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BEHAVIOR OF GASES Gases have weight Gases take up space Gases exert pressure Gases fill their containers Gases doing all of these things!

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Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are: Gases are mostly empty space The molecules in a gas are separate, very small and very far apart

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Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are: Gas molecules are in constant, chaotic motion Collisions between gas molecules are elastic (there is no energy gain or loss)

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Kinetic Theory of Gases The basic assumptions of the kinetic molecular theory are: The average kinetic energy of gas molecules is directly proportional to the absolute temperature Gas pressure is caused by collisions of molecules with the walls of the container

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Measurements of Gases To describe a gas, its volume, amount, temperature, and pressure are measured. Volume: measured in L, mL, cm 3 (1 mL = 1 cm 3 ) Amount: measured in moles (mol), grams (g) Temperature: measured in KELVIN (K) K = ºC + 273 Pressure: measured in mm Hg, torr, atm, etc. P = F / A (force per unit area)

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Moderate Force (about 100 lbs) Small Area (0.0625 in 2 ) Enormous Pressure (1600 psi) P = F / A

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Bed of Nails Large Surface Area (lots of nails) Moderate Force Small Pressure P = F / A

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Units of Pressure Units of Pressure: 1 atm = 760 mm Hg 1 atm = 760 torr 1 atm = 1.013 x 10 5 Pa 1 atm = 101.3 kPa 1 atm = 1.013 bar

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Boyle’s Law As P , V (when T and n are constant) and vice versa…. INVERSE RELATIONSHIP V 1/P P 1 V 1 = P 2 V 2 For a given number of molecules of gas at a constant temperature, the volume of the gas varies inversely with the pressure.

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Example: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant) P 1 V 1 = P 2 V 2 (1.2 atm)(12 L) = (3.6 atm)V 2 V 2 = 4.0 L

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Charles’ Law Jacques Charles (1746-1828) The volume of a given number of molecules is directly proportional to the Kelvin temperature. As T , V (when P and n are constant) and vice versa…. DIRECT RELATIONSHIP V T

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Example: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? (express answer in K and °C) V 1 / T 1 = V 2 / T 2 (117 mL) / (373 K) = (234 mL) / T 2 T 2 = 746 K T 2 = 473 ºC

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Combined gas law This is for one gas undergoing changing conditions of temp, pressure, and volume. Combining Boyle’s law (pressure-volume) with Charles’ Law (volume-temp):

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Example 1: A sample of neon gas occupies 105 L at 27°C under a pressure of 985 torr. What volume would it occupy at standard conditions? P 1 = 985 torr V 1 = 105 L T 1 = 27 °C = 300. K P 2 = 1 atm = 760 torr V 2 = ? T 2 = 0 °C = 273 K P 1 V 1 T 2 = P 2 V 2 T 1 (985 torr)(105 L)(273K) = (760torr)(V 2 )(300K) V 2 = 124 L

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Example 2: A sample of gas occupies 10.0 L at 240°C under a pressure of 80.0 kPa. At what temperature would the gas occupy 20.0 L if we increased the pressure to 107 kPa? P 1 = 80.0 kPa V 1 = 10.0 L T 1 = 240 °C = 513 K P 2 = 107 kPa V 2 = 20.0 L T 2 = ? P 1 V 1 T 2 = P 2 V 2 T 1 (80.0kPa)(10.0L)(T 2 ) = (107kPa)(20.0L)(513K) T 2 = 1372K≈ 1370K

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Example 3: A sample of oxygen gas occupies 23.2 L at 22.2 °C and 1.3 atm. At what pressure (in mm Hg) would the gas occupy 11.6 L if the temperature were lowered to 12.5 °C? P 1 = 1.3 atm = 988 mmHg V 1 = 23.2 L T 1 = 22.2 °C = 295.2 K P 2 = ? V 2 = 11.6 L T 2 = 12.5 °C = 285.5 K P 1 V 1 T 2 = P 2 V 2 T 1 (988mm Hg)(23.2L)(285.5K) = (P 2 )(11.6L)(295.2K) P 2 = 1938 mm Hg ≈ 1900 mmHg

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Gases: Standard Molar Volume & The Ideal Gas Law Avogadro’s Law: at the same temperature and pressure, equal volumes of all gases contain the same # of molecules (& moles). Standard molar volume = 22.4 L @STP This is true of “ideal” gases at reasonable temperatures and pressures,the behavior of many “real” gases is nearly ideal.

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Example: 1.00 mole of a gas occupies 36.5 L, and its density is 1.36 g/L at a given temperature & pressure. a) What is its molecular weight (molar mass)?

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Example: 1.00 mole of a gas occupies 36.5 L, and its density is 1.36 g/L at a given temperature & pressure. b) What is the density of the gas under standard conditions?

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The IDEAL GAS LAW Shows the relationship among the pressure, volume, temp. and # moles in a sample of gas. P = pressure (atm) V = volume (L) n = # moles T = temp (K) R = universal gas constant = 0.0821 The units of R depend on the units used for P, V & T

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Example 1: What volume would 50.0 g of ethane, C 2 H 6, occupy at 140 ºC under a pressure of 1820 torr? P = (1820 torr)(1 atm/760 torr) = 2.39 atm V = ? n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol T = 140 °C + 273 = 413 K PV = nRT (2.39 atm)(V) = (1.67 mol)(0.0821 L·atm/mol·K)(413 K) V = 23.6 L

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Example 2: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH 4, measured at standard conditions. P = 1.00 atm V = 8.96 L n = ? T = 273 K PV = nRT (1 atm)(8.96 L) = (n)(0.0821 L·atm/mol·K)(273 K) n = 0.400 mol (a)

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Example 2: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH 4, measured at standard conditions. Or the easier way… (a)

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Example 2: Calculate (a) the # moles in, and (b) the mass of an 8.96 L sample of methane, CH 4, measured at standard conditions. Convert moles to grams… (b)

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Example 3: Calculate the pressure exerted by 50.0 g ethane, C 2 H 6, in a 25.0 L container at 25 ºC? P = ? V = 25.0 L n = (50.0 g)(1 mol / 30.0 g) = 1.67 mol T = 25 °C + 273 = 298 K PV = nRT (P)(25.0 L) = (1.67 mol)(0.0821 L·atm/mol·K)(298 K) P = 1.63 atm

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Determining Molecular Weights & Molecular Formulas of Gases: If the mass of a volume of gas is known, we can use this info. to determine the molecular formula for a compound.

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Example 1: A 0.109 g sample of pure gaseous compound occupies 112 mL at 100. ºC and 750. torr. What is the molecular weight of the compound? First find # moles Then use mass to determine g/mol… PV = nRT (0.99 atm)(0.112 L) = (n)(0.0821 L·atm/mol·K)(373 K) n = 0.00362 mol

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Example 2: A compound that contains only C and H is 80.0% C and 20.0% H by mass. At STP, 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound? First find the empirical formula… Empirical formula = CH 3

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Example 2: A compound that contains only C and H is 80.0% C and 20.0% H by mass. At STP, 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound? Next, determine the MW of the sample… 0.024 mol

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Example 2: A compound that contains only C and H is 80.0% C and 20.0% H by mass. At STP, 546 mL of the gas has a mass of 0.732 g. What is the molecular (true) formula for the compound? Finally, determine the molecular formula… Empirical formula = CH 3 15.0 g/mol True MW = 30.0 g/mol CH 3 x 2 = C 2 H 6

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Partial Pressures and Mole Fractions In a mixture of gases each gas exerts the pressure it would exert if it occupied the volume alone. The total pressure exerted by a mixture of gases is the sum of the partial pressures of the individual gases: P total = P 1 + P 2 + P 3 + …

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Example: If 100.0 mL of hydrogen gas, measured at 25 C and 3.00 atm, and 100.0 mL of oxygen, measured at 25 C and 2.00 atm, what sould be the pressure of the mixture of gases? P total = P 1 + P 2 + P 3 + … P T = 3.00 atm + 2.00 atm P T = 5.00 atm Notice the two gases are measured at the same temp. and vol.

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Vapor Pressure of a Liquid The pressure exerted by its gaseous molecules in equilibrium with the liquid; increases with temperature

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Vapor Pressure of a Liquid P atm = P gas + P H 2 O or P gas = P atm - P H 2 O

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Vapor Pressure of a Liquid Temp. ( C) v.p. of water (mm Hg) Temp. ( C) v.p. of water (mm Hg) 1815.482118.65 1916.482219.83 2017.542321.07 See A-2 for a complete table

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Example 1: A sample of hydrogen gas was collected by displacement of water at 25 C. The atmospheric pressure was 748 mm Hg. What pressure would the dry hydrogen exert in the same conditions? P H 2 = P atm - P H 2 O P H 2 =748 mm Hg – 23.76 mm Hg P H 2 = 724.24 mm Hg P H 2 724 mm Hg

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Example 2: A sample of oxygen gas was collected by displacement of water. The oxygen occupied 742 mL at 27 C. The barometric pressure was 753 mm Hg. What volume would the dry oxygen occupy at STP? P O2 = P atm - P H2O P O2 =753 mm Hg – 26.74 mm Hg P O2 = 726 mm Hg P 1 V 1 T 2 = P 2 V 2 T 1 (726 mm Hg)(0.742 L)(273K) = (760 mm Hg)(V 2 )(300K) V 2 = 0.645 L

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Example 3: A student prepares a sample of hydrogen gas by electrolyzing water at 25 C. She collects 152 mL of H2 at a total pressure of 758 mm Hg. Calculate: (a) the partial pressure of hydrogen, and (b) the number of moles of hydrogen collected. P H 2 = P atm - P H 2 O P H 2 =758 mm Hg – 23.76 mm Hg P H 2 = 734 mm Hg

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Example 3: A student prepares a sample of hydrogen gas by electrolyzing water at 25 C. She collects 152 mL of H 2 at a total pressure of 758 mm Hg. Calculate: (a) the partial pressure of hydrogen, and (b) the number of moles of hydrogen collected. PV = nRT (0.966 atm)(0.152 L) = (n)(0.0821 L·atm/mol·K)(298 K) n = 0.00600 mol H 2

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Graham’s Law of Diffusion & Effusion Where, Rate = rate of diffusion or effusion MM=molar mass

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Stoichiometry of Gaseous Reactions A balanced equation can be used to relate moles or grams of substances taking part in a reaction. (AND VOLUME!)

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Example: Hydrogen peroxide is the active ingredient in commercial preparations for bleaching hair. What mass of hydrogen peroxide must be used to produce 1.00 L of oxygen gas at 25 C and 1.00 atm? 2H 2 O 2 O 2 + 2H 2 O PV = nRT (1.00 atm)(1.00 L) = (n)(0.0821 L·atm/mol·K)(298 K) n = 0.0409 mol O 2

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