Without the invisible blanket of air surrounding us, balloons like Mr. Peanut couldn’t float.

Mercury is used in barometers because of its density: 13.6 g/mL. A corresponding water barometer would be 3 stories high! The Journal of Chemical Education published an article about constructing one- de Grys, Hans; Thirty Feet and Rising: Constructing and Using a Water Barometer To Explore Chemical Principles; J. Chem. Educ. 2003, 80, 1156.

FIGURE 9.2 A column of air 1m2 in cross-sectional area extending from the earth’s surface through the upper atmosphere has a mass of about 10,300 kg, producing an atmospheric pressure of approximately 101,000 Pa.

FIGURE 9.1 Molecular view of a gas.

FIGURE 9.1 Molecular view of a gas.

FIGURE 9.5 Boyle’s law.

FIGURE 9.5 Boyle’s law.

FIGURE 9.5 Boyle’s law.

FIGURE 9.5 Boyle’s law.

FIGURE 9.5 Boyle’s law.

FIGURE 9.7 Charles’ law.

FIGURE 9.7 Charles’ law.

FIGURE 9.7 Charles’ law.

FIGURE 9.7 Charles’ law.

FIGURE 9.7 Charles’ law.

FIGURE 9.9 Avogadro’s law.

FIGURE 9.9 Avogadro’s law.

FIGURE 9.9 Avogadro’s law.

FIGURE 9.9 Avogadro’s law.

FIGURE 9.9 Avogadro’s law.

Chapter 5 States of Matter Gases, Liquids, and Solids Denniston Topping Caret 6th Edition Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 9/4 Gases and Gas Pressure Gas mixtures are homogeneous and compressible.

Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 9/5 Gases and Gas Pressure Pressure:

Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 9/6 Gases and Gas Pressure Pressure: Barometer

Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 9/7 Gases and Gas Pressure Barometer Pa torr mm Hg atm bar Units

Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 9/8 Gases and Gas Pressure Pa torr mm Hg atm bar (exact) Conversions Units 1 torr = 1 mm Hg 1 atm = 101 325 Pa (exact) 1 atm = 760 mm Hg (exact) 1 bar = 1 x 105 Pa

Copyright © 2008 Pearson Prentice Hall, Inc. Chapter 9/10 The Gas Laws Ideal Gas: A gas whose behavior follows the gas laws exactly. The physical properties of a gas can be defined by four variables: P pressure T temperature V volume n number of moles

5.1 The Gaseous State Kinetic Molecular Theory of Gases Gases are made up of small atoms or molecules that are in constant, random motion The distance of separation is very large compared to the size of the individual atoms or molecules Gas is mostly empty space All gas particles behave independently No attractive or repulsive forces exist between them

5.1 The Gaseous State Gas particles collide with each other and with the walls of the container without losing energy The energy is transferred from one atom or molecule to another The average kinetic energy of the atoms or molecules increases or decreases in proportion to absolute temperature As temperature goes up, particle speed goes up Kinetic Molecular Theory of Gases

5.1 The Gaseous State Kinetic Molecular Theory of Gases Explains the following statements: Gases are easily compressible – gas is mostly empty space, room for particles to be pushed together Gases will expand to fill any available volume – move freely with sufficient energy to overcome attractive forces Gases have low density – being mostly empty space; gases have low mass per unit volume

5.1 The Gaseous State Gases readily diffuse through each other – they are in continuous motion with paths readily available due to large space between adjacent particles Gases exert pressure on their containers – pressure results from collisions of gas particles with the container walls Gases behave most ideally at low pressure and high temperature Low pressure, average distance of separation is greatest, minimizing interactive forces High temperature, rapid motion overcomes interactive forces more easily

Gas Diffusion 5.1 The Gaseous State Ammonia (17.0g/mol) Ammonia diffused farther in same time, lighter moves faster Hydrogen chloride (36.5g/mol) Top: Start of expt Bottom: End of expt

Boyle’s law - volume of a gas varies inversely with the pressure exerted by the gas if the temperature and number of moles are held constant The product of pressure (P) and volume (V) is a constant Used to calculate Volume resulting from pressure change Pressure resulting from volume change PV = k1 5.1 The Gaseous State Boyle’s Law PiVi = PfVf

5.1 The Gaseous State Application of Boyle’s Law Gas occupies 10.0 L at 1.00 atm pressure Product, PV = (10.0 L) (1.00 atm) = k1 Double the pressure to 2.0 atm, decreases the volume to 5.0 L (2.0 atm)(Vx) = (10.0 L)(1.00 atm) Vx = 5.0 L

5.1 The Gaseous State Boyle’s Law Practice A 5.0 L sample of a gas at 25oC and 3.0 atm is compressed at constant temperature to a volume of 1.0 L. What is the new pressure? A 3.5 L sample of a gas at 1.0 atm is expanded at constant temperature until the pressure is 0.10 atm. What is the volume of the gas?

It is possible to relate gas volume and temperature Charles’s law - volume of a gas varies directly with the absolute temperature (K) if pressure and number of moles of gas are constant Ratio of volume (V) and temperature (T) is a constant 5.1 The Gaseous State Charles’s Law

5.1 The Gaseous State Application of Charles’s Law If a gas occupies 10.0 L at 273 K with V/T = k2 Doubling temperature to 546 K, increases volume to 20.0 L 10.0 L / 273 K = Vf / 546 K

5.1 The Gaseous State Practice with Charles’s Law A 2.5 L sample of gas at 25oC is heated to 50oC at constant pressure. Will the volume double? What would be the volume? What temperature would be required to double the volume?

If a sample of gas undergoes change involving volume, pressure, and temperature simultaneously, use the combined gas law Derived from a combination of Boyle’s law and Charles’s law 5.1 The Gaseous State Combined Gas Law

Calculate the volume of N2 resulting when 0.100 L of the gas is heated from 300. K to 350. K at 1.00 atm What do we know? Pi = 1.00 atm Pf = 1.00 atm Vi = 0.100 L Vf = ? L Ti = 300. K Tf = 350. K Vf = ViTf / Ti this is valid as Pi = Pf Vf = (0.100 L)(350. K) / 300. K = 0.117 L Note the decimal point in the temperature to indicate significance 5.1 The Gaseous State Using the Combined Gas Law

5.1 The Gaseous State Practice With the Combined Gas Law Calculate the temperature when a 0.50 L sample of gas at 1.0 atm and 25oC is compressed to 0.05 L of gas at 5.0 atm.

Avogadro’s Law - equal volumes of any ideal gas contain the same number of moles if measured under the same conditions of temperature and pressure Changes in conditions can be calculated by rewriting the equation 5.1 The Gaseous State Avogadro’s Law

5.1 The Gaseous State Using Avogadro’s Law If 5.50 mol of CO occupy 20.6 L, how many liters will 16.5 mol of CO occupy at the same temperature and pressure? What do we know? Vi = 20.6 L Vf = ? L ni = 5.50 mol nf = 16.5 mol Vf = Vinf / ni = (20.6 L)(16.5 mol) (5.50 mol) = 61.8 L CO

5.1 The Gaseous State Molar volume - the volume occupied by 1 mol of any gas STP – Standard Temperature and Pressure T = 273 K (or 0oC) P = 1 atm At STP the molar volume of any gas is 22.4 L Molar Volume of a Gas

5.1 The Gaseous State Gas Densities Density = mass / volume Calculate the density of 4.00 g He What is the mass of 1 mol of H2? 4.00 g DensityHe = 4.00g / 22.4L = 0.178 g/L at STP

Combining: Boyle’s law (relating volume and pressure) Charles’s law (relating volume and temperature) Avogadro’s law (relating volume to the number of moles) gives the Ideal Gas Law R is a constant, ideal gas constant R = 0.0821 L.Atm/mol.K If units are P in atm, V in L, n in number of moles, T in K PV=nRT 5.1 The Gaseous State The Ideal Gas Law

22.4 L 5.1 The Gaseous State Calculating a Molar Volume Demonstrate molar volume of O2 gas at STP

5.1 The Gaseous State Practice Using the Ideal Gas Law What is the volume of gas occupied by 5.0 g CH4 at 25oC and 1 atm? What is the mass of N2 required to occupy 3.0 L at 100oC and 700 mmHg?

Dalton’s law – a mixture of gases exerts a pressure that is the sum of the pressures that each gas would exert if it were present alone under the same conditions Total pressure of our atmosphere is equal to the sum of the pressures of N2 and O2 (principal components of air) Pt=p1+p2+p3+... 5.1 The Gaseous State Dalton’s Law of Partial Pressures

5.1 The Gaseous State Ideal Gases vs. Real Gases In reality there is no such thing as an ideal gas It is a useful model to explain gas behavior Nonpolar gases behave more ideally than polar gases because attractive forces are present in polar gases