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Gases Gases. Characteristics of Gases Expand to fill and assume the shape of their container Next.

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Presentation on theme: "Gases Gases. Characteristics of Gases Expand to fill and assume the shape of their container Next."— Presentation transcript:

1 Gases Gases

2 Characteristics of Gases Expand to fill and assume the shape of their container Next

3 Diffuse into one another and mix in all proportions. Particles move from an area of high concentration to an area of low concentration. Next

4 Invisible Next

5 Expand when heated

6 PropertiesProperties that determine physical behavior of a gas

7 Amount Mass or moles

8 Volume l x w x h π r²h

9 Temperature

10 Pressure The molecules in a gas are in constant motion. The purple balls represent gaseous atoms that collide with each other and the walls of the container. Gas molecules are in constant motion. "Pressure" is a measure of the collisions of the atoms with the container.

11 Pressure Force per unit area Equation: P = F/A F = force A = area Next

12 Force Pushing or pulling on something

13 Formulas For Surface Area Square: 6s² rectangle: 2ab + 2bc + 2ac Cylinder: 2πr²+ 2 πrh sphere: 4πr²

14 Units of Measurement N/m 2 Newton per square meter N/cm 2 Newton per square centimeter Pa Pascal kPa kilopascal Torr mmHg millimeters of mercury lb/in 2 pound per square inch

15 Calculating Pressure Using P = F/A Suppose that a woman weighing 135 lb and wearing high-heeled shoes momentarily places all her weight on the heel of one foot. If the area of the heel is 0.50 in²., calculate the pressure exerted on the underlying surface. P = F/A = 135 lb/0.50 in² P = 270 lb/in²

16 Liquid Pressure The pressure depends height of liquid column and density of liquid. Ρ = dgh = density x acceleration due to gravity x height

17 Calculating Pressure Exerted by a Liquid Calculate the height of a mercury column, in feet, that is equal to a column of water that is 110 ft. high. Known values d water = 1.0 g/cm 3 d mercury = 13.6 g/cm 3 g = 32 ft/s 2 Ρ = dgh P water = P mercury dgh = dgh (1.0 g/cm 3 )(32ft/s 2 )(110 ft) = (13.6 g/cm 3 )(32 ft/s 2 )(h) h = 8.1 ft. Hg

18 Barometer Barometer is a device used to measure the pressure exerted by the atmosphere. Height of mercury varies with atmospheric conditions and with altitude.

19 Aneroid barometer

20 Mercury Barometer Measurement of Gas Pressure


22 Standard Atmosphere (atm) Pressure exerted by a mercury column of exactly 760 mm in height the density of Hg = g/cm 3 (0 o C) The acceleration due to gravity (g) is m/s 2 exactly. 1 atm = 760 mmHg 760 Torr lb/in kPa Check textbook for more values

23 Converting Pressure to an Equivalent Pressure A gas is at a pressure of 1.50 atm. Convert this pressure to a.Kilopascals 1 atm = kPa 1.50 atm x kPa = 152 kPa 1 atm b.mmHg 1.50 atm x 760 mmHg = 1140 mmHg 1 atm

24 Manometers Used to compare the gas pressure with the barometric pressure. Next

25 Types of Manometers Closed-end manometer The gas pressure is equal to the difference in height (Dh) of the mercury column in the two arms of the manometer

26 Closed-end Manometer

27 Open-end Manometer The difference in mercury levels (D h ) between the two arms of the manometer gives the difference between barometric pressure and the gas pressure

28 Three Possible Relationships 1.Heights of mercury in both columns are equal if gas pressure and atmospheric pressure are equal. P gas = P bar


30 2.Gas pressure is greater than the barometric pressure. P > 0 P gas = P bar + P

31 3.Gas pressure is less than the barometric pressure. P < 0 P gas = P bar + P

32 Standard Temperature & Pressure (STP) Temperature = 0ºC Pressure = 1 atm

33 The Simple Gas Laws Boyles Law Charles Law Gay-Lussacs Law Combined Gas Law

34 Temperature & Gas Laws Three temperature scales Fahrenheit (ºF) Celsius (ºC) Kelvin K (no degree symbol) Always use K when performing calculations with the gas law equations. K = (ºC) Example: ºC = 20º K = 293 Next

35 Absolute Zero of Temperature Temperature at which the volume of a hypothetical gas becomes zero ºC Next

36 Absolute or Kelvin Scale Temperature scale that has ºC as its zero. Temperature interval of one Kelvin equals one degree celsius. 1 K = 1 ºC

37 Boyles Law For a fixed amount of gas at constant temperature, gas volume is inversely proportional to gas pressure. P 1 V 1 = P 2 V 2


39 Charles Law The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin (absolute) temperature. V 1 = V 2 T 1 T 2 or V 1 T 2 = V 2 T 1



42 Example. Charles Law A 4.50-L sample of gas is warmed at constant pressure from 300 K to 350 K. What will its final volume be? Given: V 1 = 4.50 L T 1 = 300. K T 2 = 350. K V 2 = ? Equation: V 1 = V 2 T 1 T 2 or V 1 T 2 = V 2 T 1 (4.50 L)(350. K) = V 2 (300. K) V 2 = 5.25 L

43 Gay-Lussacs Law The pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant. P 1 = P 2 T 1 T 2 or P 1 T 2 = P 2 T 1

44 On the next slide (43) The amount of gas and its volume are the same in either case, but if the gas in the ice bath (0 ºC) exerts a pressure of 1 atm, the gas in the boiling-water bath (100 ºC) exerts a pressure of 1.37 atm. The frequency and the force of the molecular collisions with the container walls are greater at the higher temperature.


46 Combined Gas Law Pressure and volume are inversely proportional to each other and directly proportional to temperature. P 1 V 1 = P 2 V 2 T 1 T 2 or P 1 V 1 T 2 = P 2 V 2 T 1

47 Example. Combined Gas Law A sample of gas is pumped from a 12.0 L vessel at 27ºC and 760 Torr pressure to a 3.5-L vessel at 52ºC. What is the final pressure? Given: P 1 = 760 Torr P 2 = ? V 1 = 12.0 L V 2 = 3.5 L T 1 = 300 K T 2 = 325 K Equation: P 1 V 1 = P 2 V 2 T 1 T 2 or P 1 V 1 T 2 = P 2 V 2 T 1 (760 Torr)(12.0 L)(325 K) = ( P 2) (3.5 L)(300 K) P 2 = 2.8 x 10³ Torr

48 Law of Combining Volumes Gay-Lussacs Law

49 Law of Combining Volumes Gases react in volumes that are related as small whole numbers. The small whole numbers are the stoichiometric coefficients.


51 Avogadros Law Volume & Moles

52 Avogadros Explanation of Gay-Lussacs Law When the gases are measured at the same temperature and pressure, each of the identical flasks contains the same number of molecules. Notice how the combining ratio: 2 volumes H 2 to 1 volume O 2 to 2 volumes H 2 O leads to a result in which all the atoms present initially are accounted for in the product.

53 Avogadros Hypothesis Different gases compared at same temperature and pressure a. Equal - equal number of Volumes molecules b. Equal number of – equal moleculess volumes

54 Avogadros Law At a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas. V = c · n V = volume c = constant n= # of moles Doubling the number of moles will cause the volume to double if T and P are constant.

55 More STP Values For Gases 1 mol of a gas = 22.4 L Number of molecules contained in 22.4 L of a gas is x 10 23

56 A Molar Volume of Gas Source: Photo by James Scherer. © Houghton Mifflin Company. All rights reserved.

57 Ideal Gas Law

58 Equation Includes all four gas variables: Volume Pressure Temperature Amount of gas Next

59 PV = nRT Gas that obeys this equation if said to be an ideal gas (or perfect gas). No gas exactly follows the ideal gas law, although many gases come very close at low pressure and/or high temperatures. Ideal gas constant, R, is R = PV nT = 1 atm x 22.4 L 1 mol x K R = L·atm/mol· K

60 Applications of the Ideal Gas Law Molar Mass Density

61 Molar Mass Determination n = number of moles Moles = mass of sample m Molar mass M Ideal Gas Equation: n = PV RT Substituting: m = PV M RT M = mRT PV

62 Example: Molar Mass The density of carbon tetrachloride vapor at 714 torr and 125ºC is 4.43 g/L. What is its molar mass? M = dRT P = (4.43 g/L)( L-atm/mol-k)(398 K) (714 torr x 1 atm/760 torr) M = 154 g/mol

63 Gas Density (d) d = m/V so V = m/d Ideal Gas Equation: PV = nRT Substituting P m = mRT d M m cancels out d = PM RT

64 Finding the Vapor Density of a Substance

65 Example. Vapor Density The mean molar mass of the atmosphere at the surface of Titan, Saturns largest moon, is 28.6 g/mol. The surface temperature is 95 K, and the pressue is 1.6 Earth atm. Assuming ideal behavior, calcuate the density of Titans atmosphere. d = PM RT = (1.6)(1 atm)(28.6 g/mol) ( L-atm/mol-K)(95 K) d = 5.9 g/L

66 Gases in Chemical Reactions Ideal Gas Law & Balanced Chemical Equation

67 Example. Reaction Stoichiometry How many moles of nitric acid can be prepared using 450 L of NO 2 at a pressure of 5.00 atm and a temperature of 295 K? Step 1: Use Ideal Gas law equation n = PV (5.00 atm)(450 L) RT ( L-atm/mol-K)(295 K) n = 92.9 mol Next ---- >

68 Step 2 3NO 2 (g) + H 2 O --- > 2HNO 3(aq) + NO (g) 92.9 mol NO 2 x 2 mol HNO 3 3 mol NO 2 = 61.9 mol HNO 3

69 Daltons Law of Partial Pressure Mixture of Gases

70 Total Pressure The total pressure of a mixture of gases is the sum of the partial pressures of the components of the mixture. P total = P A + P B + ……

71 Total Pressure: Mixture of Gases

72 Total Volume V total = V A + V B + ……. The expression for percent by volume (V A / V total ) x 100 %

73 Total Volume: Mixture of Gases

74 Mole Fraction of Compound in Mixture Fraction of all molecules in the mixture contributed by that component. Sum of all the mole fractions in a mixture is 1. Expression for mole fraction of a substance in terms of P and V n a = P a = V a n tot P tot V tot

75 Total Volume: Mixture of Gases

76 Example: Gas Mixtures & Partial Pressure A gaseous mixture made from 6.00 g O 2 and 9.00 g CH 4 is placed in a 15.0 L vessel at 0ºC. What is the partial pressure of each gas, and what is the total pressure in the vessel? Step 1: n O2 = 6.00 g O 2 x 1 mol O 2 32 g O 2 = mol O 2 Next ---- >

77 n CH4 = 9.00 g CH 4 x 1 mol CH g CH 4 = mol CH 4 Step 2: Calculate pressure exerted by each P O2 = nRT V = (0.188 mol O 2 )( L-atm/mol-K)(273 K) 15.0 L = atm

78 P CH4 = (0.563 mol)( L-atm/mol-K)(273 K) 15.0 L = atm Step 3: Add pressures P total = P O2 + P CH4 = atm atm P total = atm

79 Mole fraction: n a/ n total n O2 = mol ( ) mol n O2 = n CH4 = mol ( ) mol n CH4 = = 1.000

80 Volume of each Gas P O2 = V O2 P tot V tot atm = V O2 ( ) atm = 15 L V O2 = 3.8 L V CH4 = 15 L – 3.8 L = 11 L

81 Gas Collected Over Water Wet Gas: mixture of the desired gas and water vapor P bar = P gas + P water P gas = P bar – P water Next

82 P bar = P gas + P water P gas = P bar - P water

83 Gas Collected Over Water A sample of KClO 3 is partially decomposed, producing O 2 gas that is collected over water. The volume of gas collected is L at 26ºC and 765 torr total pressure. 2KClO 3(s) > 2 KCl (s) + 3O 2(g) Next ---- >

84 a.How many moles of O 2 are collected? Step 1: Calculate pressure of O 2 in mixture P O2 = P tot – P water vapor = 765 torr – 25 torr P O2 = 740 torr Next ---- >

85 n O2 = PV RT = (740 torr)(1atm/760 torr)(0.250 L) ( L-atm/mol-K)(299 K) n O2 = 9.92 x mol Next >

86 b.How many grams of KClO 3 were decomposed? (9.92 x mol O 2 ) x 2 mol KClO 3 x 123 g KCLO 3 3 mol O 2 1 mol KClO 3 = g KClO 3

87 c. When dry, what volume would the collected O 2 gas occupy at the same temperature and pressure? Remove water vapor. P 2 = 765 torr Temperature is the same. P 1 V 1 = P 2 V 2 V 2 = (740 torr)(0.250 L) (765 torr) V 2 = L

88 Kinetic-Molecular Theory of Gases

89 Assumptions The molecules of gases are in rapid random motion. Their average velocities (speeds) are proportional to the absolute (Kelvin) temperature. At the same temperature, the average kinetic energies of the molecules of different gases are equal. Next

90 Relationship Between Temperature and Average Kinetic Energy Higher temperature means greater motion (KE) av = 3 RT 2 Obtained from PV = RT = 2 (KE) av n 3

91 PV = RT = 2 (KE) av = ½ mv² n 3 Apply to all gases whether alone or mixed with other gases 1.Constant V a. T changes – KE changes b. T increases – pressure increases c. KE changes – speed of molecules changes 2.Constant temperature a. V increases, P decreases b. KE doesnt change – T didnt change c. Speed of molecules doesnt change-KE didnt change

92 Root-Mean Square Speed (u 2 ) u 2 represents the average of the squares of the particles velocities. The square root of u 2 is called the root mean square velocity. The unit of measurement is m/s. u 2 = 3RT Units: R = J K -1 mol -1 M or kgm 2 /s 2 K -1 mol -1 Joule = kg·m 2 /s 2 Joule is a unit of measurement for energy.

93 Example How is the root-mean square speed (rms) of F 2 molecules in a gas sample changed by a)An increase in temperature. b)An increase in volume c)Mixing with a sample of Ar at the same temperature

94 Answers a.Increases b.No effect c.No effect

95 Example. Root-Mean Square Speed Calculate the root mean square speed for the atoms in a sample of helium gas at 25ºC. Given: T = 25ºC = 298 K Known: R = J/K·mol or kgm 2 /s 2 / K·mol M He : Change g to kg 4.00 g/mol = 4.00 x kg/mol Equation: u 2 = 3RT M u 2 = 3 ( kgm 2 /s 2 /K·mol)(298K) 4.00 x kg/mol u = 1.36 x 10 3 m/s

96 Effusion The escape of gas molecules from their container through a tiny orifice or pin hole.

97 Model of Gaseous Effusion

98 Diffusion Diffusion is the moving or mixing of molecules of different substances as a result of random molecular motion.

99 Factors That Influence Effusion and Diffusion The greater the temperature the faster the ga will move or vise versa/ If the mass increases, the kinetic energy will also increase if the speed remains constant. If the mass increases, the speed decreases if the kinetic energy remains constant.

100 Example. Rate of Effusion Calculate the ratio of the effusion rates of hydrogen gas (H 2 ) and uranium hexafluoride (UF 6 ), a gas used in the enrichment process fuel for nuclear reactors. Known: Molar Masses H 2 = g/mol UF 6 = g/mol (Rate of effusion)² = M U compound M H gas = Rate of effusion = 13.21

101 Grahams Law The rate of effusion (or diffusion) of two different gases are inversely proportional to the square roots of their molar masses. (rate of effusion of A) 2 = M B (rate of effusion of B) 2 M A

102 Ratios of Rates the square root of two molar masses is also equal to the ratio Molecular speeds Effusion times Distance traveled by molecules Amount of gas effused

103 Based on Assumptions of Kinetic Molecular Theory Rates of diffusion of different gases are inversely proportional to the square roots of their densities. Rates of diffusion are inversely proportional to the square roots of their molar masses.

104 Nonideal (Real) Gases

105 Ideal vs. Nonideal (Real) Gases Ideal High pressure- compressible, volume approaches zero Force of collisions with container wall is great Nonideal High pressure – molecules are practically incompressible Force of collision with container wall is less due to attractive force among the molecules.

106 Behavior of Gases Behave ideally at High temperatures Low pressures Behave nonideally at Low temperatures High pressures

107 van der Waals Equation

108 Equation corrects for volume and intermolecular forces (P + n²a/V²)(V-nb) = nRT n²a/V² = related to intermolecular forces of attraction n²a/V² is added to P = measured pressure is lower than expected a & b have specific values for particular gases V - nb = free volume within the gas

109 Intermolecular Force of Attraction Attractive forces of the orange molecules for the purple molecule cause the purple molecule to exert less force when it collides with the wall than it would if these attractive forces did not exist.

110 Source of Sketches/problems bbooks/hillchem3/medialib/media_po rtfolio/04.html bbooks/hillchem3/medialib/media_po rtfolio/04.html Chemistry, 7 th ed. Brown, LeMay & Bursten, Prentice Hall

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