may be written as pVm = RT where Vm = volume of 1 mole of gas and R = molar gas constant 6. If the volume of n moles of ideal gas is V, then Vm = V n and pVm = RT p ( V n ) = RT pV = nRT

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 212 7. The number of moles of gas can also be expressed as n = m M or n = N NA where m = mass of gas M = molar mass of gas N = number of gas molecules NA = Avogadro number Hence, the ideal gas equation can also be written as pV = ( m M ) RT pV = ( N NA) RT 8. According to Avogadro, equal volume of different gases at the same temperature and pressure contain equal number of molecules. Hence, all gases at s.t.p. has the same number of molecules. This number is known as Avogadro number, NA. NA = 6.02 × 1023 mol–1 Example 1 Given that the molar gas constant R is 8 J K–1 mol–1, estimate the volume of a mole of gas at 300 K and pressure 1 × 105 Pa. If the volume of one gas molecule is 2 × 10–29 m3 , calculate the fraction of the gas volume which is empty space. (Avogadro number, NA = 6 × 1023 mol–1) Solution: Using pVm = RT Volume of 1 mol, Vm = RT p = 8 × 300 1 × 105 = 2.4 × 10–2 m3 Volume of gas molecules = (2 × 10–29) (6 × 1023) = 1.2 × 10–5 m3 Volume of empty space = (2.4 × 10–2) – (1.2 × 10–5) = 2.3988 × 10–2 m3 Volume of empty space Volume of gas = 2.3988 × 10–2 2.4 × 10–2 = 0.9995 Exam Tips The various forms of the ideal gas equation: pVm = RT pV = nRT pV = ( m M ) RT pV = ( N NA ) RT Exam Tips Real gases obey the ideal gas equations only when the pressure is very low, p → 0.

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 213 Example 2 The figure below shows the volume of flask X is twice that of flask Y. The flasks are connected by a narrow tube. The system is filled with an ideal gas and a steady state is established with the flasks held at 200 K and 400 K respectively. Find the ratio of the mass of gas in X to that in Y. 200 K 400 K X Y Solution: Since X and Y are connected, the pressure is the same in both flasks. Using pV = ( m M ) RT For X : p(2V) = ( mX M ) R(200) .............................. a M = mass of 1 mole of gas For Y : p(V) = ( mY M ) R(400) .............................. b a b : 2 = mX mY × 200 400 mX mY = 4 Example 3 Two vessels, X and Y of volume VX and VY are connected by a narrow tube and are kept at temperatures of TX and TY respectively. Both X and Y contain an ideal gas. If the number of molecules in X is NX, what is the number of molecules in Y? Solution: The pressure in X and Y are the same. Use the ideal gas equation pV = ( N NA) RT For X : pVX = ( NX NA ) RTX .............................. a For Y : pVY = ( NY NA ) RTY ............................. b a b : ( VX VY ) = ( NX NY ) . ( TX TY ) NY = NX( TX VY TY VX )

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 214 Example 4 A cylinder, 15 cm long and cross-sectional area S, closed at both ends, is fitted with a thermally insulated piston, so as to contain gas in each section A and B. Initially the piston is in equilibrium at a distance of 5 cm from one end of the cylinder, and the temperature of the gas in both sections are the same, 31°C. The temperature of the gas in section A is now increased to 183°C, while the gas in section B is maintained at 31°C. The piston moves through a distance x until a new equilibrium position is obtained. (a) Write an equation relating the initial and final pressures, volumes and temperatures of the gas in A. (b) Write a similar equation for the gas in B. (c) Hence, find the new of equilibrium position of the piston. Solution: (a) V1 = 5S, T1 = 31°C = 304 K, pressure = p1 V2 = (5 + x)S, T2 = 183°C = 456 K, pressure = p2 Using p1V1 T1 = p2V2 T2 p1 (5S) 304 = p2 (5 + x)S 456 .............................. a (b) V1′ = 10S, T1′ = 304 K, pressure = p1 V2′ = (10 – x)S, T2′ = T1′ = 304 K, pressure = p2 Using p1V1 T1 = p2V2 T2 p1 (10S) 304 = p2 (10 – x)S 304 10p1 = (10 – x) p2 ............................. ➁ (c) a b : p1 (5S) 304 × 1 10p1 = p2 (5 + x)S 456 × 1 (10 – x)p2 3(10 – x) = 4(5 + x) x = 1.4 cm New position of piston = 6.4 cm from one end 5 cm 10 cm A B 15 cm

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 215 Quick Check 1 1. The number of molecules in 1.0 cm3 of ideal gas at 273 K and 1 × 105 Pa is 2.7 × 1019. How many molecules are there in 1.0 cm3 of the gas at 273 K and 10–4 Pa? A 2.7 × 109 B 2.7 × 1010 C 2.7 × 1014 D 2.7 × 1019 2. Which of the following graphs best illustrates the variation of the product of (pressure × volume) for a fixed mass of an ideal gas with the thermodynamic temperature T? A C B D 3. Which one of the following is essential if the ideal gas equation pVm = RT is to be obeyed well by a real gas? A The temperature must be constant. B The volume must be small. C The pressure must be low. D The temperature must not exceed 273 K. 4. The relationship between the product pV and thermodynamic temperature T of a sample of ideal gas is (1) in the figure. pV 0 T (1) (2) The graph marked (2) will be followed by a gas if A the mass of the gas is halved. B the number of molecules is doubled. C the molecular mass is halved. D the pressure is halved. 5. A gas cylinder has a safety valve which releases gas when the pressure inside the cylinder reaches 3.0 × 105 Pa. The maximum mass of gas in this cylinder at 20°C is 16 kg. What would be the mass of gas in the cylinder at 60°C? A 5 kg C 15 kg B 14 kg D 48 kg 6. When the pressure of air in a balloon is increased by 2% by pumping air into the balloon, it is found that the number of moles of air in the balloon has increased by 5%, the thermodynamic temperature increased by 2%. By what percentage has the volume of the balloon changed? A Decreased by 3% B Decreased by 5% C Increased by 5% D Increased by 7% 7. X Y Tap The figure shows two bulbs, X and Y joined by a narrow tube fitted with a tap. The volume of X is three times that of Y. X and Y contain ideal gas at pressures of 100 kPa and 300 kPa respectively. The temperature of both X and Y are the same and is constant. What is the final pressure of the gas when the tap is opened? A 150 kPa C 250 kPa B 200 kPa D 400 kPa 8. An ideal gas exerts a pressure of 6 kPa when its temperature is 400 K and the number of molecules present per unit volume is n. Another sample of the same gas exerts a pressure of 3 kPa when its temperature is 300 K. How many molecules are present in unit volume of this second sample? A 4n 3 C 3n 4 B 3n 2 D 2n 3

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 216 9.2 Pressure of a Gas Students should be able to: • state the assumptions of the kinetic theory of an ideal gas • derive and use the equation for the pressure exerted by an ideal gas p = 1 3 ρc2 . Learning Outcomes Kinetic Theory of Gases 1. The kinetic theory of gases consists of the following assumptions: (a) A gas consists of a large number of molecules. (b) The gas molecules are in constant, free and random motion. (c) Gas molecules collide elastically with each other and with the walls of the container. The pressure of a gas is due to the collision of the gas molecules with the walls of the container. (d) The volume of the gas molecule is negligible compared to the volume of the gas. (e) Forces between molecules are negligible except during collision. (f) The time of collision between molecules is negligible compared to the time between collisions. 2. By using the kinetic theory of gases, we can derive an expression for the pressure exerted by a gas. Pressure of a Gas Molecule of mass m m m z y B A x u w v v – v c l l l Wall A Figure 9.5 1. Figure 9.5 shows an ideal gas in a cubic container of length l. The velocity of a gas molecule of mass m is c. The components of the velocity c along the x, y and z directions are u, v and w respectively. c2 = u2 + v2 + w2 2. Consider the component of velocity v towards the wall A. Since the collision of the molecule with the wall is perfectly elastic, the velocity after collision is –v. Change of momentum = mv –(–mv) = 2mv Info Physics Acrosol Spray Can The pressure inside a spray can is about twice the atmospheric pressure. Never throw aerosol spray can into a trash fire. The can will explode because the pressure inside the can increases when the temperature rises. 2009/P1/Q18, 2016/P1/Q10

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 217 3. Assuming that there is no collision between molecules, the time interval between a collision of the molecule with wall A, and the next collision with the same wall t = 2l v Hence, the rate of change of momentum due to one molecule = (2mv) 2 l v = mv 2 l 4. If the container contains N molecules moving with velocities of c1, c2, c3, …, cN with components along y directions of v1, v2, v3, …, vN respectively, then the force on the wall A due to N molecules colliding with it is F = mv1 2 l + mv2 2 l + mv3 2 l + … + mvN 2 l The pressure on wall A p = F Area = F l2 = m l3 (v1 2 + v2 2 + v3 2 + … +vN 2 ) .............................. a 5. The mean of the squares of N velocities v1, v2, v3, …, vN is written as <v2 > = v1 2 + v2 2 + v3 2 + ...... + vN 2 N Hence, equation a becomes p = m l3 N <v2 > ............................. b 6. Referring to Figure 9.6. c2 = u2 + v2 + w2 <c2 > = <u2 > + <v2 > + <w2 > As the number of molecules N is large and the molecules are in random motion <u2 > = <v2 > = <w2 > = 1 3 <c2 > Hence, from equation ➁ p = m l3 N <v2 > = 1 3 mN V <c2 > l3 = V p = 1 3 ρ <c2 > mN V = ρ, density of gas where <c2 > = c1 2 + c2 2 + c3 2 + ...... + cN 2 N is the mean of the squares of the velocities of molecules in the gas and <c2 > is known as the root-mean-square velocity of the molecules. Root-mean-square velocity cr.m.s = <c2 > = c1 2 + c2 2 + c3 2 + ...... + cN 2 N Figure 9.6 Exam Tips p = 1 3 ρ <c 2 > = 1 3 nm <c 2 > = 1 3 N ( V ) m <c 2 > m = mass of 1 molecule n = number of molecules per unit volume N = number of molecules in volume V

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 218 Example 5 Five molecules are moving with velocities as shown below. 100 m s–1 300 m s–1 300 m s–1 100 m s–1 500 m s–1 Calculate the root-mean-square velocity. Solution: Root-mean-square velocity, vr.m.s. = v1 2 + v2 2 + v3 2 + v4 2 + v5 2 5 = (100)2 + (–300)2 + (300)2 + (100)2 + (–500)2 5 = 300 m s–1 Example 6 Estimate the root-mean-square speed of air molecules at 27°C and 1.0 atmospheric pressure (1.01 × 105 Pa) if the density of air under these conditions is 1.29 kg m–3. Solution: Using p = 1 3 ρ <c2 > cr.m.s. = <c2 > = 3p ρ = 3 × 1.01 × 105 1.29 = 485 m s–1 Quick Check 2 1. Which statement is not an assumption of the kinetic theory of gases? A No loss of kinetic energy when gas molecules collide with each other B No change of momentum when a gas collides with the walls of the container C Volume of the gas molecules is negligible compared to the volume of the gas D Time of collisions between gas molecules is negligible compared to time between collisions 2. The pressure p and the volume V of a gas is related to the root-mean-square speed of its molecules by pV = 1 3 (Nm) <c2 > What does the product Nm represent? A Mass of gas in volume V B Density of gas C Number of molecules in one mole of gas D Total number of molecules in volume V 3. The kinetic theory of gases may be used to derive the expression relating the pressure p to the density ρ of a gas. p = 1 3 ρ <c2 > In this expression, <c2 > represents A root-mean-square speed of the molecules. B average of the squares of the speeds of the molecules. C square of the average speed of the molecules. D sum of the squares of the speeds of the molecules.

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 219 9.3 Molecular Kinetic Energy Students should be able to: • state and use the relationship between the Boltzmann constant and molar gas constant k = R NA • derive and use the expression for the mean translational kinetic energy of a molecule, 1 2 mc2 = 3 2 kT Learning Outcomes 1. From the equation p = 1 3 ρ <c2 > and density ρ = M Vm where M = mass of 1 mole of gas Vm = volume of 1 mole of gas then p = 1 3 ( M Vm) <c2 > pVm = 1 3 M <c2 > .............................. a 2. Comparing equation a with the ideal gas equation pVm = RT, we get 1 3 M<c2 > = RT Mass of 1 mole of gas, M = NAm where NA = Avogadro number and m = mass of 1 molecule Hence, 1 3 NA m <c2 > = RT m <c2 > = 3 ( R NA) T Mean kinetic energy of one molecule, 1 2 m <c2 > = 3 2 ( R NA) T = 3 2 kT where k = R NA is known as Boltzmann’s constant = 1.38 × 10–23 J K–1 3. The mean translational kinetic energy of the gas molecule is directly proportional to the thermodynamic temperature, T. 4. The root-mean-square speed, cr.m.s = <c2 > ∝ T 4. The speeds of four gas molecules are 10u, 9u, 5u and 6u. What is the root-mean-square speed? A 2.7u C 7.5u B 3.9u D 7.8u 5. The root-mean-square speed of the molecules of a gas in a closed container is v. If the gas is heated at constant volume until the pressure is doubled, what is the r.m.s. speed now? A 1 3 v C 2v B 1 2 v D 2v 6. A cubic container of volume 0.10 m3 contains an ideal gas at a pressure of 1.00 × 106 Pa and temperature 300 K. Assuming that the gas behaves as an ideal gas, calculate (a) the number of gas molecules in the container, (b) the mass of the gas, if its relative molecular mass is 28, (c) the density of the gas, (d) the root-mean-square speed of the molecules.

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 220 Example 7 Estimate the root-mean-square speed of helium atoms at room temperature of 27°C. Mass of helium atom is 6.6 × 10–27 kg. Solution: Using 1 2 m <c2 > = 3 2 kT cr.m.s. = <c2 > = 3kT m T = (273 + 27) K = 300 K = 3(1.38 × 10–23)(300) 6.6 × 10–27 = 1.4 × 103 m s–1 Example 8 A cylinder contains a mixture of two gases. The mass of a molecule of the gases are m1 and m2 respectively. Calculate the ratio of the root-mean-square speeds of the two gas molecules if the temperature of the mixture is constant. Solution: Suppose T = The thermodynamic temperature of the mixture Then 1 2 m1c1 2 = 3 2 kT and 1 2 m2c2 2 = 3 2 kT 1 2 m1c1 2 = 1 2 m2c2 2 c1 2 c1 2 = m2 m1 Example 9 The pressure of a fixed mass of gas in a closed container of constant volume is p. The gas is heated so that the root-mean-square speed of the gas molecule is doubled. What is the new pressure of the gas? Solution: cr.m.s. ∝ p .............................. a c′ r.m.s = 2cr.m.s. ∝ p' ............................. b b a : p’ p = 2 p’ = 4p

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 221 Example 10 The curves shown are the isotherms for a fixed mass of an ideal gas at temperature T1 and T2. (a) What is the ratio of T1/T2? (b) What is the ratio of the r.m.s. speeds of the gas molecules at T1 and T2? Solution: (a) Using the ideal gas equation: pV = nRT. At T1, the curve passes through the point (1, 1 × 105 ), hence, (1 × 105 )(1) = nRT1 .............................. a At T2, the curve passes through the point (2, 2 × 105 ) hence, (2 × 105 )(2) = nRT2 ............................. b a b : 1 4 = T1 T2 T1 T2 = 1 4 (b) <c1 2 > <c1 2 > = T1 T2 = 1 4 = 1 2 Quick Check 3 1. The r.m.s. speed of nitrogen molecule at 300 K is c. What is the temperature of the gas when the r.m.s. speed is 2c? A 425 K C 1 146 K B 600 K D 1 200 K 2. p 0 1 V –– a b The graph of pressure p against 1 V for an ideal gas undergoing isothermal changes is as shown in the above figure. V is the volume of the gas. The mean kinetic energy of a gas molecule is E. The number of gas molecules in the sample is given by A 3a 2bE C 2a 3bE B 3b 2aE D 2b 3aE 3. The figure below shows the variation of pressure p with density ρ for an ideal gas at 300 K and temperature T. What is the temperature T? A 355 K C 420 K B 400 K D 840 K p/fi105 Pa 4 3 2 1 1 2 3 4 T1 T2 V/m3 0

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 222 4. The r.m.s. speed of oxygen molecules in the atmosphere is 400 m s–1. The relative molecular masses of oxygen and helium are 32 and 4 respectively. What is the r.m.s. speed of helium molecules in the atmosphere? A 141 m s–1 C 1 130 m s–1 B 282 m s–1 D 3 200 m s–1 5. The r.m.s. speed of the molecules of an ideal gas at a certain temperature is c. The temperature of the gas is changed at constant volume such that its pressure is halved. What is the r.m.s. speed now? A c 2 C 2c B c 2 D 2c 6. The kinetic theory of gases is used to derive the expression p = 1 3 ρ <c2> relating the pressure p, density ρ and the mean-square speed <c2 > of gas molecules. (a) What are the assumptions made to derive this expression? (b) State the principles used in the derivation. 7. (a) Explain why molecules in a gas do not move with the same speed. (b) Explain the fact that the speed of sound in the air is of the same order of magnitude as the r.m.s. speed of gas molecules. 8. Use the equations p = 1 3 ρ <c2 > and pVm = RT for an ideal gas to derive an expression for the temperature T of an ideal gas in terms of the total translational kinetic energy of the gas molecules in one mole. Estimate the total translational kinetic energy of the molecules of a mole of ideal gas at room temperature of 30°C. 9. (a) The pressure p of an ideal gas is given by p = ( m M) RT V Summarise the macroscopic properties of a fixed mass of ideal gas. (b) A cylinder of volume 0.080 m3 contains oxygen at 280 K and pressure 90 kPa. Calculate (i) mass of oxygen in the cylinder, (ii) number of oxygen molecules in the cylinder, (iii) the r.m.s. speed of the oxygen molecules. (Relative molecular mass of oxygen = 32), (c) Behaviour of real gases deviates from that of an ideal gas. (i) State the condition necessary for a real gas to behave as an ideal gas. (ii) Sketch suitable graph to show the deviation in behaviour of a real gas from that of an ideal gas. 10. The kinetic theory of gases gives the expression for the pressure p as p = 1 3 nm <c2 > (a) Identify the physical quantity represented by each symbol in the expression. (b) Use this expression to derive an expression for the r.m.s. speed of molecules in a gas in terms of the emperature T of the gas, the mass of one mole M, and the molar gas constant R. (c) Calculate the r.m.s. speed of hydrogen molecules at –60°C. (Mass of 1 mole of hydrogen = 2.0 × 10–3 kg). 11. In the upper layers of the atmosphere, a large fraction of hydrogen molecules move so fast that they are able to escape completely from the Earth. There may also be some loss of helium molecules, but the loss of heavier molecules is negligible. Explain each of the following observations. (a) Fast molecules are able to escape. (b) Molecules of small mass are able to escape but more massive molecules are not. (c) Some helium molecules escape even though their r.m.s. speed is less than the required speed. (d) Large quantities of hydrogen remain on the Earth as a constituent of water. 12. (a) Use the kinetic theory of gases to explain qualitatively how molecules of a gas exert a pressure on the walls of the container. (b) Two moles of an ideal gas at 300 K are contained in a vessel. The gas is then heated to 600 K, but the pressure remains constant due to a leakage of gas from the vessel.

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 223 (i) How many mole of gas is left in the vessel at 600 K? (ii) If the mass of 1 mole of the gas is 32.0 g, calculate the r.m.s. speed of the gas molecule at 600 K. 13. (a) Molecules in a gas are in constant random motion. What is meant by random motion? (b) (i) Write down the ideal gas equation. Identify the symbols in your equation. (ii) Starting from the equation pV = 1 3 Mc 2 where p is the gas pressure, V the volume of the gas, M the mass of 1 mole of gas, and c 2 – the meansquare speed of gas molecules, derive an expression for the root-meansquare speed of the gas molecules in terms of m mass of one molecule, T the temperature of gas, and k the Boltzmann’s constant. 9.4 The r.m.s Speed of Gas Molecules Students should be able to: • distinguish between an ideal gas and a real gas • explain the concept of internal energy of an ideal gas • derive and use the relationship between the internal energy and the number of degrees of freedom Learning Outcomes 1. Gas molecules are constantly colliding elastically with each other and the walls of the container. 2. When two gas molecules collide, transfer of kinetic energy occurs. The kinetic energy of one moleculeincreases and the kinetic energy of the other decreases. 3. Hence, the speed of a gas molecule changes after each collision. 4. Molecules move with difference speeds. 5. Figure 9.7 shows the molecular speed distribution, which is a graph of n(v), number of molecules with speed (v ± 1 2 ) m s–1 against the speed v. This distribution is known as Maxwell distribution. Exam Tips v0 < vm < v – vm v1 v2 v0 v T2 T1 T1 0 ∆A Number of molecules n(v) Speed, v Figure 9.7 Molecular speed distribution 2008/P1/Q18, 2008/P1/Q19, 2010/P1/Q20, 2011/P1/Q18, 2013/P1/Q11, 2014/P1/Q12, 2015/P1/Q11, 2016/P1/Q12

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 224 6. For the speed distribution at temperature T1, the number of molecules with speed between v1 and v2 is given by the shaded area under the curve Δ A. Hence, the total area under the graph represents the total number of molecules in the sample. 7. There is no molecule with zero speed and the maximum speed is very large. 8. For each temperature, there is most probable speed v0. The most probable speed is the speed of the most number of molecules. 9. The graph is not symmetrical about the most probable speed. The area under the graph to the right of v0 (for v > v0) is greater than the area to the left of v0 (for v < v0). This means that more molecules have speed greater than the most probable speed. Hence, the mean speed vm is greater than the most probable speed. 10. The mean speed, vm = v1 + v2 + ...... + vN N where N = number of molecules in the sample The r.m.s. speed, v – = v1 2 + v2 2 + ...... + vN 2 N which is greater than vm. Hence, v0 < vm < v – 11. When the temperature of the gas is raised to T2 (T2 > T1), from the kinetic theory of gases, the r.m.s. speed increases. Thus, the values of v0 and vm also increase. 12. The distribution changes. • It is now more flat. • The most probable speed v0 (the peak) shifts to a higher value. • The peak is lower. • Area to the right of v0 increases, area to the left of v0 decreases. More molecules have speeds greater than v0. 13. Since the number of molecules in a sample remains constant, the total area under both curves at T1 and T2 are the same. 14. The molecular speed distribution is also dependent on the mass of the molecules. For molecules of smaller mass, more molecules have speed greater than the most probable speed compared to more massive molecules. This explains for the greater loss of H2 molecules at high altitudes compared to N2 molecules. Example 11 60° 100 mm A Q Molecular B beam P The figure shows a device used to obtain the speed distribution of molecules in a beam. Discs A and B are mounted on a common axle which rotates in the direction as shown. P and Q are narrow slits in the discs separated by an angle of 60°. The axle is rotated at such a rate that molecules of speed 400 m s–1 are able to pass through both slits.

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 225 (a) Explain how the device select molecules of this speed from the incident beam and find the required angular speed of rotation of the axle. (b) In fact, not only molecules with the speed of 400 m s–1 passes through both discs. Calculate two other speeds of the molecules that pass through the both discs. (c) Suggest how this weakness may be overcomed. Solution: (a) Molecules with the selected speed are able to pass through the slits in both discs A and B if the time taken to move from A to B = the time for the axle to rotate through 60° 0.100 400 = π 3 ω ω = π 3 × 4 000 = 4 000π 3 rad s–1 = 667 rev. s–1 (b) Slower molecules are able to pass through the slits in both discs if the time taken to = the time taken for the axle to rotate move from A to B (2π + π 3 ) rad or (4π + π 3 ) rad 0.100 v1 = 7π 3 × 3 4 000π v1 = 57 m s–1 0.100 v2 = ( 13 3 π) × 3 4 000π v2 = 31 m s–1 (c) The weakness can be overcomed by fixing another disc with a slit on the same axle midway between discs A and B and inclined at an angle of 30° to the slits on A and B. Quick Check 4 1. In a container, there is a mixture of equal number of molecules of H2 gas and CO2 gas at the same temperature. Which graph correctly shows the molecular speed distribution of the gases? C H2 CO2 n (v) v 0 D H2 CO2 n (v) 0 v A H2 CO2 n (v) v 0 B H2 CO2 n (v) 0 v

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 226 2. If – v, vr.m.s.and v0 are the mean speed, r.m.s. speed, and most probable speed of molecules of an ideal gas, which of the following is the correct relation between v –, vr.m.s. and v0? A v0< v – < vr.m.s. B v0< vr.m.s. < v – C v – < v0 < vr.m.s. D v – < vr.m.s. < v0 3. The most probable speed of oxygen molecules in a container is 400 m s–1. Which of the following is the most likely value for the mean speed? A 200 m s–1 B 380 m s–1 C 400 m s–1 D 480 m s–1 4. The graph below shows the distribution of molecular speeds for two temperatures, T1 and T2. Which of the following statements is correct? A The temperature T1 is higher than the temperature T2. B v1 and v2 are the mean speed at T1 and T2 respectively. C Area under the both graphs are the same. D Area to the left of v1 is less than the area to the left of v2. 5. (a) Sketch a graph to show the distribution of molecular speeds of a gas in a closed container. Mark on your graph, the most probable speed v0. On the same axes, sketch the distribution at a higher temperature. (b) To take part in a chemical reaction, molecules in a gas must have energy greater than a minimum value of E0 known as the activation energy. (i) If the mass of a gas molecule is m, write an expression for vm, the minimum speed of the molecule for chemical reaction to occur. (ii) If vm >> v0, the most probable speed, use your graph in (a) to explain whether the rate of chemical reaction increases or decreases when temperature increases. 9.5 Degree of Freedom and Law of Equipartition of Energy Students should be able to: • define the degrees of freedom of a gas molecule • identify the number of degrees of freedom of a monatomic, diatomic or polyatomic molecule at room temperature • explain the variation in the number of degrees of freedom of a diatomic molecule ranging from very low to very high temperatures • state and apply the law of equipartition of energy Learning Outcomes 1. Figure 9.8 shows that the velocity v of a gas molecule may be resolved into three mutually perpendicular components, vx, vy and vz along the x, y and z-axis respectively. Figure 9.8 2013/P1/Q12, 2014/P1/Q11, 2016/P1/Q11 VIDEO Degree of Freedom

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 227 2. The velocity components vx , vy and vz are independent from each other, that is, vx has no component along the y or z-axis and so on. 3. The gas molecule is said to have three degrees of freedom. 4. A degree of freedom is an independent mode of motion or an independent mode of acquiring kinetic energy. 5. The number of degrees of freedom of a molecule depends on whether the molecule is monatomic, diatomic or polyatomic. 6. Monatomic gases such as He, Ne, Ar and vapour of metals such as Na and Hg have three degrees of freedom which consists of linear motion along the x, y and z-axis. (a) (b) Figure 9.9 7. For diatomic gas (Figure 9.9(a)) such as H2, O2, Cl2, N2 and CO besides translational motion, the molecule can rotate about the Ox-axis, or Oz-axis. Hence, diatomic molecules have five degrees of freedom, three translational motion along Ox, Oy and Oz axis, and two rotational motion. 8. Rotational kinetic energy about Oy-axis (Figure 9.9(a)) is negligible because the moment of inertia of the gas molecule about the Oy-axis is negligible. 9. Gas molecules with more than two atoms are considered as polyatomic. Examples are H2O, CO2, O3, NH3, N2O4 and H2S as shown in Figure 9.9(b). 10. Polyatomic molecules have six or more degrees of freedom. Three degrees of freedom corresponding to the translational motion, three degrees of freedom for rotational motion about Ox, Oy and Oz-axis. At higher temperature, vibrational motion increases the number of degree of freedom. 11. The number of degrees of freedom for various types of molecule discussed applies for gases at normal temperature, for example, around room temperature. 12. For lower temperature such as 50 K, the rotational kinetic energy and vibrational kinetic energy are negligible for diatomic and polyatomic molecules. Hence, there are only three degrees of freedom – translational motion only. 13. For higher temperature, above 750 K, the number of degrees of freedom increases. Diatomic molecule has seven degrees of freedom, two more than at normal temperature. The vibrational kinetic energy is now significant. Exam Tips Degrees of freedom Monatomic gas : 3 translational Diatomic gas : 5 , 3 translational and 2 rotational Polyatomic gas : 6 , 3 translational and 3 rotational

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 228 Law of Equipartition of Energy 1. This principle states that the mean kinetic energy associated with each degree of freedom of a molecule is 1 2 kT, where k = Boltzmann’s constant T = thermodynamic temperature 2. For a molecule which has f degrees of freedom, mean kinetic energy = f 2 kT Example 12 Assuming that thermal neutrons behave like monatomic gas, estimate the r.m.s. speed of thermal neutrons at 300 K. (Mass of neutron = 1.7 × 10–27 kg, Boltzmann’s constant k = 1.4 × 10–23) Solution: For monatomic gas, number of degrees of freedom, f = 3. Hence kinetic energy, 1 2 m <c2 > = 3 2 kT cr.m.s = <c 2 > = 3kT m = 3 × (1.4 × 10–23)300 1.7 × 10–27 = 2 700 m s–1 Quick Check 5 1. The effective number of degrees of freedom for N2 gas at room temperature is A 3 C 6 B 5 D 7 2. The mean kinetic energy of CO2 gas molecules at temperature T is A 1 2 kT C 5 2 kT B 2 3 kT D 3kT 3. The mean kinetic energy of molecules of an ideal diatomic gas at temperature T is E. What is the mean kinetic energy of molecules of an ideal monatomic gas at temperature 2T? A 0.5E B 0.6E C 1.2E D 2E 4. Which is the correct statement of the law of equipartition of energy for gas molecules? A Mean kinetic energy is 1 2 kT. B Kinetic energy for each degree of freedom is 1 2 kT. C All gas molecules have the same number of degree of freedom. D Total translation kinetic energy is 1 2 kT.

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 229 9.6 Internal Energy of an Ideal Gas Students should be able to: • distinguish between an ideal gas and a real gas • explain the concept of internal energy of an ideal gas • derive and use the relationship between the internal energy and the number of degrees of freedom Learning Outcomes 1. The internal energy of a system is the sum of the kinetic energy and potential energy of the molecules in the system. The kinetic energy is due to the motion of the molecules and the potential energy is due to the forces between molecules. 2. In real gases, there are forces between molecules, especially when the pressure of the gas is high. Hence the internal energy of real gases consists of both kinetic energy and potential energy. 3. In an ideal gas, there is no forces between gas molecules. Hence the gas molecules has no potential energy. Hence the internal energy of an ideal gas is the sum of the kinetic energy of the gas molecules. 4. Internal energy of a mole of ideal gas is the sum of the kinetic energy of NA (Avogadro’s number) molecules. According to the law of equipartition of energy, at temperature T kinetic energy of 1 molecule = f 2kT (f is the degrees of freedom) Internal energy of 1 mole, U = NA( f 2kT) = f 2 (NAkT) (NA k = R, molar gas constant) = f 2RT 5. Internal energy of n moles of ideal gas, U = n ( f 2RT) 6. The table below summarises the kinetic energy of a gas molecule, and the internal energy U of a mole of gas at normal temperature. Table 9.1 Type of gas Monatomic Diatomic Polyatomic Number of degrees of freedom, f 3 5 6 Kinetic energy of one molecule = f 2 kT 3 2 kT 5 2 kT 3kT Internal energy of one mole, U = f 2 RT 3 2 RT 5 2 RT 3RT 2010/P1/Q19, 2017/P1/Q11, Q20

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 230 Example 13 (a) A vessel contains N molecules of an ideal monatomic gas at pressure p and temperature T. Write an expression for the total kinetic energy of the molecule in terms of N, T and k, Boltzmann’s constant. (b) An additional N molecules of the same gas is introduced into the vessel in such a way that the total kinetic energy remains unchanged. What happens to (i) the temperature? (ii) the pressure of the gas? Solution: (a) For monatomic gas, f = 3 Mean kinetic energy of gas molecule = 3 2 kT Total kinetic energy of N molecules = 3 2 NkT (b) (i) With an additional N molecules, total number of molecules = 2N. If T1 = new temperature, since total kinetic energy remains unchanged, (2N) ( 3 2 kT1) = 3 2 NkT T1 = T 2 (ii) At temperature T, <c2 > ∝ T At temperature T1 = T 2 , <c1 2 > ∝ T 2 <c1 2 > = 1 2 <c2 > Using p = 1 3 ρ <c2 > at temperature T At temperature, T1 = T 2 p1 = 1 3 (2ρ) <c1 2 > = 1 3 (2ρ) ( <c2 > 2 ) = 1 3 ρ <c2 > = p That is the pressure remains unchanged. (ii) Alternative method. At temperature T, pV = ( N NA) RT At temperature, T1 = T 2 , p1V = ( 2N NA ) R ( T 2 ) = ( N NA) RT = pV Hence, p1 = p

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 231 Example 14 Calculate the ratio of the total translational kinetic energy of molecules in 1.0 mole of gas at 17°C to the average kinetic energy of a sprinter of mass 70 kg running 100 m dash in 10 s. Solution: For translational motion, the number of degrees of freedom, f = 3. The total translational kinetic energy of molecules in 1.0 mole of gas, U = 3 2 RT. Average kinetic energy of sprinter, E = 1 2 mv2 = 1 2 m ( d t ) 2 Hence, U E = 3 2 RT 1 2 m ( d t ) 2 T = (273 + 17)K = 290 K = 3 × (8.31)(290) 70 × (10)2 d t = 100 10 = 10 m s–1 = 1.0 Quick Check 6 1. The internal energy of a fixed mass of ideal gas depends on A its pressure B its temperature C its volume D its pressure and volume 2. The pressure of an ideal gas in a closed container is p, and the mean kinetic energy of the molecules is E. If the mean kinetic energy increases to 9E, what is the pressure of the gas? A 3p C 27p B 9p D 81p 3. The r.m.s. speed of the molecules of an ideal gas at pressure p and volume V is c. The gas is heated until its volume is 3 2 V and pressure 6p. What is the r.m.s. speed of the molecules? A 3c C 6c B 4c D 9c 4. A cylinder X of volume V contains N molecules of an ideal gas at a pressure of 400 Pa and temperature of 200 K. Another cylinder Y is of volume 2V and contains 4N molecules of the same ideal gas at a temperature of 300 K. What is the pressure of the gas in cylinder Q? A 200 Pa C 900 Pa B 600 Pa D 1200 Pa 5. At room temperature, the r.m.s. speeds of oxygen molecules, and carbon dioxide molecules in the air around us are co and cc respectively. What is the value of the ratio co/cc? [Relative atomic mass of oxygen = 16, carbon = 12] A 0.85 C 1.32 B 1.17 D 1.66 6. What is the r.m.s. speed of oxygen molecules at 23°C? [Mass of 1 mole of oxygen = 32.0 g] A 134 m s–1 B 173 m s–1 C 483 m s–1 D 624 m s–1 7. Starting from the kinetic theory equation p = 1 3 ρ <c2> and the ideal gas equation pVm = RT, derive an expression for the internal energy of one mole of ideal monatomic gas in terms of R, the molar gas constant and T.

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 232 STPM PRACTICE 9 1. Oxygen molecules at temperature T have the same r.m.s. speed as helium molecules at 300 K. What is the temperature T? A 106 K B 600 K C 850 K D 2400 K 2. The volume of an ideal gas is V when its pressure is p. The r.m.s. speed of the gas molecules is proportional to A pV B pV C p V D p V 3. The pressure of an ideal gas in a vessel is 1.5 × 105 Pa when the temperature is 350 K. What is the number of gas molecules per unit volume? A 1.7 × 1020 m–3 B 1.4 × 1021 m–3 C 1.2 × 1022 m–3 D 3.1 × 1025 m–3 4. The pressure of an ideal gas in a vessel is p. The gas is heated at constant volume until the r.m.s. speed of the gas molecules is doubled. What is the new pressure of the gas? A 2p C 3p B 2p D 4p 5. Which of the following is not an assumption of the kinetic theory of gases? A Gas molecules move with the same speed, known as the r.m.s. speed B Gas molecules are in constant free random motion C Collisions between gas molecules are perfectly elastic D The size of a gas molecule is negligible compared to the distance between gas molecules 6. Which statement about the internal energy U of a fixed mass of an ideal gas of pressure p, volume V and temperature T is correct? A U ∝ p C U ∝ pV B U ∝ V D U ∝ T 7. The molecular speed distribution of a gas is shown in the figure below. Number of molecules with speed v Speed v vfi nʹ 0 Which of the statements is correct? A The mean speed is v’ B The r.m.s. speed is greater than v’ C The maximum number of molecules is n’ D More than half of the number of molecules have speeds less than v’ Important Formulae 1. Ideal gas equation: pVm = RT pV = nRT = ( N NA )RT 2. Pressure, P = 1 3 pc2 = 1 3 (nm)c2 3. Translational K.E. of a gas molecule, 1 2 mc2 = 3 2 kT 4. For each degree of freedom, kinetic energy of a molecule = 1 2 kT 5. Internal energy of n moles, U = f 2 nRT

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 233 13. The density of water is 1 × 103 kg m–3 and its relative molecular mass is 18. Assuming Avogadro number as 6 × 1023 mol–1, estimate the volume of one molecule of water. 14. The pressure p of an ideal gas is related to its temperature T by the equation p = nkT (a) Identify the quantities n and k in the equation. (b) Write another equation from the kinetic theory of gases relating the pressure p, n and the mean square speed <c 2 > of the gas molecules, and m the mass of one molecule. (c) A plasma contains both protons and electrons in thermal equilibrium. Given that protons and electrons behave as molecules of real gases. If the r.m.s. speed electrons is 3 × 106 m s–1, calculate (i) r.m.s. speed of protons, (ii) the temperature of the plasma. 15. (a) What is meant by an ideal gas? (b) State two conditions under which a real gas can be approximated to an ideal gas. (c) A container of fixed volume 2.5 × 10–2 m3 containing 0.20 mol of an ideal gas is heated to 600.0 °C. Calculate the pressure of the ideal gas. 16. (a) (i) State two assumptions of an ideal gas. (ii) Under what conditions does a real gas behaves close to an ideal gas. (b) A cylinder of volume 0.75 m3 contains 5.6 kg hydrogen gas at a temperature of 27o C. If the gas behaves as an ideal gas, what is the pressure of the gas? (Molecular mass of hydrogen is 0.002 kg mol-1) Discuss whether the actual pressure of the gas is greater or less than the calculated value. (c) Hydrogen gas has five degrees of freedom. (i) Explain what is meant by degrees of freedom. Name the degrees of freedom of the hydrogen gas. 8. A r.m.s. speed of air molecule is 485 m s–1 at standard temperature and pressure. This mean that at standard temperature and pressure, A most of the air molecules move at a speed of 485 m s–1 B the mean speed of air molecules is less than 485 m s–1 C all the air molecules move at a speed of 485 m s–1 D the mean speed of air molecules is 485 m s–1 9. Which statement about the degrees of freedom of a gas molecule at room temperature is correct? A Monatomic gas molecule has one degree of freedom for translational motion B Monatomic gas molecule has two degrees of freedom for rotational motion C Diatomic gas molecule has three degrees of freedom for translational motion D Diatomic gas molecule has five degrees of freedom for rotational motion 10. At a temperature T, the average kinetic energy of molecules of a monatomic gas is E. What is the average kinetic energy of molecules of a diatomic gas at temperature of 3T? A 2E B 4E C 5E D 6E 11. (a) A vessel contains N molecules of a diatomic gas at a temperature T. Use the law of equipartition of energy to derive the expression for the kinetic energy K of the N molecules in the gas. (b) A closed vessel contains 2.5 moles of oxygen gas at a temperature of 300 K. The temperature of the gas is then raised to 500 K. Find the increase in the internal energy of the gas. 12. The density of water and steam are 1.0 × 103 kg m–3 and 6.1 × 10–1 kg m–3 respectively. Calculate the ratio of mean separation between steam molecules mean separation between water molecules .

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 234 1 1. B pV = ( N NA )RT hence N ∝ p 2. A pV = nRT hence pV ∝ T 3. C Low pressure, gas molecules far apart 4. A pV = nRT Gradient ∝ n = m M 5. B pV = ( m M )RT = constant, m ∝ 1 T 6. C pV = nRT and (1.02p)V’ = (1.05n)R(1.02T) 7. A p2(3V + V) = (100)3V + (300)V 8. D pV = ( N NA )RT then n = N V ∝ p T 2 1. B 2. A 3. B 4. D (10u)2 + (9u)2 + (5u)2 + (6u)2 4 5. C V = constant, T ∝ p, vrms ∝ T 6. (a) 2.41 × 1025 pV = ( N NA ) RT (b) 1.12 kg n = N NA , mass = nM (c) 11.2 kg m–3 ρ = m V (d) 518 m s–1 p = 1 3 ρ < c 2 > 3 1. D vrms ∝ T 2. A pV = ( N NA )RT = N( R NA )T= NkT E = 3 2 kT, Gradient = a b = pV = NkT, N = ( a b )( 3 2E ) = 3a 2bE 3. C p = 1 3 ρc2 and T ∝ c2 ∝ p ρ 4. C 1 2 m0c0 2 = 3 2 kT = 1 2 mHec2 He 5. A V = constant, T ∝ p and c ∝ T ∝ p 6. (a) and (b) Refer to page 219 & 220. 7. (a) • Gas molecules collide elastically. • Transfer of kinetic energy. • Kinetic energy of molecules is different after collision. (b) • Sound energy transferred by air molecules. • Collisions between air molecules. 8. NA( 1 2 m <c2 >) = 3RT 2 3.80 × 103 J 9. (a) p ∝ 1 V if T constant V T = constant if p constant p ∝ T, if V constant (b) (i) 0.099 kg pV = ( m M ) RT (ii) 1.86 × 1024 n = m M , N = nNA (iii) 467 m s–1 p = 1 3 ρ <c2 > (c) (i) Very low pressure, p → 0 (ii) 0 Real gas(eg. H2 ) p Ideal gas pV (ii) The temperature at a height of 1000 km from the Earth’s surface is 1500 K. What is the mean r.m.s. speed of hydrogen molecules at 1500 K? (iii) The escape speed from the Earth at a height of 1000 km is 10.0 km s-1. Discuss the possibility of finding hydrogen gas at a height of 1000 km above the Earth’s surface. ANSWERS

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 235 10. (a) n = number of molecules per unit volume m = mass of 1 molecule <c2 > = mean square speed (b) c = c2 = 3RT M (Refer to Page 219) (c) 1.6 × 103 m s–1 Use c = 3RT M 11. (a) Velocity of molecules > escape velocity. (b) Translational kinetic energy of small molecule = translational kinetic energy of massive molecule at the same temperature For m small, v is large, > escape velocity. For m big, v is small, < escape velocity. (c) Random motion of gas molecules. If velocity of molecule > escape velocity, molecule escapes. (d) Water (H2O) molecule are massive. Velocity < escape velocity. 12. (a) • Molecules collide with walls of container. • Change of momentum after collision. • Force on wall = rate of change of momentum. • Pressure = force/area. (b) (i) 1 mole, pV = n1RT1 = n2RT2 (ii) 684 m s–1 <c2 > = 3RT M 13. (a) At any instant, number of molecules having a certain speed is the same for all directions. (b) (i) Refer to page 211. (ii) c2 = 3RT M Refer to page 219 4 1. B Most of CO2 (heavier) molecules have lower v. 2. A Refer to page 224. 3. D vmean > v0 most probable speed 4. C Refer to page 224. 5. (a) See graph in (b)(ii) below. (b) (i) 1 2 mvm 2 = E0 vm = 2E0 m (ii) 0 n (v ) A2 A1 T2 T1 vm v For T1, A1 represents number of molecules with v > vm (take part in chemical reaction). For T2, A2 > A1. Hence, number of molecules taking part in chemical reaction increases. Therefore, rate of chemical reaction increases. 5 1. B : Refer to page 227. 2. D : Mean K.E. = f 2 kT, f = 6 (CO2 is polyatomic) 3. C : E ∝ T 4. B : Refer to page 227. 6 1. B : Internal energy U ∝ T 2. B : V = constant, T∝ p and E ∝ T ∝ p 3. A : pV = nRT ∝ T and c ∝ T 4. D : pV = ( N NA )RT X : (400)V= ( N NA )R(200) Y : p2(2V)= ( 4N NA )R(300) p2 = 1200 Pa 5. B : crms = 3RT M , Mo = 32, Mc = (12 + 32) = 44 co cc = 44 32 = 1.17 6. C : crms = 3RT M = 3(8.31)(273 + 23) 0.0320 m s–1 = 480 m s–1 7. C : U = 3 2 RT Refer to page 219 STPM Practice 9 1. B : c = 3RT M = 3RT 0.032 = 3R(300) 0.004 T = 2400 K 2. B : pV = nRT R.m.s. speed ∝ T ∝ pV 3. A : pV = ( N NA)RT, N V = (6.02 × 1023)(1.5 × 105 ) (8.31)(350) m-3 = 3.1 × 1025 m-3 4. D: p = 1 3 ρc2 , p’ = 1 3 ρ(2c)2 = 4p

9 Physics Term 1 STPM Chapter 9 Kinetic Theory of Gases 236 5. A : Refer to page 216 6. C : U ∝ T, pV = nRT, hence U ∝ pV 7. B : r.m.s. speed > most probable speed v’ 8. B : vmean < vrms 9. C : Refer to page 237 10. C : Monatomic, E = 3 2 kT Diatomic, E2 = 5 2 k(3T) = (5)(3) 3 E = 5E 11. (a) Diatomic gas, f = 5 Kinetic energy of a molecule = 5 2 kT Total kinetic energy of N molecules K = 5 2 NkT (b) U = 5 2 nRT ∆U = 5 2 (2.5)R(500 – 300) J = 10.4 kJ 12. 12 M = ρV = ρ (Nd3 ) 13. 3 × 10–29 m3 14. (a) n : number of molecules per unit volume k : Boltzmann’s constant pVm = RT = (NAk)T p = ( NA Vm )kT = nkT (b) p = 1 3 nm <c2 > (c) (i) 7.02 × 104 m s–1 1 2 m1 <c1 2 > = 1 2 m2 <c2 2 > (ii) T = 2 × 105 K 1 2 m <c2 > = 3 2 kT 15. (a) Ideal gas: A gas that obeys the ideal gas equation for all pressures. (b) Conditions: 1) Very low pressure (2) Low temperature (c) pV = nRT p = (0.20)(8.31)(600 + 273) 2.5 × 10–2 Pa = 5.80 × 104 Pa 16. (a) (i) Assumptions: – No force between gas molecules, except during collisions. – Collisions between gas molecules, and gas molecules with the walls of the container are perfectly elastic. – Volume of gas molecules negligible compared with volume of the gas. – Internal energy of the gas consists of the kinetic energy of the gas molecules. (Any two) (ii) Conditions: – Low pressure (p → 0) – Low temperature. (b) pV = ( m M )RT p = (5.6)(8.31)(273 + 27) (0.0020)(0.75) Pa = 9.31 × 106 Pa Actual pressure is greater than 9.31 × 106 Pa. Reason: High pressure – hydrogen gas does not behave like an ideal gas. (c) (i) Degree of freedom: independent mode of motion or independent mode of acquiring kinetic energy. Three degrees of motion (f = 3) for translational motion, and f = 2 for rotational motion. (ii) Translational kinetic energy = 1 2 Mc2 = 3 2 RT r.m.s. speed, c = 3RT M = 3(8.31)(1500) 0.0020 m s–1 = 4.3 km s–1 (iii) Since r.m.s. speed of hydrogen molecules, 4.3 km s–1 < 10.0 km s–1, the escape speed, there is strong possibility of finding hydrogen molecules at a height of 1000 km.

237 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases CHAPTER 237 First Law of Thermodynamics Q = ∆U + W Work, W 0 V p W = ∫ = area, A V1 V2 V2 V1 pdV Isothermal and Adiabatic Changes Isothermal changes ∆T = 0 and pV = constant 0 Isotherms Adiabatic V p T1 T2 Adiabatic changes Q = 0 pVγ = constant TVγ–1 = constant Thermodynamics of Gases Heat Capacities CV, m = dU dT Cp, m – CV, m = R γ = Cp, m CV, m Concept Map THERMODYNAMICS OF 10 GASES Bilingual Keywords Adiabatic change: Perubahan adiabatik Diatomic: Dwiatom Internal energy: Tenaga dalam Isothermal change: Perubahan isoterma Reversible change: Perubahan berbalik Specific heat capacity: Muatan haba tentu

238 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases INTRODUCTION 1. In thermodynamics of gas, you study about the changes to the internal energy of a gas and factors affecting the internal energy. 2. Thermodynamics also involves work done and heat supplied or lost by a gas. 10.1 Heat Capacities Students should be able to: • define heat capacity, specific heat capacity and molar heat capacity • use the equations: Q = C∆θ , Q = mc∆θ, Q = nCV,m∆θ and Q = nCp.m∆θ Learning Outcome 1. The specific heat capacity, c of a material is the heat required to raise the temperature of a unit mass of the material by one kelvin (1 K). 2. Example of specific heat capacity, c of some materials: Water = 4 180 J kg–1 K–1, Copper = 400 J kg–1 K–1 3. The heat ∆Q absorbed by a material of mass m when its temperature increases by ∆T is ∆Q = mc ∆T When the temperature of the material decreases by ∆T, the heat lost is ∆Q = mc ∆T 4. The heat capacity C of a body is the heat required to raise the temperature of the body by 1 K. Heat capacity, C = mc where m = mass of body, c = specific heat capacity. 5. The molar heat capacity of a material, Cm is the heat required to raise the temperature of one mole of the material by 1 K. (Unit: J mol–1 K–1). The heat gained by n moles of a material when its temperature increases by ∆T is ∆Q = n Cm ∆T 6. The heat required to raise the temperature of a gas depends on whether the gas is allowed to expand when it is heated. 7. A gas has two heat capacities, one at constant volume, and another at constant pressure. 8. The molar heat capacity of a gas at constant volume CV,m is the heat required to raise the temperature of one mole of the gas by 1 K at constant volume. 9. The molar heat capacity of a gas at constant pressure Cp,m is the heat required to raise the temperature of one mole of the gas by 1 K at constant pressure. 10. The heat supplied to increase the temperature of n moles of gas by ∆T (a) at constant volume is ∆Q = n CV,m ∆T (b) at constant pressure is ∆Q = n Cp,m ∆T 2009/P1/Q21, 2010/P1/Q21, 2013/P1/Q13

239 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 11. Cp,m > CV,m When the temperature of one mole of gas is raised by 1 K at constant volume, there is no increase in volume, ∆V = 0. Hence, no external work is done by the gas. • The heat supplied is used only to increase the internal energy of the gas. • When the temperature of one mole of gas is raised by 1 K at constant pressure, the volume of the gas increases. • External work is done by the gas during the expansion. • The increase in internal energy in both cases, at constant volume and at constant pressure are the same because the increases in temperature are the same. • Hence, when gas is heated at constant pressure, the heat supplied is used to increase the internal energy and for the gas to do external work. Hence, Cp,m > CV,m Example 1 The molar heat capacity of hydrogen gas at constant volume is 20.5 J mol-1 K-1. Calculate (a) the specific heat capacity of hydrogen at constant volume, (b) the heat capacity of 0.50 kg of hydrogen at constant volume, (c) the heat required to raise the temperature of 0.50 kg of hydrogen from 30o C to 45o C at constant volume. (Mass of 1 mole of hydrogen = 2.0 g) Solution: (a) Molar heat capacity at constant volume, CV,m = 20.5 J mol–1 K–1 Mass of 1 mole of hydrogen, M = 2.0 g = 2.0 × 10–3 kg Specific heat capacity at constant volume, cV = CV,m M = 20.5 2.0 × 10–3 J kg–1 K–1 = 1.03 × 104 J kg–1 K–1 (b) Heat capacity of 0.50 kg at constant volume, CV = mcV = (0.50)( 1.03 × 104 ) J K–1 = 5.15 × 103 J K–1 (c) Heat required to raise the temperature of 0.50 kg hydrogen from 30o C to 45o C at constant volume. Q = CV(∆θ) = (5.15 × 103 )(45 – 30) J = 7.73 × 104 J Example 2 Explain why (a) there is a significant difference between the specific heat capacity of a gas at constant volume, and at constant pressure. (b) the difference is not significant for solid or liquid.

240 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Solution: (a) For a gas, the specific heat capacity at constant pressure, cp > cV, the specific heat capacity of constant volume. When 1 kg gas is heated at constant volume, the gas does not expand, (∆V = 0). No work is done by the gas. The heat supplied cv to raise the temperature of 1 kg of gas by 1 K is only used to increase the internal energy of the gas cv = ∆U .............................. a When 1 kg of the gas is heated at constant pressure, the gas expands. Work is done by the gas (W = p∆V). The heat supplied to 1 kg of gas cp is used to increase the internal energy of the gas, ∆U and for the gas to do work. cp = ∆U + p∆V .............................. b Comparing a and b, cp > cV .............................. c (b) For solid or liquid, the change of volume when the temperature increases is negligible. Work done by solid or liquid, p∆V is negligible (∆V = 0). Hence, cp = cV. Quick Check 1 1. The molar heat capacity of an ideal gas at constant pressure is greater than at constant volume because A the molecules acquire greater kinetic energy at constant pressure. B work is done against intermolecular forces as the gas expands. C the potential energy of the molecules increase when they are further apart. D work is done against external pressure when the gas expands. 2. The heat required to produce a certain increase in temperature for a fixed mass of gas depends on whether the gas is heated at constant volume or at constant pressure. Discuss whether this is true for a liquid. 3. A mass of helium in a closed vessel is 0.08 kg. The gas requires 7500 J of heat to raise its temperature from 320 K to 350 K. (a) What is the heat capacity of the gas at constant volume? (b) What is the specific heat capacity of the gas at constant volume? (c) What is the molar heat capacity of the gas at constant volume? (mass of 1 mole of helium = 4.0 g) 10.2 Work Done by a Gas Students should be able to: • derive and use the equation for work done by a gas W = ∫p dV Learning Outcome 1. Work is done by a gas when the gas expands. Conversely, work is done on a gas when the gas is compressed. 2. Figure 10.1 shows a gas at pressure p in a cylinder of cross-sectional area A fitted with a light frictionless piston. The force on the piston, F = pA Area, A Piston Gas dx p F Figure 10.1 2011/P1/Q19, 2015/P1/Q12, 2017/P1/Q12

241 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 3. When the gas expands by pushing the piston through a small distance dx, the work done by the gas is dW = F dx = pA dx A dx = dV = p dV Hence, when the volume of gas increase from V1 to V2, the work done by the gas is W = ∫ V V 1 2 p dV = Area under the graph of p against V (Figure 10.2). 4. When a gas is compressed from volume V1 ′ to V2 ′. Work done on the gas, W = ∫ V V 1 2′ ′ p dV Since V2′ < V1′, the value of work done is negative. The work done on the gas = shaded area under p-V curve in Figure 10.3 5. When a gas expands at constant pressure, p as shown in Figure 10.4, work done by the gas, W = ∫ V V 1 2 p dV = p ∫ V V 1 2 dV = p (V2 – V1) = p ∆V = Pressure × Increase in volume Example 3 When water changes into steam at 100°C and pressure at 1.01 × 105 Pa, 1.0 cm3 water changes to 1 600 cm3 of steam. (a) What is the work done against atmospheric pressure when 1 kg of water changes into steam? (b) Explain the difference in the value of the specific latent heat of evaporation, 2.26 × 106 J kg–1 and the work done during vaporization of 1 kg of water. Solution: (a) Volume of 1 kg of water = 1 000 cm3 Volume of 1 kg of steam = (1 600 × 1 000) cm3 Since the change in volume occurs at constant pressure, work done, W = p ∆V = (1.01 × 105 ) (1.6 × 106 – 1.0 × 103 ) × 10–6 = 1.615 × 105 J (b) Specific latent heat of vaporization, L = 2.26 × 106 J kg–1 Work done against atmospheric pressure = 1.615 × 105 J kg–1 Difference of energy = (2.26 × 106 – 1.615 × 105 ) J kg–1 = 2.099 × 106 J kg–1 The difference of energy is used to break completely the bonds between water molecules. p V1 V V2 0 Area = Work done by gas Figure 10.2 Expansion p V2' V1' 0 V Area = Work done on gas Figure 10.3 Compression Figure 10.4

242 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases Example 4 A cylinder of uniform cross-sectional area 0.10 m2 is fitted with a smooth piston of mass 1.0 kg. The cylinder contains 6.5 × 10–2 m3 air at a pressure of 1.0 × 105 Pa. A bullet of mass 32 g travelling at a speed of 240 m s–1 hits the piston and embeds in it after collision. This causes the piston to be displaced through a small displacement. Assuming that there is no loss of energy after collision, calculate the maximum displacement of the piston. State any other assumptions you made. Solution: M u m M v m Before collision After collision Using the principle of conservation of momentum, velocity of bullet and piston after collision, v is given by mu = (m + M)v v = mu (m + M) Total kinetic energy of piston and bullet = 1 2 (m + M)v 2 = 1 2 m2 u2 (m + M) By the principle of conservation of energy, Work done on gas = Loss in kinetic energy of piston and bullet Fx = 1 2 m2 u2 (m + M) F = pA x = m2 u2 2(pA) (m + M) = (32 × 10–3)2 (240)2 2(1.0 × 105 ) (0.10) (0.032 + 1.0) = 2.86 × 10–3 m Assumption: • Since the displacement of the piston is small, the change in volume is small and the pressure may be assumed to be constant at (1.0 × 105 ) Pa • Work done on gas = p ∆V since p = constant

243 10 Physics Term 1 STPM Chapter 10 Thermodynamics of Gases 1. A fixed mass of ideal gas undergoes the changes of pressure and volume W → X → Y → Z → W as shown in the p-V graph. p/105 Pa V/10–4 m3 1.0 2.0 4.0 3.0 W X Z Y What is the nett work done by the gas in a complete cycle? A 20 J C 60 J B 40 J D 80 J 2. W Z Y X A fixed mass of gas undergoes a cycle of changes WXYZW as shown. The shaded area represents A work done by gas during WXY – work done on gas during YZW. B work done on gas during YZW – work done by gas during WXY. C work done by gas during ZW – work done on gas during XY. D work done on gas during ZW – work done by gas during XY. 3. An ideal gas in a cylinder is fixed with a frictionless piston. The piston is moved from volume V1 at pressure p1 to volume V2 at pressure p2 (V2 < V1). The temperature remains constant throughout the process. The work done by the gas is A zero, because the gas obeys Boyle’s law, hence p2V2 – p1V1 = 0. B negative, because the volume decrease. C positive, because the piston has been moved. D negative, because the pressure p2 < p1. 4. When mass m of water changes into vapour at its boiling point, the volume increased by ∆V. The atmospheric pressure is p. If the specific latent heat of vaporisation of water is l, then the change in the internal energy is A ml B p∆V C ml – p∆V D ml + p∆V 5. When three moles of an ideal gas are compressed at constant pressure, its temperature increases from 300 K to 380 K. What is the work done on the gas? A 1.00 kJ B 1.50 kJ C 1.99 kJ D 2.99 kJ 6. At 100°C and pressure 1.01 × 105 Pa, 1.00 kg steam occupies a volume of 1.67 m3 . The volume of 1.00 kg water is 1.04 × 10–3 m3 . The specific latent heat of vaporisation of water at 100°C is 2.26 × 106 J kg–1. When 1.00 kg water changes into steam at 100°C and at pressure of 1.01 × 105 Pa, calculate (a) heat supplied, (b) work done by the system, (c) increase in internal energy of the system. 7. An ideal gas expands from 50 cm3 to 100 cm3 at constant pressure 1 000 N m–2. The temperature of the gas before expansion is 27°C, what is its temperature after expansion? Calculate the work done during the expansion. 8. The relative molecular mass of ethanol is 46.0 and its density is 790 kg m–3. Ethanol boils at 350 K at atmospheric pressure of 1.01 × 105 N m–2. For 1.00 kg of ethanol, calculate (a) the number of molecules, (b) work done by the system when ethanol changes to vapour at 350 K, (c) increase in internal energy per molecule of ethanol. (Specific latent heat of vaporisation of ethanol = 850 kJ kg–1). Quick Check 2