1. The kinetic theory of gases and the gas laws

2. 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;

3. 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!

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

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

6. 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)

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

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

9. 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)

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

11. 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

12. The behaviour of gases

13. 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!

14. 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)

15. 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!

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

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

18. 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)

19. 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)

20. 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

21. 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

22. 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

23. 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

24. 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

25. 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.

26. 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.
                                                                                                                                                                                                

27. 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.

28. pV = constant T

29. 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

30. 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

31. 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

32. 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

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