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Copyright © 2011 Pearson Education, Inc. Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College.

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Presentation on theme: "Copyright © 2011 Pearson Education, Inc. Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College."— Presentation transcript:

1 Copyright © 2011 Pearson Education, Inc. Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA

2 Copyright © 2011 Pearson Education, Inc. The Structure of a Gas 2Tro: Chemistry: A Molecular Approach, 2/e Gases are composed of particles that are flying around very fast in their container(s) The particles in straight lines until they encounter either the container wall or another particle, then they bounce off If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there

3 Copyright © 2011 Pearson Education, Inc. Gases Pushing 3Tro: Chemistry: A Molecular Approach, 2/e Gas molecules are constantly in motion As they move and strike a surface, they push on that surface push = force If we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting pressure = force per unit area

4 Copyright © 2011 Pearson Education, Inc. The Effect of Gas Pressure 4Tro: Chemistry: A Molecular Approach, 2/e The pressure exerted by a gas can cause some amazing and startling effects Whenever there is a pressure difference, a gas will flow from an area of high pressure to an area of low pressure the bigger the difference in pressure, the stronger the flow of the gas If there is something in the gass path, the gas will try to push it along as the gas flows

5 Copyright © 2011 Pearson Education, Inc. Air Pressure 5Tro: Chemistry: A Molecular Approach, 2/e The atmosphere exerts a pressure on everything it contacts the atmosphere goes up about 370 miles, but 80% is in the first 10 miles from the earths surface This is the same pressure that a column of water would exert if it were about 10.3 m high

6 Copyright © 2011 Pearson Education, Inc. Atmospheric Pressure Effects Differences in air pressure result in weather and wind patterns The higher in the atmosphere you climb, the lower the atmospheric pressure is around you at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 psi 6Tro: Chemistry: A Molecular Approach, 2/e

7 Copyright © 2011 Pearson Education, Inc. Pressure Imbalance in the Ear If there is a difference in pressure across the eardrum membrane, the membrane will be pushed out – what we commonly call a popped eardrum 7Tro: Chemistry: A Molecular Approach, 2/e

8 Copyright © 2011 Pearson Education, Inc. The Pressure of a Gas 8Tro: Chemistry: A Molecular Approach, 2/e Gas pressure is a result of the constant movement of the gas molecules and their collisions with the surfaces around them The pressure of a gas depends on several factors number of gas particles in a given volume volume of the container average speed of the gas particles

9 Copyright © 2011 Pearson Education, Inc. Measuring Air Pressure gravity 9Tro: Chemistry: A Molecular Approach, 2/e We measure air pressure with a barometer Column of mercury supported by air pressure Force of the air on the surface of the mercury counter balances the force of gravity on the column of mercury

10 Copyright © 2011 Pearson Education, Inc. 1.The height of the column increases because atmospheric pressure decreases with increasing altitude 2.The height of the column decreases because atmospheric pressure decreases with increasing altitude 3.The height of the column decreases because atmospheric pressure increases with increasing altitude 4.The height of the column increases because atmospheric pressure increases with increasing altitude 1.The height of the column increases because atmospheric pressure decreases with increasing altitude 2.The height of the column decreases because atmospheric pressure decreases with increasing altitude 3.The height of the column decreases because atmospheric pressure increases with increasing altitude 4.The height of the column increases because atmospheric pressure increases with increasing altitude Practice – What happens to the height of the column of mercury in a mercury barometer as you climb to the top of a mountain? Tro: Chemistry: A Molecular Approach, 2/e10

11 Copyright © 2011 Pearson Education, Inc. Common Units of Pressure 11Tro: Chemistry: A Molecular Approach, 2/e

12 Copyright © 2011 Pearson Education, Inc. Example 5.1: A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg? because mmHg are smaller than psi, the answer makes sense 1 atm = 14.7 psi, 1 atm = 760 mmHg 132 psi mmHg Check: Solution: Conceptual Plan: Relationships: Given: Find: psiatmmmHg 12Tro: Chemistry: A Molecular Approach, 2/e

13 Copyright © 2011 Pearson Education, Inc. PracticeConvert 45.5 psi into kPa Tro: Chemistry: A Molecular Approach, 2/e13

14 Copyright © 2011 Pearson Education, Inc. PracticeConvert 45.5 psi into kPa because kPa are smaller than psi, the answer makes sense 1 atm = 14.7 psi, 1 atm = kPa psi kPa Check: Solution: Conceptual Plan: Relationships: Given: Find: psiatmkPa 14Tro: Chemistry: A Molecular Approach, 2/e

15 Copyright © 2011 Pearson Education, Inc. Manometers The pressure of a gas trapped in a container can be measured with an instrument called a manometer Manometers are U-shaped tubes, partially filled with a liquid, connected to the gas sample on one side and open to the air on the other A competition is established between the pressures of the atmosphere and the gas The difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere 15Tro: Chemistry: A Molecular Approach, 2/e

16 Copyright © 2011 Pearson Education, Inc. Manometer for this sample, the gas has a larger pressure than the atmosphere, so 16Tro: Chemistry: A Molecular Approach, 2/e

17 Copyright © 2011 Pearson Education, Inc. Boyles Law Robert Boyle (1627–1691) 17Tro: Chemistry: A Molecular Approach, 2/e Pressure of a gas is inversely proportional to its volume constant T and amount of gas graph P vs V is curve graph P vs 1/V is straight line As P increases, V decreases by the same factor P x V = constant P 1 x V 1 = P 2 x V 2

18 Copyright © 2011 Pearson Education, Inc. Boyles Experiment 18Tro: Chemistry: A Molecular Approach, 2/e Added Hg to a J-tube with air trapped inside Used length of air column as a measure of volume

19 Copyright © 2011 Pearson Education, Inc. 19Tro: Chemistry: A Molecular Approach, 2/e

20 Copyright © 2011 Pearson Education, Inc. 20Tro: Chemistry: A Molecular Approach, 2/e

21 Copyright © 2011 Pearson Education, Inc. Boyles Experiment, P x V 21Tro: Chemistry: A Molecular Approach, 2/e

22 Copyright © 2011 Pearson Education, Inc. Boyles Law: A Molecular View Pressure is caused by the molecules striking the sides of the container When you decrease the volume of the container with the same number of molecules in the container, more molecules will hit the wall at the same instant This results in increasing the pressure 22Tro: Chemistry: A Molecular Approach, 2/e

23 Copyright © 2011 Pearson Education, Inc. Boyles Law and Diving Scuba tanks have a regulator so that the air from the tank is delivered at the same pressure as the water surrounding you. This allows you to take in air even when the outside pressure is large. 23Tro: Chemistry: A Molecular Approach, 2/e Because water is more dense than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm at 20 m the total pressure is 3 atm If your tank contained air at 1 atm of pressure, you would not be able to inhale it into your lungs you can only generate enough force to overcome about 1.06 atm

24 Copyright © 2011 Pearson Education, Inc. Boyles Law and Diving 24Tro: Chemistry: A Molecular Approach, 2/e If a diver holds her breath and rises to the surface quickly, the outside pressure drops to 1 atm According to Boyles law, what should happen to the volume of air in the lungs? Because the pressure is decreasing by a factor of 3, the volume will expand by a factor of 3, causing damage to internal organs. Always Exhale When Rising!!

25 Copyright © 2011 Pearson Education, Inc. P 1 V 1 = P 2 V 2 Example 5.2: A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm? because P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does V 1 =7.25 L, P 1 = 4.52 atm, P 2 = 1.21 atm V 2, L Check: Solution: Conceptual Plan: Relationships: Given: Find: V 1, P 1, P 2 V2V2 25Tro: Chemistry: A Molecular Approach, 2/e

26 Copyright © 2011 Pearson Education, Inc. Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to atm. If the volume of the balloon is now 2.78 x 10 3 mL, what was it originally? 26Tro: Chemistry: A Molecular Approach, 2/e

27 Copyright © 2011 Pearson Education, Inc. P 1 V 1 = P 2 V 2, 1 atm = 760 torr (exactly) A balloon is put in a bell jar and the pressure is reduced from 782 torr to atm. If the volume of the balloon is now 2.78x 10 3 mL, what was it originally? because P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does V 2 =2780 mL, P 1 = 762 torr, P 2 = atm V 1, mL Check: Solution: Conceptual Plan: Relationships: Given: Find: V 2, P 1, P 2 V1V1 27Tro: Chemistry: A Molecular Approach, 2/e

28 Copyright © 2011 Pearson Education, Inc. Charless Law Jacques Charles (1746–1823) 28Tro: Chemistry: A Molecular Approach, 2/e Volume is directly proportional to temperature constant P and amount of gas graph of V vs. T is straight line As T increases, V also increases Kelvin T = Celsius T V = constant x T if T measured in Kelvin

29 Copyright © 2011 Pearson Education, Inc. If the lines are extrapolated back to a volume of 0, they all show the same temperature, °C, called absolute zero If you plot volume vs. temperature for any gas at constant pressure, the points will all fall on a straight line 29Tro: Chemistry: A Molecular Approach, 2/e

30 Copyright © 2011 Pearson Education, Inc. The pressure of gas inside and outside the balloon are the same At high temperatures, the gas molecules are moving faster, so they hit the sides of the balloon harder – causing the volume to become larger The pressure of gas inside and outside the balloon are the same At low temperatures, the gas molecules are not moving as fast, so they dont hit the sides of the balloon as hard – therefore the volume is small Charless Law – A Molecular View 30Tro: Chemistry: A Molecular Approach, 2/e

31 Copyright © 2011 Pearson Education, Inc. Example 5.3: A gas has a volume of 2.57 L at 0.00 °C. What was the temperature at 2.80 L? 31 because T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does V 1 =2.57 L, V 2 = 2.80 L, t 2 = 0.00 °C t 1, K and °C Check: Solution: Conceptual Plan: Relationships: Given: Find: V 1, V 2, T 2 T1T1 Tro: Chemistry: A Molecular Approach, 2/e

32 Copyright © 2011 Pearson Education, Inc. Practice – The temperature inside a balloon is raised from 25.0 °C to °C. If the volume of cold air was 10.0 L, what is the volume of hot air? 32Tro: Chemistry: A Molecular Approach, 2/e

33 Copyright © 2011 Pearson Education, Inc. 33 The temperature inside a balloon is raised from 25.0 °C to °C. If the volume of cold air was 10.0 L, what is the volume of hot air? when the temperature increases, the volume should increase, and it does V 1 =10.0 L, t 1 = 25.0 °C L, t 2 = °C V 2, L Check: Solution: Conceptual Plan: Relationships: Given: Find: V 1, T 1, T 2 V2V2 Tro: Chemistry: A Molecular Approach, 2/e

34 Copyright © 2011 Pearson Education, Inc. Avogadros Law Amedeo Avogadro (1776–1856) 34Tro: Chemistry: A Molecular Approach, 2/e Volume directly proportional to the number of gas molecules V = constant x n constant P and T more gas molecules = larger volume Count number of gas molecules by moles Equal volumes of gases contain equal numbers of molecules the gas doesnt matter

35 Copyright © 2011 Pearson Education, Inc. Example 5.4:A mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L? because n and V are directly proportional, when the volume increases, the moles should increase, and they do V 1 = 4.65 L, V 2 = 6.48 L, n 1 = mol n 2, and added moles Check: Solution: Conceptual Plan: Relationships: Given: Find: V 1, V 2, n 1 n2n2 35Tro: Chemistry: A Molecular Approach, 2/e

36 Copyright © 2011 Pearson Education, Inc. Practice If 1.00 mole of a gas occupies 22.4 L at STP, what volume would moles occupy? 36Tro: Chemistry: A Molecular Approach, 2/e

37 Copyright © 2011 Pearson Education, Inc. Practice If 1.00 mole of a gas occupies 22.4 L at STP, what volume would moles occupy? because n and V are directly proportional, when the moles decrease, the volume should decrease, and it does V 1 =22.4 L, n 1 = 1.00 mol, n 2 = mol V2V2 Check: Solution: Conceptual Plan: Relationships: Given: Find: V 1, n 1, n 2 V2V2 37Tro: Chemistry: A Molecular Approach, 2/e

38 Copyright © 2011 Pearson Education, Inc. By combining the gas laws we can write a general equation R is called the gas constant The value of R depends on the units of P and V we will use and convert P to atm and V to L The other gas laws are found in the ideal gas law if two variables are kept constant Allows us to find one of the variables if we know the other three Ideal Gas Law 38Tro: Chemistry: A Molecular Approach, 2/e

39 Copyright © 2011 Pearson Education, Inc. Example 5.6: How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C? 1 mole at STP occupies 22.4 L, because there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas V = 3.24 L, P = 24.3 psi, t = 25 °C n, mol Check: Solution: Conceptual Plan: Relationships: Given: Find: P, V, T, Rn 39Tro: Chemistry: A Molecular Approach, 2/e

40 Copyright © 2011 Pearson Education, Inc. Standard Conditions Because the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions STP Standard pressure = 1 atm Standard temperature = 273 K 0 °C 40Tro: Chemistry: A Molecular Approach, 2/e

41 Copyright © 2011 Pearson Education, Inc. Practice – A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions? 41Tro: Chemistry: A Molecular Approach, 2/e

42 Copyright © 2011 Pearson Education, Inc. A gas occupies 10.0 L at 44.1 psi and 27 °C. What volume will it occupy at standard conditions? 1 mole at STP occupies 22.4 L, because there is more than 1 mole, we expect more than 22.4 L of gas V 1 = 10.0L, P 1 = 44.1 psi, t 1 = 27 °C, P 2 = 1.00 atm, t 2 = 0 °C V 2, L Check: Solution: Conceptual Plan: Relationships: Given: Find: P 1, V 1, T 1, RnP 2, n, T 2, RV2V2 42Tro: Chemistry: A Molecular Approach, 2/e

43 Copyright © 2011 Pearson Education, Inc. Practice Calculate the volume occupied by 637 g of SO 2 (MM 64.07) at 6.08 x 10 4 mmHg and –23 °C 43Tro: Chemistry: A Molecular Approach, 2/e

44 Copyright © 2011 Pearson Education, Inc. PracticeCalculate the volume occupied by 637 g of SO 2 (MM 64.07) at 6.08 x 10 4 mmHg and –23 °C. m SO2 = 637 g, P = 6.08 x 10 4 mmHg, t = 23 °C, V, L Solution: Conceptual Plan: Relationships: Given: Find: P, n, T, RVgn 44Tro: Chemistry: A Molecular Approach, 2/e

45 Copyright © 2011 Pearson Education, Inc. Molar Volume Solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L x molecules of gas notice: the gas is immaterial We call the volume of 1 mole of gas at STP the molar volume it is important to recognize that one mole measures of different gases have different masses, even though they have the same volume 45Tro: Chemistry: A Molecular Approach, 2/e

46 Copyright © 2011 Pearson Education, Inc. Molar Volume 46Tro: Chemistry: A Molecular Approach, 2/e

47 Copyright © 2011 Pearson Education, Inc. Practice How many liters of O STP can be made from the decomposition of g of PbO 2 ? 2 PbO 2 (s) 2 PbO(s) + O 2 (g) (PbO 2 = 239.2, O 2 = 32.00) 47Tro: Chemistry: A Molecular Approach, 2/e

48 Copyright © 2011 Pearson Education, Inc. Practice How many liters of O STP can be made from the decomposition of g of PbO 2 ? 2 PbO 2 (s) 2 PbO(s) + O 2 (g) because less than 1 mole PbO 2, and ½ moles of O 2 as PbO 2, the number makes sense 1 mol O 2 = 22.4 L, 1 mol PbO 2 = 239.2g, 1 mol O 2 2 mol PbO g PbO 2, 2 PbO 2 2 PbO + O 2 L O 2 Check: Solution: Conceptual Plan: Relationships: Given: Find: L O 2 mol PbO 2 g PbO 2 mol O 2 48Tro: Chemistry: A Molecular Approach, 2/e

49 Copyright © 2011 Pearson Education, Inc. Density at Standard Conditions Density is the ratio of mass to volume Density of a gas is generally given in g/L The mass of 1 mole = molar mass The volume of 1 mole at STP = 22.4 L 49Tro: Chemistry: A Molecular Approach, 2/e

50 Copyright © 2011 Pearson Education, Inc. Practice – Calculate the density of N 2 (g) at STP 50Tro: Chemistry: A Molecular Approach, 2/e

51 Copyright © 2011 Pearson Education, Inc. 51 Practice – Calculate the density of N 2 (g) at STP the units and number are reasonable N2,N2, d N2, g/L Check: Solution: Conceptual Plan: Relationships: Given: Find: MMd Tro: Chemistry: A Molecular Approach, 2/e

52 Copyright © 2011 Pearson Education, Inc. Gas Density 52Tro: Chemistry: A Molecular Approach, 2/e Density is directly proportional to molar mass

53 Copyright © 2011 Pearson Education, Inc. Example 5.7: Calculate the density of N 2 at 125°C and 755 mmHg because the density of N 2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is P = 755 mmHg, t = 125 °C, d N2, g/L Check: Solution: Conceptual Plan: Relationships: Given: Find: P, MM, T, Rd 53Tro: Chemistry: A Molecular Approach, 2/e

54 Copyright © 2011 Pearson Education, Inc. Practice Calculate the density of a gas at 775 torr and 27 °C if moles weighs g 54Tro: Chemistry: A Molecular Approach, 2/e

55 Copyright © 2011 Pearson Education, Inc. m=9.988g, n=0.250 mol, P=775 mmHg, t=27 °C, density, g/L m = 9.988g, n = mol, P = atm, T = 300.K density, g/L Solution: Practice Calculate the density of a gas at 775 torr and 27 °C if moles weighs g the unit is correct and the value is reasonable Check: Conceptual Plan: Relationships: Given: Find: V, mdP, n, T, RV 55Tro: Chemistry: A Molecular Approach, 2/e

56 Copyright © 2011 Pearson Education, Inc. Molar Mass of a Gas 56Tro: Chemistry: A Molecular Approach, 2/e One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law

57 Copyright © 2011 Pearson Education, Inc. m=0.311g, V=0.225 L, P=886 mmHg, t=55°C, molar mass, g/mol m=0.311g, V=0.225 L, P= atm, T=328 K, molar mass, g/mol Example 5.8: Calculate the molar mass of a gas with mass g that has a volume of L at 55°C and 886 mmHg the unit is correct, the value is reasonable Check: Solution: Conceptual Plan: Relationships: Given: Find: n, mMMP, V, T, Rn 57Tro: Chemistry: A Molecular Approach, 2/e

58 Copyright © 2011 Pearson Education, Inc. Practice What is the molar mass of a gas if 12.0 g occupies 197 L at 3.80 x 10 2 torr and 127 °C? 58Tro: Chemistry: A Molecular Approach, 2/e

59 Copyright © 2011 Pearson Education, Inc. m=12.0 g, V= 197 L, P=380 torr, t=127°C, molar mass, g/mol m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K, molar mass, g/mol Practice What is the molar mass of a gas if 12.0 g occupies 197 L at 380 torr and 127 °C? the unit is correct and the value is reasonable Check: Solution: Conceptual Plan: Relationships: Given: Find: n, mMMP, V, T, Rn 59Tro: Chemistry: A Molecular Approach, 2/e

60 Copyright © 2011 Pearson Education, Inc. Mixtures of Gases When gases are mixed together, their molecules behave independent of each other all the gases in the mixture have the same volume all completely fill the container each gass volume = the volume of the container all gases in the mixture are at the same temperature therefore they have the same average kinetic energy Therefore, in certain applications, the mixture can be thought of as one gas even though air is a mixture, we can measure the pressure, volume, and temperature of air as if it were a pure substance we can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules 60Tro: Chemistry: A Molecular Approach, 2/e

61 Copyright © 2011 Pearson Education, Inc. 61 Composition of Dry Air Tro: Chemistry: A Molecular Approach, 2/e

62 Copyright © 2011 Pearson Education, Inc. Partial Pressure The pressure of a single gas in a mixture of gases is called its partial pressure We can calculate the partial pressure of a gas if we know what fraction of the mixture it composes and the total pressure or, we know the number of moles of the gas in a container of known volume and temperature The sum of the partial pressures of all the gases in the mixture equals the total pressure Daltons Law of Partial Pressures because the gases behave independently 62Tro: Chemistry: A Molecular Approach, 2/e

63 Copyright © 2011 Pearson Education, Inc. The partial pressure of each gas in a mixture can be calculated using the ideal gas law 63Tro: Chemistry: A Molecular Approach, 2/e

64 Copyright © 2011 Pearson Education, Inc. Example 5.9: Determine the mass of Ar in the mixture the units are correct, the value is reasonable P He =341 mmHg, P Ne =112 mmHg, P tot = 662 mmHg, V = 1.00 L, T=298 K mass Ar, g Check: Solution: Conceptual Plan: Relationships: Given: Find: P tot, P He, P Ne P Ar P Ar = atm, V = 1.00 L, T=298 K mass Ar, g P Ar, V, T nArnAr mArmAr P Ar = P tot – (P He + P Ne ) 64Tro: Chemistry: A Molecular Approach, 2/e

65 Copyright © 2011 Pearson Education, Inc. Practice – Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe. 65Tro: Chemistry: A Molecular Approach, 2/e

66 Copyright © 2011 Pearson Education, Inc. Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe the unit is correct, the value is reasonable P tot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol P Ne, atm Check: Solution: Conceptual Plan: Relationships: Given: Find: n Xe, V, T, RP Xe P tot, P Xe P Ne 66Tro: Chemistry: A Molecular Approach, 2/e

67 Copyright © 2011 Pearson Education, Inc. Mole Fraction The fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes The ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, c for gases, = volume % / 100% The partial pressure of a gas is equal to the mole fraction of that gas times the total pressure 67Tro: Chemistry: A Molecular Approach, 2/e

68 Copyright © 2011 Pearson Education, Inc. Solution: Ex 5.10: Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O 2 at 298 K m He = 24.2 g, m O2 = 4.32 g, V = 12.5 L, T = 298 K He, O2, P He, atm, P O2, atm, P total, atm Conceptual Plan: Relationships: Given: Find: n tot, V, T, RP tot m gas n gas gas gas, P total P gas n He = 6.05 mol, n O2 = mol, V = 12.5 L, T = 298 K He = , O2 = , P He, atm, P O2, atm, P total, atm 68Tro: Chemistry: A Molecular Approach, 2/e

69 Copyright © 2011 Pearson Education, Inc. Practice – Find the mole fraction of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe 69Tro: Chemistry: A Molecular Approach, 2/e

70 Copyright © 2011 Pearson Education, Inc. Find the mole fraction of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe the unit is correct, the value is reasonable P tot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol P Ne, atm Check: Solution: Conceptual Plan: Relationships: Given: Find: n Xe, V, T, RP Xe P tot, P Xe P Ne P tot, P Ne Ne 70Tro: Chemistry: A Molecular Approach, 2/e

71 Copyright © 2011 Pearson Education, Inc. Collecting Gases 71Tro: Chemistry: A Molecular Approach, 2/e Gases are often collected by having them displace water from a container The problem is that because water evaporates, there is also water vapor in the collected gas The partial pressure of the water vapor, called the vapor pressure, depends only on the temperature so you can use a table to find out the partial pressure of the water vapor in the gas you collect if you collect a gas sample with a total pressure of mmHg* at 25 °C, the partial pressure of the water vapor will be mmHg – so the partial pressure of the dry gas will be mmHg Table 5.4*

72 Copyright © 2011 Pearson Education, Inc. Collecting Gas by Water Displacement 72Tro: Chemistry: A Molecular Approach, 2/e

73 Copyright © 2011 Pearson Education, Inc. 73 Vapor Pressure of Water Tro: Chemistry: A Molecular Approach, 2/e

74 Copyright © 2011 Pearson Education, Inc. 74 Ex 5.11: 1.02 L of O 2 collected over water at 293 K with a total pressure of mmHg. Find mass O 2. V=1.02 L, P=755.2 mmHg, T=293 K mass O 2, g Solution: Conceptual Plan: Relationships: Given: Find: P tot, P H2O P O2 V=1.02 L, P O2 = atm, T=293 K mass O 2, g P O2,V,Tn O2 g O2 Tro: Chemistry: A Molecular Approach, 2/e

75 Copyright © 2011 Pearson Education, Inc. Practice – 0.12 moles of H 2 is collected over water in a 10.0 L container at 323 K. Find the total pressure. 75Tro: Chemistry: A Molecular Approach, 2/e

76 Copyright © 2011 Pearson Education, Inc. V=10.0 L, n H2 = 0.12 mol, T = 323 K P total, mmHg moles of H 2 is collected over water in a 10.0 L container at 323 K. Find the total pressure. Solution: Conceptual Plan: Relationships: Given: Find: P H2, P H2O P total n H2,V,TP H2 Tro: Chemistry: A Molecular Approach, 2/e

77 Copyright © 2011 Pearson Education, Inc. Reactions Involving Gases P, V, T of Gas Amole Amole BP, V, T of Gas B 77Tro: Chemistry: A Molecular Approach, 2/e The principles of reaction stoichiometry from Chapter 4 can be combined with the gas laws for reactions involving gases In reactions of gases, the amount of a gas is often given as a volume instead of moles as weve seen, you must state pressure and temperature The ideal gas law allows us to convert from the volume of the gas to moles; then we can use the coefficients in the equation as a mole ratio When gases are at STP, use 1 mol = 22.4 L

78 Copyright © 2011 Pearson Education, Inc. Ex 5.12: What volume of H 2 is needed to make 35.7 g of CH 3 OH at 738 mmHg and 355 K? CO(g) + 2 H 2 (g) CH 3 OH(g) m CH3OH = 37.5g, P=738 mmHg, T=355 K V H2, L Solution : Conceptual Plan: Relationships: Given: Find: P, n, T, RVg CH 3 OHmol CH 3 OHmol H 2 n H2 = mol, P= atm, T=355 K V H2, L 78Tro: Chemistry: A Molecular Approach, 2/e

79 Copyright © 2011 Pearson Education, Inc. Ex 5.13: How many grams of H 2 O form when 1.24 L H 2 reacts completely with O 2 at STP? O 2 (g) + 2 H 2 (g) 2 H 2 O(g) V H2 = 1.24 L, P = 1.00 atm, T = 273 K mass H2O, g Solution : Concept Plan: Relationships: Given: Find: H 2 O = g/mol, 1 mol = 22.4 STP 2 mol H 2 O : 2 mol H 2 g H 2 OL H 2 mol H 2 mol H 2 O 79Tro: Chemistry: A Molecular Approach, 2/e

80 Copyright © 2011 Pearson Education, Inc. Practice – What volume of O 2 at atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s) 2 Hg(l) + O 2 (g) (MM HgO = g/mol) 80Tro: Chemistry: A Molecular Approach, 2/e

81 Copyright © 2011 Pearson Education, Inc. What volume of O 2 at atm and 313 K is generated by the thermolysis of 10.0 g of HgO? 2 HgO(s) 2 Hg(l) + O 2 (g) m HgO = 10.0g, P=0.750 atm, T=313 K V O2, L Solution : Conceptual Plan: Relationships: Given: Find: P, n, T, RVg HgOmol HgOmol O 2 n O2 = mol, P=0.750 atm, T=313 K V O2, L 81Tro: Chemistry: A Molecular Approach, 2/e

82 Copyright © 2011 Pearson Education, Inc. Properties of Gases 82Tro: Chemistry: A Molecular Approach, 2/e Expand to completely fill their container Take the shape of their container Low density much less than solid or liquid state Compressible Mixtures of gases are always homogeneous Fluid

83 Copyright © 2011 Pearson Education, Inc. Kinetic Molecular Theory 83Tro: Chemistry: A Molecular Approach, 2/e The particles of the gas (either atoms or molecules) are constantly moving The attraction between particles is negligible When the moving gas particles hit another gas particle or the container, they do not stick; but they bounce off and continue moving in another direction like billiard balls

84 Copyright © 2011 Pearson Education, Inc. Kinetic Molecular Theory 84Tro: Chemistry: A Molecular Approach, 2/e There is a lot of empty space between the gas particles compared to the size of the particles The average kinetic energy of the gas particles is directly proportional to the Kelvin temperature as you raise the temperature of the gas, the average speed of the particles increases but dont be fooled into thinking all the gas particles are moving at the same speed!!

85 Copyright © 2011 Pearson Education, Inc. Gas Properties Explained – Pressure Because the gas particles are constantly moving, they strike the sides of the container with a force The result of many particles in a gas sample exerting forces on the surfaces around them is a constant pressure 85Tro: Chemistry: A Molecular Approach, 2/e

86 Copyright © 2011 Pearson Education, Inc. Gas Properties Explained – Indefinite Shape and Indefinite Volume Because the gas molecules have enough kinetic energy to overcome attractions, they keep moving around and spreading out until they fill the container. As a result, gases take the shape and the volume of the container they are in. 86Tro: Chemistry: A Molecular Approach, 2/e

87 Copyright © 2011 Pearson Education, Inc. Gas Properties Explained - Compressibility Because there is a lot of unoccupied space in the structure of a gas, the gas molecules can be squeezed closer together 87Tro: Chemistry: A Molecular Approach, 2/e

88 Copyright © 2011 Pearson Education, Inc. Gas Properties Explained – Low Density Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume; the result is they have low density 88Tro: Chemistry: A Molecular Approach, 2/e

89 Copyright © 2011 Pearson Education, Inc. Density & Pressure 89Tro: Chemistry: A Molecular Approach, 2/e Pressure is the result of the constant movement of the gas molecules and their collisions with the surfaces around them When more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure also higher density

90 Copyright © 2011 Pearson Education, Inc. Gas Laws Explained – Boyles Law Boyles Law says that the volume of a gas is inversely proportional to the pressure Decreasing the volume forces the molecules into a smaller space More molecules will collide with the container at any one instant, increasing the pressure 90Tro: Chemistry: A Molecular Approach, 2/e

91 Copyright © 2011 Pearson Education, Inc. Gas Laws Explained – Charless Law Charless Law says that the volume of a gas is directly proportional to the absolute temperature Increasing the temperature increases their average speed, causing them to hit the wall harder and more frequently on average To keep the pressure constant, the volume must then increase 91Tro: Chemistry: A Molecular Approach, 2/e

92 Copyright © 2011 Pearson Education, Inc. Gas Laws Explained – Avogadros Law Avogadros Law says that the volume of a gas is directly proportional to the number of gas molecules Increasing the number of gas molecules causes more of them to hit the wall at the same time To keep the pressure constant, the volume must then increase 92Tro: Chemistry: A Molecular Approach, 2/e

93 Copyright © 2011 Pearson Education, Inc. Gas Laws Explained – Daltons Law of Partial Pressures Daltons Law says that the total pressure of a mixture of gases is the sum of the partial pressures Kinetic-molecular theory says that the gas molecules are negligibly small and dont interact Therefore the molecules behave independently of each other, each gas contributing its own collisions to the container with the same average kinetic energy Because the average kinetic energy is the same, the total pressure of the collisions is the same 93Tro: Chemistry: A Molecular Approach, 2/e

94 Copyright © 2011 Pearson Education, Inc. Daltons Law & Pressure Because the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side 94Tro: Chemistry: A Molecular Approach, 2/e

95 Copyright © 2011 Pearson Education, Inc. Kinetic Energy and Molecular Velocities Average kinetic energy of the gas molecules depends on the average mass and velocity KE = ½mv 2 Gases in the same container have the same temperature, therefore they have the same average kinetic energy If they have different masses, the only way for them to have the same kinetic energy is to have different average velocities lighter particles will have a faster average velocity than more massive particles 95Tro: Chemistry: A Molecular Approach, 2/e

96 Copyright © 2011 Pearson Education, Inc. Molecular Speed vs. Molar Mass To have the same average kinetic energy, heavier molecules must have a slower average speed 96Tro: Chemistry: A Molecular Approach, 2/e

97 Copyright © 2011 Pearson Education, Inc. Temperature and Molecular Velocities _ KE avg = ½N A mu 2 N A is Avogadros number KE avg = 1.5RT R is the gas constant in energy units, J/molK 1 J = 1 kgm 2 /s 2 Equating and solving we get N A mass = molar mass in kg/mol As temperature increases, the average velocity increases 97Tro: Chemistry: A Molecular Approach, 2/e

98 Copyright © 2011 Pearson Education, Inc. Molecular Velocities All the gas molecules in a sample can travel at different speeds However, the distribution of speeds follows a statistical pattern called a Boltzman distribution Ee talk about the average velocity of the molecules, but there are different ways to take this kind of average The method of choice for our average velocity is called the root-mean-square method, where the rms average velocity, u rms, is the square root of the average of the sum of the squares of all the molecule velocities 98Tro: Chemistry: A Molecular Approach, 2/e

99 Copyright © 2011 Pearson Education, Inc. Boltzman Distribution 99Tro: Chemistry: A Molecular Approach, 2/e

100 Copyright © 2011 Pearson Education, Inc. Temperature vs. Molecular Speed As the absolute temperature increases, the average velocity increases the distribution function spreads out, resulting in more molecules with faster speeds 100Tro: Chemistry: A Molecular Approach, 2/e

101 Copyright © 2011 Pearson Education, Inc. Ex 5.14: Calculate the rms velocity of O 2 at 25 °C O 2, t = 25 °C u rms Solution: Conceptual Plan: Relationships: Given: Find: MM, Tu rms 101Tro: Chemistry: A Molecular Approach, 2/e

102 Copyright © 2011 Pearson Education, Inc. Practice – Calculate the rms velocity of CH 4 (MM 16.04) at 25 °C 102Tro: Chemistry: A Molecular Approach, 2/e

103 Copyright © 2011 Pearson Education, Inc. The mass of O 2 (32.00) is 2x the mass of CH 4 (16.04). The rms of CH 4 (681 m/s) is 2x the rms of O 2 (482 m/s) Practice – Calculate the rms velocity of CH 4 at 25 °C CH 4, t = 25 °C u rms Solution: Conceptual Plan: Relationships: Given: Find: MM, Tu rms 103Tro: Chemistry: A Molecular Approach, 2/e

104 Copyright © 2011 Pearson Education, Inc. Mean Free Path Molecules in a gas travel in straight lines until they collide with another molecule or the container The average distance a molecule travels between collisions is called the mean free path Mean free path decreases as the pressure increases 104Tro: Chemistry: A Molecular Approach, 2/e

105 Copyright © 2011 Pearson Education, Inc. Diffusion and Effusion The process of a collection of molecules spreading out from high concentration to low concentration is called diffusion The process by which a collection of molecules escapes through a small hole into a vacuum is called effusion The rates of diffusion and effusion of a gas are both related to its rms average velocity For gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of its molar mass 105Tro: Chemistry: A Molecular Approach, 2/e

106 Copyright © 2011 Pearson Education, Inc. Effusion 106Tro: Chemistry: A Molecular Approach, 2/e

107 Copyright © 2011 Pearson Education, Inc. Grahams Law of Effusion For two different gases at the same temperature, the ratio of their rates of effusion is given by the following equation: Thomas Graham (1805–1869) 107Tro: Chemistry: A Molecular Approach, 2/e

108 Copyright © 2011 Pearson Education, Inc. 108 Ex 5.15: Calculate the molar mass of a gas that effuses at a rate times N 2 MM, g/mol Solution: Conceptual Plan: Relationships: Given: Find: rate A /rate B, MM N2 MM unknown Tro: Chemistry: A Molecular Approach, 2/e

109 Copyright © 2011 Pearson Education, Inc. Practice – Calculate the ratio of rate of effusion for oxygen to hydrogen 109Tro: Chemistry: A Molecular Approach, 2/e

110 Copyright © 2011 Pearson Education, Inc. 110 Practice – Calculate the ratio of rate of effusion for oxygen to hydrogen O 2, g/mol; H g/mol Solution: Conceptual Plan: Relationships: Given: Find: MM O2, MM H2 rateA/rateB This means that, on average, the O 2 molecules are traveling at ¼ the speed of H 2 molecules Tro: Chemistry: A Molecular Approach, 2/e

111 Copyright © 2011 Pearson Education, Inc. Ideal vs. Real Gases 111Tro: Chemistry: A Molecular Approach, 2/e Real gases often do not behave like ideal gases at high pressure or low temperature Ideal gas laws assume 1.no attractions between gas molecules 2.gas molecules do not take up space based on the kinetic-molecular theory At low temperatures and high pressures these assumptions are not valid

112 Copyright © 2011 Pearson Education, Inc. Real Gas Behavior Because real molecules take up space, the molar volume of a real gas is larger than predicted by the ideal gas law at high pressures 112Tro: Chemistry: A Molecular Approach, 2/e

113 Copyright © 2011 Pearson Education, Inc. The Effect of Molecular Volume At high pressure, the amount of space occupied by the molecules is a significant amount of the total volume The molecular volume makes the real volume larger than the ideal gas law would predict van der Waals modified the ideal gas equation to account for the molecular volume b is called a van der Waals constant and is different for every gas because their molecules are different sizes Johannes van der Waals (1837–1923) 113Tro: Chemistry: A Molecular Approach, 2/e

114 Copyright © 2011 Pearson Education, Inc. Real Gas Behavior Because real molecules attract each other, the molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures 114Tro: Chemistry: A Molecular Approach, 2/e

115 Copyright © 2011 Pearson Education, Inc. The Effect of Intermolecular Attractions At low temperature, the attractions between the molecules is significant The intermolecular attractions makes the real pressure less than the ideal gas law would predict van der Waals modified the ideal gas equation to account for the intermolecular attractions a is another van der Waals constant and is different for every gas because their molecules have different strengths of attraction 115Tro: Chemistry: A Molecular Approach, 2/e

116 Copyright © 2011 Pearson Education, Inc. van der Waals Equation Combining the equations to account for molecular volume and intermolecular attractions we get the following equation used for real gases 116Tro: Chemistry: A Molecular Approach, 2/e

117 Copyright © 2011 Pearson Education, Inc. Real Gases A plot of PV/RT vs. P for 1 mole of a gas shows the difference between real and ideal gases It reveals a curve that shows the PV/RT ratio for a real gas is generally lower than ideal for low pressures – meaning the most important factor is the intermolecular attractions It reveals a curve that shows the PV/RT ratio for a real gas is generally higher than ideal for high pressures – meaning the most important factor is the molecular volume 117Tro: Chemistry: A Molecular Approach, 2/e

118 Copyright © 2011 Pearson Education, Inc. PV/RT Plots 118Tro: Chemistry: A Molecular Approach, 2/e

119 Copyright © 2011 Pearson Education, Inc. 119 Structure of the Atmosphere The atmosphere shows several layers, each with its own characteristics The troposphere is the layer closest to the Earths surface circular mixing due to thermal currents – weather The stratosphere is the next layer up less air mixing The boundary between the troposphere and stratosphere is called the tropopause The ozone layer is a layer of high O 3 concentration located in the stratosphere Tro: Chemistry: A Molecular Approach, 2/e

120 Copyright © 2011 Pearson Education, Inc. 120 Air Pollution Air pollution is materials added to the atmosphere that would not be present in the air without, or are increased by, mans activities though many of the pollutant gases have natural sources as well Pollution added to the troposphere has a direct effect on human health and the materials we use because we come in contact with it and the air mixing in the troposphere means that we all get a smell of it! Pollution added to the stratosphere may have indirect effects on human health caused by depletion of ozone and the lack of mixing and weather in the stratosphere means that pollutants last longer before washing out Tro: Chemistry: A Molecular Approach, 2/e

121 Copyright © 2011 Pearson Education, Inc. 121 Pollutant Gases, SO x SO 2 and SO 3, oxides of sulfur, come from coal combustion in power plants and metal refining as well as volcanoes Lung and eye irritants Major contributors to acid rain 2 SO 2 + O H 2 O 2 H 2 SO 4 SO 3 + H 2 O H 2 SO 4 Tro: Chemistry: A Molecular Approach, 2/e

122 Copyright © 2011 Pearson Education, Inc. 122 Pollutant Gases, NO x NO and NO 2, oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plants as well as lightning storms NO 2 causes the brown haze seen in some cities Lung and eye irritants Strong oxidizers Major contributors to acid rain 4 NO + 3 O H 2 O 4 HNO 3 4 NO 2 + O H 2 O 4 HNO 3 Tro: Chemistry: A Molecular Approach, 2/e

123 Copyright © 2011 Pearson Education, Inc. 123 Pollutant Gases, CO CO comes from incomplete burning of fossil fuels in cars, trucks, and power plants Adheres to hemoglobin in your red blood cells, depleting your ability to acquire O 2 At high levels can cause sensory impairment, stupor, unconsciousness, or death Tro: Chemistry: A Molecular Approach, 2/e

124 Copyright © 2011 Pearson Education, Inc. 124 Pollutant Gases, O 3 Ozone pollution comes from other pollutant gases reacting in the presence of sunlight as well as lightning storms known as photochemical smog and ground-level ozone O 3 is present in the brown haze seen in some cities Lung and eye irritant Strong oxidizer Tro: Chemistry: A Molecular Approach, 2/e

125 Copyright © 2011 Pearson Education, Inc. 125 Major Pollutant Levels Government regulation has resulted in a decrease in the emission levels for most major pollutants Tro: Chemistry: A Molecular Approach, 2/e

126 Copyright © 2011 Pearson Education, Inc. 126 Stratospheric Ozone Ozone occurs naturally in the stratosphere Stratospheric ozone protects the surface of the earth from over-exposure to UV light from the Sun O 3 (g) + UV light O 2 (g) + O(g) Normally the reverse reaction occurs quickly, but the energy is not UV light O 2 (g) + O(g) O 3 (g) Tro: Chemistry: A Molecular Approach, 2/e

127 Copyright © 2011 Pearson Education, Inc. 127 Ozone Depletion Chlorofluorocarbons became popular as aerosol propellants and refrigerants in the 1960s CFCs pass through the tropopause into the stratosphere There, CFCs can be decomposed by UV light, releasing Cl atoms CF 2 Cl 2 + UV light CF 2 Cl + Cl Cl atoms catalyze O 3 decomposition and remove O atoms so that O 3 cannot be regenerated NO 2 also catalyzes O 3 destruction Cl + O 3 ClO + O 2 O 3 + UV light O 2 + O ClO + O O 2 + Cl Tro: Chemistry: A Molecular Approach, 2/e

128 Copyright © 2011 Pearson Education, Inc. 128 Ozone Holes Satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions Tro: Chemistry: A Molecular Approach, 2/e


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