2. Gases & Kinetic Molecular Theory Kinetic molecular theory Gases Physical properties Temperature Pressure Boyles Law Charles Law Gay Lussacs Law Combined Gas Law Ideal Gas Law

3. KINETIC MOLECULAR THEORY

4. Kinetic Molecular theory The Kinetic Molecular Theory is what modern day chemist’s use to explain the behaviors and characteristics of gases ◦ The word Kinetic refers to Motion ◦ The word Molecular refers to molecules

5. Kinetic Molecular theory Assumptions 1. A gas is composed of small particles 2. These particles have an insignificant volume (they are small) and are spaced far apart from one another.

6. Kinetic Molecular theory Assumptions 3. There are NO attractive or repulsive forces between particles. 4. The particles in a gas are in constant random motion 5. Particles move in straight paths and are completely independent of each other

7. Kinetic Molecular theory Assumptions 6. Particles path is only changed by colliding with another particle or the sides of its container. 7. All collisions a gas particle undergoes are perfectly elastic. 8. No energy is lost from one particle to another and the total kinetic energy remains constant.

8. Nature of Gases (physical properties) Regardless of their chemical identity, gases tend to exhibit similar physical behaviors Gases can be monoatomic (He) one atom Gases can be diatomic (N 2 ) Two atoms Gases can be polyatomic (CH 4 ) many atoms

9. Nature of Gases (physical properties) Gases have mass Gases are compressible Gases fill their containers Gases exert pressure by colliding with objects in their path ◦ The sum of all the force per unit of area caused by these collisions is equal to the pressure exerted.

10. Nature of Gases (physical properties)  Pressure of a gas is dependant on its temperature. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids.

11. Nature of Gases (physical properties) It is usually necessary to have a mole or more of gas particles to have a significant change in mass. Gases have high velocity when compared to their masses.

12. Nature of Gases (physical properties) This gives them a lot of energy and movement. This movement causes them to spread out. The movement of gases spreading out is called diffusion.

13. Ideal vs. Real Gases Ideal gasesReal gases Have no volume Have elastic collisions Are in constant random motion in a straight line Don’t attract or repel each other Have their own volume Attract each other Behaves well at ◦ low temperatures ◦ High temperatures ◦ In non-polar molecules

14. A word about temperature Temperature is defined as : the average kinetic energy of the particles that make up an object. The higher the temperature the more energy the particles have. The more energy the more impacts The more impacts then the more pressure.

15. A word about temperature Always use absolute temperature (Kelvin) when working with gases. ºF ºC K -45932212 -2730100 0273373 K = ºC + 273

16. A word about pressure The gases in the air are exerting a pressure called atmospheric pressure. This pressure is a result of the mass of the gases around us and the force of gravity.

17. A word about pressure KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi

18. STP Standard Temperature & Pressure 0°C273 K Or 1 atm101.325 kPa

19. THE GAS LAWS Boyles, Charles and Gay-Lussac Laws

20. Boyles Law Robert Boyle was among the first to note the relationship between Pressure and volume of a gas. He (actually some other dude) measured the volume of air at different pressures, and observed a pattern of behavior which led to his law. During his experiments Temperature and How much gas is used were NOT allowed to change.

21. Boyles Law PV = k

22. HOW DOES PRESSURE VS. VOLUME OF GASES RELATE GRAPHICALLY Volume Pressure PV = k Temperature, # of particles remain constant Temperature, # of particles remain constant

23. Boyles Law Another way to look at this relationship…. What if we had to change conditions …since PV = k THEN we can assume….. P 1 V 1 = P 2 V 2 Practice: a gas has a volume of 3.0 liters at 2 atm. What is the volume at 4atm.

24. Boyles Law Example: a gas has a volume of 3.0 liters at 2 atm. What is the volume at 4atm. Determine which variables you have ◦ Given:  P1 = 2 atm  V1 = 3 liters  P2 = 4 ATM  V2 = ? Determine which law is being represented : Boyles law is P and V

25. Boyles Law Example: a gas has a volume of 3.0 liters at 2 atm. What is the volume at 4atm. Rearrange the equation for the variable you don’t know. P1 V1 = V2 P2 Plug in the variables and solve:

26. Practice #2: A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg Given: Solve: Rearrange:

27. GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Practice Problem #3. A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW = P 1 x V 1 = P 2 x V 2 PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

28. Charles’s Law Jaques Charles determined the relationship between temperature and volume of a gas. He measured the volume of air at different temperatures and observed their pattern of behavior. During his experiments the pressure of the system and the volume of gas used were held constant.

29. Charles’s Law

30. HOW DOES TEMPERATURE VS. VOLUME OF GASES RELATE GRAPHICALLY Temp Volume V/T = k Pressure, # of particles remain constant Pressure, # of particles remain constant

31. Charles’s Law The volume and absolute Temperature (K) of a gas are directly related At constant mass and constant pressure V T

32. Charles Law Another way to look at this relationship…. What if we had to change conditions …since V / T = k THEN we can assume….. V 1 V 2 T 1 T 2 =

33. Charles Law Practice Problem #1: A gas has a volume Of 3.0 Liters at 127 °C. What is its volume at 227 °C? Given : T1 = 127 C V1 = 3.0 liters T2 =227 C V2 = ??? V 1 V 2 T 1 T 2 =

34. Charles Law Example: A gas has a volume of 3.0 Liters at 127 °C. What is its volume at 227 °C? Plug in the variables : 3.0L = V2 400K 500 K Re-arrange equation : (500 K) (3.0L) = V2 (400K) V2 = 3.8L V 1 V 2 T 1 T 2 =

35. Practice problem #3: A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K V 1 V 2 T 1 T 2 =

36. GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Practice Problem #2 A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW= V1/T1 = V2/T2 TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

37. Gay Lussac’s Law Lussac determined the relationship between temperature and pressure of a gas. He measured the temperature of air at different pressures and observed a pattern of behavior. He saw that the volume of a system and amount of gas were held constant

38. Gay Lussac’s Law

39. HOW DOES PRESSURE VS. TEMPERATURE OF GASES RELATE GRAPHICALLY Temp Pressure P/T = k Volume, # of particles remain constant Volume, # of particles remain constant

40. Gay Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume P T

41. Gay-Lussac’s Law Another way to look at this relationship…. What if we had to change conditions …since P / T = k THEN we can assume….. P 1 P 2 T 1 T 2 =

42. Gay Lussac’s Law Practice Problem A gas has a pressure of 3.0 atm at 127 °C. What is it’s pressure at 227°C Determine the Variables T1 = 127°C + 273 = 400 K P1 = 3.0 atm T2 = 227°C = 273 = 500K P2 = ? P 1 P 2 T 1 T 2 =

43. Gay Lussac’s Law Practice Problem A gas has a pressure of 3.0 atm at 127 °C. What is it’s pressure at 227°C Plug in the variables: 3.0 atm = P2 400K 500K P 1 P 2 T 1 T 2 =

44. GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 Practice Problem #2 A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW P1/T1= P2/T2 PP TT (765 torr)T 2 = (560. torr)(309K) T 2 = 226 K = -47°C

45. LAW RELAT- IONSHIP LAW CON- STANT Boyle’s P VP VP VP V P 1 V 1 = P 2 V 2 T, n Charles’ V TV TV TV T V 1 /T 1 = V 2 /T 2 P, n Gay- Lussac’s P TP TP TP T P 1 /T 1 = P 2 /T 2 V, n

46. = kPV PTPT VTVT T Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2

47. GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.325 kPa T 2 = 273 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 cm 3 )(273 K) =(101.325 kPa) V 2 (298 K) V 2 = 5.09 cm 3 Gas Law Problems A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

48. Ideal Gas Law Ideal Gas Law

49. PV T VnVn Ideal Gas Law = k PV nT UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R

50. B. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K PV=nRT

51. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L  atm/mol  K WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K P = 3.01 atm Ideal Gas Law Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW

52. GIVEN: V = ?V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3  kPa/mol  K Ideal Gas Law Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3 IDEAL GAS LAW

53. What is the volume (in liters) occupied by 49.8 g of HCl at STP? PV = nRT T = 0 0 C = 273.15 K P = 1 atm n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = 1 atm 1.37 mol x 0.0821 x 273.15 K Latm molK V = 30.6 L

54. Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1P1 T1T1 P2P2 T2T2 = P 1 = 1.20 atm T 1 = 291 K P 2 = ? T 2 = 358 K P 2 = P 1 x T2T2 T1T1 = 1.20 atm x 358 K 291 K = 1.48 atm