Gases and Kinetic Molecular Theory

Transcription

1 1 Gases and Kinetic Molecular Theory 1

2 CHAPTER GOALS 1. Comparison of Solids, Liquids, and Gases. Composition of the Atmosphere and Some Common Properties of Gases 3. Pressure 4. Boyle s Law: The Volume-Pressure Relationship 5. Charles Law: The Volume-Temperature Relationship; The Absolute Temperature Scale 6. Standard Temperature and Pressure 7. The Combined Gas Law Equation 8. Avogadro s Law and the Standard Molar Volume

3 CHAPTER GOALS 9. Summary of Gas Laws: The Ideal Gas Equation 10. Determination of Molecular Weights and Molecular Formulas of Gaseous Substances 11. Dalton s Law of Partial Pressures 1. Mass-Volume Relationships in Reactions Involving Gases 13. The Kinetic-Molecular Theory 14. Diffusion and Effusion of Gases 15. Real Gases: Deviations from Ideality 3

4 Comparison of Solids, Liquids, and Gases The density of gases is much less than that of solids or liquids. Densities Solid Liquid Gas (g/ml) H O CCl Gas molecules must be very far apart compared to liquids and solids. 4

5 Composition of the Atmosphere and Some Common Properties of Gases Composition of Dry Air Gas % by Volume N O 0.94 Ar 0.93 CO 0.03 He, Ne, Kr, Xe 0.00 CH H

6 Pressure Pressure is force per unit area. lb/in N/m Gas pressure as most people think of it. 6

7 Pressure Atmospheric pressure is measured using a barometer. Definitions of standard pressure 76 cm Hg 760 mm Hg 760 torr 1 atmosphere kpa Hg density 13.6 g/ml 7

8 Boyle s Law: The Volume-Pressure Relationship V 1/P or V k (1/P) or PV k P 1 V 1 k 1 for one sample of a gas. P V k for a second sample of a gas. k 1 k for the same sample of a gas at the same T. Thus we can write Boyle s Law mathematically as P 1 V 1 P V 8

9 Boyle s Law: The Volume-Pressure Relationship Example 1-1: At 5 o C a sample of He has a volume of 4.00 x 10 ml under a pressure of 7.60 x 10 torr. What volume would it occupy under a pressure of.00 atm at the same T? P 1 V 1 1 P V P 1 V 1 V P 1 ( 760 torr torr)( )( mL ml) ) torr torr ml 9

10 Boyle s Law: The Volume-Pressure Relationship Notice that in Boyle s law we can use any pressure or volume units as long as we consistently use the same units for both P 1 and P or V 1 and V. Use your intuition to help you decide if the volume will go up or down as the pressure is changed and vice versa. 10

11 Charles Law: The Volume-Temperature Relationship Gases liquefy before reaching 0K absolute zero C Volume (L) vs. Temperature (K) 11

12 Charles Law: The Volume-Temperature Relationship Charles s law states that the volume of a gas is directly proportional to the absolute temperature at constant pressure. Gas laws must use the Kelvin scale to be correct. Relationship between Kelvin and centigrade. o K C

13 Charles Law: The Volume-Temperature Relationship V T 1 1 Mathematical form of Charles law. k V V V T or or V V kt kt or or k k T T V and k however th e k's are equal T V T 1 1 V T in the most useful form so 13

14 Charles Law: The Volume-Temperature Relationship Example 1-: A sample of hydrogen, H, occupies 1.00 x 10 ml at 5.0 o C and 1.00 atm. What volume would it occupy at 50.0 o C under the same pressure? T T V1 V V1 T V T T T 1 V ml 33 K 98 K 108 ml 1 14

15 Standard Temperature and Pressure Standard temperature and pressure is given the symbol STP. It is a reference point for some gas calculations. Standard P atm or kpa Standard T K or 0.00 o C 15

16 The Combined Gas Law Equation Boyle s and Charles Laws combined into one statement is called the combined gas law equation. Useful when the V, T, and P of a gas are changing. For a Boyle' P V 1 given Boyle' s Law P V 1 1 s 1 Law P V sample P V of gas : Charles' The Charles' V T V T Law V V T Law combined gas law is : P V T k P1 V T 1 1 P V T 16

17 The Combined Gas Law Equation Example 1-3: A sample of nitrogen gas, N, occupies 7.50 x 10 ml at C under a pressure of 8.10 x 10 torr. What volume would it occupy at STP? V 750 ml P T T K P 8101 torr Solve V1 348 K T 810 torr 1 Solve for V 750 ml P V P for V 73 T K 760 torr V? 1 1 P T1 P T1 ( 810 torr )( 750 ml )( 73 K) ( 760 torr )( 348 K) 67 ml V? 73 K P 760 torr TP1 V1 T 17

18 The Combined Gas Law Equation Example 1-4 : A sample of methane, CH 4, occupies.60 x 10 ml at 3 o C under a pressure of atm. At what temperature would it occupy 5.00 x 10 ml under a pressure of 1.0 x 10 3 torr? You do it! 18

19 The Combined Gas Law Equation V P T T ml atm 380 torr 305 K T1 P V P V K ( 305 K)( 100 torr )( 500 ml ) ( 380 torr )( 60 ml ) 1580 o C V P T 500 ml 100 torr? 19

20 Avogadro s Law and the Standard Molar Volume 0

21 Avogadro s Law and the Standard Molar Volume Avogadro s Law states that at the same temperature and pressure, equal volumes of two gases contain the same number of molecules (or moles) of gas. If we set the temperature and pressure for any gas to be STP, then one mole of that gas has a volume called the standard molar volume. The standard molar volume is.4 L at STP. This is another way to measure moles. For gases, the volume is proportional to the number of moles. 11. L of a gas at STP mole 44.8 L? moles 1

22 Avogadro s Law and the Standard Molar Volume Example 1-5: One mole of a gas occupies 36.5 L and its density is 1.36 g/l at a given temperature and pressure. (a) What is its molar mass? (b) What is its density at STP?? g L 136. g mol mol L? g LSTP 49.6 g mol mol.4 L g / mol.1 g/l

23 Summary of Gas Laws: The Ideal Gas Law Boyle s Law - V 1/P (at constant T & n) Charles Law V T (at constant P & n) Avogadro s Law V n (at constant T & P) Combine these three laws into one statement V nt/p Convert the proportionality into an equality. V nrt/p This provides the Ideal Gas Law. PV nrt R is a proportionality constant called the universal gas constant. 3

24 Summary of Gas Laws: The Ideal Gas Law We must determine the value of R. Recognize that for one mole of a gas at 1.00 atm, and 73 K (STP), the volume is.4 L. Use these values in the ideal gas law. R PV nt atm. 4 L ( )( ) ( 1.00 mol )( 73 K) L atm mol K 4

25 Summary of Gas Laws: The Ideal Gas Law R has other values if the units are changed. R J/mol K Use this value in thermodynamics. R kg m /s K mol Use this later in this chapter for gas velocities. R dm 3 kpa/k mol This is R in all metric units. R cal/k mol This the value of R in calories rather than J. 5

26 Summary of Gas Laws: The Ideal Gas Law Example 1-6: What volume would 50.0 g of ethane, C H 6, occupy at 1.40 x 10 o C under a pressure of 1.8 x 10 3 torr? To use the ideal gas law correctly, it is very important that all of your values be in the correct units! 1. T K. P 180 torr (1 atm/760 torr).39 atm g (1 mol/30 g) 1.67 mol 6

27 Summary of Gas Laws: The Ideal Gas Law V n R T P L atm K mol mol K atm atm ( 1.67 mol ) ( ) ( 1.67 mol ) ( 413 K) 3.6 L 7

28 Summary of Gas Laws: The Ideal Gas Law Example 1-7: Calculate the number of moles in, and the mass of, an 8.96 L sample of methane, CH 4, measured at standard conditions. You do it! 8

29 Summary of Gas Laws: The Ideal Gas Law ( 1.00 atm )( L) n PV RT L atm mol K 16.0 g? g CH mol mol ( 73 K) 6.40 g mol CH 4 9

30 Summary of Gas Laws: The Ideal Gas Law Example 1-8: Calculate the pressure exerted by 50.0 g of ethane, C H 6, in a 5.0 L container at 5.0 o C. P You do it! n 1.67 mol and T 98 K n R T P V L atm 1.67 mol K mol K 5.0 L P 1.63 atm ( ) ( ) 30

31 Determination of Molecular Weights and Molecular Formulas of Gaseous Substances Example 1-9: A compound that contains only carbon and hydrogen is 80.0% carbon and 0.0% hydrogen by mass. At STP, 546 ml of the gas has a mass of 0.73 g. What is the molecular (true) formula for the compound? 100 g of compound contains 80 g of C and 0 g of H. 31

32 Determination of Molecular Weights and Molecular Formulas of Gaseous Substances 1mol CC? mol? mol C Catoms 80.0 g g C mol mol C C 1.0 g gc C 1mol? mol H? mol HHatoms 0.0 g H H mol mol H H 1.01g H Determine 3 the the empirical 1.01g H smallest w hole number ratio. formula 1mol is C CH 3 with mass 15 3

33 Determination of Molecular Weights and Molecular Formulas of Gaseous Substances Remember, the molar mass is the mass divided by the number of moles g g n mol mol actual mass 30.0 empirical mass 15.0 Thus the molecular formula is the empirical formula doubled. ( CH ) 3 CH6 33

34 Determination of Molecular Weights and Molecular Formulas of Gaseous Substances Example 1-10: A 1.74 g sample of a compound that contains only carbon and hydrogen contains 1.44 g of carbon and g of hydrogen. At STP 101 ml of the gas has a mass of 0.6 gram. What is its molecular formula? You do it! 34

35 Determination of Molecular Weights and Molecular Formulas of Gaseous Substances 1mol C? mol C atoms 1.44 g C 0.10 mol C 1.0 g C 1mol H? mol H atoms g H 0.97 mol H 1.01g H n PV RT C H ( 1.00 atm)( L) L atm mol K 5 with mass ( 73 K) mol 35

36 Determination of Molecular Weights and Molecular Formulas of Gaseous Substances? g mol 0.6 g mol 58.1g/mol ( C H ) C H

37 Dalton s Law of Partial Pressures Dalton s law states that the pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases. P total P A + P B + P C

38 Dalton s Law of Partial Pressures Example 1-11: If 1.00 x 10 ml of hydrogen, measured at 5.0 o C and 3.00 atm pressure, and 1.00 x 10 ml of oxygen, measured at 5.0 o C and.00 atm pressure, were forced into one of the containers at 5.0 o C, what would be the pressure of the mixture of gases? P P + P Total H O 3.00 atm +.00 atm 5.00 atm 38

39 Dalton s Law of Partial Pressures Vapor Pressure is the pressure exerted by a substance s vapor over the substance s liquid at equilibrium. 39

40 Dalton s Law of Partial Pressures Example 1-1: A sample of hydrogen was collected by displacement of water at 5.0 o C. The atmospheric pressure was 748 torr. What pressure would the dry hydrogen exert in the same container? P P + P P P P Total H H O H Total H O ( 748 4) P torr 74 torr H P from table H O 40

41 Dalton s Law of Partial Pressures Example 1-13: A sample of oxygen was collected by displacement of water. The oxygen occupied 74 ml at 7.0 o C. The barometric pressure was 753 torr. What volume would the dry oxygen occupy at STP? You do it! 41

42 Dalton s Law of Partial Pressures V 74 ml V? 1 T 300 K T 73 K 1 P torr P 760 torr ( ) 1 V ml 73 K 76 torr STP 300 K 760 torr 4

43 Mass-Volume Relationships in Reactions Involving Gases In this section we are looking at reaction stoichiometry, like in Chapter 3, just including gases in the calculations. (s) MnO & KClO KCl + 3 O 3 mol KClO 3 yields mol KCl and 3 mol O (1.6g) yields (74.6g) and 3 (3.0g) Those 3 moles of O can also be thought of as: (s) 3(.4L) or 67. L at STP (g) 43

44 Mass-Volume Relationships in Reactions Involving Gases Example 1-14: What volume of oxygen measured at STP, can be produced by the thermal decomposition of 10.0 g of KClO 3? You do it! 44

45 Mass-Volume Relationships in? L? L STP STP O O Reactions Involving Gases 10.0 g 3.9 L STP KClO O 3 1mol KClO g KClO 3 3 mol O mol KClO 3.4 LSTP O 1mol O 45

46 The Kinetic-Molecular Theory The basic assumptions of kineticmolecular theory are: Postulate 1 Gases consist of discrete molecules that are relatively far apart. Gases have few intermolecular attractions. The volume of individual molecules is very small compared to the gas s volume. Proof - Gases are easily compressible. 46

47 The Kinetic-Molecular Theory Postulate Gas molecules are in constant, random, straight line motion with varying velocities. Proof - Brownian motion displays molecular motion. 47

48 The Kinetic-Molecular Theory Postulate 3 Gas molecules have elastic collisions with themselves and the container. Total energy is conserved during a collision. Proof - A sealed, confined gas exhibits no pressure drop over time. 48

49 The Kinetic-Molecular Theory Postulate 4 The kinetic energy of the molecules is proportional to the absolute temperature. The average kinetic energies of molecules of different gases are equal at a given temperature. Proof - Brownian motion increases as temperature increases. 49

50 The Kinetic-Molecular Theory The kinetic energy of the molecules is proportional to the absolute temperature. The kinetic energy of the molecules is proportional to the absolute temperature. Displayed in a Maxwellian distribution. 50

51 The Kinetic-Molecular Theory The gas laws that we have looked at earlier in this chapter are proofs that kinetic-molecular theory is the basis of gaseous behavior. Boyle s Law P 1/V As the V increases the molecular collisions with container walls decrease and the P decreases. Dalton s Law P total P A + P B + P C +... Because gases have few intermolecular attractions, their pressures are independent of other gases in the container. Charles Law V T An increase in temperature raises the molecular velocities, thus the V increases to keep the P constant. 51

52 The Kinetic-Molecular Theory 5

53 The Kinetic-Molecular Theory The root-mean square velocity of gases is a very close approximation to the average gas velocity. Calculating the root-mean square velocity is simple: RT u rms 3 M To calculate this correctly: m The value of R kg m /s K mol And M must be in kg/mol. 53

54 The Kinetic-Molecular Theory Example 1-17: What is the root mean square velocity of N molecules at room T, 5.0 o C? u rms u rms kg m kg m sec K sec K mol 0.08 kg/mol kg / mol m / s 1159 mi / hr ( 98 ) ( 98K ) 54

55 The Kinetic-Molecular Theory Example 1-18: What is the root mean square velocity of He atoms at room T, 5.0 o C? You do it! 55

56 The Kinetic-Molecular Theory u rms kg m sec K mol kg / mol K ( ) 1363 m / s 3067 mi / hr Can you think of a physical situation that proves He molecules have a velocity that is so much greater than N molecules? What happens to your voice when you breathe He? 56

57 Diffusion and Effusion of Gases Diffusion is the intermingling of gases. Effusion is the escape of gases through tiny holes. 57

58 Diffusion and Effusion of Gases This is a demonstration of diffusion. 58

59 Diffusion and Effusion of Gases The rate of effusion is inversely proportional to the square roots of the molecular weights or densities. R R 1 R R 1 or M M 1 D D 1 59

60 Diffusion and Effusion of Gases Example 1-15: Calculate the ratio of the rate of effusion of He to that of sulfur dioxide, SO, at the same temperature and pressure. R M He SO R M SO He 641. g / mol 4.0 g / mol 16 4 R 4R He SO 60

61 Diffusion and Effusion of Gases Example 1-16: A sample of hydrogen, H, was found to effuse through a pinhole 5. times as rapidly as the same volume of unknown gas (at the same temperature and pressure). What is the molecular weight of the unknown gas? You do it! R H M unk R M M unk 5. 7 unk H Munk.0 g/mol Munk.0 g/mol 7(.0 g/mol) 54 g/mol 61

62 Real Gases: Deviations from Ideality Real gases behave ideally at ordinary temperatures and pressures. At low temperatures and high pressures real gases do not behave ideally. The reasons for the deviations from ideality are: 1. The molecules are very close to one another, thus their volume is important.. The molecular interactions also become important. 6

63 Real Gases: Deviations from Ideality van der Waals equation accounts for the behavior of real gases at low temperatures and high pressures. P + n a ( V nb) nrt V The van der Waals constants a and b take into account two things: 1. a accounts for intermolecular attraction. b accounts for volume of gas molecules At large volumes a and b are relatively small and van der Waal s equation reduces to ideal gas law at high temperatures and low pressures. 63

64 Real Gases: Deviations from Ideality What are the intermolecular forces in gases that cause them to deviate from ideality? 1. For nonpolar gases the attractive forces are London Forces.. For polar gases the attractive forces are dipole-dipole attractions or hydrogen bonds. 64

65 Real Gases: Deviations from Ideality Example 1-19: Calculate the pressure exerted by 84.0 g of ammonia, NH 3, in a 5.00 L container at 00. o C using the ideal gas law. You do it! P n 84.0 g NH3 nrt V ( 4.94 mol ) ( 473 K) P 1mol 17.0 g 4.94 mol L atm mol K 5.00 L 38.4 atm 65

66 Real Gases: Deviations from Ideality Example 1-0: Solve Example 1-19 using the van der Waal s equation. n 4.94 mol a P n a V P ( V - nb ) nrt V - nb L atm mol n a V b L mol nrt 66

67 Real Gases: Deviations from Ideality ( )( 4.94 mol )( 473K) ( 4.94 mol ) ( 4.17 ) L atm L atm mol K P L 5.00 L (4.94 mol )( mol ) ( 5.00 L) L atm P 4.07 atm (39.8 atm L P 35.7 atm which is a 7.6% difference 4.1atm) from ideal mol 67

68 Synthesis Question The lethal dose for hydrogen sulfide is 6.0 ppm. In other words, if in 1 million molecules of air there are six hydrogen sulfide molecules then that air would be deadly to breathe. How many hydrogen sulfide molecules would be required to reach the lethal dose in a room that is 77 feet long, 6 feet wide and 50. feet tall at 1.0 atm and 5.0 o C? 68

69 Synthesis Question 77 ft 1 in ft.54 cm in 347 cm 6 ft 1 in ft.54 cm in 1890 cm 50 ft 1 in ft.54 cm in 154 cm V 347 cm 1890 cm 154 cm cm cm 3 1 L 1000 cm L 69

70 Synthesis Question n PV RT L cm cm T K 6 ( 1atm)( L) ( 0.081L atm )( 98 K) mol K L 76,000 mol 6 of air Lethal molecules 9 76,000 mol molecules of air mol molecule of HS 3 dose molecules of air molecules 6 10 molecules of air of H S 70

71 Group Question Tires on a car are typically filled to a pressure of 35 psi at 3.00 x 10 K. A tire is 16 inches in radius and 8.0 inches in thickness. The wheel that the tire is mounted on is 6.0 inches in radius. What is the mass of the air in the tire? 71

72 1 Gases and Kinetic Molecular Theory 7