Title: Honors Chemistry, Chapter 11
1Chapter 11 Molecular Composition of Gases
2Gay-Lussacs Law of Combining Volumes of Gases
- hydrogen gas oxygen gas ? water vapor
- 2 volumes 1 volume ? 2 volumes
- hydrogen chlorine?hydrogen chloride gas
- 1 volume 1 volume ? 2 volumes
- Gay-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.
3Avogadros Law
- Avogadros law states that equal volumes of gases
at the same temperature and pressure contain
equal numbers of molecules. - V kn
- Where V is volume
- n is the number of moles
- k is a constant
4Avogadros Law
- Assume hydrogen, oxygen and chlorine are diatomic
molecules. - H2(g) Cl2(g) ? 2HCl(g)
- 1 vol. 1 vol. ? 2 vol.
- and
- 2H2(g) O2(g) ? 2H2O(g)
- 2 vol. 1 vol. ? 2 vol.
5Molar Volume
- The 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.)
6Sample Problem 11-1
- What 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
7Sample Problem 11-2
- What 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 SO232.0032.07 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
8Thought Problem
- Consider 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?
9Thought Problem
- How 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?
10Chapter 11, Section 1 Review
- State 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.
11Idea Gas Law
- Boyles Law V a 1/P
- Charless Law V a T
- Avogadros Law V a n
- or V a nT/P
- or V nRT/P
- or PV nRT
- Where R is the Ideal Gas Constant
12Ideal Gas Constant
- R _PV_ _(1 atm)(22.4140 L)_
- nT (1 mol)(273.15 K)
- 0.082057 L atm/(mol K)
13Numerical Values of the Ideal Gas Constant
14Sample Problem 11-3
- What 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 N2
- T 298 K
- P nRT/V 0.500 mol N2 x
- 0.0821 L atm/ (mol K) x 298 K / 10.0 L
-
1.22 atm
15Sample Problem 11-4
- What is the volume, in liters, of 0.250 mol of
oxygen at 20 oC. and 0.974 atm pressure? - P 0.974 atm
- n 0.250 mol of Oxygen
- T 20 oC. or 273.2 20 293.2 K
- V nRT/P
- 0.250 x 0.0821 (L atm)/(mol K) x 293.2 K /
0.974 atm 6.17 L O2
16Molar Mass
- PV nRT m RT / M (n m / M)
- Where m is mass and M is molar mass
- Solve for M
- M m RT/(PV)
17Gas Density
- D m/V m DV
- PV m RT/ M
- PV DV RT/M
- D M P / (RT)
18Problem 11-6
- At 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 L
- T 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
19Example Density Problem
- The 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 K
- P 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
20Chapter 11, Section 2 Review
- State 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.
21Chapter 11, Section 2 Review
- Using 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.
22Stoichiometry of Gases
- Coefficients 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
23Volume Ratios
- Volume 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
24Sample Problem 11-7
- What 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
25Volume-Mass and Mass-Volume Calculations
- We can use the ideal gas law to calculate
problems like - gas vol A ? moles A ? moles B ? mass B
- or
- mass A ? moles A ? moles B ? gas vol B
26Sample Problem 11-8
- How 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 CO2 - or
- n 5.00L CO2 / (22.4 L/mol)0.233 mol CO2
27Sample Problem 11-8 continued
Molecular 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
28Sample Problem 11-9
- How 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
29Sample Problem 11-9 Continued
V nRT/P (11.3 mol H2) x 0.0821 L atm / (mol
K) x 308 K / (0.980 atm) 292 L H2
30Chapter 11, Section 3 Review
- Explain 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.
31Diffusion and Effusion
- Diffusion 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.
32Average Kinetic Energy
- For two gases at the same temperature
- Avg kinetic energy ½ MAvA2 ½ MBvB2
- MAvA2 MBvB2
- vA2 / vB2 MB / MA
- vA MB
- --------- ------------
- vB MA
33Grahams 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
34Sample Problem 11-10
- Compare 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
35Chapter 11, Section 4 Review
- State 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.