2. Gas Laws BOYLE CHARLES AVOGADRO GAY-LUSSAC

3. What happens to the volume of a gas when you increase the pressure? (e.g. Press a syringe that is stoppered) Consider This Push!

4. What happens to the volume of a gas when you increase the pressure? (e.g. Press a syringe that is stoppered) Consider This Push!

5. What happens to the Volume of a Gas When you Increase the Pressure? (e.g. Press a syringe that is stoppered) Consider This Why? There is lots of space between gas particles. Therefore, gases are compressible!

6. Let’s investigate the relationship between pressure and volume if the quantity of gas and temperature are held constant. 100 kPa

7. Volume = 50 L 10050

8. 200 kPa Volume = 25 L 10050 20025

9. 400 kPa Volume = 12.5 L 10050 20025 40012.5

10. 800 kPa Volume = 6.25 L 10050 20025 40012.5 8006.25 What is the mathematical relationship between P and V? P x V = constant 5000

11. Boyle’s Law In the 17th Century Robert Boyle described this property as, “the spring of air”. Boyle showed that when temperature and amount of gas were constant then: P  1/V OR: PV = k

12. Boyle’s Law For a fixed quantity of gas at a constant temperature, the volume and pressure are inversely proportional. Boyle showed that when temperature and amount of gas were constant then: P  1/V OR: PV = k

13. Who Cares? Scuba Divers! At sea level air pressure = 100 kPa At 10 m deep in water pressure = 200 kPa At 20 m deep = 300 kPa At 30 m deep = 400 kPa SCUBA provides air at the same pressure

14. What will the volume of air in the diver’s lungs be at the surface (100 kPa)? What will happen to the diver? A Scuba Diver goes to a depth of 90 m and takes a breath of 3 L volume from her tank. Suddenly! A Dolphin lunges at the diver and takes the SCUBA! The Diver holds her breath and quickly returns to the surface.

15. A Scuba Diver goes to a depth of 90 m and takes a breath of 3 L volume from her tank. Assuming that T is constant we can use Boyle’s Law: PV = k At 90 m: k = P 1 V 1 At surface: k = P 2 V 2 Therefore! P 1 V 1 = P 2 V 2 (1000 kPa)(3 L) = 100 kPa(V 2 ) V 2 = 300 L Yikes Exploding lungs

16. Consider This! Two balloons are filled with equal volumes of air. What happens to the Volume of each if one is heated and the other is frozen?

17. HEATED BALLOON FROZEN BALLOON 50 o C, V=1.18 L

18. FROZEN BALLOON HEATED BALLOON 60 o C, V=1.22 L40 o C, V=1.14 L

19. FROZEN BALLOON HEATED BALLOON 70 o C, V=1.25 L30 o C, V=1.11 L

20. FROZEN BALLOON HEATED BALLOON 80 o C, V=1.29 L20 o C, V=1.07 L

21. FROZEN BALLOON HEATED BALLOON 90 o C, V=1.32 L10 o C, V=1.04 L

22. FROZEN BALLOON HEATED BALLOON 100 o C, V=1.36 L0 o C, V=1.00 L

23. To study this relationship let’s look at this data in a table. Graph this data using temperature as the independent variable

25. If this line is extended backwards the volume of 0 L of gas is found to be -273 o C

26. -273.15 o C is known as absolute zero. When using gas laws, temperature must be expressed using a temperature scale where 0 is -273.15 o C. This is called the Kelvin scale of absolute temperature. 0 K = -273.15 o C When changing o C to K simply add 273.15 What is 12.3 o C in K? 12.3 + 273.15 = 285.45 = 285.5 K

27. Now let’s look at this table with temperatures in Kelvin (K). Can you spot a mathematical relationship between T and V.

28. V/T in Kelvin is a constant. 0.0037

29. Charles’ Law In the 17th Century Jacques Charles examined the relationship between Temperature and Volume Charles showed that when Pressure and amount of gas were constant then: V  T OR: V/T = k

30. Charles’ Law For a fixed quantity of gas at a constant pressure, absolute temperature and volume are directly proportional. Charles showed that when Pressure and amount of gas were constant then: V  T OR: V/T = k

31. Combined Gas Law If V1V1 T1T1 = V2V2 T2T2 V1P1V1P1 V2P2V2P2 and V1P1V1P1 T1T1 V2P2V2P2 T2T2 = = then or V 1 P 1 T 2 =V 2 P 2 T 1 What does P 2 = ? V2T1V2T1 V2T1V2T1

32. RECAP Charles’ Law At Constant P and n: V/T = k Boyle’s Law At Constant T and n: VP = k Combined Gas Law For a fixed quantity of gas: VP/T = k

33. If 12.5 L of a gas at a pressure of 125 kPa is placed in an elastic container at 15 o C what volume would it occupy if the pressure is increased to 145 kPa? Given: V 1 = 12.5 L P 1 = 125 kPa T 1 = 15 o C V 2 = ? P 2 = 145 kPa T 2 = 15 o C V1P1V1P1 T1T1 = V2P2V2P2 T2T2 Since T 1 = T 2 cancel them to get V1P1V1P1 V2P2V2P2 = (12.5 L)(125 kPa) = V 2 (145 kPa) V 2 = (12.5 L)(125 kPa)/145 kPa V 2 = 10.8 L

34. Does this answer make sense? 12.5 L @ 125 kPa 125 kPa

35. Does this answer make sense? 165 kPa

36. Does this answer make sense? 165 kPa

37. Does this answer make sense? 165 kPa V reduced to 10.8 L Yes, as the pressure increases at constant temperature the volume decreases.

38. If 15.6 L of a gas at a pressure of 165 kPa is placed in an elastic container at 15 o C what volume would it occupy if the temperature is increased to 98 o C? Given: V 1 = 15.6 L P 1 = 165 kPa T 1 = 15 o C V 2 = ? T 2 = 98 o C P 2 = 165 kPa V1P1V1P1 T1T1 = V2P2V2P2 T2T2 Since P 1 = P 2 cancel them to get V 1 /T 1 V 2 / T 2 = (15.6 L)/(288 K) = V 2 /(371 K) V 2 = (15.6 L)(371 K) / 288K V 2 = 20.1 L 288 K 371 K

39. If 5.3 L of a gas at a pressure of 75 kPa is placed in an elastic container at 24 o C what volume would it occupy if the temperature is increased to 62 o C and pressure to 155 kPa? Given: V 1 = 5.3 L P 1 = 75 kPa T 1 = 24 o C V 2 = ? T 2 = 62 o C P 2 = 155 kPa V1P1V1P1 T1T1 = V2P2V2P2 T2T2 (5.3 L)(75 kPa)(335 K) 297 K 335 K OR V 1 P 1 T 2 =V 2 P 2 T 1 (155 kPa)(297K) V 2 = V 2 = 2.9 L

40. Ideal Gas Law An Ideal Gas is a hypothetical gas that obeys all the gas laws perfectly under all conditions. PV = nRT Where n is the number of moles of gas, P is pressure in kPa, R = 8.313 kPaL/molK and T is temperature in K. Find the mass of helium gas which would be introduced into a 0.95 L container to produce a pressure of 125 kPa at 25 o C.

41. Find the volume occupied by 25 g of chlorine gas at SATP. 4.2 g of propane gas is introduced into a 325 mL container at 45 o C. What is the pressure of the container. Propane is C 3 H 8. What is the density of NH 3 gas at STP if 1.0 mol of this gas occupies 22.4 L. At what temperature does methane gas have a density of 1.2 mg/L if its pressure is 65 kPa. Methane is CH 4.

43. This gas exerts a pressure of 100 kPa inside this container. If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container? 170 kPa

44. This gas exerts a pressure of 100 kPa inside this container. If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container? 170 kPa

45. This gas exerts a pressure of 100 kPa inside this container. If another gas is injected into the same container and it exerts a pressure of 70 kPa what is the total pressure in the container? 170 kPa

46. The total pressure of a gas mixture is the sum of the partial pressures of each of the gases in the mixture. Example If a container of air has a pressure of 100 kPa and the % of N 2 in the container is 78%, % of O 2 is 21%, what are the partial pressures of each of these gases inside this container. P N 2 = 78 kPa, P O 2 = 21 kPa

47. 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 What would the total pressure be if the gas in container 1 was injected into container 2.

48. 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 The total pressure is the sum of the pressures of gas 1 and gas 2.Since gas 1 changed volume and temperature its pressure changed.

49. 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 V 1 =5.0 L, V 2 =6.0 L, T 1 =300 K, T 2 = 400 K P 1 =125 kPa, P 2 =? V 1 P 1 T 2 =V 2 P 2 T 1 P 2 = 5.0 L x 125 kPa x 400 K 6.0 L x 300 K = 139 kPa

50. 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 Total pressure is 155 kPa + 139 kPa = 294 kPa = 2.9 x 10 2 kPa 139 kPa

51. 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 Find the total pressure in container 1 if the gas in container 2 is injected into container 1.

52. 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 V 1 =6.0 L, V 2 =5.0 L, T 1 =400 K, T 2 = 300 K P 1 =155 kPa, P 2 =? V 1 P 1 T 2 =V 2 P 2 T 1 P 2 = 6.0 L x 155 kPa x 300 K 5.0 L x 400 K = 139.5 kPa

53. 5.0 L 300 K 125 kPa 6.0 L, 400 K, 155 kPa 1 2 Total pressure is 125 kPa + 139.5 kPa = 264.5 kPa = 2.6 x 10 2 kPa 139.5 kPa