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  • 1. Thermal Physics Topic 3.2 Thermal Properties of Matter
  • 2. Heat Capacity/Thermal Capacity
    • When different substances undergo the same temperature change they can store or release different amounts of energy.
    • The temperature change that occurs when a substance absorbs heat depends on the amount of the substance. To quantify heat we must specify the amount of the substance.
    • The calorie is defined as:
      • The amount of heat required to raise the temperature of 1 g of water by 1 o C.
  • 3. Heat Capacity/Thermal Capacity
    • A kilocalorie is the heat required to raise the temperature of 1 kg of water by 1 o C.
    • The S.I. unit of heat is the same as all other forms of energy - the Joule (J). 1 calorie = 4.187 J.
    • Heat capacity =  Q /  T in JK -1
      •  Q = the change in thermal energy in joules
      •  T = the change in temperature in Kelvin
    • Defined as the amount of energy to change the temperature of a body by unit temperature
    • Applies to a specific BODY
  • 4. Heat Capacity - 2
    • A body with a high heat capacity will take in thermal energy at a slower rate than a substance with a low heat capacity because it needs more time to absorb a greater quantity of thermal energy
    • They also cool more slowly because they give out thermal energy at a slower rate
  • 5. Specific Heat Capacity
      • Why is it that you can touch the pastry on a McDonalds Apple pie but when you bite into it, the Apple burns your tongue?
      • Why is it that a pizza can be just right but the pineapple is always much hotter?
  • 6.
    • The answer lies in the fact that different substances have different capacities for storing heat. Put a litre of water in a saucepan and heat, it may take a few minutes to boil. Put a metal knife on the same hotplate and it will reach the same temperature much more quickly.
  • 7. If we were given 1g of both iron and water, we would have a different number of molecules of different type and mass in each sample. Water uses energy to increase the rotation of molecules, internal vibration and bond stretching. Iron atoms use the energy to increase the translational kinetic energy. This means it takes 8 times the amount of heat to raise 1g of water by 1 o C than it does for iron.
  • 8. Specific Heat Capacity
    • Defined as the quantity of heat required to raise the temperature of a unit mass of a substance by 1 degree is known as the specific heat capacity.
    • Unit mass is normally 1kg, and unit temperature rise is normally 1K
    • Specific Heat Capacity =  Q / (m  T)
    • in J kg -1 K -1
      • where m is the mass of the material
  • 9.
    • We use the symbol c for shc so the equation becomes:
          •  Q = mc  T
    • For an object made of 1specific material then
    • Heat Capacity = m x Specific Heat Capacity
    • Water has a specific heat of 4200 joules/kg o C. This means it takes 4200 J of energy to raise the temperature of 1kg of water by 1 o C.
  • 10. Specific Heat Capacity - 2
    • Unit masses of different substance s contain
      • different numbers of molecules
      • of different types
      • of different masses
    • If the same amount of internal energy is added to each unit mass
      • it is distributed amongst the molecules
  • 11. Specific Heat capacity - 3
    • The average energy change of each molecule will be different for each substance
    • Therefore the temperature changes will be different
    • So the specific heat capacities will be different
  • 12. Methods of finding the S.H.C
    • Two methods
      • Direct
      • Indirect
  • 13. Direct Method - Liquids
    • Using a calorimeter of known Heat Capacity
    • (or Specific Heat Capacity of the material and the mass of the calorimeter)
    • Because Heat Capacity = Mass x Specific Heat Capacity
  • 14.
    • Mechanical systems are difficult to achieve accurate results, as there is a loss of energy due to friction that cannot easily be accounted for. The most accurate measurements of the specific heat of water have been made with electrical systems.
  • 15. SHC of Liquids Thermometer Calorimeter Heating coil Liquid Insulation Stirrer To joulemeter or voltmeter and ammeter
  • 16. Calculations - Liquids
    • Electrical Energy input is equal to the thermal energy gained by the liquid and the calorimeter – this is the assumption that we are making
    • Work done = V x I x t
    • Energy gained by liquid = m l c l  T l
    • Energy gained by calorimeter = m c c c  T c
  • 17. Calculations - Liquids -2
    • Using conservation of energy
    • Electrical energy in = thermal energy gained by liquid + thermal energy gained by calorimeter
    • V I t = m l c l  T l + m c c c  T c
    • The only unknown is the specific heat capacity of the liquid
  • 18. Question
    • How much heat energy would be required to raise the temperature of 5 kg of water from 19 o C to 44 o C?
  • 19. Solution
    • m = 5 kg
    • c = 4200 J/kg o C
    • T i = 19 o C
    • T f = 44 o C
    •  Q = mc  T
    •  Q = 5 x 4200 x (44 -19)
    •  Q = 5.25 x 10 5 J
  • 20. Question 2
    • If 40 000 J of heat are provided to 4 kg of water at 20 o C, what final temperature will be achieved?
  • 21. Solution
    • Q = 40 000 J
    • m = 4 kg
    • c = 4200 J/kg o C
    • T i = 20 o C
    •  Q = mc  T
    • 4 x 10 4 = 4 x 4200 x ( T f - 20)
    • T f = (4 x 10 4 /4 x 4200) + 20
    • T f = 22.4 o C
  • 22. Direct Method - Solids
    • Using a specially prepared block of the material
    • The block is cylindrical and has 2 holes drilled in it
      • one for the thermometer and one for the heater
      • Heater hole in the centre, so the heat spreads evenly through the block
      • Thermometer hole, ½ way between the heater and the outside of the block, so that it gets the averge temperature of the block
  • 23. SHC of Solids Insulation Thermometer Heating coil Solid Insulation To joulemeter or voltmeter and ammeter
  • 24. Calculations - Solids
    • Again using the conservation of energy
    • Electrical Energy input is equal to the thermal energy gained by the solid
    • Electrical energy = V x I x t
    • Energy gained by solid = m s c s  T s
  • 25. Calculations - Solids -2
    • V x I x t = m s c s  T s
    • The only unknown is the specific heat capacity of the solid
  • 26. Question 3
    • How much water could be boiled using an immersion heater that draws a 5A current in 15 minutes from room temperature (20 o C)?
  • 27. Solution
    • VIt = mc  T
    • m = Vit/c  T
    • m = (240 x 5 x 15 x 60)/(4200 x (100 -20))
    • m = 1.08 x 10 6 /3.36 x 10 5
    • m = 3.2 kg
  • 28. Question 4
    • A person wants to make 4, 250ml cups of hot coffee. If they were to use an electric kettle 240 V that used 1500W, how long would it take to boil the minimum amount of water from 25 o C)?
  • 29. Solution
    • P = VI and VIt = mc  T
    •  Pt = mc  T
    • t = mc  T / P
    • t = (4 x 0.25) x 4200 x (100 - 25)/1500
    • t = (1 x 4200 x 75)/1500
    • t = 315000/1500
    • t = 210 s (or 3½ min)
  • 30. Indirect Method
    • Sometimes called the method of mixtures
    • In the case of solid, a known mass of solid is heated to a known temperature (usually by immersing in boiling water for a period of time)
    • Then it is transferred to a known mass of liquid in a calorimeter of known mass
  • 31. Indirect Method cont…..
    • The change in temperature is recorded and from this the specific heat capacity of the solid can be found
    • Energy lost by block = Energy gained by liquid and calorimeter
    • m b c b  T b = m w c w  T w + m c c c  T c
      • the SHC of water and the calorimeter are needed
  • 32. Apparatus Heat Thermometer Beaker Boiling Water Block Thermometer Calorimeter Water Block Insulation
  • 33. Indirect Method – cont….
    • In the case of a liquid
    • A hot solid of known specific heat capacity is transferred to a liquid of unknown specific heat capacity
    • A similar calculation then occurs
  • 34. Question 5
    • Ten silver spoons, each with a mass of 30g, are removed from a pan of boiling water, quickly dried, and then placed in a pan of water at room temperature (20 o C). The pan contains 500g of water. The temperature rises to 23 o C. What is the specific heat of silver?
  • 35. Solution
    • ( mc  T ) silver = ( mc  T ) water
    • (0.03 x 10) x c s x (100 - 23) = 0.5 x 4200 x (23 - 20)
    • c x 0.3 x 77 = 2100 x 3
    • 23.1 c = 6300
    • c = 6300/23.1
    • c = 270J/kg o C
  • 36. Question 6
    • Determine the final temperature of a 0.2 kg mass of hot coffee at 90 o C contained in a foam-insulating cup if 0.1 kg of cold water at 10 o C is poured into it? Assume the specific heat capacity of coffee is 4000 J/kg o C.
  • 37. Solution
    • c water = 4200 J/kg o C
    • c coffee = 4000 J/kg o C
    • m 2 = 0.1 kg
    • T 2i = 10 o C
    • m 1 = 0.2 kg
    • T 1i = 90 o C
    • Q gained = Q lost
    • m 1 c coffee  T 1 = m 2 c water  T 2
    • 0.2 x 4000 x (90 - T f )=0.1 x 4200 x ( T f - 10)
    • 72000 - 800 T f = 420 T f -4200
    • 1220 T f = 76200
    • T f = 62.5 o C
  • 38. Question 7
    • What will be the final temperature reached when a 250 g rod of copper ( c copper = 385 J/kg o C) is taken from a beaker of boiling water and plunged into 100g of water at 20 o C contained in another beaker?
  • 39. Solution
    • m 1 = 0.25 kg
    • c 1 = 385 J/kg o C
    • T 1 i = 100 o C
    • m 2 = 0.1 kg
    • c 2 = 4200 J/kg o C
    • T 2 i = 20 o C
    • Q gained = Q lost
    • m 1 c 1  T 1 = m 2 c 2  T 2
    • 0.25 x 385 x (100- T f )=0.1 x 4200 x ( T f - 20)
    • 9625 - 96.25 T f = 420 T f - 8400
    • 516.25 T f = 18025 
    • T f = 34.9 o C
  • 40. Phases (States) of Matter
    • Matter is defined as anything that has mass and occupies space
    • There are 4 states of matter
    • Solids, Liquids, Gases and Plasmas
    • Most of the matter on the Earth in the form of the first 3
    • Most of the matter in the Universe is in the plasma state
  • 41. Macroscopic
    • Macroscopic properties are all the observable behaviours of that material such as shape, volume, compressibility
    • The many macroscopic or physical properties of a substance can provide evidence for the nature of that substance
  • 42. Macroscopic Characteristics
  • 43.
    • These will help to explain what is happening at the atomic level, and this part of the model will be interpreted later
  • 44. Microscopic Characteristics
  • 45. Solids.
    • Before the turn of the previous century ;
      • it was thought that the content of a solid ,
      • determined its characteristics.
    • It was what made diamonds hard ;
      • lead heavy and iron magnetic.
    • The characteristics of a solid are due to ;
      • its structure ,
      • the arrangement of atoms within the material.
  • 46. Solids
    • To keep the atoms in the regular pattern (lattice) ;
      • there are forces (electrical in nature),which bind them together.
    • If the atoms get too close,
      • the force becomes repulsive
      • (between electrons in their outer shells).
  • 47. Solids
    • In solids, the thermal energy is very much smaller than ;
      • the intermolecular binding energy and so ,
      • solids have specific macroscopic properties.
    • Solids maintain ;
      • a fixed shape and ,
      • a fixed size.
    • Even if a force is applied to it it does not readily change ;
      • its shape ,
      • or volume.
    • The result is that the atoms vibrate about a fixed position.
  • 48. Arrangement of Particles - 1
    • Solids
      • Closely packed
      • Strongly bonded to neighbours
      • held rigidly in a fixed position
      • the force of attraction between particles gives them PE
  • 49. Fluids
    • Liquids
    • Gases
    • are both fluids
    • Because they FLOW
  • 50. Liquids
    • T he thermal energy (due to an increase in temperature) is greater ;
      • allowing the atoms to move farther apart.
    • The binding forces are less and the atoms are able to roll over each other.
    • This gives rise to the macroscopic properties of liquids.
    • Liquids do not maintain a fixed shape ;
      • it takes the shape of the container ,
    • Like a solid, it is not readily compressible ;
      • only a very large force can ,
      • significantly change its volume.
  • 51. Arrangement of Particles - 2
    • Liquids
      • Still closely packed
      • Bonding is still quite strong
      • Not held rigidly in a fixed position and bonds can break and reform
      • PE of the particles is higher than a solid because the distance between the particles is higher
  • 52. Gases
    • The forces of attraction are so weak and ;
      • the thermal energy is so high
        • (due to another increase in temperature),
      • the atoms do not even stay close together.
    • They move very rapidly in a random manner ;
      • filling the container and ,
      • occasionally colliding with one another.
  • 53. Gases
    • The speed at which the atoms are moving is ;
      • so fast that when they do collide,
      • the force of attraction is not strong enough to ,
      • keep them together and ,
      • they fly off in a new direction.
    • A gas has neither a fixed shape nor a fixed volume ;
      • it will expand to fill its container.
  • 54. Arrangement of Particles - 3
    • Gases
      • Widely spaced
      • Only interact significantly on closest approach or collision
      • Have a much higher PE than liquids because the particles are furthest apart
  • 55. Plasma
    • At extremely high temperatures such as those found in stars, atoms are ionised. The result is a collection of nuclei (ions) and electrons referred to as plasma.
  • 56. Changes of State
    • Add energy to ice and it turns to water. Add energy to water and it turns to steam. The state of matter depends upon its temperature and the pressure that is exerted upon it. To change state, a transfer of energy is required.
    • A substance can undergo changes of state or phase changes at different temperatures
  • 57. Changes of State - 2
    • The moving particle theory can be used to explain the microscopic behaviour of these phase changes
      • When the solid is heated the particles of the solid vibrate at an increasing rate as the temperature is increased
      • The vibrational KE of the particles increases
  • 58. Changes of State -3
      • At the melting point a temperature is reached at which the particles vibrate with sufficient thermal energy to break from their fixed positions and begin to slip over each other
      • As the solid continues to melt more and more particles gain sufficient energy to overcome the forces between the particles and over time all the solid particles are changed to a liquid
      • The PE of the system increases as the particle s move apart
  • 59. Changes in State - 4
      • As the heating continues the temperature of the liquid rises due to an increase in the vibrational, rotational and translational energy of the particles
      • At the boiling point a temperature is reached at which the particles gain sufficient energy to overcome the inter-particle forces and escape into the gaseous state. PE increases.
      • Continued heating at the boiling point provides the energy for all the particles to change
  • 60. Heating Curve Solid Liquid Gas Solid - liquid phase change Liquid - gas phase change Temp / o C Time /min
  • 61. Changes of State Freezing/solidification vaporisation condensation melting sublimation Thermal energy given out Thermal energy added GAS SOLID LIQUID
  • 62. Evaporation
    • The process of evaporation is a change from the liquid state to the gaseous state which occurs at a temperature below the boiling point
    • The Moving Particle (Kinetic) theory can be applied to understand the evaporation process
  • 63. Explanation
    • A change of state from liquid to gas that takes place at the surface of the liquid.
        •  
    • The temperature of any body is related to the average kinetic energy of its molecules. As the molecules move in a random manner, some molecules may collide. Some may lose kinetic energy and some may collide and increase kinetic energy, enough to overcome the attractive forces of their neighbouring molecules.
  • 64. So
    • As the higher kinetic energy molecules have escaped, the average kinetic energy of the liquid has been reduced. This means the liquid left behind has been cooled and there is a corresponding temperature drop.
    •  
    • This is the principle used by evaporative air conditioners and perspiration.
  • 65. Cooling
    • Condensation is the opposite process to evaporation. This is the cooling of a gas to a liquid. When water vapour molecules collide with a cold can of Coke, giving up so much kinetic energy, they condense into a liquid.
  • 66. Cool
    • Condensation is a warming process. The kinetic energy lost by the gas molecules warms the surface that they strike. A steam burn is more dangerous than a boiling water burn at the same temperature. Steam gives up energy when it condenses to the liquid that wets the skin.
  • 67. Factors Affecting The Rate
    • Evaporation can be increased by
      • Increasing temperature
      • (more particles have a higher KE)
      • Increasing surface area
      • (more particles closer to the surface)
      • Increasing air flow above the surface
      • (gives the particles somewhere to go to)
  • 68. Freezing
    • When energy is continually withdrawn from a liquid, molecular motion slows until the forces of attraction becomes enough to cause them to fuse. The molecules then vibrate about equilibrium positions and form a solid. This is called freezing.
    • If foreign material such as sugar or salt is added to water, the freezing temperature is lowered. The foreign ions get in the way of the water molecules and stop them forming ice molecules, Antifreeze is often used for this purpose.
    • Freezing is a cooling process.
  • 69. Melting
    • The opposite of freezing is melting
    • This occurs when energy is added.
    • The average kinetic energy is increased and so the temperature is increased.
  • 70. Vaporisation
    • Evaporation takes place at the surface of a liquid. A change of state from liquid to gas can also take place within the liquid.
    • The gas that forms beneath the surface occurs as bubbles which move up and out into the surrounding air. This is also called boiling.
    • The pressure of the bubbles within the bubble must be great enough to resist the pressure of the liquid water.
  • 71. Sublimation
    • This is the process whereby a solid changes directly into a vapour without passing through the liquid phase.
    • Carbon dioxide will do this at atmospheric pressure.
  • 72. Latent Heat
    • The thermal energy which a particle absorbs in melting, vaporising or sublimation or gives out in freezing, condensing or sublimating is called Latent Heat because it does not produce a change in temperature
  • 73. Latent Heat cont….
    • During a change of state, there is no change in temperature until all of the substance has changed state. If we study boiling water and steam that are both at 100 o C, they both have the same average kinetic energy.
    • The molecules in steam however, has much more potential energy as they are free to move and are not held together. When water turns to steam, no temperature rise is observed as the energy absorbed goes into increasing the potential energy.
  • 74. Water
    • Let us look at what happens when 1.0 kg of water is heated from -20 o C where it is ice, until it has become steam at 100 o C at 1 atm pressure (1.03 x 10 5 Pa).
    • As heat is added, its temperature increases at the rate of about 1 o C for every 2 kJ of heat added.
    • When the temperature reaches 0 o C, the temperature stops rising even though heat is still added. When 340kJ have been added, all the ice has turned to water and temperature is still 0 o C.
  • 75.
    • The energy required to change 1 kg of a substance from the solid to liquid state is called the latent heat of fusion (L f ) . This also refers to the amount of heat released when a liquid is turned to solid. For water, the L f = 3.34 x 10 5 J.
    • The water will now increase in temperature at the rate of 1 o C for every 4 kJ of heat added. When the temperature reaches 100 o C, the temperature again remains constant until all of the water is turned to steam.
    • The energy required to change 1 kg of a substance from the liquid to gaseous state is called the latent heat of vaporisation (L v ) . This also refers to the amount of heat released when a gas is turned to liquid.
  • 76. The heat required to change the state of a substance can also be expressed mathematically.   Q = mL
  • 77. Definition
    • The quantity of heat energy required to change one kilogram of a substance from one phase to another, without a change in temperature is called the Specific Latent Heat of Transformation
    • Latent Heat =  Q / m in J kg -1
  • 78. Types of Latent Heat
    • Fusion
    • Vaporisation
    • Sublimation
    • The latent heat of fusion of a substance is less than the latent heat of vaporisation or the latent heat of sublimation
  • 79. Questions
    • When dealing with questions think about
      • where the heat is being given out
      • where the heat is being absorbed
      • try not to miss out any part
  • 80. Question
    • How much heat is required to convert 40 g of ice to water at 0 o C?
  • 81. Solution
    • Q = mL
    • Q = (4 x 10 -2 ) x (3.34 x 10 5 )
    • Q = 1.3 x 10 4 J absorbed
  • 82. Question 2
    • Water at 95 o C is mixed with an equal mass of ice at 0 o C. Find the final temperature achieved.
  • 83. Solution
    • Energy lost by water cooling =
    • Energy gained by ice melting +
    • Energy gained by ice warming
    • mc  T = m L f + mc  T
    • m x 4200 x (95 - T f ) =
    • ( m x (3.34 x 10 5 )) + (( m x 4200 x ( T f - 0))
  • 84.
    • divide both sides by m .
    • 399000 - 4200 T f = 334000 + 4200 T f
    • 8400 T f = 65000
    • T f = 7.7 o C
  • 85. Question 3
    • A large polystyrene pot contains 2 kg of water at 20 o C. Steam at 100 o C is blown into the water and the temperature reaches 50 o C. Find the mass of steam used.
  • 86. Solution
    • m 2 = ?
    • T i = 100 o C
    • T f = 50 o C
    • L v = 2.26 x 10 6 J kg -1
    • m 1 = 2 kg
    • c = 4200 J kg -1 o C -1
    • T i = 20 o C
    • T f = 50 o C
  • 87.
    • m 1 c  T = m 2 L v + m 2 c  T
    •   2 x 4200 x 30 =
    • m 2 x (2.26 x 10 6 ) + m 2 x 4200 x 50
    • m 2 = (2 x 4200 x 30)/(2.26x10 6 + 4200 x 50)
    • m 2 = 0.1 kg
  • 88. Methods of finding Latent Heat
    • Using similar methods as for specific heat capacity
    • The latent heat of fusion of ice can be found by adding ice to water in a calorimeter
  • 89. Apparatus Block of ice Thermometer Calorimeter Water Block of ice Insulation
  • 90.
    • The change in temperature is recorded and from this the latent heat of fusion of the ice can be found
    • Energy gained by block melting = Energy lost by liquid and calorimeter
    • m b L b = m w c w  T w + m c c c  T c
    • the SHC of water and the calorimeter are needed
  • 91.
    • The latent heat of vaporisation of a liquid could be found by an electrical method
  • 92. Latent Heat of Vaporisation Insulation Thermometer Heating coil Liquid in Calorimeter To joulemeter or voltmeter and ammeter
  • 93.
    • The initial mass of the liquid is recorded
    • The change in temperature is recorded for heating the liquid to boiling
    • The liquid is kept boiling
    • The new mass is recorded
    • Energy supplied by heater = energy to raise temperature of liquid + energy use to vaporise some of the liquid
    • (The calorimeter also needs to be taken in to account.
    • V I t = m l c l  T l + m e L e + m c c c  T c
  • 94. Question 4
    • A 200W immersion heater is used to raise the temperature of water to its boiling point. The heater is left on for 4 minutes after the water boils. What mass of water will be boiled off in this time?
  • 95. Solution
    • m L v = Pt
    • m x 2.26 x 10 6 = 200 x 240
    • m = 200 x 240/(2.26 x 10 6 )
    • m = 2.12 x 10 -2 kg
  • 96. 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.
    • 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 increased
    • Also the change in momentum is greater, therefore greater force
  • 97. Pressure continued
    • When a force is applied to a piston in a cylinder containing a volume of gas
    • The particles take up a smaller volume
    • Smaller area to collide with
    • And hence collisions are more frequent with the sides leading to an increase in pressure
  • 98.
    • 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.
  • 99. Collisions
    • Because the collisions are perfectly elastic
    • There is no loss of KE as a result of the collisions
  • 100. An Ideal Gas
    • Is a theoretical gas that obeys the gas laws
    • And thus fit the ideal gas equation exactly
  • 101. 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
  • 102. 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
  • 103. 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 10 23 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.
  • 104. 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 E k . This simplification allows us to make straight forward calculations.
  • 105. 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  F g is negligible.
  • 106. 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.