Presentation is loading. Please wait.

Presentation is loading. Please wait.

Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo.

Similar presentations


Presentation on theme: "Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo."— Presentation transcript:

1 Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo Tro

2 Tro, Chemistry: A Molecular Approach2 Air Pressure & Shallow Wells water for many homes is supplied by a well less than 30 ft. deep with a pump at the surface the pump removes air from the pipe, decreasing the air pressure in the pipe the outside air pressure then pushes the water up the pipe the maximum height the water will rise is related to the amount of pressure the air exerts

3 Tro, Chemistry: A Molecular Approach3 Atmospheric Pressure pressure is the force exerted over an area on average, the air exerts the same pressure that a column of water 10.3 m high would exert 14.7 lbs./in 2 so if our pump could get a perfect vacuum, the maximum height the column could rise is 10.3 m

4 Tro, Chemistry: A Molecular Approach4 Gases Pushing 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

5 Tro, Chemistry: A Molecular Approach5 The Effect of Gas Pressure the pressure exerted by a gas can cause some amazing and startling effects whenever there is a pressure difference, a gas will flow from area of high pressure to 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

6 Tro, Chemistry: A Molecular Approach6 Atmospheric Pressure Effects differences in air pressure result in weather and wind patterns the higher up 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 rapid changes in atmospheric pressure may cause your ears to pop due to an imbalance in pressure on either side of your ear drum

7 Tro, Chemistry: A Molecular Approach7 Pressure Imbalance in 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.

8 Tro, Chemistry: A Molecular Approach8 The Pressure of a Gas 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 Tro, Chemistry: A Molecular Approach9 Measuring Air Pressure use a barometer column of mercury supported by air pressure force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury gravity

10 Tro, Chemistry: A Molecular Approach10 Common Units of Pressure UnitAverage Air Pressure at Sea Level pascal (Pa),101,325 kilopascal (kPa) atmosphere (atm)1 (exactly) millimeters of mercury (mmHg)760 (exactly) inches of mercury (inHg)29.92 torr (torr)760 (exactly) pounds per square inch (psi, lbs./in 2 )14.7

11 Example 5.1 – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg? since mmHg are smaller than psi, the answer makes sense 1 atm = 14.7 psi, 1 atm = 760 mmHg 132 psi mmHg Check: Solution: Concept Plan: Relationships: Given: Find: psiatmmmHg

12 Tro, Chemistry: A Molecular Approach12 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 pressure 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

13 Tro, Chemistry: A Molecular Approach13 Manometer for this sample, the gas has a larger pressure than the atmosphere, so

14 Tro, Chemistry: A Molecular Approach14 Boyles Law 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

15 Tro, Chemistry: A Molecular Approach15 Boyles Experiment added Hg to a J-tube with air trapped inside used length of air column as a measure of volume

16 Tro, Chemistry: A Molecular Approach16

17 Tro, Chemistry: A Molecular Approach17

18 Tro, Chemistry: A Molecular Approach18 Boyles Experiment, P x V

19 Tro, Chemistry: A Molecular Approach19 When you double the pressure on a gas, the volume is cut in half (as long as the temperature and amount of gas do not change)

20 Tro, Chemistry: A Molecular Approach20 Boyles Law and Diving since water is denser 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 pressure you would not be able to inhale it into your lungs

21 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? since 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: Concept Plan: Relationships: Given: Find: V 1, P 1, P 2 V2V2

22 Tro, Chemistry: A Molecular Approach22 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 2780 mL, what was it originally?

23 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 2780 mL, what was it originally? since 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: Concept Plan: Relationships: Given: Find: V 1, P 1, P 2 V2V2

24 Tro, Chemistry: A Molecular Approach24 Charles Law 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

25 Tro, Chemistry: A Molecular Approach25 Charles Law – A Molecular View 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 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

26 26 The data fall on a straight line. If the lines are extrapolated back to a volume of 0, they all show the same temperature, °C, called absolute zero

27 T(K) = t(°C) , Example 5.3 – A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L? since 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: Concept Plan: Relationships: Given: Find: V 1, V 2, T 2 T1T1

28 Tro, Chemistry: A Molecular Approach28 Practice – The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air?

29 T(K) = t(°C) , The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air? since T and V are directly proportional, when the temperature increases, the volume should increase, and it does V 1 =10.0 L, t 1 = 25.0°C L, t 2 = 250.0°C V 2, L Check: Solution: Concept Plan: Relationships: Given: Find: V 1, T 1, T 2 V2V2

30 Tro, Chemistry: A Molecular Approach30 Avogadros Law 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

31 mol added = n 2 – n 1, 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? since n and V are directly proportional, when the volume increases, the moles should increase, and it does V 1 =4.65 L, V 2 = 6.48 L, n 1 = mol n 2, and added moles Check: Solution: Concept Plan: Relationships: Given: Find: V 1, V 2, n 1 n2n2

32 Tro, Chemistry: A Molecular Approach32 Ideal Gas Law By combing 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 3

33 1 atm = 14.7 psi T(K) = t(°C) 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, since 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: Concept Plan: Relationships: Given: Find: P, V, T, Rn

34 Tro, Chemistry: A Molecular Approach34 Standard Conditions since 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

35 Tro, Chemistry: A Molecular Approach35 Practice – A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions?

36 1 atm = 14.7 psi T(K) = t(°C) 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, since there is more than 1 mole, we expect more than 22.4 L of gas V 1 = 10.0 L, P 1 = 44.1 psi, t 1 = 27 °C, P 2 = 1.00 atm, t 2 = 0°C V 2, L Check: Solution: Concept Plan: Relationships: Given: Find: P 1, V 1, T 1, RnP 2, n, T 2, RV2V2

37 Tro, Chemistry: A Molecular Approach37 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 of different gases have different masses, even though they have the same volume

38 Tro, Chemistry: A Molecular Approach38 Molar Volume

39 Tro, Chemistry: A Molecular Approach39 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

40 Tro, Chemistry: A Molecular Approach40 Gas Density density is directly proportional to molar mass

41 1 atm = 760 mmHg, MM = g T(K) = t(°C) Example 5.7 – Calculate the density of N 2 at 125°C and 755 mmHg since 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: Concept Plan: Relationships: Given: Find: P, MM, T, Rd

42 Tro, Chemistry: A Molecular Approach42 Molar Mass of a Gas 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

43 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 value 31.9 g/mol is reasonable m=0.311g, V=0.225 L, P=886 mmHg, t=55°C, molar mass, g/mol Check: Solution: Concept Plan: Relationships: Given: Find: n, mMMP, V, T, Rn 1 atm = 760 mmHg, T(K) = t(°C) m=0.311g, V=0.225 L, P= atm, T=328 K, molar mass, g/mol

44 Tro, Chemistry: A Molecular Approach44 Practice - Calculate the density of a gas at 775 torr and 27°C if moles weighs g

45 Calculate the density of a gas at 775 torr and 27°C if moles weighs g the value 1.65 g/L is reasonable m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C, density, g/L Check: Solution: Concept Plan: Relationships: Given: Find: V, mdP, n, T, RV 1 atm = 760 mmHg, T(K) = t(°C) m=9.988g, n=0.250 mol, P= atm, T=300. K density, g/L

46 Tro, Chemistry: A Molecular Approach46 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

47 Tro, Chemistry: A Molecular Approach47 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

48 Tro, Chemistry: A Molecular Approach48 Composition of Dry Air

49 Tro, Chemistry: A Molecular Approach49 The partial pressure of each gas in a mixture can be calculated using the ideal gas law

50 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: Concept Plan: Relationships: Given: Find: P tot, P He, P Ne P Ar P tot = P a + P b + etc., 1 atm = 760 mmHg, MM Ar = g/mol 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 )

51 Tro, Chemistry: A Molecular Approach51 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.

52 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: Concept Plan: Relationships: Given: Find: n Xe, V, T, RP Xe P tot, P Xe P Ne

53 Tro, Chemistry: A Molecular Approach53 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, for gases, = volume % / 100% the partial pressure of a gas is equal to the mole fraction of that gas times the total pressure

54 Tro, Chemistry: A Molecular Approach54 Mountain Climbing & Partial Pressure our bodies are adapted to breathe O 2 at a partial pressure of 0.21 atm Sherpa, people native to the Himalaya mountains, are adapted to the much lower partial pressure of oxygen in their air partial pressures of O 2 lower than 0.1 atm will lead to hypoxia unconsciousness or death climbers of Mt Everest carry O 2 in cylinders to prevent hypoxia on top of Mt Everest, P air = atm, so P O2 = atm

55 Tro, Chemistry: A Molecular Approach55 Deep Sea Divers & Partial Pressure its also possible to have too much O 2, a condition called oxygen toxicity P O2 > 1.4 atm oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions its also possible to have too much N 2, a condition called nitrogen narcosis also known as Rapture of the Deep when diving deep, the pressure of the air divers breathe increases – so the partial pressure of the oxygen increases at a depth of 55 m the partial pressure of O 2 is 1.4 atm divers that go below 50 m use a mixture of He and O 2 called heliox that contains a lower percentage of O 2 than air

56 Tro, Chemistry: A Molecular Approach56 Partial Pressure & Diving

57 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 = 43.2 g V = 12.5 L, T = 298 K He, O2, P He, atm, P O2, atm, P total, atm Solution: Concept Plan: Relationships: Given: Find: n tot, V, T, RP tot m gas n gas gas gas, P total P gas MM He = 4.00 g/mol MM O2 = g/mol 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

58 Tro, Chemistry: A Molecular Approach58 Collecting Gases gases are often collected by having them displace water from a container the problem is that since 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*

59 Tro, Chemistry: A Molecular Approach59 Vapor Pressure of Water

60 Tro, Chemistry: A Molecular Approach60 Collecting Gas by Water Displacement

61 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: Concept Plan: Relationships: Given: Find: P tot, P H2O P O2 1 atm = 760 mmHg, P total = P A + P B, O 2 = g/mol V=1.02 L, P O2 = mmHg, T=293 K mass O 2, g P O2,V,Tn O2 g O2

62 Tro, Chemistry: A Molecular Approach62 Practice – 0.12 moles of H 2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

63 V=10.0 L, n H2 =0.12 mol, T=323 K P total, atm 0.12 moles of H 2 is collected over water in a 10.0 L container at 323 K. Find the total pressure. Solution: Concept Plan: Relationships: Given: Find: P H2, P H2O P total 1 atm = 760 mmHg P total = P A + P B, n H2,V,TP H2

64 Tro, Chemistry: A Molecular Approach64 Reactions Involving Gases 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, 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 P, V, T of Gas Amole Amole BP, V, T of Gas B

65 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 : Concept Plan: Relationships: Given: Find: P, n, T, RV 1 atm = 760 mmHg, CH 3 OH = g/mol 1 mol CH 3 OH : 2 mol H 2 g CH 3 OHmol CH 3 OHmol H 2 n H2 = mol, P= atm, T=355 K V H2, L

66 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

67 Tro, Chemistry: A Molecular Approach67 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)

68 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 : Concept Plan: Relationships: Given: Find: P, n, T, RV 1 atm = 760 mmHg, HgO = g/mol 2 mol HgO : 1 mol O 2 g HgOmol HgOmol O 2 n O2 = mol, P=0.750 atm, T=313 K V O2, L

69 Tro, Chemistry: A Molecular Approach69 Properties of Gases 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

70 Tro, Chemistry: A Molecular Approach70 Kinetic Molecular Theory the particles of the gas (either atoms or molecules) are constantly moving the attraction between particles is negligible when the moving particles hit another particle or the container, they do not stick; but they bounce off and continue moving in another direction like billiard balls

71 Tro, Chemistry: A Molecular Approach71 Kinetic Molecular Theory there is a lot of empty space between the particles compared to the size of the particles the average kinetic energy of the 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 particles are moving at the same speed!!

72 Tro, Chemistry: A Molecular Approach72 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.

73 Tro, Chemistry: A Molecular Approach73 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

74 Tro, Chemistry: A Molecular Approach74 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

75 Tro, Chemistry: A Molecular Approach75 Density & Pressure 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

76 Tro, Chemistry: A Molecular Approach76 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

77 Tro, Chemistry: A Molecular Approach77 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 in order to keep the pressure constant, the volume must then increase

78 Tro, Chemistry: A Molecular Approach78 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 in order to keep the pressure constant, the volume must then increase

79 Tro, Chemistry: A Molecular Approach79 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 independent of each other, each gas contributing its own collisions to the container with the same average kinetic energy since the average kinetic energy is the same, the total pressure of the collisions is the same

80 Tro, Chemistry: A Molecular Approach80 Daltons Law & Pressure since the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side

81 Tro, Chemistry: A Molecular Approach81 Deriving the Ideal Gas Law from Kinetic-Molecular Theory pressure = Force total /Area F total = F 1 collision x number of collisions in a particular time interval F 1 collision = mass x 2(velocity)/time interval no. of collisions is proportional to the number of particles within the distance (velocity x time interval) from the wall F total α massvelocity 2 x Area x no. molecules/Volume Pressure α mv 2 x n/V Temperature α mv 2 P α Tn/V, PV=nRT

82 Tro, Chemistry: A Molecular Approach82 Calculating Gas Pressure

83 Tro, Chemistry: A Molecular Approach83 Molecular Velocities all the gas molecules in a sample can travel at different speeds however, the distribution of speeds follows a pattern called a Boltzman distribution we 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

84 Tro, Chemistry: A Molecular Approach84 Boltzman Distribution

85 Tro, Chemistry: A Molecular Approach85 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, 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

86 Tro, Chemistry: A Molecular Approach86 Molecular Speed vs. Molar Mass in order to have the same average kinetic energy, heavier molecules must have a slower average speed

87 Tro, Chemistry: A Molecular Approach87 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

88 Tro, Chemistry: A Molecular Approach88 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

89 T(K) = t(°C) , O 2 = g/mol Ex 5.14 – Calculate the rms velocity of O 2 at 25°C O 2, t = 25°C u rms Solution: Concept Plan: Relationships: Given: Find: MM, Tu rms

90 Tro, Chemistry: A Molecular Approach90 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

91 Tro, Chemistry: A Molecular Approach91 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 both the rates of diffusion and effusion of a gas are 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 the molar mass

92 Tro, Chemistry: A Molecular Approach92 Effusion

93 Tro, Chemistry: A Molecular Approach93 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:

94 Ex 5.15 – Calculate the molar mass of a gas that effuses at a rate times N 2 MM, g/mol Solution: Concept Plan: Relationships: Given: Find: rate A /rate B, MM N2 MM unknown N 2 = g/mol

95 95 Ideal vs. Real Gases 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

96 Tro, Chemistry: A Molecular Approach96 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

97 Tro, Chemistry: A Molecular Approach97 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

98 Tro, Chemistry: A Molecular Approach98 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 called a van der Waals constant and is different for every gas because their molecules are different sizes

99 Tro, Chemistry: A Molecular Approach99 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

100 Tro, Chemistry: A Molecular Approach100 Van der Waals Equation combining the equations to account for molecular volume and intermolecular attractions we get the following equation used for real gases a and b are called van der Waal constants and are different for each gas

101 Tro, Chemistry: A Molecular Approach101 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 ideality 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 ideality for high pressures – meaning the most important factor is the molecular volume

102 Tro, Chemistry: A Molecular Approach102 PV/RT Plots

103 Tro, Chemistry: A Molecular Approach103 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 located in the stratosphere

104 Tro, Chemistry: A Molecular Approach104 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

105 Tro, Chemistry: A Molecular Approach105 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 contributor to acid rain 2 SO 2 + O H 2 O 2 H 2 SO 4 SO 3 + H 2 O H 2 SO 4

106 Tro, Chemistry: A Molecular Approach106 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 contributor to acid rain 4 NO + 3 O H 2 O 4 HNO 3 4 NO 2 + O H 2 O 4 HNO 3

107 Tro, Chemistry: A Molecular Approach107 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

108 Tro, Chemistry: A Molecular Approach108 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 irritants strong oxidizer

109 Tro, Chemistry: A Molecular Approach109 Major Pollutant Levels government regulation has resulted in a decrease in the emission levels for most major pollutants

110 Tro, Chemistry: A Molecular Approach110 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)

111 Tro, Chemistry: A Molecular Approach111 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 removes 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

112 Tro, Chemistry: A Molecular Approach112 Ozone Holes satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions


Download ppt "Chapter 5 Gases Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA 2008, Prentice Hall Chemistry: A Molecular Approach, 1 st Ed. Nivaldo."

Similar presentations


Ads by Google