chapter 101 gases. 2 homework: 10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72
TRANSCRIPT
Chapter 10 1
GasesGases
Chapter 10Chapter 10
Chapter 10 2
Homework:10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72
Chapter 10 3
Characteristics of GasesCharacteristics of Gases
- Expand to fill a volume (expandability)- Compressible- Readily forms homogeneous mixtures with other gases- These behaviors are due to large distances between the
gas molecules.
Chapter 10 4
PressurePressurePressure - force acting on an object per unit area.
Chapter 10 5
PressurePressurePressure - force acting on an object per unit area.
AF
P
Chapter 10 6
PressurePressurePressure - force acting on an object per unit area.
AF
P
- Atmospheric pressure is measured with a barometer.
- Standard atmospheric pressure is the pressure required to support 760 mm of Hg in a column.
- There are several units used for pressure:- Pascal (Pa), N/m2
- Millimeters of Mercury (mmHg)- Atmospheres (atm)
Chapter 10 7
PressurePressure- Conversion Factors
- 1 atm = 760 mmHg - 1 atm = 760 torr - 1 atm = 1.01325 105 Pa - 1 atm = 101.325 kPa.
Chapter 10 8
The Gas LawsThe Gas Laws- There are four variables required to describe a gas:
- Amount of substance: moles- Volume of substance: volume- Pressures of substance: pressure- Temperature of substance: temperature
- The gas laws will hold two of the variables constant and see how the other two vary.
Chapter 10 9
The Gas LawsThe Gas LawsThe Pressure-Volume Relationship: Boyle’s LawBoyle’s Law - The volume of a fixed quantity of gas is
inversely proportional to its pressure.
Chapter 10 10
The Gas LawsThe Gas LawsThe Pressure-Volume Relationship: Boyle’s LawBoyle’s Law - The volume of a fixed quantity of gas is
inversely proportional to its pressure.
) and (constant 1
TnV
P
Chapter 10 11
The Gas LawsThe Gas LawsThe Pressure-Volume Relationship: Boyle’s LawBoyle’s Law - The volume of a fixed quantity of gas is
inversely proportional to its pressure.
2211
) and (constant 1
VPVP
TnV
P
Chapter 10 12
The Gas LawsThe Gas LawsThe Pressure-Volume Relationship: Boyle’s Law
Chapter 10 13
The Gas LawsThe Gas Laws
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
The Temperature-Volume Relationship: Charles’s Law
Chapter 10 14
The Gas LawsThe Gas Laws
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
The Temperature-Volume Relationship: Charles’s Law
)and(constant PnTV
Chapter 10 15
The Gas LawsThe Gas Laws
Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.
The Temperature-Volume Relationship: Charles’s Law
2
2
1
1
)and(constant
T
V
T
V
PnTV
Chapter 10 16
The Gas LawsThe Gas LawsThe Temperature-Volume Relationship:
Charles’s Law
Chapter 10 17
The Gas LawsThe Gas Laws
Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.
The Quantity-Volume Relationship: Avogadro’s Law
Chapter 10 18
The Gas LawsThe Gas Laws
Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.
) and (constant TPnV
The Quantity-Volume Relationship: Avogadro’s Law
Chapter 10 19
The Ideal Gas EquationThe Ideal Gas Equation- Combine the gas laws (Boyle, Charles, Avogadro)
yields a new law or equation.
Chapter 10 20
The Ideal Gas EquationThe Ideal Gas Equation- Combine the gas laws (Boyle, Charles, Avogadro)
yields a new law or equation.
Ideal gas equation:
PV = nRT
R = gas constant = 0.08206 L.atm/mol-K
P = pressure (atm) V = volume (L)
n = moles T = temperature (K)
Chapter 10 21
The Ideal Gas EquationThe Ideal Gas Equation- We define STP (standard temperature and pressure)
as 0C, 273.15 K, 1 atm.- Volume of 1 mol of gas at STP is 22.4 L.
Chapter 10 22
Gas Densities and Molar Mass- Rearranging the ideal-gas equation with M as molar
mass yields
Further Applications of The Ideal-Gas Further Applications of The Ideal-Gas EquationEquation
RT
Pd
M
Chapter 10 23
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
Dalton’s Law - In a gas mixture the total pressure is given by the sum of partial pressures of each component:
Pt = P1 + P2 + P3 + …
- The pressure due to an individual gas is called a partial pressure.
Dalton’s Law
Chapter 10 24
Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures
- The partial pressure of a gas can be determined if you know the mole fraction of the gas of interest and the total pressure of the system.
- Let ni be the number of moles of gas i exerting a partial pressure Pi, then
Pi = iPt
where i is the mole fraction (ni/nt).
Partial Pressures and Mole Fractions
Chapter 10 25
Kinetic-Molecular TheoryKinetic-Molecular Theory- Theory developed to explain gas behavior- To describe the behavior of a gas, we must first
describe what a gas is:– Gases consist of a large number of molecules in constant
random motion.
– Volume of individual molecules negligible compared to volume of container.
– Intermolecular forces (forces between gas molecules) negligible.
– Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature.
– Average kinetic energy of molecules is proportional to temperature.
Chapter 10 26
Kinetic-Molecular TheoryKinetic-Molecular Theory
• Pressure exerted by a gas is the result of bombardment of the walls of the container by the gas molecules.
• Pressure varies directly with the number of molecules hitting the wall per unit time.– If the volume is reduced the number of impacts is
increased; therefore, the pressure is increased.
Boyle’s Law
Chapter 10 27
Kinetic-Molecular TheoryKinetic-Molecular Theory
• The force that a gas molecule strikes the side of a container is directly proportional to the temperature.
• As the temperature is increased the kinetic energy (force) of the gas particles increases.
• For the pressure to be constant, the area the force is applied to must be increased (the volume is increased).
Charles’ Law
Chapter 10 28
Kinetic-Molecular TheoryKinetic-Molecular Theory
• There are large distances between the gas molecules.• The components of a mixture will bombard the walls
of the container with the same frequency in the presence of a mixture as it would by itself.
Dalton’s Law
Chapter 10 29
Kinetic-Molecular TheoryKinetic-Molecular Theory
• As kinetic energy increases, the velocity of the gas molecules increases.
• Root mean square speed, u, is the speed of a gas molecule having average kinetic energy.
• Average kinetic energy, , is related to root mean square speed:
= ½mu2
Molecular Speed
Chapter 10 30
Kinetic-Molecular TheoryKinetic-Molecular TheoryMolecular Speed
Chapter 10 31
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
• Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas.
• Mathematically:
Molecular Speed
Chapter 10 32
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
• Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas.
• Mathematically:
MRT
u3
Molecular Speed
Chapter 10 33
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
• Consider two gases at the same temperature: the lighter gas has a higher u than the heavier gas.
• Mathematically:
• The lower the molar mass, M, the higher the u for that gas at a constant temperature.
MRT
u3
Molecular Speed
Chapter 10 34
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Chapter 10 35
Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Effusion
Effusion – The escape of gas through a small opening.
Diffusion – The spreading of one substance through another.
Chapter 10 36
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
Graham’s Law of Effusion
Chapter 10 37
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
Graham’s Law of Effusion
1
2
2
1MM
rr
Chapter 10 38
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.
• Gas escaping from a balloon is a good example.
Graham’s Law of Effusion
1
2
2
1MM
rr
Chapter 10 39
Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Effusion
Chapter 10 40
Molecular Effusion and DiffusionMolecular Effusion and Diffusion
• Diffusion of a gas is the spread of the gas through space.
• Diffusion is faster for light gas molecules.• Diffusion is slowed by gas molecules colliding with
each other.• Average distance of a gas molecule between collisions
is called mean free path.
Diffusion and Mean Free Path
Chapter 10 41
Molecular Effusion and DiffusionMolecular Effusion and DiffusionDiffusion and Mean Free Path• At sea level, mean free path is about 6 10-6 cm.
Chapter 10 42
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior• From the ideal gas equation, we have
PV = nRT• This equation breaks-down at
– High pressure• At high pressure, the attractive and repulsive forces between gas
molecules becomes significant.
Chapter 10 43
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior
Chapter 10 44
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior• From the ideal gas equation, we have
PV = nRT• This equation breaks-down at
– High pressure• At high pressure, the attractive and repulsive forces between gas
molecules becomes significant.
– Small volume
Chapter 10 45
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior• From the ideal gas equation, we have
PV = nRT• This equation breaks-down at
– High pressure• At high pressure, the attractive and repulsive forces between gas
molecules becomes significant.
– Small volume• At small volumes, the volume due to the gas molecules is a source of
error.
Chapter 10 46
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior
• Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions.
The Van der Waals Equation
Chapter 10 47
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior
• Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions.
The Van der Waals Equation
2
2
V
annbV
nRTP
Chapter 10 48
Real Gases: Deviations from Ideal Real Gases: Deviations from Ideal BehaviorBehavior
• Two terms are added to the ideal gas equation to correct for volume of molecules and one to correct for intermolecular attractions.
• a and b are constants, determined by the particular gas.
The Van der Waals Equation
2
2
V
annbV
nRTP
Chapter 10 49
Homework:10.12, 10.28, 10.42, 10.48, 10.54, 10.66, 10.72
Chapter 10 50
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
Chapter 10 51
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
Chapter 10 52
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
2211 VPVP
Chapter 10 53
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
)89.4()50.7(988.0 2
2221
LPLatm
VPVP
Chapter 10 54
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. Calculate the pressure of the gas if its volume is decreased to 4.89L while its temperature is held constant.
--BOYLES LAW
atmL
LatmP
LPLatm
VPVP
52.189.4
)50.7(988.0
)89.4()50.7(988.0
2
2
2221
Chapter 10 55
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
Chapter 10 56
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
Chapter 10 57
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
2
2
1
1
T
V
T
V
Chapter 10 58
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
KK
V
P
V
P
30128273
kelvintoetemperaturconvert2
2
2
1
Chapter 10 59
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
x
L
K
L
V
P
V
P
00.4
301
50.72
2
2
1
Chapter 10 60
ProblemProblem
A sample of gas occupies a volume of 7.50L at 0.988 atm and 28.0oC. At what temperature in degrees Celsius is the volume of the gas 4.00L if the pressure is kept constant.
--Charles’ Law
5.1125.160
00.4
301
50.72
2
2
1
Ctoconvert o
Kx
x
L
K
L
V
P
V
P
Chapter 10 61
ProblemProblemCalcium hydride, CaH2, reacts with water to form hydrogen gas:
CaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
How many grams of CaH2 are needed to generate 10.0L of H2 gas if the pressure of H2 is 740 torr at 23oC?
Chapter 10 62
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
-Calculate moles of H2 formed
-Calculate moles of CaH2 needed
-Convert moles CaH2 to grams
Chapter 10 63
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
)(/)(08206.0
29627323
0.10
974.0760/740
KmolatmLr
KCT
LV
torrP
o
Chapter 10 64
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2 K
Latmn
KmolatmLr
KCT
LV
torrP
H
o
Chapter 10 65
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
Chapter 10 66
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
Chapter 10 67
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
molx
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
2005.0
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
Chapter 10 68
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
)/10.42(2005.0
2005.0
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
2
molgmolgCaH
molx
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
Chapter 10 69
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
g
molgmolgCaH
molx
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
44.8
)/10.42(2005.0
2005.0
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
2
Chapter 10 70
ProblemProblemCaH2(s) + 2 H2O(l) Ca(OH)2(aq) + 2 H2(g)
g
molgmolgCaH
molx
mol
x
H
CaH
moln
K
Latmn
KmolatmLr
KCT
LV
torrP
H
H
o
44.8
)/10.42(2005.0
2005.0
401.02
1
401.0
)296(08206.0
)0.10(974.0
)(/)(08206.0
29627323
0.10
974.0760/740
2
2
2
2
2