Chem 106, Prof. T. L. Heise
CHE 106: General Chemistry
CHAPTER
TEN
Copyright © Tyna L. Heise 2001
All Rights Reserved
1
Chem 106, Prof. T. L. Heise
Do Now
 Turn
in VSEPR Lab
 Make sure you have included bond angles and the
DIPOLE MOMENTS for any of the polar
molecules
 Any questions about the upcoming test?
2
Chem 106, Prof. T. L. Heise
3
Gases
How
is matter encountered?
–NOT on the atomic or molecular
scale
–As a large collection of atoms or
molecules
–Recognized as solids liquids and
gases
Chapt. 10.1
Chem 106, Prof. T. L. Heise
4
Gases
Characteristics
of Gases
–Expands to fill container completely
–Highly compressible
–Form homogeneous mixtures
regardless of identities or
proportions
Chapt. 10.1
Chem 106, Prof. T. L. Heise
5
Gases
Why
exhibit these characteristics?
–Individual particles are far apart
–Act as if they are only molecule
present
Chapt. 10.1
Chem 106, Prof. T. L. Heise
6
Properties of Gases
When measuring gases, easiest properties
are
1) Temperature - thermochemistry
2) Volume - solution chemistry
3) Pressure
Chapt. 10.2
Chem 106, Prof. T. L. Heise
7
Pressure
• - conveys idea of force
•
P= F
•
A
•
•
P = pressure
F = force
A = area
- pressure is exerted on any
surface a gas contacts
Chapt. 10.2
Chem 106, Prof. T. L. Heise
8
Atmospheric pressure
Gases of our atmosphere exert a force on
surface of the earth due to gravity.
- force exerted is equal to mass times
acceleration due to gravity
F = ma
since P = F
then P = ma
a = 9.8 m
A
A
s2
Chapt. 10.2
Chem 106, Prof. T. L. Heise
9
Atmospheric pressure
Atmospheric pressure
is measured using a
mercury barometer
•Tube is filled with
mercury
•Small portion falls
back into dish when
inverted, vacuum
exists above liquid in
column
•Column moves up or
down depending on
atmospheric force on
surface of mercury in
dish
Chapt. 10.2
Chem 106, Prof. T. L. Heise
10
Atmospheric pressure
Units of Measurement:
1 atm
760 mmHg
760 torr
1.01325 x 105 Pa
101.325 kPa
Convert 13.33 kPa into atm
13.33 kPa
1 atm
101.325 kPa
= 0.1316 atm
Chapt. 10.2
Chem 106, Prof. T. L. Heise
11
Pressures of Enclosed gases
Manometers
•Similar in operation to barometer
•Two types, closed tube and open tube
•Closed tube - measures pressures
below atmospheric
•Difference in tube heights =
pressure
Chapt. 10.2
Chem 106, Prof. T. L. Heise
12
Pressures of Enclosed gases
•Open tube - measures pressures
near atmospheric
•Difference in tube heights
relates pressure of gas to
atmospheric pressure
Chapt. 10.2
Chem 106, Prof. T. L. Heise
13
Pressures of Enclosed gases
Chapt. 10.2
Chem 106, Prof. T. L. Heise
14
Pressures of Enclosed gases
Sample Problem: A vessel connected to an open
end manometer is filled with gas to a pressure of
0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
Chapt. 10.2
Chem 106, Prof. T. L. Heise
15
Pressures of Enclosed gases
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
a) Convert atm into torr
Chapt. 10.2
Chem 106, Prof. T. L. Heise
16
Pressures of Enclosed gases
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
a) Convert atm into torr
0.835 atm
760 torr
=
1 atm
Chapt. 10.2
Chem 106, Prof. T. L. Heise
17
Pressures of Enclosed gases
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
a) Convert atm into torr
0.835 atm
760 torr
1 atm
=
634 torr
* which is less
than atmospheric
pressure
Chapt. 10.2
Chem 106, Prof. T. L. Heise
18
Pressures of Enclosed gases
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
a) Convert atm into torr
atmospheric pressure makes arm open to air lower and
arm attached to gas container higher
Chapt. 10.2
Chem 106, Prof. T. L. Heise
19
Pressures of Enclosed gases
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
b) Pgas = Patm + difference in heights
Chapt. 10.2
Chem 106, Prof. T. L. Heise
20
Pressures of Enclosed gases
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
b) Pgas = Patm + difference in heights
Pgas = 634 torr
Patm = 755 torr
634 = 755 + difference
Chapt. 10.2
Chem 106, Prof. T. L. Heise
21
Pressures of Enclosed gases
Sample Problem: A vessel connected to an open end
manometer is filled with gas to a pressure of 0.835 atm.
The atmospheric pressure is 755 torr.
a) which arm of manometer is higher?
b) what is the height difference?
b) Pgas + difference in heights = Patm
Pgas = 634 torr
Patm = 755 torr
634 + X = 755
X = 121 torr
Chapt. 10.2
Chem 106, Prof. T. L. Heise
Now
 Open
test
the floor to questions about the upcoming
22
Chem 106, Prof. T. L. Heise
23
The Gas Laws
Four variables needed to adequately describe a
gas
•Temperature (T)
•Pressure (P)
•Volume (V)
•Number of moles (n)
•Equations that express relationships
between these variables are the gas laws.
Chapt. 10.3
Chem 106, Prof. T. L. Heise
24
Boyle’s Law
Boyle investigated the relationship between volume
and pressure.
As pressure
increased,
the volume
decreased;
proves an
inverse
relationship
PV = constant
Chapt. 10.3
Chem 106, Prof. T. L. Heise
25
Charles’s Law
Charles investigated the relationship between volume
and temperature.
As temp
decreased,
the volume
decreased;
proves a
direct
relationship
V = constant
T
Chapt. 10.3
Chem 106, Prof. T. L. Heise
26
Avogadro’s Law
Avogadro investigated the relationship between volume
and amount of substance.
Avogadro’s Hypothesis - equal
volumes of gases at same temperature
and pressure contain equal numbers
of molecules
22.4 L
As number
of molecules
doubles, the
volume as
doubles proves a
direct
relationship
Chapt. 10.3
Chem 106, Prof. T. L. Heise
Check it out!
P
vs V
 T vs P
 T vs V
27
Chem 106, Prof. T. L. Heise
28
The Ideal Gas Equation
Combining the previous three laws, allows for a better
mathematical look at gases.
The term R
in the gas
equation is
called the
gas
constant.
The conditions of 0°C and
1 atm are referred to as
standard temperature and
pressure!
Chapt. 10.4
Chem 106, Prof. T. L. Heise
29
The Ideal Gas Equation
The ideal gas law allows us to isolate all
variables that are to be held constant and set
up proportionalities.
P1V1 = P2V2
V 1 = V2
T1 T2
P1 = P2
T1 T2
Chapt. 10.4
Chem 106, Prof. T. L. Heise
30
The Ideal Gas Equation
Sample Exercise:
Tennis balls are usually filled with air or N2
gas to a pressure above atmospheric pressure to
increase their “bounce”. If a particular tennis
ball has a volume of 144 cm3 and contains 0.33 g
of N2 gas, what is the pressure inside the ball at
24°C?
1) PV = nRT
Chapt. 10.4
Chem 106, Prof. T. L. Heise
31
The Ideal Gas Equation
Sample Exercise:
Tennis balls are usually filled with air or N2 gas
to a pressure above atmospheric pressure to increase their
“bounce”. If a particular tennis ball has a volume of
144 cm3 and contains 0.33 g of N2 gas, what is the pressure
inside the ball at 24°C?
2) PV = nRT
P=x
R = 0.08206 L-atm/mol-K
V = 144 cm3
T = 24°C
n = 0.33 g N2
Chapt. 10.4
Chem 106, Prof. T. L. Heise
32
The Ideal Gas Equation
Sample Exercise:
Tennis balls are usually filled with air or N2 gas
to a pressure above atmospheric pressure to increase their
“bounce”. If a particular tennis ball has a volume of
144 cm3 and contains 0.33 g of N2 gas, what is the pressure
inside the ball at 24°C?
3) PV = nRT
P=x
R = 0.08206 L-atm/mol-K
V = 144 cm3 = 144 mL = 0.144 L
T = 24°C = 297 K
n = 0.33 g N2
1 mol N2 = 0.012 mol N2
28 g N2
Chapt. 10.4
Chem 106, Prof. T. L. Heise
33
The Ideal Gas Equation
Sample Exercise:
Tennis balls are usually filled with air or N2 gas
to a pressure above atmospheric pressure to increase their
“bounce”. If a particular tennis ball has a volume of
144 cm3 and contains 0.33 g of N2 gas, what is the pressure
inside the ball at 24°C?
4) P = nRT = 0.012 mol N2 (0.08206 L-atm/mol-K) (297 K)
V
0.144 L
Chapt. 10.4
Chem 106, Prof. T. L. Heise
34
The Ideal Gas Equation
Sample Exercise:
Tennis balls are usually filled with air or N2 gas
to a pressure above atmospheric pressure to increase their
“bounce”. If a particular tennis ball has a volume of
144 cm3 and contains 0.33 g of N2 gas, what is the pressure
inside the ball at 24°C?
2) P = nRT = 0.012 mol N2 (0.08206 L-atm/mol-K) (297 K)
V
0.144 L
P = 2.0 atm
Chapt. 10.4
Chem 106, Prof. T. L. Heise
35
The Ideal Gas Equation
Sample Exercise:
A large natural gas storage tank is arranged
so that the pressure is maintained at 2.20 atm.
On a cold December day when the temperature is
-15°C, the volume of gas in the tank is 28,500 ft3.
What is the volume of the same quantity of gas on
a warm July day when the temperature is 31°C?
Chapt. 10.4
Chem 106, Prof. T. L. Heise
36
The Ideal Gas Equation
Sample Exercise:
A large natural gas storage tank is arranged so that the
pressure is maintained at 2.20 atm. On a cold December day when
the temperature is -15°C, the volume of gas in the tank is 28,500
ft3. What is the volume of the same quantity of gas on a warm
July day when the temperature is 31°C?
1) V1 = V2
T1 T2
Chapt. 10.4
Chem 106, Prof. T. L. Heise
37
The Ideal Gas Equation
Sample Exercise:
A large natural gas storage tank is arranged so that the
pressure is maintained at 2.20 atm. On a cold December day when
the temperature is -15°C, the volume of gas in the tank is 28,500
ft3. What is the volume of the same quantity of gas on a warm
July day when the temperature is 31°C?
1) V1 = V2
T1 T2
V1 = 28,500 ft3
T1 = -15°C
V2 = x
T2 = 31°C
Chapt. 10.4
Chem 106, Prof. T. L. Heise
38
The Ideal Gas Equation
Sample Exercise:
A large natural gas storage tank is arranged so that the
pressure is maintained at 2.20 atm. On a cold December day when
the temperature is -15°C, the volume of gas in the tank is 28,500
ft3. What is the volume of the same quantity of gas on a warm
July day when the temperature is 31°C?
2) V1 = V2
T1 T2
V1 = 28,500 ft3 (12)3 in3 16.4 cm3
1 ft3
1 in3
1 L
= 8.08 x 105 L
103 cm3
Chapt. 10.4
Chem 106, Prof. T. L. Heise
39
The Ideal Gas Equation
Sample Exercise:
A large natural gas storage tank is arranged so that the
pressure is maintained at 2.20 atm. On a cold December day when
the temperature is -15°C, the volume of gas in the tank is 28,500
ft3. What is the volume of the same quantity of gas on a warm
July day when the temperature is 31°C?
3) V1 = V2
T1 T2
V1 = 8.08 x 105 L
T1 = -15°C = 258 K
V2 = x
T2 = 31°C = 304 K
Chapt. 10.4
Chem 106, Prof. T. L. Heise
40
The Ideal Gas Equation
Sample Exercise:
A large natural gas storage tank is arranged so that the
pressure is maintained at 2.20 atm. On a cold December day when
the temperature is -15°C, the volume of gas in the tank is 28,500
ft3. What is the volume of the same quantity of gas on a warm
July day when the temperature is 31°C?
4) V1 = V2
T1 T2
8.08 x 105 L =
258 K
X
304 K
Chapt. 10.4
Chem 106, Prof. T. L. Heise
41
The Ideal Gas Equation
Sample Exercise:
A large natural gas storage tank is arranged so that the
pressure is maintained at 2.20 atm. On a cold December day when
the temperature is -15°C, the volume of gas in the tank is 28,500
ft3. What is the volume of the same quantity of gas on a warm
July day when the temperature is 31°C?
4) V1 = V2
T1 T2
8.08 x 105 L =
258 K
X
304 K
X = 9.52 x 105 L
Chapt. 10.4
Chem 106, Prof. T. L. Heise
42
Further Applications of Law
Ideal gas equation can be used to determine
- density of gas
- molar mass of gas
- volumes of gases formed or consumed
Chapt. 10.5
Chem 106, Prof. T. L. Heise
43
Further Applications of Law
Gas Densities and Molar Mass
n = moles
V
L
* multiply both sides by molar mass, M
n M = PM
V
RT
Chapt. 10.5
Chem 106, Prof. T. L. Heise
44
Further Applications of Law
Gas Densities and Molar Mass
nM
V
= moles
L
g
moles
* multiplying both sides by molar mass, M,
will give us g
L
,
so that
Density = PM
RT
Chapt. 10.5
Chem 106, Prof. T. L. Heise
45
Further Applications of Law
Gas Densities and Molar Mass
Density = PM
RT
this equation can algebraically
be rearranged to solve for
molar mass
Molar Mass = d R T
P
Chapt. 10.5
Chem 106, Prof. T. L. Heise
46
Further Applications of Law
Sample Exercise:
Calculate the average molar mass of dry air if it has a
density of 1.17 g/L at 21°C and 740.0 torr
1) M = d R T
P
Chapt. 10.4
Chem 106, Prof. T. L. Heise
47
Further Applications of Law
Sample Exercise:
Calculate the average molar mass of dry air if it has a
density of 1.17 g/L at 21°C and 740.0 torr
2) M = d R T
P
d = 1.17 g/L
R = 62.36 L-torr/mol-K
T = 21°C = 294 K
P = 740.0 torr
Chapt. 10.4
Chem 106, Prof. T. L. Heise
48
Further Applications of Law
Sample Exercise:
Calculate the average molar mass of dry air if it has a
density of 1.17 g/L at 21°C and 740.0 torr
3) M = d R T
d = 1.17 g/L
P
R = 62.36 L-torr/mol-K
T = 21°C = 294 K
P = 740.0 torr
= 1.17 g/L ( 62.36 L-torr/mol-K) (294 K)
740.0 torr
Chapt. 10.4
Chem 106, Prof. T. L. Heise
49
Further Applications of Law
Sample Exercise:
Calculate the average molar mass of dry air if it has a
density of 1.17 g/L at 21°C and 740.0 torr
4) M = d R T
d = 1.17 g/L
P
R = 62.36 L-torr/mol-K
T = 21°C = 294 K
P = 740.0 torr
= 1.17 g/L ( 62.36 L-torr/mol-K) (294 K)
740.0 torr
= 29.0 g/L
Chapt. 10.4
Chem 106, Prof. T. L. Heise
50
Further Applications of Law
Sample Exercise:
The mean molar mass of the atmosphere at the surface of
Titan, Saturn’s largest moon, is 28.6 g/mol. The surface
temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming
ideal behavior, calculate the density of Titan’s atmosphere.
1) d = P M
RT
Chapt. 10.4
Chem 106, Prof. T. L. Heise
51
Further Applications of Law
Sample Exercise:
The mean molar mass of the atmosphere at the surface of
Titan, Saturn’s largest moon, is 28.6 g/mol. The surface
temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming
ideal behavior, calculate the density of Titan’s atmosphere.
2) d = P M
RT
Pressure = 1.6 atm
M = 28.6 g/mol
R = 0.08206 L-atm/mol-K
T = 95 K
Chapt. 10.4
Chem 106, Prof. T. L. Heise
52
Further Applications of Law
Sample Exercise:
The mean molar mass of the atmosphere at the surface of
Titan, Saturn’s largest moon, is 28.6 g/mol. The surface
temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming
ideal behavior, calculate the density of Titan’s atmosphere.
3) d = P M
RT
=
1.6 atm(28.6 g/mol)
0.08206 L-atm/mol-K ( 95 K)
Chapt. 10.4
Chem 106, Prof. T. L. Heise
53
Further Applications of Law
Sample Exercise:
The mean molar mass of the atmosphere at the surface of
Titan, Saturn’s largest moon, is 28.6 g/mol. The surface
temperature is 95 K, and the pressure is 1.6 Earth atm. Assuming
ideal behavior, calculate the density of Titan’s atmosphere.
4) d = P M
RT
=
1.6 atm(28.6 g/mol)
0.08206 L-atm/mol-K ( 95 K)
= 5.9 g/L
Chapt. 10.5
Chem 106, Prof. T. L. Heise
54
Further Applications of Law
Volumes of Gases in Chemical Reactions
n = PV
RT
Gas
Data
A knowledge of gases is often
important because gases are often
reactants or products in chemical
reactions
Moles
of
Gas A
- coefficients in balanced
equations is again going to be
very important
Moles
g of
of
Gas B
Gas B
Chapt. 10.5
Chem 106, Prof. T. L. Heise
55
Further Applications of Law
Sample Exercise:
In the first step of the industrial process for making nitric
acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the
presence of a suitable catalyst. The following reaction occurs:
4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to
react with 1.00 mol of O2 in this reaction?
Chapt. 10.5
Chem 106, Prof. T. L. Heise
56
Further Applications of Law
Sample Exercise:
In the first step of the industrial process for making nitric
acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the
presence of a suitable catalyst. The following reaction occurs:
4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to
react with 1.00 mol of O2 in this reaction?
1) No need to convert into moles of O2, we were given that
information
Chapt. 10.5
Chem 106, Prof. T. L. Heise
57
Further Applications of Law
Sample Exercise:
In the first step of the industrial process for making nitric
acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the
presence of a suitable catalyst. The following reaction occurs:
4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to
react with 1.00 mol of O2 in this reaction?
2) Convert from mol O2 into mol NH3
1.00 mol O2
4 mol NH3 = 0.800 mol NH3
5 mol O2
Chapt. 10.5
Chem 106, Prof. T. L. Heise
58
Further Applications of Law
Sample Exercise:
In the first step of the industrial process for making nitric
acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the
presence of a suitable catalyst. The following reaction occurs:
4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to
react with 1.00 mol of O2 in this reaction?
3) Use ideal gas equation and given info to solve for final variable
PV = nRT
P = 5.00 atm
n = 0.800 mol NH3
T = 850°C = 1123 K
V=X
R = 0.08206 L-atm/mol-K
Chapt. 10.5
Chem 106, Prof. T. L. Heise
59
Further Applications of Law
Sample Exercise:
In the first step of the industrial process for making nitric
acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the
presence of a suitable catalyst. The following reaction occurs:
4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to
react with 1.00 mol of O2 in this reaction?
4) Use ideal gas equation and given info to solve for final variable
PV = nRT
V = nRT = 0.800 mol ( 0.08206) (1123 K)
P
5.00 atm
Chapt. 10.5
Chem 106, Prof. T. L. Heise
60
Further Applications of Law
Sample Exercise:
In the first step of the industrial process for making nitric
acid, ammonia reacts with oxygen at 850°C and 5.00 atm in the
presence of a suitable catalyst. The following reaction occurs:
4NH3 + 5O2 --> 4NO + 6H20
How many liters of NH3 at 850°C and 5.00 atm are required to
react with 1.00 mol of O2 in this reaction?
4) Use ideal gas equation and given info to solve for final variable
PV = nRT
V = nRT = 0.800 mol ( 0.08206) (1123 K)
P
V = 14.7 L
5.00 atm
Chapt. 10.5
Chem 106, Prof. T. L. Heise
61
Gas Mixtures and Partial Pressures
John Dalton’s work with air allowed him to make
the following observation:
total pressure of a mixture equals the sum of
the pressures that each gas would exert if it were
present alone
Chapt. 10.6
Chem 106, Prof. T. L. Heise
62
Gas Mixtures and Partial Pressures
This equation implies that each gas behaves
independently, so each gas has a unique mole
quantity, and total moles equals sums of each
individual amount...
nt = n1 + n2 + n3 + . . .
Chapt. 10.6
Chem 106, Prof. T. L. Heise
63
Gas Mixtures and Partial Pressures
Sample Exercise:
What is the total pressure exerted by a
mixture of 2.00 g of H2 and 8.00 g of N2 at 273 K
in a 10.0 L vessel?
1) Solve for H2
PV = nRT
Chapt. 10.6
Chem 106, Prof. T. L. Heise
64
Gas Mixtures and Partial Pressures
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of
H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
2) Solve for H2
PV = nRT
P=X
V = 10.0 L
n = 2.00 g H2
1 mol H2
2 g H2
=
+ 8.00 g N2 1 mol N2 =
28 g N2
R = 0.08206 L-atm/mol-K
T = 273 K
Chapt. 10.6
Chem 106, Prof. T. L. Heise
65
Gas Mixtures and Partial Pressures
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of
H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
3) Solve for H2
PV = nRT
P=X
V = 10.0 L
P = nRT
V
n = 1.28 moles total
= 1.28 mol (0.08206) ( 273 K)
10.0 L
R = 0.08206 L-atm/mol-K
= 2.88 atm
T = 273 K
Chapt. 10.6
Chem 106, Prof. T. L. Heise
66
Gas Mixtures and Partial Pressures
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of
H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
4) Solve for N2
PV = nRT
Chapt. 10.6
Chem 106, Prof. T. L. Heise
67
Gas Mixtures and Partial Pressures
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of
H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
5) Solve for N2
PV = nRT
P=X
V = 10.0 L
n = 8.00 g N2
1 mol N2
28 g N2
R = 0.08206 L-atm/mol-K
T = 273 K
Chapt. 10.6
Chem 106, Prof. T. L. Heise
68
Gas Mixtures and Partial Pressures
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of
H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
6) Solve for N2
PV = nRT
P=X
P = nRT
V = 10.0 L
n = 8.00 g N2
V
1 mol N2
28 g N2
R = 0.08206 L-atm/mol-K
T = 273 K
= 0.286 mol (0.08206) ( 273 K)
10.0 L
= 0.641 atm
Chapt. 10.6
Chem 106, Prof. T. L. Heise
69
Gas Mixtures and Partial Pressures
Sample Exercise:
What is the total pressure exerted by a mixture of 2.00 g of
H2 and 8.00 g of N2 at 273 K in a 10.0 L vessel?
7) Sum individual pressures to identify total pressure
Pt = P(H2) + P(N2)
= 2.24 atm + 0.641 atm
= 2.88 atm
Chapt. 10.6
Chem 106, Prof. T. L. Heise
70
Gas Mixtures and Partial Pressures
Partial pressures and Mole Fractions
- each gas behaves independently and so it is easy to relate
the amount of a given gas to its partial pressure
- since P = nRT
V
then P1 = n1RT
V
and P1 = n1
Pt
nt
Pt = ntRT
V
- the ratio n1/nt is denoted the mole fraction of gas 1
- mole fractions are unitless values expressing ratio
Chapt. 10.6
Chem 106, Prof. T. L. Heise
71
Gas Mixtures and Partial Pressures
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated
the composition of the atmosphere of Titan, Saturn’s largest
moon. The total pressure on the surface of Titan is 1220 torr. The
atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol %
CH4. Calculate the partial pressure of each gas.
1) Solve for N2
Chapt. 10.6
Chem 106, Prof. T. L. Heise
72
Gas Mixtures and Partial Pressures
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated
the composition of the atmosphere of Titan, Saturn’s largest
moon. The total pressure on the surface of Titan is 1220 torr. The
atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol %
CH4. Calculate the partial pressure of each gas.
2) Solve for N2
Total pressure = 1220 torr
Mole fraction = 82/100 = 0.82
PN = 0.82 (1220 torr) = 1.0 x 103 torr
Chapt. 10.6
Chem 106, Prof. T. L. Heise
73
Gas Mixtures and Partial Pressures
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated
the composition of the atmosphere of Titan, Saturn’s largest
moon. The total pressure on the surface of Titan is 1220 torr. The
atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol %
CH4. Calculate the partial pressure of each gas.
3) Solve for Ar
Chapt. 10.6
Chem 106, Prof. T. L. Heise
74
Gas Mixtures and Partial Pressures
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated
the composition of the atmosphere of Titan, Saturn’s largest
moon. The total pressure on the surface of Titan is 1220 torr. The
atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol %
CH4. Calculate the partial pressure of each gas.
4) Solve for Ar
Total pressure = 1220 torr
Mole fraction = 12/100 = 0.12
PAr = 0.12 (1220 torr) = 150 torr
Chapt. 10.6
Chem 106, Prof. T. L. Heise
75
Gas Mixtures and Partial Pressures
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated
the composition of the atmosphere of Titan, Saturn’s largest
moon. The total pressure on the surface of Titan is 1220 torr. The
atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol %
CH4. Calculate the partial pressure of each gas.
5) Solve for CH4
Chapt. 10.6
Chem 106, Prof. T. L. Heise
76
Gas Mixtures and Partial Pressures
Sample Exercise:
From data gathered by Voyager 1, scientists have estimated
the composition of the atmosphere of Titan, Saturn’s largest
moon. The total pressure on the surface of Titan is 1220 torr. The
atmosphere consists of 82 mol % N2, 12 mol % Ar and 6.0 mol %
CH4. Calculate the partial pressure of each gas.
2) Solve for CH4
Total pressure = 1220 torr
Mole fraction = 6.0/100 = 0.060
PCH4 = 0.060(1220 torr) = 73 torr
Chapt. 10.6
Chem 106, Prof. T. L. Heise
77
Gas Mixtures and Partial Pressures
Collecting gases over water
- when determining moles of produced gases, the best way
to collect a gas sample is over water
Chapt. 10.6
Chem 106, Prof. T. L. Heise
78
Gas Mixtures and Partial Pressures
Collecting gases over water
- when determining moles of produced gases, the best way
to collect a gas sample is over water
- the volume of gas is measured by raising and lowering
collecting container until water levels are equal inside and
out…when this occurs, atmospheric pressures are equal
- total pressure inside is equal to sum of pressure of
gas collected and pressure of water vapor
Ptotal = Pgas + PH2O
Chapt. 10.6
Chem 106, Prof. T. L. Heise
79
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
1) Calculate partial pressure of N2
Chapt. 10.6
Chem 106, Prof. T. L. Heise
80
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
1) Calculate partial pressure of N2
Pt = PN + Pwater
Pt = 745 torr
Pwater = 25.21 torr
Chapt. 10.6
Chem 106, Prof. T. L. Heise
81
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
1) Calculate partial pressure of N2
Pt = PN + Pwater
Pt = 745 torr
Pwater = 25.21 torr
745 torr = X + 25.21 torr
Chapt. 10.6
Chem 106, Prof. T. L. Heise
82
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
1) Calculate partial pressure of N2
Pt = PN + Pwater
Pt = 745 torr
Pwater = 25.21 torr
745 torr = X + 25.21 torr
PN = 720. torr
Chapt. 10.6
Chem 106, Prof. T. L. Heise
83
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
Chapt. 10.6
Chem 106, Prof. T. L. Heise
84
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
PV = nRT
Chapt. 10.6
Chem 106, Prof. T. L. Heise
85
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
PV = nRT
P = 720. Torr
V = 0.511 L
n=X
R = 62.36 L-torr/mol-K
T = 26°C = 299 K
Chapt. 10.6
Chem 106, Prof. T. L. Heise
86
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
PV = nRT
P = 720. torr
V = 0.511 L
n=X
n = PV
RT
= 720. torr(0.511 L)
62.36 (299 K)
R = 62.36 L-torr/mol-K
T = 26˚C = 299 K
Chapt. 10.6
Chem 106, Prof. T. L. Heise
87
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
2) Use Ideal Gas Equation to determine moles of N2
PV = nRT
P = 720. Torr
V = 0.511 L
n=X
R = 62.36 L-torr/mol-K
T = 26˚C = 299 K
n = PV
RT
= 720. torr(0.511 L)
62.36 (299 K)
= 0.0197 mol N2
Chapt. 10.6
Chem 106, Prof. T. L. Heise
88
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
3) Convert from moles of N2 to g NH4NO2
Chapt. 10.6
Chem 106, Prof. T. L. Heise
89
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
3) Convert from moles of N2 to g NH4NO2
0.0197 mol N2
1 mol NH4NO2
64.0 g NH4NO2
1 mol N2
1 mol NH4NO2
Chapt. 10.6
Chem 106, Prof. T. L. Heise
90
Gas Mixtures and Partial Pressures
Sample exercise:
When a sample of NH4NO2 is decomposed in a test tube,
511 mL of N2 gas is collected over water at 26°C and 745 torr total
pressure. How many grams of NH4NO2 were decomposed?
3) Convert from moles of N2 to g NH4NO2
0.0197 mol N2
1 mol NH4NO2
64.0 g NH4NO2
1 mol N2
1 mol NH4NO2
= 1.26 g NH4NO2
Chapt. 10.6
Chem 106, Prof. T. L. Heise
91
Kinetic Molecular Theory
Ideal gas equation describes how gases behave, but not
why they do . . .
Kinetic molecular theory explains why
- gases consist of large numbers of molecules
that are in continuous, random motion
- volume of the gas molecules themselves is
negligible when compared to the volume of the
gas as a whole
- attractive and repulsive forces between gases
molecules is negligible
Chapt. 10.7
Chem 106, Prof. T. L. Heise
92
Kinetic Molecular Theory
Ideal gas equation describes how gases behave, but not
why they do . . .
Kinetic molecular theory explains why
- energy can be transferred between molecules
during collisions, but the average kinetic energy
of the molecules does not change with time, as
long as the temperature is remaining constant i.e. collisions are elastic
- average kinetic energy is proportional to the
absolute temperature
Chapt. 10.7
Chem 106, Prof. T. L. Heise
93
Kinetic Molecular Theory
This theory is helpful in explaining the pressure and
temperature at a molecular level:
pressure of gas is caused
by the collisions of the
molecules against the wall
of a container
magnitude of pressure
is determined by both
how often and how
forcefully the molecules
strike the wall
Chapt. 10.7
Chem 106, Prof. T. L. Heise
94
Kinetic Molecular Theory
This theory is helpful in explaining the pressure and
temperature at a molecular level:
absolute temperature of
a gas is the measure of
the average kinetic energy
of its molecules
molecular motion
increases with
increasing temperature
this is on average,
individual molecules
have individual speeds
Chapt. 10.7
Chem 106, Prof. T. L. Heise
95
Kinetic Molecular Theory
Applications of the Gas Laws
1. Effect of a volume increase at
constant temp
- molecules must move a longer
distance between collisions
- fewer collisions per unit time
with wall of container
- pressure decreases
Chapt. 10.7
Chem 106, Prof. T. L. Heise
96
Kinetic Molecular Theory
Applications of the Gas Laws
2. Effect of a temperature increase
at constant volume
- increase in speed and u
- more collisions per unit time
with wall of container
- pressure increases
Chapt. 10.7
Chem 106, Prof. T. L. Heise
97
Molecular Effusion and Diffusion
According to the kinetic-molecular theory, the
average kinetic energy of any collection of gases,
u = 3RT
M
R = gas constant
T = temp
M = molar mass
Chapt. 10.8
Chem 106, Prof. T. L. Heise
98
Molecular Effusion and Diffusion
has a specific value of u at any given temperature
-two gases at same temp have same avg.
kinetic energy
-if masses are different, than the speed of
particles will be different because of the inclusion
of M in formula
Chapt. 10.8
Chem 106, Prof. T. L. Heise
99
Molecular Effusion and Diffusion
The dependence of speed on mass,
has several implications:
Effusion - the escape of a
molecule through a tiny hole
Effusion rate is inversely
proportional to the square
root of molar mass
Normally rates are compared
as a ratio
Chapt. 10.8
Chem 106, Prof. T. L. Heise
100
Molecular Effusion and Diffusion
The dependence of speed on mass, has several implications:
Diffusion - spreading of one substance throughout a
space or through another substance
Diffusion is also related to size of particle, however,
molecule collisions make diffusion much more difficult
- average distance traveled by a molecule as it
diffuses is called the mean free path, which
varies with pressure
Chapt. 10.8
Chem 106, Prof. T. L. Heise
101
Deviations from an Ideal Gas
Although the ideal gas equation is useful, all real gases fail to
obey the relationships to some degree
- deviation from an ideal gas occurs most at high
pressure and low temperature
- to ensure as much compliance as possible to the ideal
gas equation, a real gas should be considered when it is
at high temperatures and low pressure
WHY?
Real gases DO
- have molecular attractions
- lose energy when they collide
- have volume
Chapt. 10.9
Chem 106, Prof. T. L. Heise
102
Deviations from an Ideal Gas
Chapt. 10.9
Chem 106, Prof. T. L. Heise
103
Deviations from an Ideal Gas
Van der Waals’ Equation: takes into account volume
and molecular attraction
P = nRT - n2a
V - nb
V2
* constants (a) and (b) are different for each gas and must
be identified using a table
Chapt. 10.9
Chem 106, Prof. T. L. Heise
Chapter Ten; Review
 Characteristics
of gases
 Pressure
 Boyle’s,
Charles’s, & Avogadro’s Laws
 Ideal gas equation
 Gas densities and Molar mass
 Dalton’s Law of partial pressures
 Mole fractions
 Kinetic Molecular theory
 Effusion, Diffusion, u and mean free path
 Deviations from ideal gas behavior
 Van der Waals equation
104