1. The kinetic model of a gas Unit 4: Module 3 – Kinetic Theory There are two ways of describing a gas: macroscopic and microscopic. Macroscopic: Considers a gas on a large scale. It has mass and volume. A change in state/phase occurs once a liquid has reached its boiling point. Microscopic: Considers a large number of moving molecules that collide with one another and the walls of the container holding them.

2. Kinetic Theory of an Ideal Gas What is it? It explains the movement of gas molecules by relating macroscopic and microscopic descriptions. Real gases do NOT meet the conditions assumed for an ideal gas. An ideal gas has internal energy ONLY in the form of random kinetic energy Unit 4: Module 3 – Kinetic Theory

3. All gas molecules are perfectly elastic. They bounce off each other and the walls of the container without losing any KE. Molecules never come to a stop or settle at the bottom of a container. All gases consist of identical molecules in continuous random motion (Brownian Motion). Assumptions of an Ideal Gas Unit 4: Module 3 – Kinetic Theory

4. Assumptions of an Ideal Gas (Continued) Molecules exert forces on each other only when they collide. Therefore they must be a relatively long way apart. The gravitational force on the molecules is negligible Molecules are so tiny that they take up no space at all. Volume of the molecule is negligible compared to the volume of the container. Unit 4: Module 3 – Kinetic Theory

5. Kinetic Theory Unit 4: Module 3 – Kinetic Theory A gas exerts pressure on the walls of a container due to its collisions. Consider ONE gas molecule of mass, m travelling at a velocity, v towards the wall of a container. As it collides with the container wall it elastically collides. This means its change in momentum is 2mv. d When the molecule next collides with the wall it would have travelled a distance of 2d. As speed = distance/time. Time between collisions = distance / speed Using the above info: Time between collisions = 2d / v

6. As we know Force is proportional to the rate of change of momentum Therefore F ∝ change in momentum/ time F ∝ 2mv/(2d/v) F ∝ mv 2 /d If the mass of the molecule decreases then the force exerted on the walls of the container decrease. This is hard to consider for just ONE molecule but remember a real gas would have millions of molecules each exerting a force - PRESSURE Unit 4: Module 3 – Kinetic Theory Kinetic Theory Cont.

7. Unit 4: Module 3 – Kinetic Theory Kinetic Theory Cont.(again) If the force due to one molecule is: F ∝ mv 2 /d Consider n molecules: F ∝ nmv 2 /d The pressure on a face of the cube container (of area d 2 ) is Pressure = nmv 2 /d 3 (as pressure = force/area) or Pressure = nmv 2 /V (where V is volume) As these particles can move in 3 dimensions, the total number of molecules in the box is 3n = N P = Nmv 2 /3V = Mv 2 /3V = 1/3 ρv 2 Where M = total mass of gas and ρ its density d

8. Molecular Speed The molecules in a gas do not all travel at the same speed. The average value of molecular speed, c is not the mean but a mean square speed. We write this as c 2 Therefore we write: P = 1/3 ρv 2 as P = 1/3 ρc 2 The square root of the average (mean) of the squares of the speeds is known as the r.m.s. speed. Unit 4: Module 3 – Kinetic Theory

9. Worked Example Example Qu: Four molecules have speeds of 300, 400, 500 and 600ms -1. Calculate a) the mean (average) speed and b) the mean square speed, c 2 c) the r.m.s. speed. a)Mean speed = 300+400+500+600 = 450ms -1. 4 b)Mean square speed = 300 2 + 400 2 + 500 2 + 600 2 = 215000 m 2 s -2 4 c) r.m.s. speed (square root of mean speed) =  ( 215000) = 460 ms -1 Unit 4: Module 3 – Kinetic Theory

10. An Ideal Gas – The Theoretical Equation The total mass of gas in the box = N x m as density = mass/ volume density = N x m V If we replace ρ in P = 1/3 ρv2 ρv2 with Nm/V We can rewrite the equation as: pV = 1 Nm 3 Unit 4: Module 3 – Kinetic Theory

11. An Ideal Gas – The Theoretical Equation Using Kinetic Theory it is possible to derive the equations that describe and explain things like pressure and temperature in terms of the movement of individual molecules. pV = 1 Nm 3 Where : p = pressure V = volume N = No of atoms/molecules m = mass of one atom/molecule = mean square speed of the atoms Unit 4: Module 3 – Kinetic Theory

12. Kinetic Energy We now know: pV = 1 Nm and pV = nRT 3 If we equate both (for 1 mole of gas): 1 N A m = RT 3 (rearrange and multiply both sides by ½) We get: 1 m = 3RT Where K =R 2 2N A N A K = Boltzmann constant (JK -1 ) As 1 m = Mean Kinetic Energy of a gas molecule 2 E k = 3 KT Where E k = kinetic energy 2 K = Boltzmann Constant T = Kelvin temperature Unit 4: Module 3 – Kinetic Theory