Chapter 101 Gases

2 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72

Chapter 103 Characteristics of Gases -Expand to fill a volume (expandability) -Compressible -Readily forms homogeneous mixtures with other gases -These behaviors are due to large distances between the gas molecules.

Chapter 104 Pressure Pressure - force acting on an object per unit area.

Chapter 105 Pressure Pressure - force acting on an object per unit area.

Chapter 106 Pressure Pressure - force acting on an object per unit area. -Atmospheric pressure is measured with a barometer. -Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column. -There are several units used for pressure: -Pascal (Pa), N/m 2 -Millimeters of Mercury (mmHg) -Atmospheres (atm)

Chapter 107 Pressure -Conversion Factors -1 atm = 760 mmHg -1 atm = 760 torr -1 atm =  10 5 Pa -1 atm = kPa.

Chapter 108 The Gas Laws -There are four variables required to describe a gas: -Amount of substance: moles -Volume of substance: volume -Pressures of substance: pressure -Temperature of substance: temperature -The gas laws will hold two of the variables constant and see how the other two vary.

Chapter 109 The Gas Laws The Pressure-Volume Relationship: Boyle’s Law Boyle’s Law - The volume of a fixed quantity of gas is inversely proportional to its pressure.

Chapter 1010 The Gas Laws The Pressure-Volume Relationship: Boyle’s Law Boyle’s Law - The volume of a fixed quantity of gas is inversely proportional to its pressure.

Chapter 1011 The Gas Laws The Pressure-Volume Relationship: Boyle’s Law Boyle’s Law - The volume of a fixed quantity of gas is inversely proportional to its pressure.

Chapter 1012 The Gas Laws The Pressure-Volume Relationship: Boyle’s Law

Chapter 1013 The Gas Laws Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases. The Temperature-Volume Relationship: Charles’s Law

Chapter 1014 The Gas Laws Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases. The Temperature-Volume Relationship: Charles’s Law

Chapter 1015 The Gas Laws Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases. The Temperature-Volume Relationship: Charles’s Law

Chapter 1016 The Gas Laws The Temperature-Volume Relationship: Charles’s Law

Chapter 1017 The Gas Laws Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas. The Quantity-Volume Relationship: Avogadro’s Law

Chapter 1018 The Gas Laws Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas. The Quantity-Volume Relationship: Avogadro’s Law

Chapter 1019 The Ideal Gas Equation -Combine the gas laws (Boyle, Charles, Avogadro) yields a new law or equation.

Chapter 1020 The Ideal Gas Equation -Combine the gas laws (Boyle, Charles, Avogadro) yields a new law or equation. Ideal gas equation: PV = nRT R = gas constant = L.atm/mol-K P = pressure (atm)V = volume (L) n = molesT = temperature (K)

Chapter 1021 The Ideal Gas Equation -We define STP (standard temperature and pressure) as 0  C, K, 1 atm. -Volume of 1 mol of gas at STP is 22.4 L.

Chapter 1022 Gas Densities and Molar Mass -Rearranging the ideal-gas equation with M as molar mass yields Further Applications of The Ideal-Gas Equation

Chapter 1023 Gas Mixtures and Partial Pressures Dalton’s Law - In a gas mixture the total pressure is given by the sum of partial pressures of each component: P t = P 1 + P 2 + P 3 + … - The pressure due to an individual gas is called a partial pressure. Dalton’s Law

Chapter 1024 Gas Mixtures and Partial Pressures -The partial pressure of a gas can be determined if you know the mole fraction of the gas of interest and the total pressure of the system. -Let n i be the number of moles of gas i exerting a partial pressure P i, then P i =  i P t where  i is the mole fraction (n i /n t ). Partial Pressures and Mole Fractions

Chapter 1025 Kinetic-Molecular Theory -Theory developed to explain gas behavior -To describe the behavior of a gas, we must first describe what a gas is: –Gases consist of a large number of molecules in constant random motion. –Volume of individual molecules negligible compared to volume of container. –Intermolecular forces (forces between gas molecules) negligible. –Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature. –Average kinetic energy of molecules is proportional to temperature.

Chapter 1026 Kinetic-Molecular Theory Pressure exerted by a gas is the result of bombardment of the walls of the container by the gas molecules. Pressure varies directly with the number of molecules hitting the wall per unit time. –If the volume is reduced the number of impacts is increased; therefore, the pressure is increased. Boyle’s Law

Chapter 1027 Kinetic-Molecular Theory The force that a gas molecule strikes the side of a container is directly proportional to the temperature. As the temperature is increased the kinetic energy (force) of the gas particles increases. For the pressure to be constant, the area the force is applied to must be increased (the volume is increased). Charles’ Law

Chapter 1028 Kinetic-Molecular Theory There are large distances between the gas molecules. The components of a mixture will bombard the walls of the container with the same frequency in the presence of a mixture as it would by itself. Dalton’s Law

Chapter 1029 Kinetic-Molecular Theory As kinetic energy increases, the velocity of the gas molecules increases. Root mean square speed, u, is the speed of a gas molecule having average kinetic energy. Average kinetic energy, , is related to root mean square speed:  = ½mu 2 Molecular Speed

Chapter 1030 Kinetic-Molecular Theory Molecular Speed

Chapter 1031 Molecular Effusion and Diffusion Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas. Mathematically: Molecular Speed

Chapter 1032 Molecular Effusion and Diffusion Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas. Mathematically: Molecular Speed

Chapter 1033 Molecular Effusion and Diffusion Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas. Mathematically: The lower the molar mass, M, the higher the u for that gas at a constant temperature. Molecular Speed

Chapter 1034 Molecular Effusion and Diffusion

Chapter 1035 Molecular Effusion and Diffusion Graham’s Law of Effusion Effusion – The escape of gas through a small opening. Diffusion – The spreading of one substance through another.

Chapter 1036 Molecular Effusion and Diffusion Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass. Graham’s Law of Effusion

Chapter 1037 Molecular Effusion and Diffusion Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass. Graham’s Law of Effusion

Chapter 1038 Molecular Effusion and Diffusion Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass. Gas escaping from a balloon is a good example. Graham’s Law of Effusion

Chapter 1039 Molecular Effusion and Diffusion Graham’s Law of Effusion

Chapter 1040 Molecular Effusion and Diffusion Diffusion of a gas is the spread of the gas through space. Diffusion is faster for light gas molecules. Diffusion is slowed by gas molecules colliding with each other. Average distance of a gas molecule between collisions is called mean free path. Diffusion and Mean Free Path

Chapter 1041 Molecular Effusion and Diffusion Diffusion and Mean Free Path At sea level, mean free path is about 6  cm.

Chapter 1042 Real Gases: Deviations from Ideal Behavior From the ideal gas equation, we have PV = nRT This equation breaks-down at –High pressure At high pressure, the attractive and repulsive forces between gas molecules becomes significant.

Chapter 1043 Real Gases: Deviations from Ideal Behavior

Chapter 1044 Real Gases: Deviations from Ideal Behavior From the ideal gas equation, we have PV = nRT This equation breaks-down at –High pressure At high pressure, the attractive and repulsive forces between gas molecules becomes significant. –Small volume

Chapter 1045 Real Gases: Deviations from Ideal Behavior From the ideal gas equation, we have PV = nRT This equation breaks-down at –High pressure At high pressure, the attractive and repulsive forces between gas molecules becomes significant. –Small volume At small volumes, the volume due to the gas molecules is a source of error.

Chapter 1046 Real Gases: Deviations from Ideal Behavior Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions. The Van der Waals Equation

Chapter 1047 Real Gases: Deviations from Ideal Behavior Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions. The Van der Waals Equation

Chapter 1048 Real Gases: Deviations from Ideal Behavior Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions. a and b are constants, determined by the particular gas. The Van der Waals Equation

Chapter 1049 Homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72

Chapter 1050 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.

Chapter 1051 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant. --BOYLES LAW

Chapter 1052 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant. --BOYLES LAW

Chapter 1053 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant. --BOYLES LAW

Chapter 1054 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant. --BOYLES LAW

Chapter 1055 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.

Chapter 1056 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant. --Charles’ Law

Chapter 1057 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant. --Charles’ Law

Chapter 1058 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant. --Charles’ Law

Chapter 1059 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant. --Charles’ Law

Chapter 1060 Problem A sample of gas occupies a volume of 7.50L at atm and 28.0 o C. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant. --Charles’ Law

Chapter 1061 Problem Calcium hydride, CaH 2, reacts with water to form hydrogen gas: CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g) How many grams of CaH 2 are needed to generate 10.0L of H 2 gas if the pressure of H 2 is 740 torr at 23 o C?

Chapter 1062 Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g) -Calculate moles of H 2 formed -Calculate moles of CaH 2 needed -Convert moles CaH 2 to grams

Chapter 1063 Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

Chapter 1064 Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

Chapter 1065 Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

Chapter 1066 Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

Chapter 1067 Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

Chapter 1068 Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

Chapter 1069 Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)

Chapter 1070 Problem CaH 2 (s) + 2 H 2 O(l)  Ca(OH) 2 (aq) + 2 H 2 (g)