1. Kinetic Molecular Theory & Gases An Honors/AP Chemistry Presentation

2. Kinetic Molecular Theory FKinetic means motion FSo the K.M.T. studies the motions of molecules. FSolids - vibrate a little FLiquids - vibrate, rotate, and translate (a little) FGases - vibrate, rotate, and translate (a lot)! FKinetic means motion FSo the K.M.T. studies the motions of molecules. FSolids - vibrate a little FLiquids - vibrate, rotate, and translate (a little) FGases - vibrate, rotate, and translate (a lot)!

3. Basic Assumptions of KMT FGases consist of large numbers of molecules in continuous random motion. FThe volume of the molecules is negligible compared to the total volume. FGases consist of large numbers of molecules in continuous random motion. FThe volume of the molecules is negligible compared to the total volume. FIntermolecular interactions are negligible. FWhen collisions occur, there is a transfer of kinetic energy, but no loss of kinetic energy. FThe average kinetic energy is proportional to the absolute temperature. FIntermolecular interactions are negligible. FWhen collisions occur, there is a transfer of kinetic energy, but no loss of kinetic energy. FThe average kinetic energy is proportional to the absolute temperature.

4. Gas Properties FVolume - amount of space (L or mL) FTemperature - relative amount of molecular motion (K) FPressure - the amount of force molecules exert over a given area (atm, Torr, Pa, psi, mm Hg) FMoles - the number of molecules (mol) FVolume - amount of space (L or mL) FTemperature - relative amount of molecular motion (K) FPressure - the amount of force molecules exert over a given area (atm, Torr, Pa, psi, mm Hg) FMoles - the number of molecules (mol)

5. Temperature Conversions FC = 5/9(F-32) FF = 9/5C + 32 FK = C + 273 FSo what is the absolute temperature (K) of an object at -40 o F? FC = 5/9(F-32) FF = 9/5C + 32 FK = C + 273 FSo what is the absolute temperature (K) of an object at -40 o F?

6. Answer to Temperature Conversion F-40 o F = -40 o C F-40 o C = 233 K or 230 K F-40 o F = -40 o C F-40 o C = 233 K or 230 K

7. Pressure Conversions F1 atm = 760 mm Hg = 760 Torr = 101,325 Pa = 14.7 psi FHow many atmospheres is 12.0 psi? FHow many Torr is 1.25 atm? FHow many Pascals is 720 mm Hg? F1 atm = 760 mm Hg = 760 Torr = 101,325 Pa = 14.7 psi FHow many atmospheres is 12.0 psi? FHow many Torr is 1.25 atm? FHow many Pascals is 720 mm Hg?

8. Answers to Pressure Conversions F12.0 psi =.816 atm F1.25 atm =950. Torr F720 mm Hg = 96000 Pa F12.0 psi =.816 atm F1.25 atm =950. Torr F720 mm Hg = 96000 Pa

9. A Barometer FA mercury barometer measures air pressure by allowing atmospheric pressure to press on a bath of mercury, forcing mercury up a long tube. The more pressure, the higher the column of mercury.

10. More on the barometer FAlthough American meteorologists will sometimes measure the height in inches, typically this pressure is measured in mm Hg. F1 mm Hg = 1 Torr FAlthough American meteorologists will sometimes measure the height in inches, typically this pressure is measured in mm Hg. F1 mm Hg = 1 Torr

11. S.T.P. FWhen making comparisons we often use benchmarks or standards to compare against. FIn chemistry Standard Temperature is 0 o C (273K) and Standard Pressure is 1 atm. FWhen making comparisons we often use benchmarks or standards to compare against. FIn chemistry Standard Temperature is 0 o C (273K) and Standard Pressure is 1 atm.

12. Boyle’s Law FIf the amount and temperature of the gas are held constant, then the volume of a gas is inversely proportional to the pressure it exerts. FMathematically this means that the pressure times the volume is a constant. FP*V = k FP 1 V 1 =P 2 V 2 FIf the amount and temperature of the gas are held constant, then the volume of a gas is inversely proportional to the pressure it exerts. FMathematically this means that the pressure times the volume is a constant. FP*V = k FP 1 V 1 =P 2 V 2

13. Boyle’s Law in Action

14. Sample Questions FThe volume of a balloon is 852 cm 3 when the air pressure is 1.00 atm. What is the volume if the pressure drops to.750 atm? FA gas is trapped in a 2.20 liter space beneath a piston exerting 25.0 psi. If the volume expands to 2.75 L, what is the new pressure? FThe volume of a balloon is 852 cm 3 when the air pressure is 1.00 atm. What is the volume if the pressure drops to.750 atm? FA gas is trapped in a 2.20 liter space beneath a piston exerting 25.0 psi. If the volume expands to 2.75 L, what is the new pressure?

15. The Answers are… FP 1 V 1 =P 2 V 2 F(1atm)(852cm 3 )= (.750atm)*V 2 V 2 =1140cm 3 F(25.0psi)(2.20L)=P 2 (2.75L) P 2 = 20.0 psi FP 1 V 1 =P 2 V 2 F(1atm)(852cm 3 )= (.750atm)*V 2 V 2 =1140cm 3 F(25.0psi)(2.20L)=P 2 (2.75L) P 2 = 20.0 psi

16. Charles’ Law FIf the amount and the pressure of a gas are held constant, then the volume of a gas is directly proportional to its absolute temperature. FMathematically, this means that the volume divided by the temperature is a constant. FV/T = k FV 1 /T 1 =V 2 /T 2 FIf the amount and the pressure of a gas are held constant, then the volume of a gas is directly proportional to its absolute temperature. FMathematically, this means that the volume divided by the temperature is a constant. FV/T = k FV 1 /T 1 =V 2 /T 2

17. Charles Law in Action

18. Sample Questions FThe volume of a balloon is 5.00 L when the temperature is 20.0 o C. If the air is heated to 40.0 o C, what is the new volume? F3.00 L of air are held under a piston at 0.00 o C. If the air is allowed to expand at constant pressure to 4.00 L, what is the new Celsius temperature of the gas? FThe volume of a balloon is 5.00 L when the temperature is 20.0 o C. If the air is heated to 40.0 o C, what is the new volume? F3.00 L of air are held under a piston at 0.00 o C. If the air is allowed to expand at constant pressure to 4.00 L, what is the new Celsius temperature of the gas?

19. The Answers Are… FV 1 /T 1 =V 2 /T 2 F5.00L/293K = V 2 /313K V 2 =5.34L F273K/3.00L = T 2 /4.00L T 2 =364K=91 o C FV 1 /T 1 =V 2 /T 2 F5.00L/293K = V 2 /313K V 2 =5.34L F273K/3.00L = T 2 /4.00L T 2 =364K=91 o C

20. The Gay-Lussac Law FIf the amount and volume of the gas are held constant, then the pressure exterted by the gas is directly proportional to its absolute temperature. FMathematically this means that the pressure divided by the temperature is a constant. FP/T = k FP 1 /T 1 =P 2 /T 2 FIf the amount and volume of the gas are held constant, then the pressure exterted by the gas is directly proportional to its absolute temperature. FMathematically this means that the pressure divided by the temperature is a constant. FP/T = k FP 1 /T 1 =P 2 /T 2

21. The Gay-Lussac Law in Action

22. Sample Questions FA tank of oxygen is stored at 3.00 atm and -20 o C. If the tank is accidentally heated to 80 o C, what is the new pressure in the tank? FA piston is trapped in place at a temperature of 25 o C and a pressure of 112 kPa. At what celcius temperature is the pressure 102 kPa? FA tank of oxygen is stored at 3.00 atm and -20 o C. If the tank is accidentally heated to 80 o C, what is the new pressure in the tank? FA piston is trapped in place at a temperature of 25 o C and a pressure of 112 kPa. At what celcius temperature is the pressure 102 kPa?

23. The Answers are… FP 1 /T 1 =P 2 /T 2 F(3atm)/(253 K)= P 2 / (353 K) P 2 =4.19 atm F(298 K)/(112 kPa)=T 2 /(102kPa) T 2 = 271 K = -2 o C FP 1 /T 1 =P 2 /T 2 F(3atm)/(253 K)= P 2 / (353 K) P 2 =4.19 atm F(298 K)/(112 kPa)=T 2 /(102kPa) T 2 = 271 K = -2 o C

24. Avogadro’s Law FIf the temperature and the pressure of a gas are held constant, then the volume of a gas is directly proportional to the amount of gas. FMathematically, this means that the volume divided by the # of moles is a constant. FV/n = korV/m = k FV 1 /n 1 =V 2 /n 2 or V 1 /m 1 =V 2 /m 2 FIf the temperature and the pressure of a gas are held constant, then the volume of a gas is directly proportional to the amount of gas. FMathematically, this means that the volume divided by the # of moles is a constant. FV/n = korV/m = k FV 1 /n 1 =V 2 /n 2 or V 1 /m 1 =V 2 /m 2

25. Avogadro’s Law in Action

26. Sample Questions FThe volume of a balloon is 5.00 L when there is.250 mol of air. If 1.25 mol of air is added to the balloon, what is the new volume? F3.00 L of air has a mass of about 4.00 grams. If more air is added so that the volume is now 24.0 L, what is the mass of the air now? FThe volume of a balloon is 5.00 L when there is.250 mol of air. If 1.25 mol of air is added to the balloon, what is the new volume? F3.00 L of air has a mass of about 4.00 grams. If more air is added so that the volume is now 24.0 L, what is the mass of the air now?

27. The Answers Are… FV 1 /n 1 =V 2 /n 2 or V 1 /m 1 =V 2 /m 2 F5.00L/.250 mol = V 2 /1.50 mol V 2 =30.0 L F4.00g/3.00L = m 2 /24.0L m 2 =32.0 g FV 1 /n 1 =V 2 /n 2 or V 1 /m 1 =V 2 /m 2 F5.00L/.250 mol = V 2 /1.50 mol V 2 =30.0 L F4.00g/3.00L = m 2 /24.0L m 2 =32.0 g

28. The Combined Gas Law FThis law combines the inverse proportion of Boyle’s Law with the direct proportions of Charles’, Gay-Lussac’s, and Avogadro’s Laws. FP 1 V 1 /(n 1 T 1 ) = P 2 V 2 /(n 2 T 2 ) For FP 1 V 1 /T 1 = P 2 V 2 /T 2 FThis law combines the inverse proportion of Boyle’s Law with the direct proportions of Charles’, Gay-Lussac’s, and Avogadro’s Laws. FP 1 V 1 /(n 1 T 1 ) = P 2 V 2 /(n 2 T 2 ) For FP 1 V 1 /T 1 = P 2 V 2 /T 2

29. Four Gas Laws in One FThe combined gas law could be used in place of any of the previous 4 gas laws. FFor example, in Boyle’s Law, we assume that the amount and temperature are constant. So if we cross them off of the combined gas law: FP 1 V 1 /(n 1 T 1 ) = P 2 V 2 /(n 2 T 2 ) FP 1 V 1 = P 2 V 2 FThe combined gas law could be used in place of any of the previous 4 gas laws. FFor example, in Boyle’s Law, we assume that the amount and temperature are constant. So if we cross them off of the combined gas law: FP 1 V 1 /(n 1 T 1 ) = P 2 V 2 /(n 2 T 2 ) FP 1 V 1 = P 2 V 2

30. Another Example FA sample of hydrogen has a volume of 12.8 liters at 104 o F and 2.40 atm. What is the volume at STP?

31. The answer is: FP 1 V 1 /(n 1 T 1 ) = P 2 V 2 /(n 2 T 2 ) FP 1 =2.40atm,V 1 =12.8L, T 1 =104 o F=40 o C=313K, T 2 =273K, P 2 =1atm, n 1 =n 2 F(2.4atm)(12.8L)/(313K) = (1atm)V 2 /(273K) FV 2 = 26.8 L FP 1 V 1 /(n 1 T 1 ) = P 2 V 2 /(n 2 T 2 ) FP 1 =2.40atm,V 1 =12.8L, T 1 =104 o F=40 o C=313K, T 2 =273K, P 2 =1atm, n 1 =n 2 F(2.4atm)(12.8L)/(313K) = (1atm)V 2 /(273K) FV 2 = 26.8 L

32. The Ideal Gas Law FIf, P 1 V 1 /(n 1 T 1 ) = P 2 V 2 /(n 2 T 2 ) FThen PV/(nT) = constant FThat constant is R, the ideal gas law constant. FR =.0821 L*atm/(mol*K) FR = 8.314 J/(mol*K) FSo, PV=nRT FIf, P 1 V 1 /(n 1 T 1 ) = P 2 V 2 /(n 2 T 2 ) FThen PV/(nT) = constant FThat constant is R, the ideal gas law constant. FR =.0821 L*atm/(mol*K) FR = 8.314 J/(mol*K) FSo, PV=nRT

33. But what about…  Since n = m/ M, we can substitute into PV = nRT and get  PV M = mRT FSince D = m/V, we can substitute in again and get  P M = DRT  Since n = m/ M, we can substitute into PV = nRT and get  PV M = mRT FSince D = m/V, we can substitute in again and get  P M = DRT

34. So which one is it? FLike a good carpenter, it is good to have many tools so that you can choose the right tool for the right job.  If I am solving a gas problem with density, I use P M = DRT. FIf I am solving a gas problem with moles, I use PV = nRT.  If I am solving a gas problem with mass, I use PV M = mRT. FLike a good carpenter, it is good to have many tools so that you can choose the right tool for the right job.  If I am solving a gas problem with density, I use P M = DRT. FIf I am solving a gas problem with moles, I use PV = nRT.  If I am solving a gas problem with mass, I use PV M = mRT.

35. Such as…. FUnder what pressure would oxygen have a density of 8.00 g/L at 300 K?  P M = DRT FP(32 g/mol) = (8 g/L)(.0821 latm/molK)(300 K) FP = 6.16 atm FUnder what pressure would oxygen have a density of 8.00 g/L at 300 K?  P M = DRT FP(32 g/mol) = (8 g/L)(.0821 latm/molK)(300 K) FP = 6.16 atm

36. An Important Number FWhat is the volume of 1 mole of a gas at STP? FPV = nRT V = nRT/P FV = (1mol)*(.0821Latm/molK)(273K)/ (1atm) FV = 22.4 L FThis is called the standard molar volume of an ideal gas. FWhat is the volume of 1 mole of a gas at STP? FPV = nRT V = nRT/P FV = (1mol)*(.0821Latm/molK)(273K)/ (1atm) FV = 22.4 L FThis is called the standard molar volume of an ideal gas.

37. Gas Stoichiometry FWe had said that stoichiometry implied a ratio of molecules, or moles. Up until now we only used mole ratios. FHowever Avogadro said that the volume is directly proportional to the number of molecules. FThis means that we can do stoichiometry with volume or moles. FWe had said that stoichiometry implied a ratio of molecules, or moles. Up until now we only used mole ratios. FHowever Avogadro said that the volume is directly proportional to the number of molecules. FThis means that we can do stoichiometry with volume or moles.

38. Example 1 of Gas Stoichiometry FWhat volume of hydrogen is needed to synthesize 6.00 liters of ammonia? FN 2 (g) + 3 H 2 (g) --> 2 NH 3 (g) F6.00 L H 2 x (2 NH 3 /3 H 2 ) = 4.00 L NH 3 FWhat volume of hydrogen is needed to synthesize 6.00 liters of ammonia? FN 2 (g) + 3 H 2 (g) --> 2 NH 3 (g) F6.00 L H 2 x (2 NH 3 /3 H 2 ) = 4.00 L NH 3

39. Example 2 of Gas Stoichiometry FWhat mass of nitrogen is needed to synthesize 20.0 L of ammonia at 1.50 atm and 25 o C? FN 2 (g) + 3 H 2 (g) --> 2 NH 3 (g) F20.0 L NH 3 x (1 N 2 /2 NH 3 ) = 10.0 L N 2  PV M = mRT F(1.5 atm)(10 L)(28 g/mol) = m(.0821Latm/molK)(298K) Fm = 17.2 g N 2 FWhat mass of nitrogen is needed to synthesize 20.0 L of ammonia at 1.50 atm and 25 o C? FN 2 (g) + 3 H 2 (g) --> 2 NH 3 (g) F20.0 L NH 3 x (1 N 2 /2 NH 3 ) = 10.0 L N 2  PV M = mRT F(1.5 atm)(10 L)(28 g/mol) = m(.0821Latm/molK)(298K) Fm = 17.2 g N 2

40. Dalton’s Law FWhen we talk about air pressure, we need to understand that air is not oxygen. FAir is a solution of nitrogen (78.09%), oxygen (20.95%), argon (.93%), and CO 2 (.03%). FSo when we talk about air pressure, which gas are we talking about? FWhen we talk about air pressure, we need to understand that air is not oxygen. FAir is a solution of nitrogen (78.09%), oxygen (20.95%), argon (.93%), and CO 2 (.03%). FSo when we talk about air pressure, which gas are we talking about?

41. ALL OF THEM! FDalton’s Law of Partial Pressures states that the total pressure of a system is equal to the sum of the partial (or individual) pressures of each component. FP total = P 1 + P 2 + … P x FSo if air pressure is 1 atm, then we can assume that the N 2 is.78 atm, the O 2 is.21 atm, and the Ar is about.01 atm. FDalton’s Law of Partial Pressures states that the total pressure of a system is equal to the sum of the partial (or individual) pressures of each component. FP total = P 1 + P 2 + … P x FSo if air pressure is 1 atm, then we can assume that the N 2 is.78 atm, the O 2 is.21 atm, and the Ar is about.01 atm.

42. A Corollary FIf we extend Boyle’s Law and Avogadro’s Law, we could infer that, at constant temperature and volume, the pressure of a gas is directly proportional to its pressure. FP 1 /P total = n 1 /n total FIf we extend Boyle’s Law and Avogadro’s Law, we could infer that, at constant temperature and volume, the pressure of a gas is directly proportional to its pressure. FP 1 /P total = n 1 /n total

43. An important example FA sample of CaCO 3 is heated, releasing CO 2, which is collected over water (a typical practice). FThe pressure in the collection bottle is the sum of the pressure of the CO 2 plus the pressure of the water vapor (since some water always evaporates). FP total = P CO2 + P H2O

44. Sample Water Vapor Pressures

45. So in our example FIf a total pressure of 365 Torr is collected at 25 o C in a 100 ml collection bottle: FWhat is the partial pressure of CO 2 ? FWhat mass of CaCO 3 decomposed? FIf a total pressure of 365 Torr is collected at 25 o C in a 100 ml collection bottle: FWhat is the partial pressure of CO 2 ? FWhat mass of CaCO 3 decomposed?

46. Here’s how it works FP total = P CO2 + P H2O F365 Torr = P CO2 + 23.8 Torr FP CO2 = 341.2 Torr =.449 atm  PV M = mRT F(.449 atm)(.100 L)(44.0 g/mol) = m(.0821Latm/molK)(298K) Fm =.0807 g CO 2 FP total = P CO2 + P H2O F365 Torr = P CO2 + 23.8 Torr FP CO2 = 341.2 Torr =.449 atm  PV M = mRT F(.449 atm)(.100 L)(44.0 g/mol) = m(.0821Latm/molK)(298K) Fm =.0807 g CO 2

47. Corollary Problem FA gas collection bottle contains.25 mol of He,.50 mol Ar, and.75 mol of Ne. If the partial pressure of Helium is 200 Torr: FWhat is the total pressure in the system? FWhat are the partial pressures of Ne and Ar? FA gas collection bottle contains.25 mol of He,.50 mol Ar, and.75 mol of Ne. If the partial pressure of Helium is 200 Torr: FWhat is the total pressure in the system? FWhat are the partial pressures of Ne and Ar?

48. The answers are… Fn He =.25 mol, n Ar =.50 mol, n Ne =.75 mol, P He = 200 Torr. Fn total =1.50 mol FP total /P he = n total /n He FP total /200Torr = 1.50 mol/.25 mol FP total = 1200 Torr FP Ar /P total = n Ar /n total FP ar /1200 =.50 mol/1.50 mol FP Ar = 400 Torr FP Ne = 1200 Torr - 400 Torr - 200 Torr FP ne = 600 Torr Fn He =.25 mol, n Ar =.50 mol, n Ne =.75 mol, P He = 200 Torr. Fn total =1.50 mol FP total /P he = n total /n He FP total /200Torr = 1.50 mol/.25 mol FP total = 1200 Torr FP Ar /P total = n Ar /n total FP ar /1200 =.50 mol/1.50 mol FP Ar = 400 Torr FP Ne = 1200 Torr - 400 Torr - 200 Torr FP ne = 600 Torr

49. Temperature and Kinetic Energy FEarlier, I stated that temperature is a relative measure of molecular motion. FBy definition, Kinetic energy is a measure of the energy of motion. FPretty similar right? FEarlier, I stated that temperature is a relative measure of molecular motion. FBy definition, Kinetic energy is a measure of the energy of motion. FPretty similar right?

50. Yes they are FKE av = 3/2*R*T FThe average kinetic energy depends only on the absolute temperature. FR, the Ideal Gas Law Constant, should be 8.314 J/molK, since we will want the energy in the proper SI unit of Joules. FKE av = 3/2*R*T FThe average kinetic energy depends only on the absolute temperature. FR, the Ideal Gas Law Constant, should be 8.314 J/molK, since we will want the energy in the proper SI unit of Joules.

51. A Thought Question FWhich of the following ideal gases would have the largest average kinetic energy at 25 o C? He, N 2, CO, or H 2

52. They are all the same! FSince Ke av = 3/2*R*T, the mass does not make a difference (ideally). FKE = 3/2*(8.314J/molK)*(298K) FKE = 3716 J/mol FSince Ke av = 3/2*R*T, the mass does not make a difference (ideally). FKE = 3/2*(8.314J/molK)*(298K) FKE = 3716 J/mol

53. Speed vs Kinetic Energy In physics, you learned that KE = 1/2*m*v 2. The velocity, v, describes the speed of an object in a specific direction. If the mass, m, is measured in kg and the velocity is measured in m/s, then the kinetic energy would be measured in Joules.

54. Physics to Chemistry FRewriting the physics version, we could say that v=√(2*KE/m). FIn chemistry, the Kinetic energy is measured in J/mol, so the mass would have to be measured in Kg/mol which is essentially molar mass. FRewriting the physics version, we could say that v=√(2*KE/m). FIn chemistry, the Kinetic energy is measured in J/mol, so the mass would have to be measured in Kg/mol which is essentially molar mass.

55. Root Mean Square Speed FIn Chemistry, we are not worried about velocities in multiple directions. We want an average speed independent of direction. FWe call this V rms - the root mean square speed.  V rms = √(3RT/ M ) FIn Chemistry, we are not worried about velocities in multiple directions. We want an average speed independent of direction. FWe call this V rms - the root mean square speed.  V rms = √(3RT/ M )

56. A Thought Question Revisited FWhich of the following ideal gases would have the largest root mean square speed at 25 o C? He, N 2, CO, or H 2

57. This Time They Are Different  V rms = √(3RT/ M )  For He, V rms = √(3RT/ M ) = √(3*8.314J/molK*298K)/4g/mol = 1363 m/s  For N 2, V rms = √(3RT/ M ) = √(3*8.314J/molK*298K)/28g/mol = 515 m/s FSince the molar mass is the same for N 2 and CO, their V rms would be the same, 515 m/s.  For H 2, V rms = √(3RT/ M ) = √(3*8.314J/molK*298K)/2g/mol = 1928 m/s FBecause H 2 is the lightest, it moves the fastest.  V rms = √(3RT/ M )  For He, V rms = √(3RT/ M ) = √(3*8.314J/molK*298K)/4g/mol = 1363 m/s  For N 2, V rms = √(3RT/ M ) = √(3*8.314J/molK*298K)/28g/mol = 515 m/s FSince the molar mass is the same for N 2 and CO, their V rms would be the same, 515 m/s.  For H 2, V rms = √(3RT/ M ) = √(3*8.314J/molK*298K)/2g/mol = 1928 m/s FBecause H 2 is the lightest, it moves the fastest.

58. And this leads us to… FGraham’s Law FThe rate of effusion (or diffusion) is inversely proportional to the square root of the molar mass. FEffusion is the process of a gas escaping from one container through a small opening. FDiffusion is the process of a gas spreading out in a large container. FGraham’s Law FThe rate of effusion (or diffusion) is inversely proportional to the square root of the molar mass. FEffusion is the process of a gas escaping from one container through a small opening. FDiffusion is the process of a gas spreading out in a large container. Rate 1 *√ M 1 = Rate 2 *√ M 2

59. Rate vs. Speed FWhen we say rate, we are talking about an amount of gas (moles, grams, or even liters) per unit of time. FThis is not the same as speed which is distance over time. FHowever, the main idea is the same; lighter gases move/effuse faster. FWhen we say rate, we are talking about an amount of gas (moles, grams, or even liters) per unit of time. FThis is not the same as speed which is distance over time. FHowever, the main idea is the same; lighter gases move/effuse faster.

60. For example FUnder a given set of conditions, oxygen diffuses at 10 L/hr. A different gas diffuses at 20 L/hr under the same conditions. What is the molar mass of this gas?

61. 2 ways to solve this FBy the equation:  Rate 1 *√ M 1 = Rate 2 *√ M 2  10 L/hr(√32g/mol) = 40 L/hr *√ M 2  M 2 = 2 g/mol FBy Logic: FIf the rate of the unknown gas is 4 times faster, it must be 4 2, or 16, times lighter. F32 g/mol divided by 16 is 2 g/mol. FBy the equation:  Rate 1 *√ M 1 = Rate 2 *√ M 2  10 L/hr(√32g/mol) = 40 L/hr *√ M 2  M 2 = 2 g/mol FBy Logic: FIf the rate of the unknown gas is 4 times faster, it must be 4 2, or 16, times lighter. F32 g/mol divided by 16 is 2 g/mol.

62. Real vs. Ideal FAt the start of the presentation, we talked about the major assumptions of the Kinetic Molecular Theory. FIf a gas obeys the KMT, it is ideal. FIf it doesn’t obey the KMT, it is real. FAt the start of the presentation, we talked about the major assumptions of the Kinetic Molecular Theory. FIf a gas obeys the KMT, it is ideal. FIf it doesn’t obey the KMT, it is real.

63. So What does that Mean? FThe molecules of an ideal gas do not interact with one another, except to collide elastically. FThe molecules of a real gas will interact, to some degree. FSince no gases are always ideal, the trick is to make a real gas behave ideally. FThe molecules of an ideal gas do not interact with one another, except to collide elastically. FThe molecules of a real gas will interact, to some degree. FSince no gases are always ideal, the trick is to make a real gas behave ideally.

64. Real Gases Behaving Ideally FIf we don’t want the molecules attracting or repelling one another, the first issue is to use a nonpolar gas. FIf we use smaller amounts of the gas, there are less chances of them interacting. FIf we don’t want the molecules attracting or repelling one another, the first issue is to use a nonpolar gas. FIf we use smaller amounts of the gas, there are less chances of them interacting.

65. Real Gases Behaving Ideally FIf we put the gas in larger volumes, the molecules will not interact as much. FLikewise, if we keep the gas under low pressure, the molecules will not interact as much. FThis could also be stated by having molecules that have low densities. FIf we put the gas in larger volumes, the molecules will not interact as much. FLikewise, if we keep the gas under low pressure, the molecules will not interact as much. FThis could also be stated by having molecules that have low densities.

66. Real Gases Behaving Ideally FThe smaller the molecules, the less likely they are to interact. FLastly, at higher temperatures the molecules are moving too fast to actually interact with one another - they are more likely to collide elastically. FThe smaller the molecules, the less likely they are to interact. FLastly, at higher temperatures the molecules are moving too fast to actually interact with one another - they are more likely to collide elastically.

67. Phase Diagrams FA phase diagram shows how the different states of matter exist based on the pressure and temperature.

68. A Typical Phase Diagram

69. Water is not Typical…

70. …and Helium is weird!