1. The Gas Laws Do Now read pages 70-71

2. The Gas Laws What happens if the Pressure and Volume are changed and constant temperature

3. 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 )

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

5. 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

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

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

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

9. Explaining the behaviour of gases When we compress (reduce the volume) a gas at constant temperature, the pressure increases. Why?

10. Explaing the behaviour of gases When we compress (reduce the volume) a gas at constant temperature, the pressure increases. Why? A smaller volume increases the likelihood of a particle colliding with the container walls. Boyle’s Law

11. The Gas Laws Do Q1-3 page 71

12. The behaviour of gases- Pressure Law http://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp http://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp When we heat a gas at constant volume, what happens to the pressure? Why? Let’s do it!

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

14. Boyle’s Law w States that the pressure of a fixed mass of gas is inversely proportional to its volume at constant temperature w P  1/V or PV = constant w When the conditions are changed w P 1 V 1 = P 2 V 2

15. What to do A column of trapped dry air in a sealed tube by the oil The pressure on this volume of air can be varied by pumping air in or out of the oil reservoir to obtain different pressures Wait to allow the temperature to return to room temperature

16. Charles’ Law w States that the volume of a fixed mass of gas is directly proportional to its absolute temperature at constant pressure w V  T or V/T = constant w When the conditions are changed w V 1 /T 1 = V 2 /T 2

17. The Experiment Tap 1 Tap 2 Tap 3 Water reservoir Fixed mass of gas Mercury in U tube

18. What to do w Fill the mercury column with mercury using the right hand tube (tap 1 open, tap 2 closed) w With tap 1 open drain some mercury using tap 2, then close tap 1 and 2. To trap a fixed mass of gas w Fill the jacket with water (make sure tap 3 is closed)

19. and then w Change the temperature of the water by draining some water from tap 3 and adding hot water w Equalise the pressure by leveling the columns using tap 2 w Read the volume from the scale

20. The Results V T K V T o C A value for absolute zero

21. The Results P V P 1/ V PV P

22. The Charles’ Law copy 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 Charles’ law If T is in Kelvin

23. Explaing the behaviour of gases When we heat a gas a constant pressure, the volume increases. Why?

24. Explaining the behaviour of gases When we heat a gas a constant pressure, the volume increases. Why? Increasing the volume reduces the chance of particles colliding with the container walls, opposing the effect of the particles increased kinetic energy. Charles Law

25. Explaing the behaviour of gases When we heat a gas a constant pressure, the volume increases. Why?

26. Explaining the behaviour of gases When we heat a gas a constant pressure, the volume increases. Why? Increasing the volume reduces the chance of particles colliding with the container walls, opposing the effect of the particles increased kinetic energy. Charles Law

27. The Pressure law 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

28. Explaining the behaviour of gases http://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp http://phet.colorado.edu/sims/ideal-gas/gas-properties.jnlp When we heat a gas at constant volume, the pressure increases. Why?

29. Explaining the behaviour of gases When we heat a gas at constant volume, the pressure increases. Why? Increased average kinetic energy of the particles means there are more collisions with the container walls in a period of time and the collisions involve a greater change in momentum. Pressure Law

30. Absolute Zero and the Kelvin Scale w Charles’ Law and the Pressure Law suggest that there is a lowest possible temperature that substances can go w This is called Absolute Zero w The Kelvin scale starts at this point and increases at the same scale as the Celsius Scale

31. w Therefore -273 o C is equivalent to 0 K w ∆1 o C is the same as ∆1 K w To change o C to K, add 273 w To change K to o C, subtract 273

32. The equation of state By combining these three laws pV = constant V/T = constant p/T = constant We get pV/T = constant Or p 1 V 1 =p 2 V 2 T 1 T 2 Remember, T must be in Kelvin

33. An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At sea level these values are 300K and 1.0 x 10 5 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

34. An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 m 3. Therefore at sea level p 1 = 1.0 x 10 5 Pa, V 1 = 0.83 m 3, T 1 = 300K.

35. An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 p 1 = 1.0 x 10 5 Pa, V 1 = 0.83 m 3, T 1 = 300K. At the top of Mount Everest p 2 = 3.3 x 10 4 Pa, V 2 = ? m 3, T 1 = 250K.

36. An example At the top of Mount Everest the temperature is around 250K, with atmospheric pressure around 3.3 x 10 4 Pa. At seas level these values are 300K and 1.0 x 10 5 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 p 1 = 1.0 x 10 5 Pa, V 1 = 0.83 m 3, T 1 = 300K. At the top of Mount Everestp 2 = 3.3 x 10 4 Pa, V 2 = ? m 3, T 1 = 250K. p 1 V 1 /T 1 = p 2 V 2 /T 2 (1.0 x 10 5 Pa x 0.83 m 3 )/300K = (3.3 x 10 4 Pa x V 2 )/250K V 2 = 2.1 m 3, This is the volume of 1kg of air on Everest Density = mass/volume = 1/2.1 = 0.48 kg.m -3.

37. pV= constant T

38. The equation of state of an ideal gas Experiment has shown us that pV = nR T p - pressure (Pa) V - volume (m 3 ) n - number of mols R - molar gas constant ( 8.31 J mol -1 K -1 ) T - Temperature (K) Remember, T must be in Kelvin

39. Sample question A container of hydrogen of volume 0.1m 3 and temperature 25°C contains 3.20 x 10 23 molecules. What is the pressure in the container? K.A.Tsokos “Physics for the IB Diploma” 5 th Edition

40. Sample question A container of hydrogen of volume 0.1m 3 and temperature 25°C contains 3.20 x 10 23 molecules. What is the pressure in the container? # moles = 3.20 x 10 23 /6.02 x 10 23 = 0.53 K.A.Tsokos “Physics for the IB Diploma” 5 th Edition

41. Sample question A container of hydrogen of volume 0.1m 3 and temperature 25°C contains 3.20 x 10 23 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 10 4 N.m -2 K.A.Tsokos “Physics for the IB Diploma” 5 th Edition

42. An Ideal Gas w Is a theoretical gas that obeys the gas laws w And thus fit the ideal gas equation exactly

43. Real Gases w Real gases conform to the gas laws under certain limited conditions w But they condense to liquids and then solidify if the temperature is lowered w Furthermore, there are relatively small forces of attraction between particles of a real gas w This is not the case for an ideal gas

44. Questions! Questions Lots of questions. Homework questions due 24 th January