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Gases.

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Presentation on theme: "Gases."— Presentation transcript:

1 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 that determine physical behavior of a gas
Properties that determine physical behavior of a gas

7 Amount Mass or moles

8 Volume l x w x h πr²h

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+ 2πrh sphere: 4πr²

14 Units of Measurement N/m2 Newton per square meter
N/cm Newton per square centimeter Pa Pascal kPa kilopascal Torr Torr mmHg millimeters of mercury lb/in 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 dwater = 1.0 g/cm dmercury = 13.6 g/cm g = 32 ft/s2 Ρ = dgh Pwater = Pmercury dgh = dgh (1.0 g/cm3)(32ft/s2)(110 ft) = (13.6 g/cm3)(32 ft/s2)(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 Measurement of Gas Pressure
Mercury Barometer

21

22 Standard Atmosphere (atm)
Pressure exerted by a mercury column of exactly 760 mm in height the density of Hg = g/cm3 (0oC) The acceleration due to gravity (g) is m/s2 exactly. 1 atm = 760 mmHg 760 Torr lb/in2 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 Kilopascals atm = kPa 1.50 atm x kPa = 152 kPa 1 atm mmHg 1.50 atm x mmHg = 1140 mmHg

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 (Dh) between the two arms of the manometer gives the difference between barometric pressure and the gas pressure

28 Three Possible Relationships
Heights of mercury in both columns are equal if gas pressure and atmospheric pressure are equal. Pgas = Pbar

29 Source of sketch: http://cwx. prenhall

30 Gas pressure is greater than the barometric pressure.
Pgas = Pbar + ∆P

31 Gas pressure is less than the barometric pressure.
Pgas = Pbar + ∆P

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

33 Boyle’s Law Charles’ Law Gay-Lussac’s Law Combined Gas Law
The Simple Gas Laws Boyle’s Law Charles’ Law Gay-Lussac’s 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 Boyle’s Law For a fixed amount of gas at constant temperature, gas volume is inversely proportional to gas pressure. P1V1 = P2V2

38

39 Charles’ Law The volume of a fixed amount of gas at constant pressure is directly proportional to the Kelvin (absolute) temperature. V1 = V2 T T2 or V1T2 = V2T1

40

41

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: V1 = 4.50 L T1 = 300. K T2 = 350. K V2 = ? Equation: V1 = V2 T T2 or V1T2 = V2T1 (4.50 L)(350. K) = V2 (300. K) V2 = 5.25 L

43 Gay-Lussac’s Law The pressure of a sample of gas is directly proportional to the absolute temperature when volume remains constant. P1 = P2 T T2 or P1T2 = P2T1

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.

45

46 Combined Gas Law Pressure and volume are inversely proportional to each other and directly proportional to temperature. P1V1 = P2V2 T T2 or P1V1T2 = P2V2T1

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: P1 = 760 Torr P2 = ? V1 = 12.0 L V2 = 3.5 L T1 = 300 K T2 = 325 K Equation: P1V1 = P2V T T2 or P1V1T2 = P2V2T1 (760 Torr)(12.0 L)(325 K) = ( P2)(3.5 L)(300 K) P2 = 2.8 x 10³ Torr

48 Law of Combining Volumes
Gay-Lussac’s 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.

50

51 Avogadro’s Law Volume & Moles

52 Avogadro’s Explanation of Gay-Lussac’s 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 H2 to 1 volume O2 to 2 volumes H2O leads to a result in which all the atoms present initially are accounted for in the product.

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

54 Avogadro’s 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 1023

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, Saturn’s 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 Titan’s 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 NO2 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 3NO2 (g) + H2O --- > 2HNO3(aq) + NO(g) 92.9 mol NO2 x 2 mol HNO3 3 mol NO2 = 61.9 mol HNO3

69 Dalton’s 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. Ptotal = PA + PB + ……

71 Total Pressure: Mixture of Gases

72 Total Volume Vtotal = VA + VB + …….
The expression for percent by volume (VA / Vtotal) 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 na = Pa = Va ntot Ptot Vtot

75 Total Volume: Mixture of Gases

76 Example: Gas Mixtures & Partial Pressure
A gaseous mixture made from 6.00 g O2 and 9.00 g CH4 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: nO2 = 6.00 g O2 x 1 mol O2 32 g O2 = mol O2 Next ---- >

77 Step 2: Calculate pressure exerted by each PO2 = nRT V
nCH4 = 9.00 g CH4 x mol CH4 16.0 g CH4 = mol CH4 Step 2: Calculate pressure exerted by each PO2 = nRT V = (0.188 mol O2)( L-atm/mol-K)(273 K) 15.0 L = atm

78 PCH4 = (0.563 mol)(0.0821 L-atm/mol-K)(273 K)
Step 3: Add pressures Ptotal = PO2 + PCH4 = atm atm Ptotal = atm

79 Mole fraction: na/ntotal
nO2 = mol ( ) mol nO2 = 0.250 nCH4 = mol ( ) mol nCH4 = 0.750 = 1.000

80 Volume of each Gas PO2 = VO2 Ptot Vtot 0.281 atm = VO2
( ) atm = L VO2 = 3.8 L VCH4 = 15 L – 3.8 L = 11 L

81 Gas Collected Over Water
Wet Gas: mixture of the desired gas and water vapor Pbar = Pgas + Pwater Pgas = Pbar – Pwater Next

82 Pbar = Pgas + Pwater Pgas = Pbar - Pwater

83 Gas Collected Over Water
A sample of KClO3 is partially decomposed, producing O2 gas that is collected over water. The volume of gas collected is L at 26ºC and 765 torr total pressure. 2KClO3(s) > 2 KCl(s) + 3O2(g) Next ---- >

84 How many moles of O2 are collected?
Step 1: Calculate pressure of O2 in mixture PO2 = Ptot – Pwater vapor = 765 torr – 25 torr PO2 = 740 torr Next ---- >

85 nO2 = PV RT = (740 torr)(1atm/760 torr)(0.250 L) ( L-atm/mol-K)(299 K) nO2 = x 10-3 mol Next >

86 How many grams of KClO3 were decomposed?
(9.92 x 10-3 mol O2) x 2 mol KClO3 x g KCLO3 3 mol O mol KClO3 = g KClO3

87 c. When dry, what volume would the collected O2 gas occupy at the same temperature and pressure?
Remove water vapor. P2 = 765 torr Temperature is the same. P1V1 = P2V2 V2 = (740 torr)(0.250 L) (765 torr) V2 = 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

91 Apply to all gases whether alone or mixed with other gases
PV = RT = 2 (KE)av = ½ mv² n Apply to all gases whether alone or mixed with other gases Constant V a. T changes – KE changes b. T increases – pressure increases c. KE changes – speed of molecules changes Constant temperature a. V increases, P decreases b. KE doesn’t change – T didn’t change c. Speed of molecules doesn’t change-KE didn’t 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-1mol-1 M or kgm2/s2 K-1mol-1 Joule = kg·m2/s2 Joule is a unit of measurement for energy.

93 Example How is the root-mean square speed (rms) of F2 molecules in a gas sample changed by An increase in temperature. An increase in volume Mixing with a sample of Ar at the same temperature

94 Answers Increases 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 kgm2/s2 / K·mol MHe : Change g to kg 4.00 g/mol = 4.00 x 10-3 kg/mol Equation: u 2 = 3RT M u 2= 3 ( kgm2/s2 /K·mol)(298K) 4.00 x 10-3 kg/mol u = 1.36 x 103 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 (H2) and uranium hexafluoride (UF6), a gas used in the enrichment process fuel for nuclear reactors. Known: Molar Masses H2 = g/mol UF6 = g/mol (Rate of effusion)² = MU compound MH gas = 2.016 Rate of effusion = 13.21

101 Graham’s 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 = MB (rate of effusion of B)2 MA

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
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 van der Waals Equation 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
Chemistry, 7th ed. Brown, LeMay & Bursten, Prentice Hall


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