Honors Chemistry Chapter 5 Gases 5 1 Gases

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Honors Chemistry Chapter 5 Gases

Honors Chemistry Chapter 5 Gases

5. 1 Gases • Temperature vs. Intermolecular attraction • Atomic Gases • Noble Gases,

5. 1 Gases • Temperature vs. Intermolecular attraction • Atomic Gases • Noble Gases, H 2, N 2, O 2, F 2, Cl 2 • Molecular Gases • Usually light molecules with weak attraction forces • Eg: HCl, CO 2, NH 3, H 2 S, NO 2 • Ionic Compounds • Strong forces; not normally gases

5. 2 Pressure • • • Force per unit area P = F/A N/m

5. 2 Pressure • • • Force per unit area P = F/A N/m 2 unit defined as Pascal (Pa) Standard air pressure = 101. 325 k. Pa Also called 1 atmosphere (atm) Measured by unequal mercury levels Manometers and barometers Common unit called mm. Hg (or Torr) Standard air pressure = 760 mm. Hg

5. 2 Dimensional Analysis • Convert 75. 0 k. Pa to mm. Hg •

5. 2 Dimensional Analysis • Convert 75. 0 k. Pa to mm. Hg • 75. 0 k. Pa 760 mm. Hg ------ x -------- = 563 mm. Hg 1 101. 325 k. Pa • Try this one • Convert 1. 25 atm to k. Pa

5. 3 Boyle’s Law • • Pressure is inversely proportional to volume Hold temperature

5. 3 Boyle’s Law • • Pressure is inversely proportional to volume Hold temperature and amount of gas constant V a 1/P V = k x (1/P) PV = k Best used with changing conditions P 1 V 1 = P 2 V 2

5. 3 Boyle’s Law Problems • A 175 m. L sample of methane is

5. 3 Boyle’s Law Problems • A 175 m. L sample of methane is stored at 125 k. Pa. What pressure is needed to compress the gas to a volume of 50. 0 m. L? • P 1 V 1 = P 2 V 2 • (125 k. Pa) (175 m. L) = P 2 (50. 0 m. L) • P 2 = 438 k. Pa • Try this one • A sample of argon occupies 476 m. L at 650 Torr. Find the volume at 975 Torr.

5. 3 Charles’ Law • • Also credited to Gay-Lussac Volume is directly proportional

5. 3 Charles’ Law • • Also credited to Gay-Lussac Volume is directly proportional to temperature Hold pressure and amount of gas constant Va. T V = k. T Linear relationship Must use Kelvins! V 1 V 2 --- = --T 1 T 2

5. 3 Charles’ Law Problems • A 5. 00 L helium balloon is heated

5. 3 Charles’ Law Problems • A 5. 00 L helium balloon is heated from 20 o. C to 75 o. C. Find its new volume. • V 1/T 1 = V 2/T 2 • 5. 00 L V 2 ----- = -------293 K 348 K • V 2 = 5. 94 L • Try this one • A 670 m. L sample of chlorine is stored at 50 o. C. At what temperature will its volume be 450 m. L?

5. 3 More Gas Laws • Another form of Charles’ Law • Pressure is

5. 3 More Gas Laws • Another form of Charles’ Law • Pressure is directly proportional to temperature • P = k. T • P 1/T 1 = P 2/T 2 • Avogadro’s Law • Volume is directly proportional to the amount of gas present • Van • Volume relationships in chemical reactions

5. 3 Avogadro’s Law Problems • How many liters of hydrogen are needed to

5. 3 Avogadro’s Law Problems • How many liters of hydrogen are needed to completely react with 1 liter of oxygen? • 2 H 2 + O 2 2 H 2 O • 2 mol hydrogen react with 1 mol oxygen • V a n, so…. • 2 L hydrogen react with 1 L oxygen • Try this one • How many liters of ammonia are formed when 1 L of hydrogen reacts with excess nitrogen?

5. 4 The Ideal Gas Equation • Ideal Gas • No intermolecular attraction forces

5. 4 The Ideal Gas Equation • Ideal Gas • No intermolecular attraction forces • Particles have no volume • Combine Boyle’s, Charles, and Avogadro’s Laws • PV = n. RT • STP = 1 atm, 273 K • Molar volume of a gas = 22. 414 L at STP • R = 0. 0821 atm L / mol K

5. 4 Ideal Gas Equation Problems • A sample of fluorine occupies 3. 65

5. 4 Ideal Gas Equation Problems • A sample of fluorine occupies 3. 65 L at 45 o. C and 2. 50 atm. How many moles of fluorine are present? • PV = n. RT • (2. 50 atm)(3. 65 L) = n (0. 0821)(318 K) • n = 0. 350 mol • Try this one • A 0. 500 mol sample of propane occupies 2. 15 L. If the temperature is 28 o. C, find the pressure.

5. 4 Gas Density • • • Since n = m/M…. PV = (m/M)

5. 4 Gas Density • • • Since n = m/M…. PV = (m/M) RT MPV = m. RT Divide by V to get density (m/V) MP = r. RT Gas density expressed in g/L

5. 4 Gas Density Problems • • • Find the density of nitrous oxide

5. 4 Gas Density Problems • • • Find the density of nitrous oxide at STP. First, find molecular mass of N 2 O MP = r. RT (44. 0 g/mol)(1. 00 atm) = r (0. 0821)(273 K) r = 1. 96 g/L Try this one…. • A gas is found to have a density of 2. 54 g/L at 15 o. C and 1. 50 atm. Find its molecular mass.

5. 5 Gas Stoichiometry • • Mass-Mass problems (review) Volume-Volume problems Volume is proportional

5. 5 Gas Stoichiometry • • Mass-Mass problems (review) Volume-Volume problems Volume is proportional to moles, so…. Mol relationship from reaction can be used directly • No conversions needed!

5. 5 Volume-Volume Problem • 2 H 2 + O 2 2 H 2

5. 5 Volume-Volume Problem • 2 H 2 + O 2 2 H 2 O • If 3. 25 L of oxygen react, how many liters of water vapor are formed? • 3. 25 L O 2 2 L H 2 O ------- x ------ = 6. 50 L H 2 O 1 1 L O 2 • Volume-Volume is just Avogadro’s Law!

5. 5 Mass-Volume Problems • • Key step – get to moles! Mass conversion

5. 5 Mass-Volume Problems • • Key step – get to moles! Mass conversion – use molecular mass Volume conversion – use gas equation Need to know temperature and pressure conditions

5. 5 Mass-Volume Problems • 25. 0 g of sodium react with excess water

5. 5 Mass-Volume Problems • 25. 0 g of sodium react with excess water at STP. How many liters of hydrogen are produced? • 2 Na + 2 H 2 O 2 Na. OH + H 2 • 25. 0 g Na 1 mol H 2 ------ x ----------- = 0. 543 mol 1 23. 0 g Na 2 mol Na • Now use ideal gas equation to get volume

5. 5 Mass-Volume Problems • • PV = n. RT (1. 00 atm) V

5. 5 Mass-Volume Problems • • PV = n. RT (1. 00 atm) V = (0. 543 mol)(0. 0821)(273 K) V = 12. 2 L Try this one • Potassium chlorate decomposes into potassium chloride and oxygen gas. How many grams of KCl. O 3 are needed to produce 5. 00 L of oxygen at 0. 750 atm and 18 o. C? • Hint: This one is backwards!

5. 6 Dalton’s Law • Partial pressure – the pressure of an individual gas

5. 6 Dalton’s Law • Partial pressure – the pressure of an individual gas in a mixture of gases • Total pressure of a mixture equals the sum of the partial pressures of each gas • Pt = P 1 + P 2 + P 3 +. . . • Partial pressure is proportional to the mol fraction (X 1 = n 1 / nt) • P 1 = X 1 P t

5. 6 Dalton’s Law • 2. 00 mol He is mixed with 1. 00

5. 6 Dalton’s Law • 2. 00 mol He is mixed with 1. 00 mol Ar. Find the partial pressure of each at 1. 75 atm pressure. • XHe = 2. 00 mol / 3. 00 mol = 0. 667 • XAr = 1. 00 mol / 3. 00 mol = 0. 333 • PHe = (0. 667) (1. 75 atm) = 1. 17 atm • PAr = (0. 333) (1. 75 atm) = 0. 583 atm • Try this. . . • Find the partial pressure of oxygen in air if it makes up 21% of the Earth’s atmosphere by volume. (Note: The volume gives you the mole ratio because of Avogadro’s law. )

5. 7 Kinetic Molecular Theory • Explains gas behavior in terms of molecular motion

5. 7 Kinetic Molecular Theory • Explains gas behavior in terms of molecular motion • Energy • Work done by a moving object • Measured in SI unit Joule (J) • Kinetic energy • Energy due to motion • K = ½ mv 2 • KMT is a simplification of reality (ideal gas)

5. 7 Kinetic Molecular Theory • Gas molecules are separated by great distances •

5. 7 Kinetic Molecular Theory • Gas molecules are separated by great distances • They can be treated as “point masses” • Gas molecules are in constant random motion • Frequent elastic collisions (no energy lost) • No attractive or repulsive forces • Average K is proportional to Temperature

5. 7 Distribution of Molecular Speeds • Maxwell-Boltzmann Distribution • Molecular speeds distributed around

5. 7 Distribution of Molecular Speeds • Maxwell-Boltzmann Distribution • Molecular speeds distributed around average • Peak velocity depends on temperature and on molec. mass • Root Mean Square Speed • _____ vrms = √ 3 RT/M • Rate of diffusion

5. 8 Deviations from Ideal Behavior • We made approximations! • Point masses •

5. 8 Deviations from Ideal Behavior • We made approximations! • Point masses • No intermolecular forces • These approximations become bad at. . . • High pressure • Low temperature • Liquefaction • van der Waals Equation • (P + an 2/V 2) (V – nb) = n. RT