Chapter 11 Molecular Composition of Gases

Gay-Lussacs Law of Combining Volumes of Gaseshydrogen gas + oxygen gas water vapor 2 volumes + 1 volume 2 volumeshydrogen + chlorinehydrogen chloride gas1 volume + 1 volume 2 volumesGay-Lussacs law of combining volumes of gases states that at constant temperature and pressure , the volumes of gaseous reactants and products can be expressed as ratios of small whole numbers.

Avogadros LawAvogadros law states that equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. V = knWhere V is volumen is the number of molesk is a constant

Avogadros LawAssume hydrogen, oxygen and chlorine are diatomic molecules.H2(g) + Cl2(g) 2HCl(g)1 vol. + 1 vol. 2 vol. and2H2(g) + O2(g) 2H2O(g)2 vol. + 1 vol. 2 vol.

Molar VolumeThe volume occupied by one mole of a gas at STP is known as the standard molar volume of a gas and has a volume of 22.41410 L (22.4 L for our purposes).(STP Standard Temperature and Pressure: 0oC. and 1 atm.)

Sample Problem 11-1What is the volume of 0.0680 mole of oxygen gas at STP?

moles of O2 volume of O2 in liters

0.068 mol O2 x 22.4 L/mol = 1.52 L O2

Sample Problem 11-2What is the mass in grams of 98.0 mL of SO2 at STP?Vol. of SO2 in liters mol of SO2 mass of SO2 mol mass of SO2=32.00+32.07= 64.07 g/mol

98.0 mL x 1L/1000 mL x 1 mol SO2/ 22.4 L x 64.07 g SO2/ mol SO2 = 0.280 g SO2

Thought ProblemConsider a container with mass of 1 kg and an internal volume of 22.4 L. :

If the container is filled with air, how much water would the container displace if placed in a container of water?

Thought ProblemHow much would the container weigh when filled with air?How much would the container weigh when evacuated?How much would the container weigh when filled with nitrogen? oxygen? hydrogen?

Chapter 11, Section 1 ReviewState the law of combining volumes.State Avogadros law and explain its significance.Define standard molar volume of a gas, and use it to calculate gas masses and volumes. Use standard molar volume to calculate the molar mass of a gas.

Idea Gas LawBoyles Law: V a 1/PCharless Law: V a TAvogadros Law: V a nor V a nT/PorV = nRT/Por PV = nRTWhere R is the Ideal Gas Constant

Ideal Gas Constant

R = _PV_ = _(1 atm)(22.4140 L)_ nT (1 mol)(273.15 K)

= 0.082057 L atm/(mol K)

Numerical Values of the Ideal Gas Constant

Numerical Value of RUnits0.082(L atm)/(mol K)62.4[L (mm of Hg)]/(mol K)8.314(L kPa) / (mol K)8.314( J ) / (mol K)Note: 1L atm = 101.325 J 1 J = 1 Pa m3

Sample Problem 11-3What is the pressure in atmospheres exerted by 0.500 mol of nitrogen gas in a 10.0 L container at 298 K?V = 10.0 L n = 0.500 mol of N2T = 298 KP = nRT/V = 0.500 mol N2 x 0.0821 L atm/ (mol K) x 298 K / 10.0 L = 1.22 atm

Sample Problem 11-4What is the volume, in liters, of 0.250 mol of oxygen at 20 oC. and 0.974 atm pressure?P = 0.974 atmn = 0.250 mol of OxygenT = 20 oC. or 273.2 + 20 = 293.2 KV = nRT/P = 0.250 x 0.0821 (L atm)/(mol K) x 293.2 K / 0.974 atm = 6.17 L O2

Molar MassPV = nRT = m RT / M (n = m / M) Where m is mass and M is molar mass

Solve for M M = m RT/(PV)

Gas DensityD = m/V m = DVPV = m RT/ MPV = DV RT/MD = M P / (RT)

Problem 11-6At 28oC. and 0.974 atm, 1.00 L of a gas has a mass of 5.16 g. What is the molar mass of this gas? What is the gas?P = 0.974 atm V = 1.00 LT = 28oC. + 273 = 301 K m = 5.16 g.M = m RT/PV = 5.16 g x [0.0821 L atm / (mol K)] x 301 K / ( 0.974 atm x 1.00 L) = 131 g/mol

Example Density ProblemThe density of a gas was found to be 2.0 g/L at 1.50 atm and 27oC. What is the molar mass of the gas? What is the gas?D = 2.0 g/L T = 27oC. + 273 = 300 KP = 1.50 atm D = M P / (RT)M = DRT / P = 2.0 g/L x [0.0821 L atm / (mol K)] x 300 K / 1.50 atm = 33 g/mol

Chapter 11, Section 2 ReviewState the ideal gas law.Derive the gas constant and discuss the units.Using the ideal gas law, calculate pressure, volume, temperature or amount of gas when the other three quantities are known.

Chapter 11, Section 2 ReviewUsing the ideal gas law, calculate the molar mass or density of a gas.Reduce the ideal gas law to Boyles law, Charless law, and Avogadros law. Describe the conditions under which each applies.

Stoichiometry of GasesCoefficients indicate molecule, mole, and volume ratios in gas reactions.For Example: 2CO(g) + O2(g) 2CO2(g) 2 molecules 1 molecule 2 molecules 2 mol 1 mol 2 mol 2 volumes 1 volume 2 volumes

Volume RatiosVolume Ratios from the CO + O2 Reaction

2 vol CO / 1 vol O2 or 1 vol O2 / 2 vol CO

2 vol CO / 2 vol CO2 or 2 vol CO2 / 2 vol CO

1 vol O2 / 2 vol CO2 or 2 vol CO2 / 1 vol O2

Sample Problem 11-7What is the volume, in liters, of oxygen required for the complete combustion of 0.350 L of propane? What will the volume of CO2 be? (Assume constant T and P.)C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

0.350L C3H8 x 5 vol O2 / 1 vol C3H8 = 1.75 L O2

0.350L C3H8 x 3 vol CO2 / 1vol C3H8 =1.05 L CO2

Volume-Mass and Mass-Volume CalculationsWe can use the ideal gas law to calculate problems like:

gas vol A moles A moles B mass Bormass A moles A moles B gas vol B

Sample Problem 11-8How many grams of calcium carbonate must be decomposed to produce 5.00 L of carbon dioxide at STP?CaCO3(s) D CaO(s) + CO2(g)n = PV/RT = (1atm)(5.00 L CO2) / [ 0.0821 L atm /(mol K)]/(273 K) = 0.223 mol CO2orn = 5.00L CO2 / (22.4 L/mol)=0.233 mol CO2

Sample Problem 11-8 continuedMolecular Weight of CaCO3?100.09 g CaCO3/mol

0.223 mol CO2 x 1 mol CaCO3 / (1 mol CO2) x 100.09 g CaCO3/1 mol CaCO3= 22.4 g CaCO3

Sample Problem 11-9How many liters of hydrogen at at 35oC. and 0.980 atm are needed to completely react with 875 grams of tungsten oxide?WO3(s) + 3H2(g) W(s) + 3H2O(l)Molar Mass of WO3? 231.84 g/mol

875 g WO3 x 1 mol WO3 / (231.84 g WO3) x 3 mol H2 / (1 mol WO3) = 11.3 mol H2

Sample Problem 11-9 ContinuedV = nRT/P = (11.3 mol H2) x [0.0821 L atm / (mol K)] x 308 K / (0.980 atm) = 292 L H2

Chapter 11, Section 3 ReviewExplain how Gay-Lussacs law and Avogadros law apply to volumes of gases in chemical reactions.Use a chemical equation to specify volume ratios for gaseous reactants or products, or both.Use volume ratios and the gas laws to calculate volumes, masses, or molar amounts of gaseous reactants or products.

Diffusion and EffusionDiffusion is the process where two gases gradually mix spontaneously due to the constant motion of the gas molecules.

Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container.

Average Kinetic EnergyFor two gases at the same temperature:Avg kinetic energy = MAvA2 = MBvB2MAvA2 = MBvB2

vA2 / vB2 = MB / MA

vA MB --------- = ------------ vB MA

Grahams Law of Effusion

Grahams law of effusion states that the rates of effusion of gases at the same temperature and pressure are inversely proportional to the square roots of their molar masses.Rate of effusion of A MB DensityB---------------------------- = ------- = -------------Rate of effusion of B MA DensityA

Sample Problem 11-10Compare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.Rate of effusion of H2 MO2 32.00 g/mol---------------------------- = ------- = -------------Rate of effusion of O2 MH2 2.02 g/mol

= 3.98

Chapter 11, Section 4 ReviewState Grahams law of effusion.Determine the relative rates of effusion of two gases of know molar masses.State the relationship between the molecular velocities of two gases and their molar masses.