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Properties of Gases: The Air We Breathe

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## Properties of Gases: The Air We Breathe

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**6**Properties of Gases: The Air We Breathe**Chapter Outline**• 6.1 The Gas Phase • Properties of Gases • 6.2 Atmospheric Pressure • 6.3 The Gas Laws • 6.4 The Ideal Gas Law • 6.5 Gases in Chemical Reactions • 6.6 Gas Density • 6.7 Dalton’s Law and Mixtures of Gases • 6.8 The Kinetic Molecular Theory of Gases • 6.9 Real Gases**Properties of a Gas**• Neither definite shape nor volume • Uniformly fills any container • Exerts pressure on surroundings • Changes volume with temperature and pressure • Miscible–mix together in any proportion • Much less dense than solids or liquids**Chapter Outline**• 6.1 The Gas Phase • 6.2 Atmospheric Pressure • What is pressure and how is it measured? • Units of Pressure • 6.3 The Gas Laws • 6.4 The Ideal Gas Law • 6.5 Gases in Chemical Reactions • 6.6 Gas Density • 6.7 Dalton’s Law and Mixtures of Gases • 6.8 The Kinetic Molecular Theory of Gases • 6.9 Real Gases**Atmospheric Pressure**• The atmosphere: • Layer of gases 50 km thick • Composition is fairly consistent. • Properties vary with location • Pressure, density**Pressure**• Pressure: • Force/unit area (P = F/A) • Atmospheric pressure = force exerted by gases surrounding Earth on Earth’s surface • Units of Pressure: • SI units = Newton/meter2 = 1 Pascal (Pa) • 1 standard atmosphere (1 atm) = 101,325 Pa • 1 atm = 760 mm Hg = 760 torr**Measurement of Pressure**• Barometer: measures atmospheric pressure • Height of Hg column based on balance of forces: • Gravity (pulls Hg down) • Atmospheric pressure (pushes Hg up into evacuated tube)**Elevation and Atmospheric Pressure**0.35 atm 0.62 atm 0.83 atm Sea level (1 atm)**Measuring Pressure: Manometer**Closed-end: used to measure P of a gas sample less than Patm. (a) h = Patm (b)h = Psample**Manometer (cont.)**Open-end: designed to measure P of a gas greater than Patm. (c) h = 0 (d) h = Psample- Patm**Chapter Outline**• 6.1 The Gas Phase • 6.2 Atmospheric Pressure • 6.3 The Gas Laws • Factors Affecting Gas Properties: Boyle’s Law, Charles’s Law, Avogadro’s Law, Amonton’s Law • 6.4 The Ideal Gas Law • 6.5 Gases in Chemical Reactions • 6.6 Gas Density • 6.7 Dalton’s Law and Mixtures of Gases • 6.8 The Kinetic Molecular Theory of Gases • 6.9 Real Gases**Boyle’s Law: Relating P and V**• Gases are compressible. • Volume ↓ as pressure ↑ • Boyle’s Law: P 1/V (T and n fixed) • or, PV = constant; or P1V1 = P2V2**Practice: Boyle’s Law**• A balloon is filled with hydrogen to a pressure of 1.35 atm and has a volume of 2.54 L. If temperature remains constant, what will the volume be when the pressure is increased to 2.50 atm? • Collect and Organize: P1 = 1.35 atm, V1 = 2.54 L, P2 = 2.50 atm, V2 = ?**Practice: Boyle’s Law**• A balloon is filled with hydrogen to a pressure of 1.35 atm and has a volume of 2.54 L. If temperature remains constant, what will the volume be when the pressure is increased to 2.50 atm? • Analyze: From Boyle’s Law, we know that P1V1 = P2V2. We can insert the known values for P1, V1, and P2, and rearrange to find V2.**Practice: Boyle’s Law**• A balloon is filled with hydrogen to a pressure of 1.35 atm and has a volume of 2.54 L. If temperature remains constant, what will the volume be when the pressure is increased to 2.50 atm? • Solve: P1V1 = P2V2P1V1/P2 = V2; (1.35 atm)(2.54 L)/(2.50 atm) = 1.37 L**Practice: Boyle’s Law**• A balloon is filled with hydrogen to a pressure of 1.35 atm and has a volume of 2.54 L. If temperature remains constant, what will the volume be when the pressure is increased to 2.50 atm? • Think about it: We would expect the volume to decrease as pressure increases, consistent with our calculated answer.**Charles’s Law: Relating V and T**• Charles’s Law: • V T (P, n fixed) • or, • Volume of a gas extrapolates to zero at 0 kelvin or -273 C**Avogadro’s Law: Relating V and n**Volume is directly proportional to the number of moles of gas, Vn (T, P fixed).**Amonton’s Law**PT (n, V fixed) P/T = constant**Chapter Outline**• 6.1 The Gas Phase • 6.2 Atmospheric Pressure • 6.3 The Gas Laws • 6.4 The Ideal Gas Law • Combining the Gas Laws • Standard Temperature and Pressure (STP) • 6.5 Gases in Chemical Reactions • 6.6 Gas Density • 6.7 Dalton’s Law and Mixtures of Gases • 6.8 The Kinetic Molecular Theory of Gases • 6.9 Real Gases**Ideal Gas Law**Ideal Gas Law: PV = nRT R = universal gas constant= 0.08206 L • atm K-1 mol-1 P = pressure (in atm) V = volume (in liters) n = moles T = temperature (in kelvin) Allows the calculation of gas properties when more than one variable is changing.**Combined Gas Law**Boyle’s Law: PV = constant Charles’s Law: V/T = constant Avogadro’s Law: V/n = constant Combining the gas laws: If n is constant, then PV/T = constant, and**Practice: Combined Gas Law**• A sample of oxygen gas is at 0.500 atm and occupies a volume of 11.2 L at 0 C. What volume will the gas occupy at 6.00 atm and room temperature (25 C)? • Collect and Organize: P1 = 0.500 atm, V1 = 11.2 L, T1 = 0.0 C (273 K); P2 = 6.00 atm, T2 = 25 C (298 K), V2 = ?**Practice: Combined Gas Law**• A sample of oxygen gas is at 0.500 atm and occupies a volume of 11.2 L at 0 C. What volume will the gas occupy at 6.00 atm and room temperature (25 C)? • Analyze: Using the combined gas law, we know five of the six variables and can rearrange to solve for V2.**Practice: Combined Gas Law**• A sample of oxygen gas is at 0.500 atm and occupies a volume of 11.2 L at 0 C. What volume will the gas occupy at 6.00 atm and room temperature (25 C)? • Solve: P1V1/T1 = P2V2/T2 (P1V1T2)/(T1P2) = V2 (0.500 atm)(11.2 L)(298 K)/(273 K)(6.00 atm) = 1.02 L**Practice: Combined Gas Law**• A sample of oxygen gas is at 0.500 atm and occupies a volume of 11.2 L at 0 C. What volume will the gas occupy at 6.00 atm and room temperature (25 C)? • Think about it: The temperature increases by about 10%, which should cause the volume to increase by 10%. However, the pressure increases by a factor of 12, so volume should decrease by a factor of 12. The effect of pressure in this case is much greater. So we would expect the final volume to be smaller by a factor of ~10-12 in order to be consistent with our calculated result.**Practice: Ideal Gas Law**• Calculate the pressure of 1.2 mol of methane gas in a 3.3 L container at 25 C. • Collect and Organize: n = 1.2 mol, V = 3.3 L, T = 25 C (298 K), P = ?**Practice: Ideal Gas Law**• Calculate the pressure of 1.2 mol of methane gas in a 3.3 L container at 25 C. • Analyze: We know three of the four variables in the ideal gas law and the value of the gas constant R, which enables us to rearrange to solve for P.**Practice: Ideal Gas Law**• Calculate the pressure of 1.2 mol of methane gas in a 3.3 L container at 25 C. • Solve: PV = nRTP = nRT/V = (1.2 mol)(0.08206 L • atm/(mol • K))(298 K)/(3.3 L) P = 8.9 atm**Practice: Ideal Gas Law**• Calculate the pressure of 1.2 mol of methane gas in a 3.3 L container at 25 C. • Think about it: The units of our calculated result are consistent with pressure, and the value appears to be reasonable.**Practice: Ideal Gas Law (2)**• An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98 °C and 754 torr pressure. Calculate the molar mass of the gas. (Hint: M = m/n) • Collect and Organize: V = 127 mL, P = 754 torr, T = 98 C, mass = 0.495 g, M = ?**Practice: Ideal Gas Law (2)**• An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98 °C and 754 torr pressure. Calculate the molar mass of the gas. (Hint: M = m/n) • Analyze: We know three of the four variables in the ideal gas law, and can rearrange to solve for n. We can calculate M = m/n. Before we calculate n, we must convert the known variables into appropriate units for the ideal gas law: V = 0.127 L, P = (754 torr )(1 atm/760 torr) = 0.9921 atm, T = 98 + 273 = 371 K**Practice: Ideal Gas Law (2)**• An experiment shows that a 0.495 g sample of an unknown gas occupies 127 mL at 98 °C and 754 torr pressure. Calculate the molar mass of the gas. (Hint: M = m/n) • Solve: PV = nRTn = PV/RT = [(0.9921 atm)(0.127 L)]/[(0.08206 L • atm/(mol • K)](371 K) = 0.00414 mol M = m/n = (0.495 g)/(0.00414 mol) = 120. g/mol**Reference Points for Gases**• Standard Temperature and Pressure (STP): • P = 1 atm; T = 273 K (0.0 C) • Molar Volume: • Volume occupied by one mole of an ideal gas at STP • V = 22.4 L (calculated from the ideal gas law)**Chapter Outline**• 6.1 The Gas Phase • 6.2 Atmospheric Pressure • 6.3 The Gas Laws • 6.4 The Ideal Gas Law • 6.5 Gases in Chemical Reactions • Stoichiometric Calculations: Converting Volumes to Moles • 6.6 Gas Density • 6.7 Dalton’s Law and Mixtures of Gases • 6.8 The Kinetic Molecular Theory of Gases • 6.9 Real Gases**Stoichiometry Calculations: Gases**• Stoichiometric Calculations: • Depend on mole/mole ratios of reactants and/or products • Moles of gas can be calculated from ideal gas law if P, V, and T are known.**Practice: Gas Stoichiometry**• Automobile air bags inflate during a crash or sudden stop by the rapid generation of N2(g)from sodium azide: 2 NaN3(s) → 2 Na(s) + 3 N2(g) How many grams of sodium azide are needed to produce sufficient N2(g) to fill a 45 x 45 x 25 cm bag to a pressure of 1.20 atm at 15 C? • Collect and Organize: We are trying to calculate the amount of a reagent (sodium azide) needed to produce enough N2(g) to fill an air bag of known dimensions at a given T and P. We know T = 15 C (288 K), P = 1.20 atm, V = (45 × 45 × 25) = 50625 cm3 = 50.625 L**Practice: Gas Stoichiometry**• Automobile air bags inflate during a crash or sudden stop by the rapid generation of N2(g)from sodium azide: 2 NaN3(s) → 2 Na(s) + 3 N2(g) How many grams of sodium azide are needed to produce sufficient N2(g) to fill a 45 x 45 x 25 cm bag to a pressure of 1.20 atm at 15 C? • Analyze: Knowing P, V, and T we can calculate n from the ideal gas law. Once we have n, we can calculate the amount of sodium azide needed stoichiometrically from the balanced chemical equation. Once the number moles of sodium azide is known, we can convert moles to grams using the molar mass.**Practice: Gas Stoichiometry**• Automobile air bags inflate during a crash or sudden stop by the rapid generation of N2(g)from sodium azide: 2 NaN3(s) → 2 Na(s) + 3 N2(g) How many grams of sodium azide are needed to produce sufficient N2(g) to fill a 45 x 45 x 25 cm bag to a pressure of 1.20 atm at 15 C? • Solve: n = PV/RT = (1.20 atm)(50.625 L)/(0.0821 L • atm/(mol • K))(288 K) = 2.57 mol From the balanced equation: (2.57 mol N2) (2 mol NaN3/3 mol N2) = 1.713 mol (1.713 mol NaN3 ) (65.01 g/mol) = 111 g sodium azides**Chapter Outline**• 6.1 The Gas Phase • 6.2 Atmospheric Pressure • 6.3 The Gas Laws • 6.4 The Ideal Gas Law • 6.5 Gases in Chemical Reactions • 6.6 Gas Density • Calculating the Density of a Gas • Density and Buoyancy • 6.7 Dalton’s Law and Mixtures of Gases • 6.8 The Kinetic Molecular Theory of Gases • 6.9 Real Gases**Gas Density**• Can be calculated from Molar Mass (M) and molar volume (V/n) • From the ideal gas law: • PV = nRT→ • Density: • When P in atm, T in kelvin, d = g/L.**Buoyancy: Gas Densities**• Buoyancy depends on differences in gas densities • Depends on: • Molar Masses He(g) = 0.169 g/L* N2(g) = 1.19 g/L* • Temperature Charles’s Law: Density as Temp * At 15ºC and 1 atm**Practice: Gas Densities**• When HCl(aq) and NaHCO3(aq) are mixed together, a reaction takes place in which a gas is one of the products. The gas has a d = 1.81 g/L at 1.00 atm and 23 C. What is the molar mass and identity of the gas? • Collect and Organize: We know density, d = 1.81 g/L, P = 1.00 atm, T = 23 C (296 K); We need to determine the molar mass and identity of the gas.**Practice: Gas Densities**• When HCl(aq) and NaHCO3(aq) are mixed together, a reaction takes place in which a gas is one of the products. The gas has a d = 1.81 g/L at 1.00 atm and 23 C. What is the molar mass and identity of the gas? • Analyze: Knowing T, P, and density, we can rearrange the ideal gas law to solve for M: M = (m/V)RT/P**Practice: Gas Densities**• When HCl(aq) and NaHCO3(aq) are mixed together, a reaction takes place in which a gas is one of the products. The gas has a d = 1.81 g/L at 1.00 atm and 23 C. What is the molar mass and identity of the gas? • Solve: M = (1.81 g/L)(0.0821 L • atm/(mol • K))(296 K)/(1.00 atm) = 43.99 = 44.0 g/mol**Practice: Gas Densities**• When HCl(aq) and NaHCO3(aq) are mixed together, a reaction takes place in which a gas is one of the products. The gas has a d = 1.81 g/L at 1.00 atm and 23 C. What is the molar mass and identity of the gas? • Think about it: The result of the calculation (44.0 g/mol) is consistent with CO2, which is the expected product of the reaction between a carbonate and an acid.**Chapter Outline**• 6.1 The Gas Phase • 6.2 Atmospheric Pressure • 6.3 The Gas Laws • 6.4 The Ideal Gas Law • 6.5 Gases in Chemical Reactions • 6.6 Gas Density • 6.7 Dalton’s Law and Mixtures of Gases • Partial Pressure and Mole Fraction • Determining Partial Pressure of Gases Collected over Water • 6.8 The Kinetic Molecular Theory of Gases • 6.9 Real Gases**Dalton’s Law of Partial Pressures**• For a mixture of gases in a container: • PTotal = P1 + P2 + P3 + . . . • Depends on ntotal, not the identity of the gas.**Mole Fraction & Partial Pressure**• Mole fraction: • Ratio of the # of moles of a component in a mixture to the total # of moles in a mixture • Mole fraction in terms of pressure: • n1 = P1V/RT; when V, T are constant, P1n1