The kinetic theory of gases and the gas laws

Kinetic theory/ideal gas We can understand the behaviour of gases using a very simple model, that of an “ideal” gas. The model makes a few simple assumptions;

You should know these by now! Ideal gas assumptions The particles of gas (atoms or molecules) obey Newton’s laws of motion. You should know these by now!

Ideal gas assumptions The particles in a gas move with a range of speeds

Ideal gas assumptions The volume of the individual gas particles is very small compared to the volume of the gas

Ideal gas assumptions The collisions between the particles and the walls of the container and between the particles themselves are elastic (no kinetic energy lost)

Ideal gas assumptions There are no forces between the particles (except when colliding). This means that the particles only have kinetic energy (no potential)

Ideal gas assumptions The duration of a collision is small compared to the time between collisions.

P = F/A Pressure – A reminder Pressure is defined as the normal (perpendicular) force per unit area P = F/A It is measured in Pascals, Pa (N.m-2)

Pressure – A reminder What is origin of the pressure of a gas?

Pressure – A reminder Collisions of the gas particles with the side of a container give rise to a force, which averaged of billions of collisions per second macroscopically is measured as the pressure of the gas Change of momentum

The behaviour of gases

The behaviour of gases http://phet. colorado When we heat a gas at constant volume, what happens to the pressure? Why? Let’s do it!

The behaviour of gases http://phet. colorado When we heat a gas at constant volume, what happens to the pressure? Why? P α T (if T is in Kelvin)

The behaviour of gases When we compress (reduce the volume) a gas at constant temperature, what happens to the pressure? Why? Let’s do it!

The behaviour of gases When we compress (reduce the volume) a gas at constant temperature, what happens to the pressure? Why? pV = constant

The behaviour of gases When we heat a gas a constant pressure, what happens to its volume? Why?

The behaviour of gases When we heat a gas a constant pressure, what happens to its volume? Why? V α T (if T is in Kelvin)

Explaining the behaviour of gases In this way we are explaining the macroscopic behaviour of a gas (the quantities that can be measured like temperature, pressure and volume) by looking at its microscopic behaviour (how the individual particles move)

p α 1/V or pV = constant The gas laws We have found experimentally that; At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume. p α 1/V or pV = constant This is known as Boyle’s law

The gas laws At constant pressure, the volume of a fixed mass of gas is proportional to its temperature; V α T or V/T = constant This is known as Charle’s law If T is in Kelvin

The gas laws At constant volume, the pressure of a fixed mass of gas is proportional to its temperature; p α T or p/T = constant This is known as the Pressure law If T is in Kelvin

T1 T2 The equation of state By combining these three laws pV = constant V/T = constant p/T = constant We get pV/T = constant Or p1V1 = p2V2 T1 T2 Remember, T must be in Kelvin

An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest? “Physics”, Patrick Fullick, Heinemann

An example Take 1kg of air at sea level At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest? Take 1kg of air at sea level Volume = mass/density = 1/1.2 = 0.83 m3. Therefore at sea level p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K.

An example Therefore at sea level At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest? Therefore at sea level p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K. At the top of Mount Everest p2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K.                                                                                                                                                                                                 

An example Density = mass/volume = 1/2.1 = 0.48 kg.m-3. At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 104 Pa. At seas level these values are 300K and 1.0 x 105 Pa respectively. If the density of air at sea level is 1.2 kg.m-3, what is the density of the air on Mount Everest? Therefore at sea level p1 = 1.0 x 105 Pa, V1 = 0.83 m3, T1 = 300K. At the top of Mount Everest p2 = 3.3 x 104 Pa, V2 = ? m3, T1 = 250K. p1V1/T1 = p2V2/T2 (1.0 x 105 Pa x 0.83 m3)/300K = (3.3 x 104 Pa x V2)/250K V2 = 2.1 m3, This is the volume of 1kg of air on Everest Density = mass/volume = 1/2.1 = 0.48 kg.m-3.

pV = constant T

T The equation of state Experiment has shown us that pV = nR Where n = number of moles of gas and R = Gas constant (8.31J.K-1.mol-1) Remember, T must be in Kelvin

Sample question A container of hydrogen of volume 0.1m3 and temperature 25°C contains 3.20 x 1023 molecules. What is the pressure in the container? K.A.Tsokos “Physics for the IB Diploma” 5th Edition

Sample question A container of hydrogen of volume 0.1m3 and temperature 25°C contains 3.20 x 1023 molecules. What is the pressure in the container? # moles = 3.20 x 1023/6.02 x 1023 = 0.53 K.A.Tsokos “Physics for the IB Diploma” 5th Edition

Sample question A container of hydrogen of volume 0.1m3 and temperature 25°C contains 3.20 x 1023 molecules. What is the pressure in the container? # moles = 3.20 x 1023/6.02 x 1023 = 0.53 P = RnT/V = (8.31 x 0.53 x 298)/0.1 = 1.3 x 104 N.m-2 K.A.Tsokos “Physics for the IB Diploma” 5th Edition

Questions! Page 84 Questions 15,16,17 and 18 Page 85 Question 19 Answers:Answers