### The Other Gas Law

```Avogadro’s Law: Volume and
Moles
General, Organic, and
Biological Chemistry
Fourth Edition
Karen Timberlake
The Other Gas
Law
the volume of a gas is
directly related to the
number of moles (n) of
gas.
Volume and Moles
Relationship
T and P are constant.
Chapter 7, Section 7
2
Chapter 7, Section 7
Learning Check
Solution
If 0.75 mole of helium gas occupies a volume of 1.5 L,
what volume will 1.2 moles of helium occupy at the
same temperature and pressure?
A. 0.94 L
B. 1.8 L
C. 2.4 L
If 0.75 mole of helium gas occupies a volume of 1.5 L,
what volume will 1.2 moles of helium occupy at the same
temperature and pressure?
Step 1 Organize the data in a table of initial and final
conditions.
Analyze the Problem.
Conditions 1 Conditions 2
V1 = 1.5 L
Chapter 7, Section 7
3
Predict
V increases
n1 = 0.75 mole n2 = 1.2 moles
Know
V2 = ?
n increases
Chapter 7, Section 7
Solution
STP
If 0.75 mole of helium gas occupies a volume of 1.5 L,
what volume will 1.2 moles of helium occupy at the same
temperature and pressure?
To make comparisons between different gases, we
use arbitrary conditions called standard temperature
(273 K) and standard pressure (1 atm). Standard
temperature and pressure is abbreviated as STP.
Step 2 Rearrange the gas law equation to solve for
the unknown quantity.
4
Standard temperature (T) = 0 °C or 273 K
Step 3 Substitute values into the gas law equation
and calculate.
Standard pressure (P) = 1 atm (760 mmHg)
Chapter 7, Section 7
5
Chapter 7, Section 7
6
1
Molar Volume as a Conversion
Factor
Molar Volume
The molar volume of a gas measured at STP (standard
temperature and pressure) is 22.4 L for 1 mole of any
gas.
The molar volume at STP
can be used to form 2 conversion factors.
and
Avogadro’s law indicates that 1 mole of any gas at STP has a
volume of 22.4 L.
Chapter 7, Section 7
7
Guide to Using Molar Volume
Chapter 7, Section 7
8
Using Molar Volume
What is the volume occupied by 2.75 moles of N2 gas
at STP?
Step 1 State the given and needed quantities.
Analyze the Problem.
Given
Need
2.75 moles of N2 at STP
Liters of N2
Step 2 Write a plan to calculate the needed
quantity.
moles N2
Chapter 7, Section 7
9
molar
volume
liters N2
Chapter 7, Section 7
Using Molar Volume
Gases in Chemical Reactions
What is the volume occupied by 2.75 moles of N2 gas
at STP?
Step 3 Write the equalities and conversion factors
including 22.4 L/mole at STP.
The volume or amount of a gas at STP in a chemical
reaction can be calculated from
10
STP conditions.
mole−mole factors from the balanced equation.
Step 4 Set up the problem with factors to cancel
units.
Chapter 7, Section 7
11
Chapter 7, Section 7
12
2
Gases in Equations at STP
Guide to Reactions Involving Gases
What volume (L) of O2 gas is needed to completely
react with 15.0 g of aluminum at STP?
Step 1 State the given and needed quantities.
Analyze the Problem.
Chapter 7, Section 7
13
Chapter 7, Section 7
14
Gases in Equations at STP
Gases in Equations at STP
What volume (L) of O2 gas is needed to completely
react with 15.0 g of aluminum at STP?
What volume (L) of O2 gas is needed to completely
react with 15.0 g of aluminum at STP?
Step 2 Write a plan to calculate the needed quantity.
Step 3 Write the equalities and conversion factors
including the molar volume.
mole-mole
grams molar
molar moles mole−mole
moles molar
molar liters
volume
mass of Al factor
factor
of O2 volume
of O2
of Al mass
Chapter 7, Section 7
15
Chapter 7, Section 7
16
General, Organic, and
Biological Chemistry
Gases in Equations at STP
Fourth Edition
Karen Timberlake
What volume (L) of O2 gas is needed to completely
react with 15.0 g of aluminum at STP?
Chapter 7
The Ideal Gas Law
Step 4 Set up the problem and calculate
Chapter 7, Section 7
Chapter 7, Section 8
3
Ideal Gas Law
R, Ideal Gas Constant
The four properties used in the measurement of a gas,
pressure (P),
volume (V),
temperature (T), and
amount (n),
can be combined to give a single expression called the
ideal gas law. PV = nRT
Rearranging the ideal gas law equation shows that the
four gas properties equal a constant, R.
Chapter 7, Section 8
To calculate the value of R, we substitute the STP
conditions for molar volume into the expression:
19
Chapter 7, Section 8
Learning Check
Solution
Another value for the universal gas constant, R, is
obtained using mmHg for the STP pressure. What is
the value of R when a pressure of 760 mmHg rather
than 1.00 atm is used?
Another value for the universal gas constant, R, is
obtained using mmHg for the STP pressure. What
is the value of R when a pressure of 760 mmHg
rather than 1.00 atm is used?
Chapter 7, Section 8
21
Unit Summary for R, the Ideal
Gas Constant
Chapter 7, Section 8
Chapter 7, Section 8
20
22
Guide to Using the Ideal Gas Law
23
Chapter 7, Section 8
24
4
Learning Check
Solution
Dinitrogen oxide (N2O), laughing gas, is used by
dentists as an anesthetic. If a 20.0 L tank of laughing
gas contains 2.86 moles of N2O at 23 °C, what is the
pressure (mmHg) in the tank?
If a 20.0 L tank of laughing gas contains 2.86 moles of
N2O at 23 ˚C, what is the pressure (mmHg) in the tank?
Step 1 State the given and needed quantities.
Analyze the Problem.
Chapter 7, Section 8
25
Chapter 7, Section 8
26
Solution
Solution
If a 20.0 L tank of laughing gas contains 2.86 moles of
N2O at 23 ˚C, what is the pressure (mmHg) in the tank?
If a 20.0 L tank of laughing gas contains 2.86 moles of
N2O at 23 ˚C, what is the pressure (mmHg) in the tank?
Step 2 Rearrange the ideal gas law equation to solve
for the needed quantity.
Step 3 Substitute the gas data into the equation and
calculate the needed quantity.
Chapter 7, Section 8
27
Ideal Gas Law and Molar Mass
Chapter 7, Section 8
28
Learning Check
A cylinder contains 5.0 L of an unknown gas at 20.0 ˚C
and 0.85 atm. If the mass of the gas in the cylinder is
5.8 g, what is the molar mass of the gas?
Chapter 7, Section 8
29
Chapter 7, Section 8
30
5
Solution
Solution
A cylinder contains 5.0 L of an unknown gas at 20.0 ˚C
and 0.85 atm. If the mass of the gas in the cylinder is
5.8 g, what is the molar mass of the gas?
Step 2 Rearrange the ideal gas law equation to solve
for the number of moles.
A cylinder contains 5.0 L of an unknown gas at 20.0 ˚C
and 0.85 atm. If the mass of the gas in the cylinder is
5.8 g, what is the molar mass of the gas?
Step 1 State the given and needed quantities.
Analyze the Problem
Step 3 Obtain the molar mass by dividing the given
number of grams by the number of moles.
Chapter 7, Section 8
31
Chemical Reactions and the Ideal
Gas Law
Chapter 7, Section 8
32
Learning Check
Nitrogen gas reacts with hydrogen gas to produce
ammonia (NH3) gas. How many liters of NH3 can be
produced at 0.93 atm and 24 ˚C from a 16.0-g sample
of nitrogen gas and an excess of hydrogen gas?
Chapter 7, Section 8
33
Chapter 7, Section 8
34
Solution
Solution
How many liters of NH3 can be produced at 0.93 atm and
24 ˚C from a 16.0-g sample of nitrogen gas and an
excess of hydrogen gas?
How many liters of NH3 can be produced at 0.93 atm and
24 ˚C from a 16.0-g sample of nitrogen gas and an
excess of hydrogen gas?
Step 1 State the given and needed quantities.
Analyze the Problem.
Step 2 Write a plan to convert the given quantity to
the needed moles.
grams
of N2
Chapter 7, Section 8
35
molar
mass
moles mole−mole moles
of N2
factor
of NH3
Chapter 7, Section 8
36
6
Solution
Solution
How many liters of NH3 can be produced at 0.93 atm and
24 ˚C from a 16.0-g sample of nitrogen gas and an
excess of hydrogen gas?
How many liters of NH3 can be produced at 0.93 atm and
24 ˚C from a 16.0-g sample of nitrogen gas and an
excess of hydrogen gas?
Step 3 Write the equalities for molar mass and mole–
mole factors.
Step 4 Set up the problem to calculate moles of
needed quantity.
Chapter 7, Section 8
37
Chapter 7, Section 8
38
General, Organic, and
Biological Chemistry
Solution
Fourth Edition
Karen Timberlake
How many liters of NH3 can be produced at 0.93 atm and
24 ˚C from a 16.0-g sample of nitrogen gas and an
excess of hydrogen gas?
Dalton’s Law of
Partial
Pressures
Step 5 Convert the moles of needed to volume using
the ideal gas law equation.
Chapter 7, Section 8
39
Chapter 7, Section 9
Partial Pressure, Dalton’s Law
Dalton’s Law of Partial Pressures
In a gas mixture, each gas exerts its partial pressure,
which is the pressure it would exert if it were the only
gas in the container.
Dalton’s Law of Partial Pressures indicates that
pressure depends on the total number of gas particles.
Dalton’s law states that the total pressure of a gas
mixture is the sum of the partial pressures of the gases
in the mixture.
Chapter 7, Section 9
41
Chapter 7, Section 9
42
7
Total Pressure
Composition of Air
For example, at STP, one mole of a pure gas will
exert the same pressure as one mole of a gas
mixture in a 22.4 L container.
V = 22.4 L
Gas mixtures
Air is a mixture of gases, including nitrogen, oxygen,
carbon dioxide, argon, and water gases.
1.0 mole N2
0.4 mole O2
0.6 mole He
1.0 mole
1.0 atm
1.0 atm
0.5 mole O2
0.3 mole He
0.2 mole Ar
1.0 mole
1.0 atm
Chapter 7, Section 9
43
Guide to Solving for Partial
Pressure
Chapter 7, Section 9
44
Learning Check
A scuba tank contains a
mixture of He and O2 gases
at 7.00 atm. If the pressure
of O2 gas is 1140 mmHg,
what is the partial pressure
of He gas in the tank?
Chapter 7, Section 9
45
Chapter 7, Section 9
46
Solution
Solution
A scuba tank contains a mixture of He and O2 gases at
7.00 atm. If the pressure of O2 gas is 1140 mmHg, what
is the partial pressure of He gas in the tank?
Step 1 Write the equation for partial pressures.
A scuba tank contains a mixture of He and O2 gases at
7.00 atm. If the pressure of O2 gas is 1140 mmHg, what
is the partial pressure of He gas in the tank?
Step 3 Substitute known pressures into the
equation and calculate the unknown partial
pressure.
Step 2 Rearrange the equation to solve for the
unknown pressure.
Chapter 7, Section 9
47
Chapter 7, Section 9
48
8
Blood Gases
Blood Gases
In the lungs, O2 enters the blood, at the same time
that CO2 is released.
In the tissues, O2 enters the cells, which release CO2
into the blood.
In the body,
O2 flows into the tissues because the partial pressure of
O2 is higher in blood and lower in the tissues.
CO2 flows out of the tissues because the partial
pressure of CO2 is higher in the tissues and lower in the
blood.
Partial Pressures (mmHg) in Blood and Tissue
Oxygenated Deoxygenated
Tissues
Blood
Blood
O2 100 mmHg
40 mmHg
30 mmHg or less
CO2 40 mmHg
46 mmHg
50 mmHg or greater
Gas