A Gas Uniformly fills any container. Easily compressed. Mixes completely with any other gas. Exerts pressure on its surroundings. Copyright © Cengage Learning. All rights reserved

Pressure SI units = Newton/meter2 = 1 Pascal (Pa) 1 standard atmosphere = 101,325 Pa 1 standard atmosphere = 1 atm = 760 mm Hg = 760 torr Copyright © Cengage Learning. All rights reserved

Barometer Device used to measure atmospheric pressure. Mercury flows out of the tube until the pressure of the column of mercury standing on the surface of the mercury in the dish is equal to the pressure of the air on the rest of the surface of the mercury in the dish. Copyright © Cengage Learning. All rights reserved

Manometer Device used for measuring the pressure of a gas in a container. Copyright © Cengage Learning. All rights reserved

Pressure Conversions: An Example The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals. Copyright © Cengage Learning. All rights reserved

Liquid Nitrogen and a Balloon Copyright © Cengage Learning. All rights reserved

Liquid Nitrogen and a Balloon What happened to the gas in the balloon? A decrease in temperature was followed by a decrease in the pressure and volume of the gas in the balloon. Copyright © Cengage Learning. All rights reserved

Liquid Nitrogen and a Balloon This is an observation (a fact). It does NOT explain “why,” but it does tell us “what happened.” Copyright © Cengage Learning. All rights reserved

Gas laws can be deduced from observations like these. Mathematical relationships among the properties of a gas (Pressure, Volume, Temperature and Moles) can be discovered. Copyright © Cengage Learning. All rights reserved

Boyle’s Law: Dependence of volume (V) on pressure (P) Gas Laws The volume of a gas depends on its Temperature, its pressure, & the amount of gas (moles of gas). The gas laws define the dependence of the volume on each of the factors individually and in combinations. Boyle’s Law: Dependence of volume (V) on pressure (P) Charles’ Law: Dependence of volume (V) on temperature (T) Avogadro’s Law: Dependence of Volume (V) on amount of gas (n) Combined Gas Law: Dependence of Volume (V) on P & T Ideal Gas Law: Dependence of Volume on P, T and n

Boyle’s Law: The volume of a fixed amount of gas at constant T varies inversely as its pressure. When pressure increases volume decreases and vice versa. Prediction of Boyle’s Law: volume = 0 when pressure = infinity. Reality: When the pressure is increased, the volume decreases and then the gas condenses to become a liquid ( as long as the temperature is below critical temperature). Boyle’s law is not valid for real gases at very high pressures.

Boyle’s Law Pressure and volume are inversely related (constant T, temperature, and n, # of moles of gas). PV = k (k is a constant for a given sample of air at a specific temperature) Copyright © Cengage Learning. All rights reserved

Boyle’s Law V a 1/P P x V = constant P1 x V1 = P2 x V2 5.3

EXERCISE! A sample of helium gas occupies 12.4 L at 23°C and 0.956 atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant? 9.88 L (0.956 atm)(12.4 L) = (1.20 atm)(V2) The new volume is 9.88 L. Copyright © Cengage Learning. All rights reserved

Charles’s Law Volume and Temperature (in Kelvin) are directly related (constant P and n). V=bT (b is a proportionality constant) K = °C + 273 0 K is called absolute zero. Copyright © Cengage Learning. All rights reserved

Charles’s law predicts that the volume of a gas increases with increasing temperature and decreases with decreasing temperature. Charles’ law predicts that the volume of a gas = 0 when its temperature = 0K = -273.15 C Reality: As the gas cooled at constant pressure, its volume decreases up to the point when it condenses. Gas laws only apply to gases. Charles’ law fails at low temperatures.

A sample of carbon monoxide gas occupies 3. 20 L at 125 0C A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? 5.3

EXERCISE! Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)? 0.849 L The new volume will become 0.849 L. Remember to convert the temperatures to Kelvin. (1.30 L) / (24.7+273) = V2 / (-78.5 + 273) Copyright © Cengage Learning. All rights reserved

Avogadro’s Law Volume and number of moles are directly related (constant T and P). V = an (a is a proportionality constant) Copyright © Cengage Learning. All rights reserved

EXERCISE! If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure? 76.3 L The new volume is 76.3 L. (2.45 mol) / (89.0 L) = (2.10 mol) / (V2) Copyright © Cengage Learning. All rights reserved

We can bring all of these laws together into one comprehensive law: V = bT (constant P and n) V = an (constant T and P) V = (constant T and n) PV = nRT (where R = 0.08206 L·atm/mol·K, the universal gas constant) Copyright © Cengage Learning. All rights reserved

EXERCISE! An automobile tire at 23°C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire. 3.27 mol The number of moles of air is 3.27 mol. (3.18 atm)(25.0 L) = n(0.08206)(23+273) Copyright © Cengage Learning. All rights reserved

EXERCISE! What is the pressure in a 304.0 L tank that contains 5.670 kg of helium at 25°C? 114 atm The pressure is 114 atm. Remember to convert kg of helium into moles of helium before using the ideal gas law. (P)(304.0) = ([5.670×1000]/4.003)(0.0821)(25+273) Copyright © Cengage Learning. All rights reserved

EXERCISE! At what temperature (in °C) does 121 mL of CO2 at 27°C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm? 696°C The new temperature is 696°C. (1.05 atm)(121 mL) / (27+273) = (1.40 atm)(293 mL) / (T2 + 273) Copyright © Cengage Learning. All rights reserved

Molar Volume of an Ideal Gas For 1 mole of an ideal gas at 0°C and 1 atm, the volume of the gas is 22.42 L. STP = standard temperature and pressure 0°C and 1 atm Therefore, the molar volume is 22.42 L at STP. Copyright © Cengage Learning. All rights reserved

EXERCISE! A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O2 are present? 3.57 g 3.57 g of O2 are present. (1.00 atm)(2.50 L) = n(0.08206)(273) n = 0.112 mol O2 × 32.00 g/mol = 3.57 g O2 Copyright © Cengage Learning. All rights reserved

Molar Mass of a Gas d = density of gas T = temperature in Kelvin P = pressure of gas R = universal gas constant Copyright © Cengage Learning. All rights reserved

What is the density of F2 at STP (in g/L)? 1.70 g/L EXERCISE! What is the density of F2 at STP (in g/L)? 1.70 g/L d = (MM)(P) / (R)(T) d = (19.00g/mol × 2)(1.00 atm) / (0.08206)(273 K) d = 1.70 g/L Copyright © Cengage Learning. All rights reserved

Exercise If 1.84 g of a gas occupies 500 mL at 15 C and 825 torr, what is the molar mass? 80.1 g/mol

Gas Stoichiometry What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l) 5.60 g C6H12O6 = 0.187 mol CO2 5.5

1. Calculate the volume of NH3(g) needed at 20. 0C and 25 1. Calculate the volume of NH3(g) needed at 20.0C and 25.0 atm to react with 150 kg of H2SO4 2NH3(g) + H2SO4(aq)  (NH4)2SO4(aq)

The balanced equation for the reaction between Al and O2 forming Aluminum oxide, Al2O3 is 4 Al(s) + 3O2(g)  2Al2O3(s) Calculate the volume of O2(g) at a pressure of 782 torr and a temperature of 25°C required to completely react with 54.0 g of Al.

For a mixture of gases in a container, PTotal = P1 + P2 + P3 + . . . The total pressure exerted is the sum of the pressures that each gas would exert if it were alone. Copyright © Cengage Learning. All rights reserved

Copyright © Cengage Learning. All rights reserved

1. A mixture of 4. 05g N2, 6. 05g Kr are placed in a 6 1. A mixture of 4.05g N2, 6.05g Kr are placed in a 6.10L vessel at 25.0C. Assume that the gases behave ideally a. Calculate the partial pressures of each of the gases in the mixture. b. Calculate the total pressure due to the mixture of gases. 2. A gas mixture contains 2.80 g N2(g) and 0.800 g O2(g) in a 2.00 L container at 23.0°C. Calculate the pressure due to this mixture.

Pressure of a component gas A in a container of Volume V & temperature T, PA = nART/V Total pressure of the mixture of gases in the same container and at the same temperature of which A is a component, Ptotal = ntotalRT/V PA/Ptotal = nA/nTotal = mole fraction of A in mixture = XA. PA/Ptotal = XA = nA/ntotal Partial pressure of a component/Total pressure of gas mixture = mole fraction of the component gas = moles of component gas/moles of all gases in the mixture.

A gas mixture contains 12. 47 g N2, 8. 15 g NH3 and 1. 98 g H2 A gas mixture contains 12.47 g N2, 8.15 g NH3 and 1.98 g H2. If the total pressure due to this mixture is 2.35 atm, calculate the partial pressure of NH3 in this mixture. What are the partial pressures of each component in a mixture of 60.67 g N2, 12.45g H2, 2.38g NH3. if the total pressure of the mixture is 25.4 atm?

Calculate the new partial pressure of oxygen. 6.13 atm EXERCISE! 27.4 L of oxygen gas at 25.0°C and 1.30 atm, and 8.50 L of helium gas at 25.0°C and 2.00 atm were pumped into a tank with a volume of 5.81 L at 25°C. Calculate the new partial pressure of oxygen. 6.13 atm Calculate the new partial pressure of helium. 2.93 atm Calculate the new total pressure of both gases. 9.06 atm For oxygen: P1V1=P2V2 (1.30 atm)(27.4 L) = (P2)(5.81 L) P2 = 6.13 atm For helium: (2.00 atm)(8.50 L) = (P2)(5.81 L) P2 = 2.93 atm The total pressure is therefore 6.13 + 2.93 = 9.06 atm. Copyright © Cengage Learning. All rights reserved

Postulates of the Kinetic Molecular Theory 1) The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). Copyright © Cengage Learning. All rights reserved

Postulates of the Kinetic Molecular Theory 2) The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. Copyright © Cengage Learning. All rights reserved

Postulates of the Kinetic Molecular Theory 3) The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other. Copyright © Cengage Learning. All rights reserved

Postulates of the Kinetic Molecular Theory The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. Average kinetic energy (KE) for 1 mole of gas at a temperature T(K) = (3/2) RT; R = 8.314 J/K.mol Predictions: Irrespective of the gas, average KE is the same for all gases at the same T Higher the T, higher the average KE Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! He H2 You are holding two balloons of the same volume. One contains helium, and one contains hydrogen. Complete each of the following statements with “different” or “the same” and be prepared to justify your answer. Copyright © Cengage Learning. All rights reserved

He H2 The pressures of the gas in the two balloons are __________. CONCEPT CHECK! He H2 The pressures of the gas in the two balloons are __________. the same Copyright © Cengage Learning. All rights reserved

He H2 The temperatures of the gas in the two balloons are __________. CONCEPT CHECK! He H2 The temperatures of the gas in the two balloons are __________. the same Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! He H2 The numbers of moles of the gas in the two balloons are __________. the same Copyright © Cengage Learning. All rights reserved

He H2 The densities of the gas in the two balloons are __________. CONCEPT CHECK! He H2 The densities of the gas in the two balloons are __________. different Copyright © Cengage Learning. All rights reserved

You then cool the gas sample. CONCEPT CHECK! You have a sample of nitrogen gas (N2) in a container fitted with a piston that maintains a pressure of 6.00 atm. Initially, the gas is at 45°C in a volume of 6.00 L. You then cool the gas sample. Copyright © Cengage Learning. All rights reserved

The pressure of the gas increases. The volume of the gas increases. CONCEPT CHECK! Which best explains the final result that occurs once the gas sample has cooled? The pressure of the gas increases. The volume of the gas increases. The pressure of the gas decreases. The volume of the gas decreases. Both volume and pressure change. d) is correct. Copyright © Cengage Learning. All rights reserved

CONCEPT CHECK! The gas sample is then cooled to a temperature of 15°C. Solve for the new condition. (Hint: A moveable piston keeps the pressure constant overall, so what condition will change?) 5.43 L The new volume is 5.43 L (use Charles’s law to solve for the new volume). (6.00 L) / (45+273) = (V2) / (15+273) Copyright © Cengage Learning. All rights reserved

Root Mean Square Velocity R = 8.3145 J/K·mol (J = joule = kg·m2/s2) T = temperature of gas (in K) M = mass of a mole of gas in kg Final units are in m/s. Copyright © Cengage Learning. All rights reserved

Diffusion – the mixing of gases. Effusion – describes the passage of a gas through a tiny orifice into an evacuated chamber. Rate of effusion measures the speed at which the gas is transferred into the chamber. Copyright © Cengage Learning. All rights reserved

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Graham’s Law of Effusion M1 and M2 represent the molar masses of the gases. Copyright © Cengage Learning. All rights reserved

Nickel forms a gaseous compound of the formula Ni(CO)x What is the value of x given that under the same conditions methane (CH4) effuses 3.3 times faster than the compound? M2 M1  r1 r2 t2 t1 = = M2 = r1 r2 ( ) 2 x M1 r1 = 3.3 x r2 = (3.3)2 x 16 = 174.2 M1 = 16 g/mol 58.7 + x • 28 = 174.2 x = 4.1 ~ 4 5.7

Compare the rates of diffusion of He (molar mass = 4g/mol) and methane, CH4 (molar mass = 16 g/mol): calculate rHe/rCH4. Which gas diffuses faster He diffuses 4 times faster than a certain gas X. Calculate the molar mass of that gas & identify the gas from the following list: (a) CH4, (b) CO2 (c) SO2 (d) Ar

We must correct for non-ideal gas behavior when: An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law. We must correct for non-ideal gas behavior when: Pressure of the gas is high. Temperature is low. Under these conditions: Concentration of gas particles is high. Attractive forces become important. Copyright © Cengage Learning. All rights reserved

Plots of PV/nRT Versus P for Several Gases (200 K) Copyright © Cengage Learning. All rights reserved

Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures Copyright © Cengage Learning. All rights reserved

Real Gases (van der Waals Equation) corrected pressure Pideal corrected volume Videal Copyright © Cengage Learning. All rights reserved

For a real gas, the actual observed pressure is lower than the pressure expected for an ideal gas due to the intermolecular attractions that occur in real gases. Copyright © Cengage Learning. All rights reserved

Values of the van der Waals Constants for Some Gases The value of a reflects how much of a correction must be made to adjust the observed pressure up to the expected ideal pressure. A low value for a reflects weak intermolecular forces among the gas molecules. Copyright © Cengage Learning. All rights reserved

Air Pollution Two main sources: Transportation Production of electricity Combustion of petroleum produces CO, CO2, NO, and NO2, along with unburned molecules from petroleum. Copyright © Cengage Learning. All rights reserved

Nitrogen Oxides (Due to Cars and Trucks) At high temperatures, N2 and O2 react to form NO, which oxidizes to NO2. The NO2 breaks up into nitric oxide and free oxygen atoms. Oxygen atoms combine with O2 to form ozone (O3). Copyright © Cengage Learning. All rights reserved

Concentration for Some Smog Components vs. Time of Day Copyright © Cengage Learning. All rights reserved

Sulfur Oxides (Due to Burning Coal for Electricity) Sulfur produces SO2 when burned. SO2 oxidizes into SO3, which combines with water droplets in the air to form sulfuric acid. Copyright © Cengage Learning. All rights reserved

Sulfur Oxides (Due to Burning Coal for Electricity) Sulfuric acid is very corrosive and produces acid rain. Use of a scrubber removes SO2 from the exhaust gas when burning coal. Copyright © Cengage Learning. All rights reserved

A Schematic Diagram of a Scrubber Copyright © Cengage Learning. All rights reserved