Ch. 11 Molecular Ch. 11 Molecular Composition of GasesComposition of Gases

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

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)

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

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

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

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?

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

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?

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

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

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

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

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

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

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?

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

Sample Problem 11-7 volume-Sample Problem 11-7 volume-volumevolume

Sample problem 11-8 volume-massSample problem 11-8 volume-mass

Sample problem 11-9Sample problem 11-9

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.

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

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

Diffusion Quicklab pg. 353Diffusion Quicklab pg. 353