Kinetic Theory (Essay)

1. close particles distant particles weak attraction strongattraction Kinetic Theory (Essay) • All matter is made up of very tiny particles, which are constantly in motion. • The molecules repel other strongly when they are forced too close together. • When they are slightly apart, they attract each other. • When they are far apart, they hardly attract or repel each other at all.

2. Interpreting temperature by kinetic theory • The motion of the particles in a body depends on its temperature. • When the temperature rises, particles vibrate more rapidly or move faster. Hence, the temperature of the body is a measure of the average kinetic energy of the particles. • Two bodies have the same temperature if particles in each body have the same average kinetic energy.

3. Particles move around freely at high speed container A gas model (Essay) • Gases have no fixed volume and shape. • They can quickly fill any space available. • In a gas, particles are very far apart. Therefore, the attraction among them is weak.

4. Particles move around freely at high speed container A gas model • The particles move at random at very high speeds (500 ms-1 at room temperature!). • They collide with each other and bombard the walls of the container. • When gas molecules hit the walls of container, they produce forces on the surface, which give rise to gas pressure. video

5. Gas pressure Pressure is defined as the perpendicular force exerted per unit area. Pressure is measured in N m-2 or pascal (Pa). 1 Pa = 1 N m-2. Normal atmospheric pressure or 1 standard pressure = 1.01 x 105 Pa

6. 10 cm 5 cm Example 1 Find the pressure exerted on the 5 kg block in each of the following cases. (a) Solution: P = F / A = 50 / (0.05 x 0.1) = 10000 Pa

7. 10 cm 5 cm 30o Example 1 Find the pressure exerted on the 5 kg block in each of the following cases. (b) Solution: P = F / A (F = normal reaction) = 50 cos30o / (0.05 x 0.1) = 8660 Pa

8. Example 2 0.2 kg • The figure shows a cylinder with a piston. If the cross section area of the cylinder is 20 cm2, and find the pressure of the gas. Given that the atmospheric pressure is 1.01 x 105 Pa. • Answer: 2.05 x 105 Pa

9. Measuring gas pressure • A Bourdon gauge is used to measure gas pressures.

10. Measuring gas pressure • A bourdon gauge works in the same way as a paper tube whistle. When a paper tube whistle is blown, it uncoils. • In a Bourdon gauge, the gas pressure makes the curved metal tube stretch out slightly. This causes the pointer to move round the scale. The higher the gas pressure, the more the pointer moves.

11. Pressure, volume and temperature • When a gas is heated, its pressure P, volume V and temperature T will change. We are going to investigate these relations.

12. Boyle’s law (P-V relationship) • In 1660, Robert Boyle investigated the relationship between the volume of a fixed amount of gas and its pressure, keeping the temperature of the gas constant. Video

13. Experiment • Press the air pump gently to change the pressure and the volume of the trapped gas in the glass tube without changing its temperature. • Measure the pressure and the volume of the gas. • Repeat the above step to get more sets of pressure and volume.

15. Results P∝ 1/V or PV = constant or P1V1 = P2V2 Boyle’s law: For a fixed mass of gas at a constant temperature, the product of its pressure and volume is constant.

16. Example 2 • The air pressure inside an inflated ball is 2 x 105 Pa. The ball is now squeezed to one-third of its original volume. Find the new pressure inside the ball. • Solution: • Let V be the original volume of the ball. • By P1V1 = P2V2 2 x 105V = P2(V/3) P2 = 6 x 105 Pa Thus, the new pressure is 6 x 105 Pa

17. Charles’ law (V – T relationship) • In this experiment, we keep the pressure of the gas constant and try to investigate the relation between its volume and temperature.

18. Gas column Experiment • Heat the water until its temperature is increased by 10oC. • Record the volume and temperature of the trapped gas. • Repeat the above step to get more sets of volume and temperature. • The mercury thread should be able to move up and down the capillary tube freely. • The friction between the mercury thread and the capillary tube should be small so that the pressure of the trapped gas is equal the atmospheric pressure.

19. Gas column Charles’ law (V – T relationship) • Heat the water until its temperature is increased by 10oC. • Record the volume and temperature of the trapped gas. • Repeat the above step to get more sets of volume and temperature. • The mercury thread should be able to move up and down the capillary tube freely. • The friction between the mercury thread and the capillary tube should be small so that the pressure of the trapped gas is equal the atmospheric pressure. simulation

20. Absolute zero • The straight line cuts the axis at -273oC which is known as the absolute zero of temperature. • Absolute zero is the theoretical lowest attainable temperature. • Theoretically, the gas at absolute zero occupies zero volume and exerts no pressure. However, in practice, it will liquefy before it can reach this temperature.

21. volume V Temperature T (K) 0 Kelvin scale • When the temperature is measured in Kelvin (K), the volume of the gas is directly proportional to its temperature shown in the figure below. • Note: Absolute temperature (K) = Celsius temperature + 273 i.e. melting point of ice: 0oC = 273 K room temperature: 27oC = (27 + 273) K = 300 K boiling point of water: 100oC = (100 + 273) K = 373 K

22. volume V or or Temperature T (K) 0 Charles’s law: For a fixed mass of gas at a constant pressure, its volume is directly proportional to its absolute temperature.

23. Example 2An inflated balloon contains 4 x 10-3 m3 of air at 27oC. It is put into a large tank of liquid nitrogen at -173 oC. What is the new volume of the balloon? • Solution: • By Charles’ law, V1/T1 = V2/T2 4 x 10-3 / (27 + 273) = V2 / (-173 + 273) V2 = 1.33 x 10-3 m3 Thus, the new volume of the balloon is 1.33 x 10-3 m3

24. Pressure law (P – T relationship) • Experiment • Heat the water unit its temperature increases by about 10oC. Stir the water thoroughly and wait for several minutes to allow the air in the flask to reach the temperature of the water. • Measure the temperature of the air by the thermometer and its pressure by the Bourdon gauge. • Repeat the above steps to get more sets of pressure and temperature.

25. or or pressure P 0 Temperature T (K) Results • The pressure of the gas increases linearly with its temperature. When the temperature is measured in Kelvin (K), the pressure is directly proportional to its temperature shown in the figure below. Pressure law: For a fixed mass of gas at a constant volume, its pressure is directly proportional to its absolute temperature.

26. Example 3When the flask used in the experiment is put into a bath of melting ice, the pressure is 8.7 x 104 Pa. What is the pressure when the flask is put into a bath of boiling water? Solution: • Temperature of melting ice = 0 oC = 273 K Temperature of boiling water = 100 oC = 373 K • By pressure law, P1/ T1 = P2 / T2 8.7 x 104 / 273 = P2 / 373 P2 = 1.19 x 105 Pa • Thus, the new pressure is 1.19 x 105 Pa

27. or General gas law • Boyle’s law: PV = constant if T is constant • Charles’ law: V/T = constant if P is constant • Pressure law: P/T = constant if V is constant • All the gas laws can be summarized by a general equation:

28. By Example 4A weather balloon contains 5 m3 of helium at the normal atmospheric pressure of 100 k Pa and at temperature of 27 C. What will be its volume when it rises to an altitude where the pressure is 80 k Pa and the temperature is 7 C? • Solution: (100 k)(5)/(27 + 273) = (80 k)(V2) / (7 + 273) V2 = 5.83 m3 Thus, its new volume is 5.83 m3

29. Equation of state and molar gas constant For 1 mole of gas (i.e. 6.02 x 1023 gas particles), (same for all gases) where R is called the universal molar gas constant. R = 8.31 J mole-1 K-1 For n moles of gas,

30. If m is the mass of a gas and M is the mass of 1 mole of gas, the number of mole • If N is the number of molecules in a gas and NA = Avogadro number = 6.02 x 1023, the number of mole From , we have , we have From Equation of state and molar gas constant .

31. R = 8.31 J K-1 mol-1

32. Example 5Show that the volume of 1 mole of gas occupied at s.t.p. is 22.4 litres.Note: s.t.p. stands for standard temperature and pressure. Standard temperature = 0oC or 273 K Standard pressure = 1 atmospheric pressure = 1.01 x 105 Pa • Solution • By PV = nRT • (1.01 x 105)V = (1)(8.31)(273) • V = 0.0224 m-3 or 22.4 litres

33. Ideal gases • An ideal gas is a gas that obeys Boyle’s law (PV is constant) for all pressures and temperatures. • However, real gases such as hydrogen, oxygen, nitrogen or carbon dioxide are not ideal gases but they behave like an ideal gas at high temperature and low pressure.

34. Example 6 • Two insulated gas cylinders A and B are connected by a tube of negligible volume as shown. • Each cylinder has an internal volume of 2 x 10-2 m3. Initially, the tap is closed and cylinder A contains 1.2 mol of an ideal gas at a temperature of 37oC. Cylinder B contains the same ideal gas at pressure 1.2 x 105 Pa and temperature 37 oC. (a) Calculate the amount, in mol, of the gas in cylinder B. Solution: Consider the gas in cylinder B By PV = nRT (1.2 x 105 )(2 x 10-2) = n(8.31)(37 + 273) n = 0.932 The amount of gas in cylinder B is 0.932 mol

35. Example 6 • Two insulated gas cylinders A and B are connected by a tube of negligible volume as shown. • Each cylinder has an internal volume of 2 x 10-2 m3. Initially, the tap is closed and cylinder A contains 1.2 mol of an ideal gas at a temperature of 37oC. Cylinder B contains the same ideal gas at pressure 1.2 x 105 Pa and temperature 37 oC. (a) Calculate the pressure of the gas in cylinder A. Solution: Consider the gas in cylinder A By PV = nRT P(2 x 10-2) = (1.2)(8.31)(37 + 273) P = 1.55 x 105 Pa The pressure of gas in cylinder A is 1.55 x 105 Pa

36. Example 6 • Two insulated gas cylinders A and B are connected by a tube of negligible volume as shown. • Each cylinder has an internal volume of 2 x 10-2 m3. Initially, the tap is closed and cylinder A contains 1.2 mol of an ideal gas at a temperature of 37oC. Cylinder B contains the same ideal gas at pressure 1.2 x 105 Pa and temperature 37 oC. (b) determine the final pressure of the gas in the cylinder. 1.2 mol 0.962 mol Solution By PV = nRT P(4 x 10-2) = (1.2 + 0.962)(8.31)(37 + 273) P = 1.37 x 105 Pa

37. Example 7There are two containers X and Y filled with the same type of ideal gas as shown. They are connected by a tube. A steady state is obtained with X held at 100 K and Y at 400 K. If the volume of X is half that of Y and the mass of gas in X is m, what is the mass of gas in Y, in terms of m? • Solution: Let P be the pressure of the gas and V be the volume of container X For the gas in container Y, P(2V) = nYR(400) --- (1) For the gas in container X, P(V) = nXR(100) --- (2) • (1)/(2): 2 = 4nY/nX nY = ½ nX Mass of gas in Y = ½ mass of gas in X = ½ m

38. Example 7 root mean squareFind (a) the mean (b) mean square, and (c) the root mean square of 1, 3, 5, 7 and 9. Solution: • Mean = (1 + 3 + 5 + 7 + 9) / 5 = 5 • Mean square = (12 + 32 + 52 + 72 + 92)/5 = (1 + 9 + 25 + 49 + 81) / 5 = 33 • Root mean square = [(12 + 32 + 52 + 72 + 92)/5] ½ = 5.74

39. Example 8 It is given that the density of hydrogen gas at s.t.p. is 0.09 kg m-3. Find the r.m.s. speed of hydrogen molecules. Solution: • By P = ⅓(rcr.m.s.2) 1.01 x 105 = ⅓(0.09cr.m.s.2) cr.m.s. = 1835 ms-1

40. Particles move around freely at high speed container Distribution of molecular speeds • A gas contains a large number of molecules in rapid motions. • In each collision between molecules, some molecules gain energy while others lose energy. • As a result, each molecule has a different speed. Simulation 1 Simulation 2

41. Maxwell distribution • Note: • Maxwell distribution is not symmetric. • When temperature increases, the distribution curve flattens out but the area under the curve remains unchanged.

42. Mean speed cm: mean speed of molecules Number of molecules with speed c co cm cr.m.s. c/m s-1 0 Most probable speed c0: speed possessed by the most number of molecules Maxwell distribution Root mean square speed cr.m.s.: It is found that c0 < cm <cr.m.s.

43. From and we have ⇒ ⇒ Temperature, molar mass and root mean square speed • For 1 mole of gas, n = 1, N = NA and the mass of the gas = molar mass Mm = mNA

44. By Note: • cr.m.s.increases with temperature. • cr.m.s.decreases with the mass of the molecule or molar mass of the gas.

45. Example 8Find the root mean square speed of hydrogen molecules at 27oC. It is given that R = 8.31 J K-1 mol-1. • Solution: • By • cr.m.s. = (3 x 8.31 x 300 / 0.002)½ = 1934 ms-1

46. Example 9Find the ratio of the root mean square speed of hydrogen to that of oxygen. • By • For hydrogen: ch = (3RT/0.002) ½ • For oxygen: co = (3RT/0.032) ½ • ch / co = (0.032/0.002)½ = 4 • the ratio of the root mean square speed of hydrogen to that of oxygen is 4.

47. Translational K.E. and Rotational K.E. • Translational motion • Rotational motion • Translational + rotational motion

48. Monoatomic and diatomic gases • Monoatomic means having only one atom in the molecule. All of the noble gases are monoatomic. • Diatomic means having two atoms in the molecule. Examples are oxygen, nitrogen

49. kinetic energy of a monoatomic gas • Translational K.E. = ½ mc2 • Rotational motion = 0 (neglected)

50. From and we have ⇒ where Translational K.E. of a monoatomic gas • For 1 mole of gas, n = 1, N = NA and the mass of the gas = molar mass Mm = mNA • Average translational kinetic energy of a molecule = which is known as Boltzmann constant.