1 The Gaseous State of Matter Chapter 12 Hein and Arena Dr. Eugene Passer Chemistry Department Bronx Community College © John Wiley and Company Version 1.1

2 Chapter Outline 12.2 The Kinetic MolecularTheoryThe KineticMolecularTheory 12.3 Measurement of Pressure of GasesMeasurement of Pressure ofGases 12.4 Dependence of Pressure on Number of Molecules and TemperatureDependence of Pressure on Number of Molecules and Temperature Avogadro’s LawAvogadro’s Law Mole-Mass-Volume Relationships of GasesMole-Mass-VolumeRelationships of Gases 12.5 Boyle’s LawBoyle’s Law 12.6 Charles’ LawCharles’ Law Dalton’s Law of Partial PressuresDalton’s Law of PartialPressures Density of GasesDensity of Gases Ideal Gas EquationIdeal Gas Equation 12.7 Gay Lussac’s LawGay Lussac’s Law 12.8 Standard Temperature and PressureStandard Temperature andPressure 12.9 Combined Gas LawsCombined Gas Laws Gas StoichiometryGas Stoichiometry Real GasesReal Gases

3 The Kinetic- Molecular Theory

4 KMT is based on the motions of gas particles. A gas that behaves exactly as outlined by KMT is known as an ideal gas. While no ideal gases are found in nature, real gases can approximate ideal gas behavior under certain conditions of temperature and pressure.

5 Principle Assumptions of the KMT 1.Gases consist of tiny subatomic particles. 2.The distance between particles is large compared with the size of the particles themselves. 3.Gas particles have no attraction for one another.

6 4.Gas particles move in straight lines in all directions, colliding frequently with one another and with the walls of the container. 5.No energy is lost by the collision of a gas particle with another gas particle or with the walls of the container. All collisions are perfectly elastic. Principle Assumptions of the KMT

7 6.The average kinetic energy for particles is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin temperature. Principle Assumptions of the KMT

8 Kinetic Energy

9 All gases have the same average kinetic energy at the same temperature. As a result lighter molecules move faster than heavier molecules. mHmH 2 = 2 mOmO 2 = 32 vHvH 2 vOvO 2 = 1 4 Kinetic Energy

10 Diffusion The ability of two or more gases to mix spontaneously until they form a uniform mixture. Stopcock closed No diffusion occurs Stopcock open Diffusion occurs

11 Effusion A process by which gas molecules pass through a very small orifice from a container at higher pressure to one at lower pressure.

12 Graham’s Law of Effusion The rates of effusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their densities, or molar masses.

13 What is the ratio of the rate of effusion of CO to CO 2 ?

14 Measurement of Pressure of Gases

15 Pressure equals force per unit area.

16 The pressure resulting from the collisions of gas molecules with the walls of the balloon keeps the balloon inflated.

17 The pressure exerted by a gas depends on the Number of gas molecules present. Temperature of the gas. Volume in which the gas is confined.

18 Mercury Barometer A tube of mercury is inverted and placed in a dish of mercury. The barometer is used to measure atmospheric pressure.

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20 Average Composition of Dry Air N2N % O2O % Ar0.93% CO % %Ne He0.0005% GasVolume Percent CH % Kr Xe, H 2, and N 2 OTrace % GasVolume Percent

21 Dependence of Pressure on Number of Molecules and Temperature

22 Pressure is produced by gas molecules colliding with the walls of a container. At a specific temperature and volume, the number of collisions depends on the number of gas molecules present. For an ideal gas the number of collisions is directly proportional to the number of gas molecules present.

23 V = 22.4 L T = O o C The pressure exerted by a gas is directly proportional to the number of molecules present.

24 Dependence of Pressure on Temperature The pressure of a gas in a fixed volume increases with increasing temperature. When the pressure of a gas increases, its kinetic energy increases.kinetic energy The increased kinetic energy of the gas results in more frequent and energetic collisions of the molecules with the walls of the container.

25 Kinetic Energy

26 The pressure of a gas in a fixed volume increases with increasing temperature. Lower T Lower P Higher T Higher P Increased pressure is due to more frequent and more energetic collisions of the gas molecules with the walls of the container at the higher temperature.

27 Boyle’s Law

28 At constant temperature (T), the volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P).

29 Graph of pressure versus volume. This shows the inverse PV relationship of an ideal gas.

30 The effect of pressure on the volume of a gas.

31 Method A. Conversion Factors Step 1. Determine whether volume is being increased or decreased. Initial volume= 8.00 L Final volume= 3.00 L volume decreases  pressure increases An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

32 Step 2. Multiply the original pressure by a ratio of volumes that will result in an increase in pressure. new pressure = original pressure x ratio of volumes An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

33 V 1 = 8.00 LP 1 = 500 torr V 2 = 3.00 L P 2 = ? An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant). Method B. Algebraic Equation Step 1. Organize the given information:

34 Step 2. Write and solve the equation for the unknown. An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

35 Step 3. Put the given information into the equation and calculate. An 8.00 L sample of N 2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

36 Charles’ Law

37 Absolute Zero on the Kelvin Scale If a given volume of any gas at 0 o C is cooled by 1 o C the volume of the gas decreases by. If a given volume of any gas at 0 o C is cooled by 20 o C the volume of the gas decreases by.

38 Absolute Zero on the Kelvin Scale If a given volume of any gas at 0 o C is cooled by 273 o C the volume of the gas decreases by o C (more precisely – o C) is the zero point on the Kelvin scale. It is the temperature at which an ideal gas would have 0 volume.

39 Volume-temperature relationship of methane (CH 4 ).

40 Charles’ Law At constant pressure the volume of a fixed mass of gas is directly proportional to the absolute temperature.

41 Effect of temperature on the volume of a gas. Pressure is constant at 1 atm. When temperature increases at constant pressure, the volume of the gas increases.

42 75 o C = 348 K 250 o C = 523 K A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C? Method A. Conversion Factors Step 1. Change o C to K: o C = K

43 Step 2: Multiply the original volume by a ratio of Kelvin temperatures that will result in an increase in volume: A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?

44 Method B. Algebraic Equation Step 1. Organize the information (remember to make units the same): V 1 = 255 mLT 1 = 75 o C = 348 K V 2 = ?T 2 = 250 o C = 523 K A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?

45 Step 2. Write and solve the equation for the unknown: A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?

46 Step 3. Put the given information into the equation and calculate : V 1 = 255 mLT 1 = 75 o C = 348 K V 2 = ?T 2 = 250 o C = 523 K A 255 mL sample of nitrogen at 75 o C is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250 o C?

47 Gay-Lussac’s Law

48 The pressure of a fixed mass of gas, at constant volume, is directly proportional to the Kelvin temperature.

49 40 o C = 313 K At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen? Method A. Conversion Factors Step 1. Change o C to K: o C = K 100 o C = 373 K temperature increases  pressure increases Determine whether temperature is being increased or decreased.

50 Step 2: Multiply the original pressure by a ratio of Kelvin temperatures that will result in an increase in pressure: At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?

51 A temperature ratio greater than 1 will increase the pressure At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?

52 Method B. Algebraic Equation Step 1. Organize the information (remember to make units the same): P 1 = 21.5 atmT 1 = 40 o C = 313 K P 2 = ?T 2 = 100 o C = 373 K At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?

53 Step 2. Write and solve the equation for the unknown: At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen?

54 Step 3. Put the given information into the equation and calculate: At a temperature of 40 o C an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100 o C what will be the pressure of the oxygen? P 1 = 21.5 atmT 1 = 40 o C = 313 K P 2 = ?T 2 = 100 o C = 373 K

55 Standard Temperature and Pressure

56 StandardStandard Temperature and Pressure Standard Conditions Standard Temperature and Pressure STP K or 0.00 o C 1 atm or 760 torr or 760 mm Hg Selected common reference points of temperature and pressure.

57 Standard (def.) Something set up and established by authority as a rule for the measure of quantity, weight, extent, value, or quality. (Merriam Webster)

58 Combined Gas Laws

59 A combination of Boyle’s and Charles’ Law. Used when pressure and temperature change at the same time. Solve the equation for any one of the 6 variables

60 final volume = ratio of pressures ratio of temperatures initial volume calculate final volume

61 final volume = ratio of pressures ratio of temperatures initial volume increases or decreases volume

62 final volume = ratio of pressures ratio of temperatures initial volume increases or decreases volume

63 A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume? o C = K 0 o C = 273 K -15 o C = 258 K Step 1. Organize the given information, putting temperature in Kelvins:

64 Step 1. Organize the given information, putting temperature in Kelvins: P 1 = 760 torrP 2 = 950 torr V 1 = 465 mLV 2 = ? T 1 = 273 KT 2 = 258 K A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?

65 Method A Conversion Factors Step 2. Set up ratios of T and P A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?

66 Step 3. Multiply the original volumes by the ratios: A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume? P 1 = 760 torrP 2 = 950 torr V 1 = 465 mLV 2 = ? T 1 = 273 KT 2 = 258 K

67 Method B Algebraic Equation A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume? Step 2. Write and solve the equation for the unknown V 2.

68 Step 2 Put the given information into the equation and calculate. A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15 o C, what would be the new volume?

69 Dalton’s Law of Partial Pressures

70 Each gas in a mixture exerts a pressure that is independent of the other gases present. The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture. P total = P a + P b + P c + P d + ….

71 A container contains He at a pressure of 0.50 atm, Ne at a pressure of 0.60 atm, and Ar at a pressure of 1.30 atm. What is the total pressure in the container? P total = P He + P Ne + P Ar P total = 0.5 atm atm atm = 2.40 atm

72 The pressure in the collection container is equal to the atmospheric pressure. The pressure of the gas collected plus the pressure of water vapor at the collection temperature is equal to the atmospheric pressure. Collecting a Gas Sample Over Water

73 Oxygen collected over water.

74 A sample of O 2 was collected in a bottle over water at a temperature of 25 o C when the atmospheric pressure was 760 torr. The vapor pressure of water at 25 o C is 23.8 torr.

75 Gay Lussac’s Law of Combining Volumes N2N2 1 volume H 2 3 volumes → → 2 NH 3 2 volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

76 Gay Lussac’s Law of Combining Volumes When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

77 Avogadro’s Law

78 Avogadro’s Law Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

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80 There are 2 molecules of hydrogen chloride. 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules 1 mol 1 mol 2 mol hydrogen + chlorine  hydrogen chloride Each molecule of hydrogen chloride contains at least 1 atom of hydrogen and 1 atom of chlorine.

81 1 volume 1 volume 2 volumes 1 molecule 1 molecule 2 molecules 1 mol 1 mol 2 mol hydrogen + chlorine → hydrogen chloride Each molecule of hydrogen and each molecule of chlorine must contain at least 2 atoms. H 2 + Cl 2 → 2 HCl

82 Mole-Mass-Volume Relationships

83 Volume of one mole of any gas at STP = 22.4 L.STP 22.4 L at STP is known as the molar volume of any gas.

L at STP is known as the molar volume of any gas.

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86 The density of neon at STP is g/L. What is the molar mass of neon?

87 Density of Gases

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89 Density of Gases liters grams

90 Density of Gases depends on T and P

91 The molar mass of SO 2 is g/mol. Determine the density of SO 2 at STP. 1 mole of any gas occupies 22.4 L at STP

92 Ideal Gas Equation

93 V  P nT

94 V  P nT atmospheres

95 V  P nT liters

96 V  P nT moles

97 V  P nT Kelvin

98 V  P nT Ideal Gas Constant

99 A balloon filled with 5.00 moles of helium gas is at a temperature of 25 o C. The atmospheric pressure is 750. torr. What is the balloon’s volume? Step 1. Organize the given information. Convert temperature to kelvins. K = o C K = 25 o C = 298K Convert pressure to atmospheres.

100 Step 2. Write and solve the ideal gas equation for the unknown. Step 3. Substitute the given data into the equation and calculate. A balloon filled with 5.00 moles of helium gas is at a temperature of 25 o C. The atmospheric pressure is 750. torr. What is the balloon’s volume?

101 Determination of Molecular Weights Using the Ideal Gas Equation

102 Calculate the molar mass of an unknown gas, if g occupies 250 mL at a temperature of 305 K and a pressure of atm. V = 250 mL = Lg = g T = 305 K P = atm

103 Gas Stoichiometry

104 All calculations are done at STP.STP Gases are assumed to behave as ideal gases. A gas not at STP is converted to STP.

105 Definition Stoichiometry: The area of chemistry that deals with the quantitative relationships among reactants and products in a chemical reaction.

106 Gas Stoichiometry Primary conversions involved in stoichiometry.

107 Mole-Volume Calculations Mass-Volume Calculations

108 Step 1 Write the balanced equation 2 KClO 3  2 KCl + 3 O 2 Step 2 The starting amount is mol KClO 3. The conversion is moles KClO 3  moles O 2  liters O 2 What volume of oxygen (at STP) can be formed from mol of potassium chlorate?

109 Step 3. Calculate the moles of O 2, using the mole-ratio method. What volume of oxygen (at STP) can be formed from mol of potassium chlorate? Step 4. Convert moles of O 2 to liters of O 2 2 KClO 3  2KCl + 3 O 2

110 The problem can also be solved in one continuous calculation. What volume of oxygen (at STP) can be formed from mol of potassium chlorate? 2 KClO 3  2KCl + 3 O 2

111 2 Al(s) + 6 HCl(aq)  2AlCl 3 (aq) + 3 H 2 (g) Step 1 Calculate moles of H 2. grams Al  moles Al  moles H 2 What volume of hydrogen, collected at 30.0 o C and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

112 Convert o C to K: 30. o C = 303 K Convert torr to atm: 2 Al(s) + 6 HCl(aq)  2AlCl 3 (aq) + 3 H 2 (g) Step 2 Calculate liters of H 2. What volume of hydrogen, collected at 30.0 o C and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

113 What volume of hydrogen, collected at 30.0 o C and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid? PV = nRT Solve the ideal gas equation for V

114 Volume-Volume Calculations

115 H 2 (g) + Cl 2 (g)  2HCl(g) 1 mol H 2 1 mol Cl 2 2 mol HCl 22.4 L STP 22.4 L STP 2 x 22.4 L STP 1 volume 2 volumes Y volume 2Y volumes For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as mole-mole relationships.

116 What volume of nitrogen will react with 600. mL of hydrogen to form ammonia? What volume of ammonia will be formed? N 2 (g) + 3H 2 (g)  2NH 3 (g)

117 Real Gases

118 Ideal Gas An ideal gas obeys the gas laws. –The volume the molecules of an ideal gas occupy is negligible compared to the volume of the gas. This is true at all temperatures and pressures. –The intermolecular attractions between the molecules of an ideal gas are negligible at all temperatures and pressures.

119 Real Gases Deviations from the gas laws occur at high pressures and low temperatures. –At high pressures the volumes of the real gas molecules are not negligible compared to the volume of the gas –At low temperatures the kinetic energy of the gas molecules cannot completely overcome the intermolecular attractive forces between the molecules.

120