General Chemistry
Principles and Modern Applications
Petrucci • Harwood • Herring
8th Edition
Chapter 6: Gases
Philip Dutton
University of Windsor, Canada
N9B 3P4
Prentice-Hall © 2002
(modified 2003 by Dr. Paul Root and 2005 by Dr. David Tramontozzi)
Contents
6-1 Properties of Gases: Gas Pressure
6-2 The Simple Gas Laws
6-3 Combining the Gas Laws:
The Ideal Gas Equation and
The General Gas Equation
6-4 Applications of the Ideal Gas Equation
6-5 Gases in Chemical Reactions
6-6 Mixtures of Gases
Contents
6-6 Mixtures of Gases
6-7 Kinetic—Molecular Theory of Gases
6-8 Gas Properties Relating to the
Kinetic—Molecular Theory
6-9 Non-ideal (real) Gases
Focus on The Chemistry of Air-Bag
Systems
Standard Conditions (STP)
Pressure
1 atm
760 mm Hg
760 torr
101.325 kPa
1.01325 bar
Temperature
0°C
273.15 K
Volume
1.0 mol of gas occupies 22.4 L at STP
The Gas Constant
0.082057 L atm mol-1 K-1
8.3145 m3 Pa mol-1 K-1
8.3145 J mol-1 K-1
62.364 L Torr mol-1 K-1
6-3 Combining the Gas Laws: The Ideal
Gas Equation and the General Gas
Equation
• Boyle’s law
• Charles’s law
• Avogadro’s law
V α 1/P
VαT
Vαn
nT
Vα
P
Making an equation we get:
Rearranging we get:
knT
V=
P
PV = nRT
The IDEAL GAS
EQUATION
Problem Solving Strategy
1. Collect the information that is given
2. Use the known conversion factors to convert
data to the correct units.
3. Identify the variable to be determined.
4. Rearrange PV=nRT to solve for the desired
quantity
PV
PV
nRT
nRT
n=
T=
V=
P=
RT
nR
P
V
5. Carry units through the calculation as check.
Example 6-7
Calculating a Gas Volume with the Ideal Gas Equation. What is
the volume occupied by 13.7 g Cl2(g) at 45°C and 745 mmHg?
Strategy:
Collect data and convert to the proper units.
n = 13.7 g Cl2 x 1 mol Cl2 = 0.193 mol Cl2
70.19 g Cl2
P = 745 mmHg x
1 atm
760 mmHg
T = 45°C + 273.15 = 318.15 K
R = 0.08206 L atm K-1 mol-1
= 0.980 atm
V = ?????
Example 6-7
Use the proper rearranged ideal gas equation and sub in all of
the data that you have figured out
nRT
V=
P
(0.193 mol Cl2)(0.08206 L atm K-1 mol-1)(318.15 K)
V=
0.980 atm
Crossing out the units that are common will leave us with the
proper units of volume
V = 5.14 L
The General Gas Equation
P1V1
P2V2
=
n1T1
n2T2
True since R is the same during
initial conditions (1) and final
conditions (2)
Combined Gas Law
P1V1
P2V2
=
T1
T2
Since we compare the same amount
of gas at the end that we had at the
beginning we can say n1=n2
The General Gas Equation
P1V1
P2V2
R=
=
n1T1
n2T2
If we hold the amount and volume constant:
P1
T1
=
P2
T2
P1T2
P2 =
T1
(1 atm)(373.15 K)
P2 =
= 1.37 atm
(273.15 K)
6-4 Applications of the Ideal Gas Equation
Molar Mass Determination
PV = nRT
and
n=
m
RT
PV =
M
Rearranging for Molar Mass
m RT
M=
PV
m
M
Example 6-10
Determining a Molar Mass with the Ideal Gas Equation.
Polypropylene is an important commercial chemical. It is used
in the synthesis of other organic chemicals and in plastics
production. A glass vessel weighs 40.1305 g when clean, dry
and evacuated; it weighs 138.2410 when filled with water at
25°C (δ=0.9970 g cm-3) and 40.2959 g when filled with
propylene gas at 740.3 mm Hg and 24.0°C. What is the molar
mass of polypropylene?
Strategy:
Determine Vflask  Determine mgas  Use the Gas Equation.
Example 6-10
Determine Vflask:
Vflask = mH2O / dH2O = (138.2410 g – 40.1305 g) / (0.9970 g cm-3)
= 98.41 cm3 = 0.09841
L
Determine mgas:
mgas = mfilled - mempty = (40.2959 g – 40.1305 g)
= 0.1654 g
Example 6-10
5-6
Use the Gas Equation:
PV = nRT
M=
m
RT
PV =
M
m RT
M=
PV
(0.6145 g)(0.08206 L atm mol-1 K-1)(297.15 K)
(0.9741 atm)(0.09841 L)
M = 42.08 g/mol
Gas Densities
PV = nRT
and
m
m
, n=
d=
M
V
m
RT
PV =
M
Rearranging to obtain density we get:
m
MP
=d=
V
RT
Example 6-11
Using the Ideal Gas Equation to Calculate a Gas Density.
What is the density of oxygen gas (O2) at 298 K and 0.987 atm?
Given :
T = 298 K
P = 0.987 atm
R = 0.082057 L atm mol-1 K-1
Strategy
Knowing
MP
d=
RT
and MO is 32 g mol-1
2
(32 g mol-1)(0.987 atm)
-1
=
1.29
g
L
d=
(0.082057 L atm mol-1 K-1)(298 K)