Properties of Gases
1
Key objectives: Be able to state (i.e. define) and
use in practical calculations, the following.
2
Key objectives: Be able to state (i.e. define) and
use in practical calculations, the following.
1. Concept of force and pressure.
3
Key objectives: Be able to state (i.e. define) and
use in practical calculations, the following.
1. Concept of force and pressure.
2. Boyle’s Law
4
Key objectives: Be able to state (i.e. define) and
use in practical calculations, the following.
1. Concept of force and pressure.
2. Boyle’s Law
3. Charles’ Law
5
Key objectives: Be able to state (i.e. define) and
use in practical calculations, the following.
1. Concept of force and pressure.
2. Boyle’s Law
3. Charles’ Law
4. Avogadro’s Hypothesis
6
Key objectives: Be able to state (i.e. define) and
use in practical calculations, the following.
1. Concept of force and pressure.
2. Boyle’s Law
3. Charles’ Law
4. Avogadro’s Hypothesis
5. Ideal gas law
7
Key objectives: Be able to state (i.e. define) and
use in practical calculations, the following.
1. Concept of force and pressure.
2. Boyle’s Law
3. Charles’ Law
4. Avogadro’s Hypothesis
5. Ideal gas law
6. Dalton’s law of partial pressures
8
Key objectives: Be able to state (i.e. define) and
use in practical calculations, the following.
1. Concept of force and pressure.
2. Boyle’s Law
3. Charles’ Law
4. Avogadro’s Hypothesis
5. Ideal gas law
6. Dalton’s law of partial pressures
7. Graham’s law of effusion
9
Key objectives: Be able to state (i.e. define) and
use in practical calculations, the following.
1. Concept of force and pressure.
2. Boyle’s Law
3. Charles’ Law
4. Avogadro’s Hypothesis
5. Ideal gas law
6. Dalton’s law of partial pressures
7. Graham’s law of effusion
8. Kinetic theory and real gases
10
Gas: Refers to a substance that is entirely gaseous at
ordinary temperatures and pressures.
11
Gas: Refers to a substance that is entirely gaseous at
ordinary temperatures and pressures.
The 4 important measurable properties of gases are:
12
Gas: Refers to a substance that is entirely gaseous at
ordinary temperatures and pressures.
The 4 important measurable properties of gases are:
1. pressure
13
Gas: Refers to a substance that is entirely gaseous at
ordinary temperatures and pressures.
The 4 important measurable properties of gases are:
1. pressure
2. volume
14
Gas: Refers to a substance that is entirely gaseous at
ordinary temperatures and pressures.
The 4 important measurable properties of gases are:
1. pressure
2. volume
3. temperature
15
Gas: Refers to a substance that is entirely gaseous at
ordinary temperatures and pressures.
The 4 important measurable properties of gases are:
1.
2.
3.
4.
pressure
volume
temperature
mass (or moles)
16
Pressure: Gases exert pressure on any surface
with which they come into contact. That is, all
gases expand uniformly to occupy whatever
space is available.
17
Pressure: Gases exert pressure on any surface
with which they come into contact. That is, all
gases expand uniformly to occupy whatever
space is available.
Pressure is define by the equation:
force
pressure =
area
18
Pressure: Gases exert pressure on any surface
with which they come into contact. That is, all
gases expand uniformly to occupy whatever
space is available.
Pressure is define by the equation:
force
pressure =
area
or
P =
F
A
19
Units of pressure
20
Units of pressure
distance
velocity =
time
21
Units of pressure
distance
velocity =
time
velocity
acceleration =
time
22
Units of pressure
distance
velocity =
time
velocity
acceleration =
time
i.e.
acceleration =
distance
time2
23
To find the SI units of force:
24
To find the SI units of force:
force = mass x acceleration
25
To find the SI units of force:
force = mass x acceleration
distance
force = mass x
time2
26
To find the SI units of force:
force = mass x acceleration
distance
force = mass x
time2
Now plug in the SI units for each quantity on the
right-hand side of the equation. The SI unit of
force is the newton (named after Newton) and
abbreviated N.
27
To find the SI units of force:
force = mass x acceleration
distance
force = mass x
time2
Now plug in the SI units for each quantity on the
right-hand side of the equation. The SI unit of
force is the newton (named after Newton) and
abbreviated N.
1 N = 1 kg m s-2
28
The SI units of pressure are: Nm-2
29
The SI units of pressure are: Nm-2
The SI unit of pressure is called the pascal
(named after Pascal) and is abbreviated Pa.
30
The SI units of pressure are: Nm-2
The SI unit of pressure is called the pascal
(named after Pascal) and is abbreviated Pa.
1 Pa = 1 Nm-2 = 1 kg m-1 s-2
31
The SI units of pressure are: Nm-2
The SI unit of pressure is called the pascal
(named after Pascal) and is abbreviated Pa.
1 Pa = 1 Nm-2 = 1 kg m-1 s-2
The most common unit of pressure is the nonSI unit, the atmosphere, abbreviated atm.
32
1 atm = 76 cm Hg
33
1 atm = 76 cm Hg
= 760 mm Hg
34
1 atm = 76 cm Hg
= 760 mm Hg
= 760 torr (named after Torricelli)
35
1 atm =
=
=
=
76 cm Hg
760 mm Hg
760 torr (named after Torricelli)
101325 Pa (by definition)
36
1 atm =
=
=
=
76 cm Hg
760 mm Hg
760 torr (named after Torricelli)
101325 Pa (by definition)
1 bar = 10 5 Pa
37
1 atm =
=
=
=
76 cm Hg
760 mm Hg
760 torr (named after Torricelli)
101325 Pa (by definition)
1 bar = 10 5 Pa
Hence
1atm = 1.01325 bar
38
Atmospheric pressure is measured by a barometer.
39
The difference between a gas and a vapor:
40
The difference between a gas and a vapor:
A gas is a substance normally in the gaseous
state at ordinary temperatures and pressures.
41
The difference between a gas and a vapor:
A gas is a substance normally in the gaseous
state at ordinary temperatures and pressures.
A vapor is the gaseous form of any substance
that is a liquid or a solid at normal
temperatures and pressures.
42
The difference between a gas and a vapor:
A gas is a substance normally in the gaseous
state at ordinary temperatures and pressures.
A vapor is the gaseous form of any substance
that is a liquid or a solid at normal
temperatures and pressures.
Thus, at room temperature and 1 atm, we
speak of water vapor and helium gas.
43
Temperature: For the gas equations, the
temperature is measured on the Kelvin scale – the
units are K (kelvin).
44
Temperature: For the gas equations, the
temperature is measured on the Kelvin scale – the
units are K (kelvin).
The absolute lowest temperature is zero degrees
kelvin.
45
Temperature: For the gas equations, the
temperature is measured on the Kelvin scale – the
units are K (kelvin).
The absolute lowest temperature is zero degrees
kelvin.
The temperature in K is determined from the
temperature measured in oC by:
degrees K = degrees oC + 273.15
46
The Gas Laws
47
The Gas Laws
The first systematic and quantitative study of
gas behavior was carried out by Boyle.
48
The Gas Laws
The first systematic and quantitative study of
gas behavior was carried out by Boyle.
Boyle noticed that when the temperature and
the amount of gas are held constant, the
volume of a gas is inversely proportional to
the applied pressure acting on the gas.
49
50
V 1
P
(const. T and n)
This is a statement of Boyle’s Law.
(V is the volume, P is the pressure, T is the
temperature, and n is the moles of gas).
51
V 1
P
(const. T and n)
This is a statement of Boyle’s Law.
(V is the volume, P is the pressure, T is the
temperature, and n is the moles of gas).
V =
C
P
(const. T and n)
52
V 1
P
(const. T and n)
This is a statement of Boyle’s Law.
(V is the volume, P is the pressure, T is the
temperature, and n is the moles of gas).
V =
C
P
(const. T and n)
(C is the proportionality constant).
53
V 1
P
(const. T and n)
This is a statement of Boyle’s Law.
(V is the volume, P is the pressure, T is the
temperature, and n is the moles of gas).
V =
C
P
(const. T and n)
(C is the proportionality constant).
This is also a statement of Boyle’s Law.
54
We can write Boyle’s law as:
PV = C
(const. T and n)
55
We can write Boyle’s law as:
PV = C
(const. T and n)
If the initial values of pressure and volume are
Pi and Vi and if the conditions are changed to a
final pressure Pf and final volume Vf then we can
write:
56
We can write Boyle’s law as:
PV = C
(const. T and n)
If the initial values of pressure and volume are
Pi and Vi and if the conditions are changed to a
final pressure Pf and final volume Vf then we can
write:
initial conditions PiVi = C
57
We can write Boyle’s law as:
PV = C
(const. T and n)
If the initial values of pressure and volume are
Pi and Vi and if the conditions are changed to a
final pressure Pf and final volume Vf then we can
write:
initial conditions PiVi = C
final conditions PfVf = C
58
Hence:
PiVi = PfVf
(const. T and n)
This is also a statement of Boyle’s Law.
59
Hence:
PiVi = PfVf
(const. T and n)
This is also a statement of Boyle’s Law.
Problem Example: If 0.100 liters of a gas,
originally at 760.0 torr, is compressed to a
pressure of 800.0 torr, at a constant
temperature, what is the final volume of the
gas?
60
No mention is made about the moles of gas – so
make the assumption that n remains constant.
Since temperature is constant, we can use
Boyle’s Law.
PiVi = PfVf
61
No mention is made about the moles of gas – so
make the assumption that n remains constant.
Since temperature is constant, we can use
Boyle’s Law.
PiVi = PfVf
This can be rearranged to read:
Pi
Vf =
Vi
Pf
62
No mention is made about the moles of gas – so
make the assumption that n remains constant.
Since temperature is constant, we can use
Boyle’s Law.
PiVi = PfVf
This can be rearranged to read:
Pi
Vf =
Vi
Pf
760.0 torr
Vf =
x 0.100 l
800.0 torr
Vf = 0.0950 l
63
Charles and Gay-Lussac’s Law
64
Charles and Gay-Lussac’s Law
Charles (1787) and Gay-Lussac (1802) were the
first investigators to study the effect of
temperature on gas volume. Their studies
showed that at constant pressure, and a fixed
number of moles of gas, that the volume of a
gas is directly proportional to the temperature
of the gas.
65
Charles and Gay-Lussac’s Law
Charles (1787) and Gay-Lussac (1802) were the
first investigators to study the effect of
temperature on gas volume. Their studies
showed that at constant pressure, and a fixed
number of moles of gas, that the volume of a
gas is directly proportional to the temperature
of the gas.
V  T (const. P and n)
This is a statement of Charles’ Law.
66
67
The proportionality can be turned into an
equality:
V = C’ T
(const. P and n)
68
The proportionality can be turned into an
equality:
V = C’ T
(const. P and n)
This is a statement of Charles’ Law.
(The prime is used to signify that this constant
is different to the constant appearing in
Boyle’s law).
69
If the initial values of temperature and volume are Ti
and Vi and if the conditions are changed to a final
temperature Tf and final volume Vf then we can write:
70
If the initial values of temperature and volume are Ti
and Vi and if the conditions are changed to a final
temperature Tf and final volume Vf then we can write:
initial conditions
Vi
= C’
Ti
71
If the initial values of temperature and volume are Ti
and Vi and if the conditions are changed to a final
temperature Tf and final volume Vf then we can write:
initial conditions
final conditions
Vi
= C’
Ti
Vf
= C’
Tf
72
If the initial values of temperature and volume are Ti
and Vi and if the conditions are changed to a final
temperature Tf and final volume Vf then we can write:
initial conditions
final conditions
Vi
= C’
Ti
Vf
= C’
Tf
Hence:
Vi
Vf
=
(const. P, n)
Ti
Tf
This is a statement of Charles’ Law.
73
Problem Example: A gas occupying 4.50 x 102 ml
is heated from 22.0 oC to 187 oC at constant
pressure. What is the final volume of the gas?
74
Problem Example: A gas occupying 4.50 x 102 ml
is heated from 22.0 oC to 187 oC at constant
pressure. What is the final volume of the gas?
There is no mention of the moles of gas
changing, so assume that n is constant. Since
pressure is constant, apply Charles’ law:
75
Problem Example: A gas occupying 4.50 x 102 ml
is heated from 22.0 oC to 187 oC at constant
pressure. What is the final volume of the gas?
There is no mention of the moles of gas
changing, so assume that n is constant. Since
pressure is constant, apply Charles’ law:
Vi
Vf
=
Ti
Tf
76
Problem Example: A gas occupying 4.50 x 102 ml
is heated from 22.0 oC to 187 oC at constant
pressure. What is the final volume of the gas?
There is no mention of the moles of gas
changing, so assume that n is constant. Since
pressure is constant, apply Charles’ law:
Vi
Vf
=
Ti
Tf
Note that it is necessary to convert the
temperatures given to units of K.
77
Ti = 22.0 + 273.15 = 295.2 K
Tf = 187 + 273.15 = 4.60 x 102 K
78
Ti = 22.0 + 273.15 = 295.2 K
Tf = 187 + 273.15 = 4.60 x 102 K
Tf
Vf =
Vi
Ti
4.60 x 102 K
2 ml
=
4.50
x
10
2.952 x 102 K
= 701 ml
79
Avogadro’s Law
80
Avogadro’s Law
In 1811 Avogadro published a hypothesis
which stated that at the same temperature
and pressure, equal volumes of gases contain
the same number of molecules (or atoms if
the gas is monatomic).
81
Avogadro’s Law
In 1811 Avogadro published a hypothesis
which stated that at the same temperature
and pressure, equal volumes of gases contain
the same number of molecules (or atoms if
the gas is monatomic).
It follows that the volume of any given gas
must be proportional to the number of
molecules present.
82
It is more convenient to work in terms of moles. A
mole contains 6.02214 x 1023 items.
83
It is more convenient to work in terms of moles. A
mole contains 6.02214 x 1023 items.
For example, one mole of argon contains 6.02214
x 1023 atoms of argon, a mole of dinitrogen
contains 6.02214 x 1023 molecules of N2.
84
It is more convenient to work in terms of moles. A
mole contains 6.02214 x 1023 items.
For example, one mole of argon contains 6.02214
x 1023 atoms of argon, a mole of dinitrogen
contains 6.02214 x 1023 molecules of N2.
A mole of apples is really, really, ...really big!
85
It is more convenient to work in terms of moles. A
mole contains 6.02214 x 1023 items.
For example, one mole of argon contains 6.02214
x 1023 atoms of argon, a mole of dinitrogen
contains 6.02214 x 1023 molecules of N2.
A mole of apples is really, really, ...really big!
The mole is a convenient “lab-sized” unit for
amount of substance.
86
Avogadro’s law can be written as:
Vn
(const. P and T)
87
Avogadro’s law can be written as:
Vn
(const. P and T)
The proportionality can be replaced by an
equality:
V = C’’n
(const. P and T)
88
Avogadro’s law can be written as:
Vn
(const. P and T)
The proportionality can be replaced by an
equality:
V = C’’n
(const. P and T)
This is a statement of Avogadro’s Law.
89
Avogadro’s law can be written as:
Vn
(const. P and T)
The proportionality can be replaced by an
equality:
V = C’’n
(const. P and T)
This is a statement of Avogadro’s Law.
(The double prime is used to signify that this
constant is different to the constants appearing
in Boyle’s law and Charles’ law).
90
If the initial values of volume and moles are Vi
and ni and if the conditions are changed to a
final volume Vf and final moles nf then we can
write:
91
If the initial values of volume and moles are Vi
and ni and if the conditions are changed to a
final volume Vf and final moles nf then we can
write:
initial conditions
Vi
= C’’
ni
92
If the initial values of volume and moles are Vi
and ni and if the conditions are changed to a
final volume Vf and final moles nf then we can
write:
initial conditions
final conditions
Vi
= C’’
ni
Vf
= C’’
nf
93
If the initial values of volume and moles are Vi
and ni and if the conditions are changed to a
final volume Vf and final moles nf then we can
write:
initial conditions
final conditions
Hence:
Vi
= C’’
ni
Vf
= C’’
nf
Vi
Vf
=
(const. P, T)
ni
nf
94
If the initial values of volume and moles are Vi
and ni and if the conditions are changed to a
final volume Vf and final moles nf then we can
write:
initial conditions
final conditions
Vi
= C’’
ni
Vf
= C’’
nf
Hence:
Vi
Vf
=
(const. P, T)
ni
nf
This is a statement of Avogadro’s Law.
95
Avogadro’s law means that, when two gases
react with each other, their volumes have a
simple ratio to each other. If the product is a
gas, its volume is related to the volume of the
reactants by a simple ratio.
96
Avogadro’s law means that, when two gases
react with each other, their volumes have a
simple ratio to each other. If the product is a
gas, its volume is related to the volume of the
reactants by a simple ratio.
Example:
3 H2(g) + N2(g)
2 NH3(g)
97
Avogadro’s law means that, when two gases
react with each other, their volumes have a
simple ratio to each other. If the product is a
gas, its volume is related to the volume of the
reactants by a simple ratio.
Example:
3 H2(g) + N2(g)
3 volumes 1 volume
2 NH3(g)
2 volumes
98
Ideal Gas Law
99
Ideal Gas Law
Summary so far:
100
Ideal Gas Law
Summary so far:
Boyle’s law:
1
V 
P
(const. T and n)
101
Ideal Gas Law
Summary so far:
Boyle’s law:
Charles’ law
1
V 
P
V T
(const. T and n)
(const. P and n)
102
Ideal Gas Law
Summary so far:
Boyle’s law:
Charles’ law
1
V 
P
V T
Avogadro’s law V  n
(const. T and n)
(const. P and n)
(const. P and T)
103
Math Aside:
104
Math Aside:
If
fX
105
Math Aside:
If f  X
and f  Y
106
Math Aside:
If f  X
and f  Y
and f  Z-1
107
Math Aside:
If f  X
and f  Y
and f  Z-1
Then
f  X Y Z-1
108
Using the result from the math aside, we can put
the expressions for Boyle’s law, Charles’ law, and
Avogadro’s law together, so that:
nT
V 
P
109
Using the result from the math aside, we can put
the expressions for Boyle’s law, Charles’ law, and
Avogadro’s law together, so that:
nT
V 
P
The proportionality can be replaced by an
equation by inserting a proportionality constant.
In this case the constant is represented by the
symbol R, and is called the gas constant. Hence:
110
Using the result from the math aside, we can put
the expressions for Boyle’s law, Charles’ law, and
Avogadro’s law together, so that:
nT
V 
P
The proportionality can be replaced by an
equation by inserting a proportionality constant.
In this case the constant is represented by the
symbol R, and is called the gas constant. Hence:
PV = n R T
Ideal gas equation
111
Ideal gas: Is defined to be a gas that satisfies the ideal
gas equation.
112
Ideal gas: Is defined to be a gas that satisfies the ideal
gas equation.
Evaluation of the gas constant: Experiments show
that 1 mol of any ideal gas at 0 OC and 1 atm
pressure occupies 22.414 liters. The conditions 0 OC
and 1 atm are called standard temperature and
pressure, abbreviated STP.
113
Ideal gas: Is defined to be a gas that satisfies the ideal
gas equation.
Evaluation of the gas constant: Experiments show
that 1 mol of any ideal gas at 0 OC and 1 atm
pressure occupies 22.414 liters. The conditions 0 OC
and 1 atm are called standard temperature and
pressure, abbreviated STP.
(1 atm) (22.414 l)
R =
( 1 mol) (273.15 K)
= 0.082057 l atm K-1 mol-1
114
In the SI unit system,
R = 8.3145 J K-1 mol-1
115
In the SI unit system,
R = 8.3145 J K-1 mol-1
Here, J stands for joule, the SI unit of energy.
116
In the SI unit system,
R = 8.3145 J K-1 mol-1
Here, J stands for joule, the SI unit of energy.
1 J = 1 Nm = 1 kg m2 s-2
117
In the SI unit system,
R = 8.3145 J K-1 mol-1
Here, J stands for joule, the SI unit of energy.
1 J = 1 Nm = 1 kg m2 s-2
Exercise:
Try to convert R = 0.082057 l atm K-1 mol-1 to the
value given in SI units. (It’s a factor-label exercise).
118
Problem Example, Ideal Gas Law: Calculate the
volume occupied by 0.168 mol of CO2 at STP.
Assume CO2 can be treated as an ideal gas.
119
Problem Example, Ideal Gas Law: Calculate the
volume occupied by 0.168 mol of CO2 at STP.
Assume CO2 can be treated as an ideal gas.
Given data: n = 0.168 mol, P = 1 atm (exact value)
T = 0 oC = 273.15 K (exact value)
120
Problem Example, Ideal Gas Law: Calculate the
volume occupied by 0.168 mol of CO2 at STP.
Assume CO2 can be treated as an ideal gas.
Given data: n = 0.168 mol, P = 1 atm (exact value)
T = 0 oC = 273.15 K (exact value)
nRT
V =
P
121
Problem Example, Ideal Gas Law: Calculate the
volume occupied by 0.168 mol of CO2 at STP.
Assume CO2 can be treated as an ideal gas.
Given data: n = 0.168 mol, P = 1 atm (exact value)
T = 0 oC = 273.15 K (exact value)
nRT
V =
P
(0.168 mol) (0.08206 l atm mol-1 K-1) (273 K)
V =
(1 atm)
= 3.76 l
122
Combined Gas Law
123
Combined Gas Law
If the initial values of the pressure, volume, temperature,
and moles are Pi, Vi, Ti, and ni, and if the conditions are
changed to a final pressure, volume, temperature, and
moles Pf, Vf, Tf, and nf respectively, then we can write:
124
Combined Gas Law
If the initial values of the pressure, volume, temperature,
and moles are Pi, Vi, Ti, and ni, and if the conditions are
changed to a final pressure, volume, temperature, and
moles Pf, Vf, Tf, and nf respectively, then we can write:
initial conditions
Pi Vi
ni Ti
= R
125
Combined Gas Law
If the initial values of the pressure, volume, temperature,
and moles are Pi, Vi, Ti, and ni, and if the conditions are
changed to a final pressure, volume, temperature, and
moles Pf, Vf, Tf, and nf respectively, then we can write:
initial conditions
final conditions
Pi Vi
ni Ti
Pf Vf
nf Tf
= R
= R
126
Combined Gas Law
If the initial values of the pressure, volume, temperature,
and moles are Pi, Vi, Ti, and ni, and if the conditions are
changed to a final pressure, volume, temperature, and
moles Pf, Vf, Tf, and nf respectively, then we can write:
initial conditions
final conditions
Hence:
Pi Vi
ni Ti
Pf Vf
nf Tf
Pi Vi
ni Ti
= R
= R
=
Pf Vf
nf Tf
127
Combined Gas Law
If the initial values of the pressure, volume, temperature,
and moles are Pi, Vi, Ti, and ni, and if the conditions are
changed to a final pressure, volume, temperature, and
moles Pf, Vf, Tf, and nf respectively, then we can write:
initial conditions
final conditions
Hence:
Pi Vi
ni Ti
Pf Vf
nf Tf
Pi Vi
= R
= R
=
Pf Vf
ni Ti
nf Tf
This is the Combined Gas Law.
128
Problem Example, Combined Gas Law: A scuba diver
carries three tanks of air. Each has a capacity of 7.00
l and is at a pressure of 1.50 x 102 atm at 25.0 oC.
What volume of air does this correspond to at STP?
129
Problem Example, Combined Gas Law: A scuba diver
carries three tanks of air. Each has a capacity of 7.00
l and is at a pressure of 1.50 x 102 atm at 25.0 oC.
What volume of air does this correspond to at STP?
Given initial data: Pi = 1.50 x 102 atm
Vi = 21. 0 l
Ti = 298.2 K (25.0 + 273.15)
130
Problem Example, Combined Gas Law: A scuba diver
carries three tanks of air. Each has a capacity of 7.00
l and is at a pressure of 1.50 x 102 atm at 25.0 oC.
What volume of air does this correspond to at STP?
Given initial data: Pi = 1.50 x 102 atm
Vi = 21. 0 l
Ti = 298.2 K (25.0 + 273.15)
Given final data: Pf = 1 atm
Tf = 273.15 K
131
Pi Vi
Pf Vf
=
ni Ti
nf Tf
Assume that the number of moles of gas is fixed, so
the combined gas law simplifies to:
Pi Vi
Pf Vf
=
Ti
Tf
132
Pi Vi
Pf Vf
=
ni Ti
nf Tf
Assume that the number of moles of gas is fixed, so
the combined gas law simplifies to:
Pi Vi
Pf Vf
=
Ti
Tf
The equation can be rearranged, so that
Pi Tf
Vf =
Vi
Pf Ti
133
(1.50 x 102 atm) (273.15 K)
Vf =
(21.0 l)
(1 atm) (298.2 K)
= 2.89 x 103 l at STP
134
Problem Example: Molar mass of a gas. Calculate the molar
mass of methane if 279 ml of the gas measured at 31.3 OC
and 492 torr has a mass 0f 0.116 g. Two steps: first find moles
of gas, then determine the molar mass.
135
Problem Example: Molar mass of a gas. Calculate the molar
mass of methane if 279 ml of the gas measured at 31.3 OC
and 492 torr has a mass 0f 0.116 g. Two steps: first find moles
of gas, then determine the molar mass.
From PV = nRT,
n=
PV
RT
136
Problem Example: Molar mass of a gas. Calculate the molar
mass of methane if 279 ml of the gas measured at 31.3 OC
and 492 torr has a mass 0f 0.116 g. Two steps: first find moles
of gas, then determine the molar mass.
From PV = nRT,
n=
=
PV
RT
1 atm
1l
(492 torr)(
)( 279 ml)(
)
760 torr
1000 ml
(0.08206 l atm mol-1 K-1)(304.5 K)
= 7.23 x 10-3 mol
137
The molar mass is given by:
molar mass  mass in grams
moles
138
The molar mass is given by:
molar mass  mass in grams
moles
0.116 g
=
7.23 x 10-3 mol
139
The molar mass is given by:
molar mass  mass in grams
moles
0.116 g
=
7.23 x 10-3 mol
= 16.0 g mol-1
140
Dalton’s Law of Partial Pressures
141
Dalton’s Law of Partial Pressures
Dalton’s Law of partial pressures: In a mixture
of gases, each component exerts the same
pressure as it would, if it were alone and
occupied the same volume.
142
Dalton’s Law of Partial Pressures
Dalton’s Law of partial pressures: In a mixture
of gases, each component exerts the same
pressure as it would, if it were alone and
occupied the same volume.
Consider a simple case: A mixture of two gases
A and B in a container of volume V and
temperature T.
143
The pressure exerted by gas A – called the
partial pressure of gas A – is given by:
144
The pressure exerted by gas A – called the
partial pressure of gas A – is given by:
nART
PA =
V
145
The pressure exerted by gas A – called the
partial pressure of gas A – is given by:
nART
PA =
V
Similarly for gas B,
nBRT
PB =
V
146
Now the total pressure PTotal is
PTotal
nTotal RT
=
V
147
Now the total pressure PTotal is
nTotal RT
PTotal =
V
Now nTotal = nA + nB so that:
148
Now the total pressure PTotal is
nTotal RT
PTotal =
V
Now nTotal = nA + nB so that:
PTotal
(nA + nB )RT
nA RT
=
=
+
V
V
nB RT
V
149
Now the total pressure PTotal is
nTotal RT
PTotal =
V
Now nTotal = nA + nB so that:
PTotal
(nA + nB )RT
nA RT
=
=
+
V
V
nB RT
V
Hence,
PTotal = PA + PB
150
Now the total pressure PTotal is
nTotal RT
PTotal =
V
Now nTotal = nA + nB so that:
PTotal
(nA + nB )RT
nA RT
=
=
+
V
V
nB RT
V
Hence,
PTotal = PA + PB
This is Dalton’s Law of partial pressures: The total
pressure is the sum of the partial pressures.
151
The preceding result can be generalized to any
number of components:
PTotal = PA + PB + PC + PD + …
where PA, PB, PC, PD, etc. are the partial
pressures of the individual gases.
152
Problem Example: Assume that 1.00 moles of
air contain 0.78 moles of dinitrogen, 0.21
moles of dioxygen, and 0.01 moles of argon.
Calculate the partial pressures of the three
gases when the air pressure is at 1.0 atm.
153
Problem Example: Assume that 1.00 moles of
air contain 0.78 moles of dinitrogen, 0.21
moles of dioxygen, and 0.01 moles of argon.
Calculate the partial pressures of the three
gases when the air pressure is at 1.0 atm.
Pair  nRT  (nN  nO  nAr) RT
V
V
2
2
154
Problem Example: Assume that 1.00 moles of
air contain 0.78 moles of dinitrogen, 0.21
moles of dioxygen, and 0.01 moles of argon.
Calculate the partial pressures of the three
gases when the air pressure is at 1.0 atm.
Pair  nRT  (nN  nO  nAr) RT
V
V
2
2
PN  nN RT
V
2
2
155
RT
n
N
PN 
n
N
V

Pair (nN  nO  nAr) RT (nN  nO  nAr)
V
2
2
2
2
2
2
2
156
RT
n
N
PN 
n
N
V

Pair (nN  nO  nAr) RT (nN  nO  nAr)
V
2
2
2
2
2
2
2
n
N
PN 
Pair
(nN  nO  nAr)
2
2
2
2
157
RT
n
N
PN 
n
N
V

Pair (nN  nO  nAr) RT (nN  nO  nAr)
V
2
2
2
2
2
2
2
n
N
PN 
Pair
(nN  nO  nAr)
2
2
2
2
But
(nN  nO  nAr)  1.00 mol
2
2
158
PN  0.78 mol 1.0 atm  0.78 atm
1.00 mol
2
Similar calculations give
PO  0.21atm and PAr  0.01 atm
2
159
Dalton’s law has a practical application when
calculating the volume of gases collected
over water. For a gas collected over water,
the measured pressure is given by
Ptotal  Pgas  PH O
2
where PH O denotes the vapor pressure of
water.
2
160
161
(p. 217)
162
Problem example: O2 generated in the
decomposition of KClO3 is collected over water.
The volume of the gas collected at 24 oC and at an
atmospheric pressure of 762 torr is 128 ml.
Calculate the number of moles of O2 obtained. The
vapor pressure of H2O at 24 oC is 22.4 torr.
First step: Calculate the partial pressure of O2.
PTotal  Pgas  PH O (Dalton’s law)
2
163
Pgas  PTotal  PH O
2
= 762 torr – 22.4 torr
= 739.6 torr (extra sig. fig)
 739.6 torr 1 atm
760 torr
= 0.973 atm
164
From the ideal gas equation PV = nRT,
PV
n=
RT
(0.973 atm) (0.128 l)
=
(0.08206 l atm mol-1K-1)(297 K)
= 0.00511 mols
165
Graham’s Law of Effusion
166
Graham’s Law of Effusion
Diffusion: A process by which one gas gradually
mixes with another. The term is also used for
solutes mixing with a solvent.
167
Graham’s Law of Effusion
Diffusion: A process by which one gas gradually
mixes with another. The term is also used for
solutes mixing with a solvent.
Effusion: The process by which a gas under
pressure escapes from one compartment of a
container to another by passing through a
small opening.
168
Graham’s law: The rate of effusion of a gas is
inversely proportional to the square root of its
density when the pressure and temperature are
held constant.
Effusion rate: abbreviated R (don’t get this
confused with the gas constant).
169
Graham’s law: The rate of effusion of a gas is
inversely proportional to the square root of its
density when the pressure and temperature are
held constant.
Effusion rate: abbreviated R (don’t get this
confused with the gas constant).
Gas density: abbreviated d.
170
Graham’s law: The rate of effusion of a gas is
inversely proportional to the square root of its
density when the pressure and temperature are
held constant.
Effusion rate: abbreviated R (don’t get this
confused with the gas constant).
Gas density: abbreviated d.
R
1
(const. P, T)
d
171
Graham’s law: The rate of effusion of a gas is
inversely proportional to the square root of its
density when the pressure and temperature are
held constant.
Effusion rate: abbreviated R (don’t get this
confused with the gas constant).
Gas density: abbreviated d.
R
1
(const. P, T)
d
This is Graham’s law of effusion.
172
The proportionality can be turned into an equality:
R d = C
(const. P and T)
This is a statement of Graham’s law of effusion.
(The constant is unrelated to any of the previous
constants in Boyle’s law, etc.).
173
For two gases A and B, the relative rates of effusion can
be evaluated as follows:
174
For two gases A and B, the relative rates of effusion can
be evaluated as follows:
For gas A:
RA d A = C
175
For two gases A and B, the relative rates of effusion can
be evaluated as follows:
For gas A:
RA d A = C
For gas B:
RB dB = C
176
For two gases A and B, the relative rates of effusion can
be evaluated as follows:
For gas A:
RA d A = C
For gas B:
RB dB = C
Hence: RA d A = RB dB
177
For two gases A and B, the relative rates of effusion can
be evaluated as follows:
For gas A:
RA d A = C
RB dB = C
For gas B:
Hence: RA d A = RB dB
RA
RB
=
dB
dA
(const. P, T)
178
For two gases A and B, the relative rates of effusion can
be evaluated as follows:
For gas A:
RA d A = C
RB dB = C
For gas B:
Hence: RA d A = RB dB
RA
RB
=
dB
dA
(const. P, T)
This is a statement of Graham’s law of effusion Law.
179
Problem example, Graham ’s law: Under conditions for
which the density of carbon dioxide is 1.96 g/l and that of
dinitrogen is 1.25 g/l, which gas will effuse more rapidly?
What is the ratio of the rates of effusion of dinitrogen to
carbon dioxide?
180
Problem example, Graham ’s law: Under conditions for
which the density of carbon dioxide is 1.96 g/l and that of
dinitrogen is 1.25 g/l, which gas will effuse more rapidly?
What is the ratio of the rates of effusion of dinitrogen to
carbon dioxide? Let A be N2 and B be CO2.
181
Problem example, Graham ’s law: Under conditions for
which the density of carbon dioxide is 1.96 g/l and that of
dinitrogen is 1.25 g/l, which gas will effuse more rapidly?
What is the ratio of the rates of effusion of dinitrogen to
carbon dioxide? Let A be N2 and B be CO2.
RN

RCO
2
2
dCO
dN
2
2
182
Problem example, Graham ’s law: Under conditions for
which the density of carbon dioxide is 1.96 g/l and that of
dinitrogen is 1.25 g/l, which gas will effuse more rapidly?
What is the ratio of the rates of effusion of dinitrogen to
carbon dioxide? Let A be N2 and B be CO2.
RN

RCO
2
2
dCO
dN
2
2
 1.96 g/l
1.25 g/l
183
Problem example, Graham ’s law: Under conditions for
which the density of carbon dioxide is 1.96 g/l and that of
dinitrogen is 1.25 g/l, which gas will effuse more rapidly?
What is the ratio of the rates of effusion of dinitrogen to
carbon dioxide? Let A be N2 and B be CO2.
RN

RCO
2
2
dCO
dN
2
2
 1.96 g/l
1.25 g/l
= 1.25
184
Problem example, Graham ’s law: Under conditions for
which the density of carbon dioxide is 1.96 g/l and that of
dinitrogen is 1.25 g/l, which gas will effuse more rapidly?
What is the ratio of the rates of effusion of dinitrogen to
carbon dioxide? Let A be N2 and B be CO2.
RN

RCO
2
2
dCO
dN
2
2
 1.96 g/l
1.25 g/l
= 1.25
So RN = 1.25 x RCO , that is, N2 effuses 1.25 times
faster than CO2.
2
2
185
An alternative form can be developed for Graham’s law of
effusion, where the ratio of the rates is determined from
the molar masses of the two gases.
186
An alternative form can be developed for Graham’s law of
effusion, where the ratio of the rates is determined from
the molar masses of the two gases.
m
m
d =
M =
V
n
187
An alternative form can be developed for Graham’s law of
effusion, where the ratio of the rates is determined from
the molar masses of the two gases.
m
m
d =
M =
V
n
so
nM
d =
V
188
An alternative form can be developed for Graham’s law of
effusion, where the ratio of the rates is determined from
the molar masses of the two gases.
m
m
d =
M =
V
n
so
nM
d =
V
and PV = n RT , so (n/V) = P/(RT)
189
An alternative form can be developed for Graham’s law of
effusion, where the ratio of the rates is determined from
the molar masses of the two gases.
m
m
d =
M =
V
n
so
nM
d =
V
and PV = n RT , so (n/V) = P/(RT)
d =
PM
RT
190
Now plug this result for d into the expression for
Graham’s law.
PMB
R A  RT
RB
PMA
RT
191
Now plug this result for d into the expression for
Graham’s law.
PMB
R A  RT
RB
PMA
RT
R A  MB
That is
(const. P, T)
RB
MA
192
Now plug this result for d into the expression for
Graham’s law.
PMB
R A  RT
RB
PMA
RT
R A  MB
That is
(const. P, T)
RB
MA
This is a statement of Graham’s law of effusion Law.
193
Exercise, Graham ’s law: What is the ratio of the rates
of effusion of dinitrogen to carbon dioxide?
194
Exercise, Graham ’s law: What is the ratio of the rates
of effusion of dinitrogen to carbon dioxide? Note: In
this example, no density information is given, so use
the formula involving the molar masses.
195
Kinetic Theory of Gases
196
The kinetic theory of gases provides a “working
model” of a gas. It is an attempt to interpret
experimental observations at the molecular
level.
197
The kinetic theory of gases provides a “working
model” of a gas. It is an attempt to interpret
experimental observations at the molecular
level.
Basic Postulates of Kinetic Theory:
198
The kinetic theory of gases provides a “working
model” of a gas. It is an attempt to interpret
experimental observations at the molecular
level.
Basic Postulates of Kinetic Theory:
1. A gas consists of an extremely large number
of tiny particles that are in constant random
motion.
199
2. The particles are separated by distances far
greater than their own dimensions. The
molecules can be considered as point-like,
that is, they posses mass but have negligible
volume (compared with the volume of the
container).
200
2. The particles are separated by distances far
greater than their own dimensions. The
molecules can be considered as point-like,
that is, they posses mass but have negligible
volume (compared with the volume of the
container).
3. All molecular collisions are elastic, that is,
the sum of the kinetic energy of the colliding
molecules remains unchanged before and
after the collision.
201
It is found: The average kinetic energy of a collection of
gas molecules is directly proportional to the absolute
temperature.
KE = ½ m v2  T
The bar over KE means average, and over v2 means the
average of v2.
202
It is found: The average kinetic energy of a collection of
gas molecules is directly proportional to the absolute
temperature.
KE = ½ m v2  T
The bar over KE means average, and over v2 means the
average of v2.
v2 is called the mean square velocity.
203
It is found: The average kinetic energy of a collection of
gas molecules is directly proportional to the absolute
temperature.
KE = ½ m v2  T
The bar over KE means average, and over v2 means the
average of v2.
v2 is called the mean square velocity.
If there are N molecules
v2 =
v12 + v22 + … + vN2
N
204
Molecules exert neither attractive nor
repulsive forces on one another. This ties in
directly with postulate 3.
205
Molecules exert neither attractive nor
repulsive forces on one another. This ties in
directly with postulate 3.
Compressibility: Since gas molecules are
separated by large distances (relative to the
molecular size), they can be compressed easily
to occupy smaller volumes.
206
Pressure – volume connection
(p. 215):
207
Pressure – volume connection: Increase the
external pressure on the piston at constant
temperature, then more molecules strike the
container walls, so the force per unit area
increases, and hence the pressure of the gas
increases.
208
Pressure – volume connection: Increase the
external pressure on the piston at constant
temperature, then more molecules strike the
container walls, so the force per unit area
increases, and hence the pressure of the gas
increases. That is, a smaller volume implies a
larger pressure. Detailed arguments lead to
P  V-1.
209
Pressure – temperature connection: An increase
in temperature increases the average velocity
of the gas molecules.
210
Pressure – temperature connection: An increase
in temperature increases the average velocity
of the gas molecules.
At higher velocities, the gas molecules strike
the container walls more frequently and with
greater force.
211
Pressure – temperature connection: An increase
in temperature increases the average velocity
of the gas molecules.
At higher velocities, the gas molecules strike
the container walls more frequently and with
greater force.
If the volume of the container is kept
constant, the area being struck is the same, so
the force per unit area, that is the pressure,
increases.
212
Pressure – temperature connection: An increase
in temperature increases the average velocity
of the gas molecules.
At higher velocities, the gas molecules strike
the container walls more frequently and with
greater force.
If the volume of the container is kept
constant, the area being struck is the same, so
the force per unit area, that is the pressure,
increases. Expect P T (const. V, n).
213
Temperature – volume connection: Increase the
temperature at constant pressure.
214
Temperature – volume connection: Increase the
temperature at constant pressure. An increase
in temperature increases the average velocity
of the gas molecules.
215
Temperature – volume connection: Increase the
temperature at constant pressure. An increase
in temperature increases the average velocity
of the gas molecules. At higher velocities, the
gas molecules strike the container walls more
frequently and with greater force.
216
Temperature – volume connection: Increase the
temperature at constant pressure. An increase
in temperature increases the average velocity
of the gas molecules. At higher velocities, the
gas molecules strike the container walls more
frequently and with greater force. This would
increase the pressure, but if this is held
constant, it would be necessary for the
volume of the container to increase to
maintain this constant pressure.
217
Temperature – volume connection: Increase the
temperature at constant pressure. An increase
in temperature increases the average velocity
of the gas molecules. At higher velocities, the
gas molecules strike the container walls more
frequently and with greater force. This would
increase the pressure, but if this is held
constant, it would be necessary for the
volume of the container to increase to
maintain this constant pressure.
Expect V T (const. P, n).
218
Graham’s law of effusion: Consider two gases
(call them A and B) at the same temperature.
219
Graham’s law of effusion: Consider two gases
(call them A and B) at the same temperature.
Kinetic theory indicates that the average
kinetic energy for both gases must be the
same.
220
Graham’s law of effusion: Consider two gases
(call them A and B) at the same temperature.
Kinetic theory indicates that the average
kinetic energy for both gases must be the
same.
gas A
KEA = ½ mA vA2
gas B
KEB = ½ mB vB2
221
Graham’s law of effusion: Consider two gases
(call them A and B) at the same temperature.
Kinetic theory indicates that the average
kinetic energy for both gases must be the
same.
gas A
KEA = ½ mA vA2
gas B
KEB = ½ mB vB2
If gas B molecules have higher mass, then the
only way the average KE can be the same is
222
that the average velocity of the B molecules is
smaller than gas A.
223
that the average velocity of the B molecules is
smaller than gas A. The molecules moving
more quickly are more likely to hit the
opening, and hence the gas with the
molecules having the lower mass will effuse
faster.
224
Maxwell Distribution of Molecular
Speeds
225
Maxwell Distribution of Molecular
Speeds
At a given instant, how many molecules are
moving at a particular speed?
226
Maxwell Distribution of Molecular
Speeds
At a given instant, how many molecules are
moving at a particular speed?
The answer to this question is provided by
Maxwell’s distribution of speeds curve.
227
Maxwell Distribution of Molecular
Speeds
(p. 205)
228
One important point to note is that although
there are always some slow-moving
molecules, there are fewer slow-moving
molecules at higher temperatures.
229
One important point to note is that although
there are always some slow-moving
molecules, there are fewer slow-moving
molecules at higher temperatures. This is
important for understanding the rates (i.e.
speeds) of gas phase chemical reactions.
230