Book 1 Section 5.2 The kinetic theory Smelling durian Explanation of the gas laws in terms of molecular motion Check-point 6 Assumptions of the kinetic theory of ideal gases p-V relationship due to molecular motion Temperature and molecular motion Check-point 7 1 2 3 4 Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory Smelling durian Do you know why the smell can spread without the help of wind? Related to the movement of gas particles. Book 1 Section 5.2 The kinetic theory

1 Explanation of the gas laws in terms of molecular motion a Random motion of gas particles Gas laws tell you what will happen to a gas under certain conditions, but do not tell you why. Kinetic theory explains the reasons behind. Expt 5d Observing random motion of molecules Book 1 Section 5.2 The kinetic theory

a Random motion of gas particles The smoke particles move about constantly along zigzag paths. Einstein proposed that: bombarded by a large number of air molecules from all directions but not in equal number  Random motion Simulation 5.4 Random motion of molecules Book 1 Section 5.2 The kinetic theory

1 Explanation of the gas laws in terms of molecular motion b Kinetic theory model The three dimensional kinetic theory model consists of a transparent tube containing a large number of ball-bearings. Tube: closed by a movable polystyrene piston Motor-driven vibrator: sets the ball-bearings in motion Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory b Kinetic theory model The ball bearings represent gas molecules. Physical quantity Kinetic theory model Weight of the piston (and cardboard discs) Pressure Voltage applied to the motor Temperature Volume Height of the piston The three-dimensional kinetic theory model Expt 5e Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory b Kinetic theory model Step 1 of the expt illustrates the following: Condition Result Interpretation voltage – weight of piston  temp constant, pressure   volume  Simulating Boyle’s law height of piston  height of piston – voltage  volume constant, temp   pressure  Simulating pressure law weight of piston  Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory b Kinetic theory model Condition Result Interpretation weight of piston – voltage  pressure constant, temp   volume  Simulating Charles’ law height of piston  pressure and temp constant, number of molecules   volume  weight of piston – voltage – number of balls  height of piston  In step 2, the big ball moves in a jerky manner.  illustrates the random motion of smoke particles Book 1 Section 5.2 The kinetic theory

1 Explanation of the gas laws in terms of molecular motion c Qualitative explanation using kinetic theory According to the kinetic theory, a gas is made up of a huge number of tiny particles, each of which has a mass. All the particles are in constant, random motion. They collide with each other and the walls of the container continually. Book 1 Section 5.2 The kinetic theory

c Qualitative explanation using kinetic theory The theory can be used to explain p andT: p: force exerted by the molecules colliding on the container walls. T : ave KE of the molecules. p  if the molecules hit the walls more often or with a greater ∆ momentum. T   molecules gain energy  move faster Simulation 5.5 Kinetic theory and gas laws Book 1 Section 5.2 The kinetic theory

c Qualitative explanation using kinetic theory Interpretation of Boyle’s law: When a gas is compressed, the molecules have less volume to move in. They hit the walls more often  greater pressure Book 1 Section 5.2 The kinetic theory

c Qualitative explanation using kinetic theory Interpretation of Pressure law: When temp , the molecules move faster. Volume fixed  molecules hit the walls more often.  speed  ∆ momentum in each collision  ∴ greater pressure Book 1 Section 5.2 The kinetic theory

c Qualitative explanation using kinetic theory Interpretation of Charles’ law: Temp   molecules move faster Frequency of collisions and ∆ momentum in each collision  To keep pressure constant  volume must  (so that the frequency of collisions ) Book 1 Section 5.2 The kinetic theory

c Qualitative explanation using kinetic theory Relationship between volume and number of molecules: Number of molecules   frequency of collisions  To keep pressure constant, volume must  Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory Check-point 6 – Q1 In the kinetic theory model, state what each of the following represents: (a) ball bearings Gas molecules. (b) weight of the piston Pressure. (c) voltage applied Temperature. (d) height of the piston Volume. Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory Check-point 6 – Q2 When pushing an inverted empty glass into water, why does water not rise into it? A The air molecules inside the glass bombard the water surface. B There is no space in the air inside the glass. C The weight of the air inside the glass is acting on the water surface. Book 1 Section 5.2 The kinetic theory

2 Assumptions of the kinetic theory of ideal gases All the molecules are identical with the same mass. All the molecules are in constant, random motion. The size of each molecule is negligible. The duration of a collision is negligible. All collisions are perfectly elastic. Intermolecular forces are negligible. A real gas does not satisfy these assumptions.  Good approximation for high T and low p Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion Consider a cubic container of length l containing N identical molecules in random motion. Suppose a molecule of mass m travels with a velocity v perpendicular to wall A. It collides elastically with the wall and bounces backwards with the same speed. Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion ∆ momentum of the molecule = –mv – mv = –2mv By conservation of momentum, total momentum of the molecule and the wall should be the same before and after collision, i.e. total ∆ momentum = 0  ∆ momentum of the wall due to this collision = +2mv Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion This molecule will be back and collide with wall A again after travelling through a round-trip of length 2l. 2l v for every time interval of , the wall will experience ∆ momentum of +2mv. Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion Average force exerted on the wall by this molecule: change in momentum time interval for the change 2mv 2l v = mv 2 l = F = Pressure exerted on the wall by this molecule: mv 2 l l 2 = force area mv 2 l 3 = mv 2 V = p = Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion Most molecules collide with the wall at an angle  need to resolve the velocity vectors into components. No change in the y-direction  ∆ momentum in the x-direction only Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion For an elastic collision, ∆ momentum = –mvx – mvx = –2mvx 2l vx Round-trip time =  pressure exerted on wall A by the molecule: mvx 2 V = p Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion There are N molecules in the container.  total pressure exerted on the wall: m V = (vx1 2 + vx2 2 + … + vxN 2) p m V = N vx 2 ̅ ......................... (1) = vx1 2 + vx2 2 + … + vxN 2 N vx 2 ̅ where Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion vx 2 ̅ vy 2 vz 2 = By symmetry, ..................... (2) (due to random motion) By Pythagoras’ theorem, c 2 = vx 2 + vy 2 + vz 2 Mean square value of velocity: c 2 = vx 2 + vy 2 + vz 2 ̅ ..................... (3) Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion c 2 = 3vx 2 ̅ Combining (2) and (3): m V = p N c 2 3 ̅ Substituting this into (1): pV 1 3 = Nmc 2 ̅ The equation is true for containers of all shapes. Book 1 Section 5.2 The kinetic theory

3 p -V relationship due to molecular motion Mean square speed and pressure Example 6 Book 1 Section 5.2 The kinetic theory

4 Temperature and molecular motion a Molecular interpretation of temperature pV 1 3 = Nmc 2 ̅ and pV = nRT 1 3 Nmc 2 ̅ = nRT  1 2 mc 2 ̅ 3RT 2 n N = 3RT 2NA =  3RT 2NA = KEaverage Book 1 Section 5.2 The kinetic theory

a Molecular interpretation of temperature 3RT 2NA = KEaverage  T (in Kelvin)  KEaverage 3RT 2NA 3 2 nRT = For n moles of gas, KEtotal = nNA × 3 2 nRT = ∴KEtotal of a gas = total internal energy Book 1 Section 5.2 The kinetic theory

4 Temperature and molecular motion b Root-mean-square speed of molecules Distribution of speed of gas molecules depends on temp. c 2 ̅ Typical speed = root of  root-mean-square speed 3pV Nm = 3RT mNA = c 2 ̅ c rms = Nm : total mass of the gas; mNA : molar mass Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory Check-point 7 – Q1 Find the velocity of a particle if the velocity components along x, y and z directions are 3 m s−1, 4 m s−1 and 5 m s−1 respectively. v = v 2 = vx 2 + vy 2 + vz 2 = 32 + 42 + 52 = 50 = 7.07 m s–1 Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory Check-point 7 – Q2 Air pressure inside a container of 40 cm3 = 120 kPa Total mass of the air = 6.5 × 10−5 kg Mean square speed of the air molecules = ? pV = 3 Nmc 2 ̅ By , 3pV Nm = 3  120  103  40  10–6 6.5 × 10−5 c 2 = ̅ = 2.22  105 m s–1 Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory Check-point 7 – Q3 Temp of gas X = 100 °C Mass of a molecule of gas X = 3.34  10–27 kg Root-mean-square speed of the molecules = ? (R = 8.31 J mol–1 K–1, NA = 6.02  1023 mol–1) Root-mean-square speed 3RT mNA = = 3  8.31  (273 + 100) 3.34  10−27  6.02  1023 = 2150 m s–1 Book 1 Section 5.2 The kinetic theory

4 Temperature and molecular motion Diffusion Example 7 Book 1 Section 5.2 The kinetic theory

4 Temperature and molecular motion Air in the can Example 8 Book 1 Section 5.2 The kinetic theory

Book 1 Section 5.2 The kinetic theory The End Book 1 Section 5.2 The kinetic theory