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# Thermal 3.2

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### Thermal 3.2

1. 1. Thermal Physics IBO DP Topic 3.2 Modelling a gas
2. 2. Pressure What is pressure? How can a magician lie or stand on a bed of nails?
3. 3. Pressure Pressure of a gas is due to the particles bombarding the walls of the container. Each collision with the wall causes a momentum change (as there is a change in direction). The force on the wall by one molecule = average rate of change of pressure.
4. 4. Pressure If the temperature increases, the average KE of the particles increases. The increase in velocity of the particles leads to a greater rate of collisions and hence the pressure of the gas increases as the collisions with the side have increase. Also the change in momentum is greater, therefore greater force.
5. 5. When a force is applied to a piston in a cylinder containing a volume of gas the particles take up a smaller volume.
6. 6. Smaller area to collide with and hence collisions are more frequent with the sides leading to an increase in pressure. Also, as the piston is being moved in. It gives the particles colliding with it more velocity.Therefore they have more KE. Therefore the temperature of the gas rises because the collisions are perfectly elastic. There is no loss of KE as a result of the collisions.
7. 7. An Ideal Gas An ideal gas is a theoretical gas that obeys the gas laws and thus fit the ideal gas equation exactly. How does this fit with your shared knowledge? TOK
8. 8. Real Gases Real gases conform to the gas laws under certain limited conditions. But they condense to liquids and then solidify if the temperature is lowered. Furthermore, there are relatively small forces of attraction between particles of a real gas. This is not the case for an ideal gas
9. 9. The Kinetic Theory of Gases When the moving particle theory is applied to gases it is generally called the kinetic theory. The kinetic theory relates the macroscopic behaviour of an ideal gas to the microscopic behaviour of its molecules or atoms.
10. 10. The Postulates 1. A gas consists of a large number of identical particles in continual random motion. Evidence 1g of hydrogen atoms contain 6.023 x 1023 particles Smoke particles move in a haphazard path which are bombarded by gas molecules which in turn have velocities that are random in magnitude and direction. Gas fills a volume Only gravity holds the atmosphere in.
11. 11. The Postulates continued 2. Collisions between particles and the wall of the container are elastic. Evidence A gas left at constant p, V & T do not change properties 3. The particles in a gas are considered to be point masses with no rotational Ek. This simplification allows us to make straight forward calculations.
12. 12. The Postulates continued 4. Volume occupied by the gas particles themselves is assumed to be negligible compared with the volume of the container as you can compress a gas. 5. There are no forces acting between the particles themselves or between the particles and the wall of the container except during collisions. There are large distances between particles  Fg is negligible.
13. 13. Macroscopic Behaviour 6. Time between collisions is large compared with the time for collisions. There are no forces between particles  particles travel in straight lines and constant v and then collide elastically and then changes v. Forces act on particles only during collision  collision time is small.
14. 14. Proof of ideal gas law Consider, a particle of mass m, in a box of side L, moving with velocity v As particle collides with wall there is a change in momentum of -mv – mv = -2mv This is only in 1 of the 3 dimensions so we need to denote this by saying x Fx = change in p/t = 2mvx/t (force of molecule on wall)
15. 15. t is the time between collisions so we know that Distance = 2L = vx * t therefore t = 2L/ vx So, Fx = 2mvx/t now becomes Fx = 2mvx/ (2L/ vx ) Fx = mvx 2/L
16. 16. Fx = mvx 2/L Now, this is only 1 particle of N so there are a range of N speeds. In addition the mean speed in 3D can be shown as vx 2 + vy 2 + vz 2 = v 2 as vx = vy = vz so vx 2 = v2 /3 Giving, F = Nmv2/3L
17. 17. F = Nmv2/3L Now, pressure, p = F/A where A = L2 So, p = Nmv2 /3L * L2 = Nmv2/3L3 and L3 = V Giving, p = Nmv2/3V Nm/V is density so this equations becomes p = 1/3ᵨ v2
18. 18. Show pV =nRT and p =1/3ᵨ v2 leads to U = 3/2 NkT
19. 19. Thermal Properties of Gases A large volume of gas can be described by bulk characteristics - pressure (p), volume (V) & temperature (T). All three terms are inter-related. Investigations involved the measurement of • Pressure • Volume • Temperature These experiments used these macroscopic properties of a gas to formulate a number of gas laws.
20. 20. Units Temperature is always measured in K Volume is usually in m3 Pressure can be different units as long as you are consistent But 1 atm = 1.01 x 105 Nm-2 = 101.3 kPa = 760 mmHg
21. 21. Gas Laws 1. Boyle’s Law: Variation of volume with pressure (const T). eg Bike pump.
22. 22. Gas Laws Boyle’s experiment can be repeated by using a flexible J-tube containing mercury, and a barometer to measure atmospheric pressure. Air was trapped in one end and varied the pressure on it, by raising the other end of the tube as shown on right:
23. 23. Gas Laws The gas pressure of the enclosed air is equal to the atmospheric pressure plus,the pressure resulting from, the height difference of Hg in the two arms. Taking a number of pressure and volume readings the graph looks like:
24. 24. A more revealing graph is obtained by plotting p vs 1/V As a straight line is obtained that passes through the origin, we can say mathematically: p  1/V
25. 25. Gas Laws If the same experiment is performed at different temperatures, a curve is still obtained but at different positions.
26. 26. The curves are called isotherms as the temp at each point on the curve is the same.
27. 27. Gas Laws We have assumed that there is a fixed amount of gas, maintained at a constant temperature. We can also write, p = k/V or pV = k k is a constant of proportionality and is equal to the slope.
28. 28. Gas Laws For different points on the graph; p1V1 = k and p2V2 = k  p1V1 = p2V2 (for const. temp)
29. 29. Gas Laws Boyle’s law states at constant temperature, the volume of a fixed mass of gas, varies inversely with its pressure. P  1/V or PV = constant When the conditions are changed P1V1 = P2V2
30. 30. The Experiment Air from foot pump Bourdon Pressure gauge Volume of dry air oil
31. 31. What to do A column of trapped dry air in a sealed tube by the oil. The pressure on this volume of air can be varied by pumping air in or out of the oil reservoir to obtain different pressures. Wait to allow the temperature to return to room temperature.
32. 32. The Results P V P 1/ V PV P
33. 33. Gas Laws 2. Charles’ Law: Variation of volume with temperature (const p). To verify this a capillary tube is sealed at one end, and a small mercury piston is included, trapping a sample of air. A thermometer is strapped to the tube and both are placed in glycerol. The pressure is fixed and is equal to atmospheric pressure.
34. 34. Gas Laws The glycerol is heated and the length of the air column is noted, at various temperatures. The length of the column can be read as the volume in arbitrary units, as the cross section is uniform.
35. 35. The results obtained can be graphed as below:
36. 36. Gas Laws This graph does not go through the origin. When extrapolated it cuts the temp axis at - 273oC
37. 37. Gas Laws This creates a new scale defined as The absolute temperature scale, where zero degrees kelvin (0K) = -273oC. A 1K rise = 1oC rise. ToC =T K -273. A real gas would liquefy before it reaches 0K and so no gas is ideal.
38. 38. Gas Laws Any extrapolation should not continue past the liquefaction point. No gas can reach a temperature below 0K as no gas could have a negative volume. Graphing for two different pressures:
39. 39. Gas Laws Using the Kelvin scale we can say mathematically: V  absolute T V = kT or k = V/T where k is the gradient of the graph. For two points, (V1,T1) and (V2,T2) V1/T1 = V2/T2
40. 40. Gas Laws Charles’ law states at constant pressure, the volume of a fixed mass of gas, varies directly with its absolute temperature. V  T or V/T = constant When the conditions are changed V1/T1 = V2/T2
41. 41. The Experiment Tap 1 Tap 2 Tap 3 Water reservoir Fixed mass of gas Mercury in U tube
42. 42. What to do Fill the mercury column with mercury using the right hand tube (tap 1 open, tap 2 closed). With tap 1 open drain some mercury using tap 2, then close tap 1 and 2. To trap a fixed mass of gas. Fill the jacket with water (make sure tap 3 is closed).
43. 43. Change the temperature of the water by draining some water from tap 3 and adding hot water. Equalise the pressure by leveling the columns using tap 2. Read the volume from the scale.
44. 44. The Results V T K V T oCA value for absolute zero
45. 45. Gas Laws 3. Gay-Lussac’s Law: (Law of pressures) Variation of pressure with temperature (const V). A bulb enclosing a fixed volume of gas is placed in a beaker of water, and attached to a mercury manometer, as shown below:
46. 46. Gas Laws
47. 47. Gas Laws The water is warmed the air expands in the bulb and tube, and causes the Hg, to be forced up the manometer. One arm is then raised to increase pressure, and compress the air, back to the original volume. The height difference is recorded.
48. 48. Gas Laws
49. 49. As the graph does not pass through the origin it must be extrapolated.
50. 50. Gas Laws We can now mathematically say: p  absolute T p = kT or k = p/T; where k is the gradient of the graph. Using two points on the graph; p1/T1 = p2/T2
51. 51. Gas Laws The law of pressures state at a constant volume, the pressure of a given mass of gas, varies directly with the Kelvin temperature. P  T or P/T = constant When the conditions are changed P1/T1 = P2/T2
52. 52. The Experiment Bourdon gauge Ice Water Fixed Mass of gas Heat
53. 53. What to do Change the temperature of the water by heating it. Record the pressure of the gas.
54. 54. The Results P T K P T oCA value for absolute zero
55. 55. Absolute Zero and the Kelvin Scale Charles’ Law and the Pressure Law suggest that there is a lowest possible temperature that substances can go This is called Absolute Zero The Kelvin scale starts at this point and increases at the same scale as the Celsius Scale
56. 56. Therefore -273oC is equivalent to 0 K ∆1oC is the same as ∆1 K To change oC to K, add 273 To change K to oC, subtract 273
57. 57. GENERAL GAS EQUATION All three laws studied above can be combined into one. From Boyle’s law:
58. 58. The curves plotted from the above situations are shown:
59. 59. Gas Laws Going from (1) to (2) V by raising T at constant pressure. V1/T1 = V*/T2 (Charles’ Law)  Going from (2) to (3) temp constant, the piston is withdrawn causing V & P. p1V* = p2V2 (Boyle’s Law)  Multiplying u and v
60. 60. Gas Laws (V1 x p1V*)/T1 = (V* x p2V2)/T2 (p1V1)/T1 = (p2V2)/T2 pV/T = constant This is called the ideal gas law and is accurate except, at low temperatures, and high pressures.
61. 61. Gas Laws The only gas that approximates the ideal gas law, over a wide range of conditions, is Helium, to 4.2K.
62. 62. Gas Laws p1V1/T1 = C C depends on the number of moles, n, and; Universal gas constant, R (8.315 J mol-1 K-1 if Pressure is in kilopascals(kPa) but R = 0.0821 L atm K-1 mol-1 if Pressure is in atmospheres(atm) For one mole of gas, pV = RT For n moles of gas, pV = nRT
63. 63. Gas Laws One mole of a substance contains; 6.023 x 1023 particles. This value is called Avogadro’s number and is given the symbol NA. To find the moles present (n) divide the total number of molecules N, by Avogadro’s number.
64. 64. Gas Laws n = N/NA m grams of a substance contains m/mA moles, where the molar mass is expressed in grams. n = m/mA
65. 65. The Mole The mole is the amount of substance which contains the same number of elementary particles as there are in 12 grams of carbon-12 Experiments show that this is 6.02 x 1023 particles A value denoted by NA and called the Avogadro Constant (units mol-1)
66. 66. Molar Mass Molar mass is the mass of one mole of the substance SI units are kg mol-1
67. 67. Example Molar mass of Oxygen gas is 32 x10-3 kg mol-1 If I have 20g of Oxygen, how many moles do I have and how many molecules? 20 x 10-3 kg / 32 x10-3 kg mol-1  0.625 mol  0.625 mol x 6.02 x 1023 molecules  3.7625 x 1023 molecules
68. 68. Use your periodic table to… Find out how many atoms are in 1. 46g of sodium 2. 64mg of copper 3. 180g of water Find out how many moles is 1. 10g of Argon 2. 1kg of Iron
69. 69. What is the molar mass of… Nitrogen Chlorine Neon Water Sulfuric acid (H2SO4) Calcium carbonate (CaCO3)
70. 70. Combining the Laws The gas laws can be combined to give a single equation For a fixed mass of gas its pressure times its volume divided by its absolute temperature is a constant PV/T = k So that P1V1/T1 = P2V2/T2
71. 71. The Ideal Gas Equation PV = nRT Where n is the number of moles R is the universal gas constant 8.31 J mol- 1 K-1
72. 72. Real vs Ideal A real gas can be considered an ideal gas when at low pressure, moderate temperature and low density. When is a model good enough? TOK