ch. 11 molecular composition of gases
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Ch. 11 Molecular Composition of Gases. 11-1 Volume-Mass Relationships of Gases. Gay-Lussac’s law of combining volumes of gases-at constant temperature and pressure, the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers Hydrogen + oxygen ->water vapor - PowerPoint PPT PresentationTRANSCRIPT
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Ch. 11 Molecular Ch. 11 Molecular Composition of GasesComposition of Gases
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11-1 Volume-Mass Relationships of 11-1 Volume-Mass Relationships of GasesGases
Gay-Lussac’s law of combining Gay-Lussac’s law of combining volumes of gases-at constant volumes of gases-at constant temperature and pressure, the temperature and pressure, the volumes of gaseous reactants and volumes of gaseous reactants and products can be expressed as ratios of products can be expressed as ratios of small whole numberssmall whole numbers
Hydrogen + oxygen ->water vaporHydrogen + oxygen ->water vapor 2L 1L 2L2L 1L 2L 2 volumes 1 volume2 volumes 1 volume 2 volumes 2 volumes
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Avogadro’s LawAvogadro’s Law
Equal volumes of gases at the same Equal volumes of gases at the same temperature and pressure contain temperature and pressure contain equal numbers of molecules (doesn’t equal numbers of molecules (doesn’t matter which gas) Fig. 11-1matter which gas) Fig. 11-1
He discovered that some molecules He discovered that some molecules can have 2 atoms or more (diatomic can have 2 atoms or more (diatomic molecules)molecules)
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Avogadro’s law also indicates gas Avogadro’s law also indicates gas volume (L) directly proportional to volume (L) directly proportional to the amount of a gas (n)the amount of a gas (n)
V = knV = kn
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Standard molar volume of a gasStandard molar volume of a gas
Avogadro’s constant = 6.022 X 10Avogadro’s constant = 6.022 X 1023 23
molecules = 1 molemolecules = 1 mole
Standard molar volume of a gas-the Standard molar volume of a gas-the volume occupied by one mole of a volume occupied by one mole of a gas at STP (22.4 L)gas at STP (22.4 L)
Fig. 11-3 1 mole of each gas Fig. 11-3 1 mole of each gas occupies 22.4 L but different massesoccupies 22.4 L but different masses
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Avogadro’s Law Sample problem Avogadro’s Law Sample problem 11-111-1
A chemical reaction produces 0.0680 A chemical reaction produces 0.0680 mol of oxygen gas. What volume in mol of oxygen gas. What volume in liters is occupied by this gas sample liters is occupied by this gas sample at STP?at STP?
0.0680 mol X 0.0680 mol X 22.4 L22.4 L = 1.52 L = 1.52 L
1 mol1 mol
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Avogadro’s Law PracticeAvogadro’s Law Practice
A sample of hydrogen gas occupies A sample of hydrogen gas occupies 14.1 L at STP. How many moles of 14.1 L at STP. How many moles of the gas are present?the gas are present?
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Converting to gramsConverting to grams
Sample problem 11-2Sample problem 11-2 A chemical reaction produced 98.0 A chemical reaction produced 98.0
mL of sulfur dioxide gas, SOmL of sulfur dioxide gas, SO22, at STP. , at STP. What was the mass in grams of the What was the mass in grams of the gas produced?gas produced?
.098 L X .098 L X 1 mol1 mol X X 64.07 g SO64.07 g SO22 = = 0.280 g0.280 g
22.4 L 1 mol22.4 L 1 mol
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Converting to grams practiceConverting to grams practice
What is the volume of 77 g of What is the volume of 77 g of nitrogen dioxide gas at STP?nitrogen dioxide gas at STP?
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11-2: The Ideal Gas Law11-2: The Ideal Gas Law
Mathematical relationship among Mathematical relationship among pressure, volume, temperature, and pressure, volume, temperature, and the number of moles of a gas.the number of moles of a gas.
Combination of Boyle’s, Charles’s, Combination of Boyle’s, Charles’s, Gay-Lussac’s, and Avogadro’s LawsGay-Lussac’s, and Avogadro’s Laws
PV = nRTPV = nRT
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Ideal gas constant (R), is derived by Ideal gas constant (R), is derived by plugging in all standard values into plugging in all standard values into the equation:the equation:
R = R = PVPV = 0.0821 = 0.0821
nTnT
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Ideal gas law sampleIdeal gas law sample
What is the pressure in atmospheres What is the pressure in atmospheres exerted by a 0.500 mol sample of exerted by a 0.500 mol sample of nitrogen gas in a 10 L container at 298 nitrogen gas in a 10 L container at 298 K?K?
Answer = 1.22 atmAnswer = 1.22 atm
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More ideal gas law practiceMore ideal gas law practice
What is the volume, in liters, of 0.250 What is the volume, in liters, of 0.250 mol of oxygen gas at 20mol of oxygen gas at 20°C and 0.974 °C and 0.974 atm pressure?atm pressure?
Answer = 6.17 LAnswer = 6.17 L
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Sample problem 11-5Sample problem 11-5
What mass of chlorine gas, ClWhat mass of chlorine gas, Cl22, in , in grams, is contained in a 10 L tank at grams, is contained in a 10 L tank at 2727°C and 3.5 atm of pressure?°C and 3.5 atm of pressure?
Answer = 101 gAnswer = 101 g
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Finding molar mass or densityFinding molar mass or density
PV = PV = mRTmRT
MM
M = M = mRTmRT M = M = DRTDRT
PV P PV P
D = D = MPMP
RTRT
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Sample ProblemSample Problem
At 28At 28°C and 0.974 atm, 1.00 L of gas °C and 0.974 atm, 1.00 L of gas has a mass of 5.16 g. What is the has a mass of 5.16 g. What is the molar mass of this gas?molar mass of this gas?
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11-3 Stoichiometry of Gases11-3 Stoichiometry of Gases
Coefficients can be used as volume Coefficients can be used as volume ratios:ratios:
2CO + O2CO + O22 -> 2CO -> 2CO22
2 volumes CO2 volumes CO
1 volume O1 volume O22
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Sample Problem 11-7 volume-Sample Problem 11-7 volume-volumevolume
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Sample problem 11-8 volume-massSample problem 11-8 volume-mass
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Sample problem 11-9Sample problem 11-9
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11-4 Effusion and Diffusion11-4 Effusion and Diffusion
Graham’s Law of Effusion-rates of Graham’s Law of Effusion-rates of diffusion and effusion depend on the diffusion and effusion depend on the relative velocities of gas moleculesrelative velocities of gas molecules
Rates of effusion of gases at the Rates of effusion of gases at the same temperature and pressure are same temperature and pressure are inversely proportional to the square inversely proportional to the square roots of their molar masses.roots of their molar masses.
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Graham’s Law formulaGraham’s Law formula
Rate of effusion ARate of effusion A = = √M√MBB
Rate of effusion B √MRate of effusion B √MAA
Molar masses can also be replaced by Molar masses can also be replaced by densities of the gases:densities of the gases:
Rate of effusion ARate of effusion A = = √density√densityBB
Rate of effusion B √denistyRate of effusion B √denistyAA
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Graham’s Law ProblemGraham’s Law Problem
Sample problem 11-10Sample problem 11-10 Compare the rates of effusion of Compare the rates of effusion of
hydrogen and oxygen at the same hydrogen and oxygen at the same temperature and pressure. (smaller temperature and pressure. (smaller molar mass gas will diffuse faster-molar mass gas will diffuse faster-how much faster?)how much faster?)
Smaller molar mass goes on bottomSmaller molar mass goes on bottom
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Diffusion Quicklab pg. 353Diffusion Quicklab pg. 353