Chapter 11
Gases
Properties of Gases
• There are 5 important properties of gases:
– Confined gases exerts pressure on the wall of a
container uniformly
– Gases have low densities
– Gases can be compressed
– Gases can expand to fill their contained uniformly
– Gases mix completely with other gases in the same
container
Chapter 11
2
Kinetic Molecular Theory of Gases
•
The Kinetic Molecular Theory of Gases is the model used
to explain the behavior of gases in nature.
1. Gas particles (atoms or molecules) move constantly in
random directions (in straight lines) with rapid velocities.
2. Gases are made up of very tiny atoms or molecules meaning
that gases are mostly empty space!
• This fact is what allows for the compression of a gas.
3.
Gas molecules or atoms have no attraction for one another.
• This is the result of the distance between the particles!
• They undergo elastic collisions with one another
4.
The average kinetic energy of a gas is proportional to the
temperature in Kelvins.
Chapter 11
3
Properties that Describe a Gas
• These properties are all related to one another.
• When one variable changes, it causes the other
three to react in a predictable manner.
Chapter 11
4
Gas Pressure (P)
• Gas pressure (P) is the result of constantly moving
gas molecules striking the inside surface of their
container.
Chapter 11
5
Atmospheric Pressure
• Atmospheric pressure is the pressure exerted by the air
on the earth.
• Evangelista Torricelli invented
the barometer in 1643 to
measure atmospheric pressure.
• Atmospheric pressure is 760
mm of mercury or 1
atmosphere (atm) at sea level.
What happens to the atmospheric
pressure as you go up in elevation?
Chapter 11
6
Units of Pressure
• Standard pressure is the atmospheric pressure at
sea level, 760 mm of mercury.
– Here is standard pressure expressed in other units:
Chapter 11
7
Gas Pressure Conversions
• The barometric pressure is 697 torr. What is the
barometric pressure in atmospheres?
– Step 1: Determine what you have: 697 torr
– Step 2: Determine what you want: ??? atm
– Step 3: Write out your plan to convert torr to atm
– Step 4: Select conversion factor(s): 1 atm = 760 torr
697 torr ×
Chapter 11
1 atm
= 0.917 atm
760 torr
8
Properties that Describe a Gas
• These properties are all related to one another.
• When one variable changes, it causes the other
three to react in a predictable manner.
Chapter 11
9
Gas Law Problems
Chapter 11
10
Boyle’s Law (V and P)
• Boyle’s Law states that the volume of a gas is inversely
proportional to the pressure at constant temperature.
• Mathematically, we write:
Pα 1.
V
• For a before and after
situation:
P1V1 = P2V2
Chapter 11
11
Boyle’s Law Problem
• A 1.50 L sample of methane gas exerts a pressure of 1650
mm Hg. What is the final pressure if the volume changes to
7.00 L?
– Step 1: Organize the data in a table with initial (1) and final (2)
conditions.
– Step 2: Rearrange the gas law to solve for the unknown (here, P2)
– Step 3: Plug your numbers into Boyle’s Law and solve for P2
P1V1 = P2V2 rearranges to
Chapter 11
P1 V1
= P2
V2
(1650 mm Hg )(1.50 L)
= 354 mm Hg
7.00 L
12
Charles’ Law (V and T)
• In 1783, Jacques Charles discovered (while hot air ballooning) that
the volume of a gas is directly proportional to the temperature in
Kelvin.
• Mathematically, we write:
TαV
• For a before and after situation:
V1
=
T1
V2
T2
Chapter 11
13
Charles’ Law Problem
• A 275 L helium balloon is heated from 20°C to 40°C. What is the final volume
at constant P?
– Step 1: Temperatures must be in Kelvin. Convert from °C to K if needed:
20°C + 273 = 293 K and 40°C + 273 = 313 K
– Step 2: Organize the data in a table with initial (1) and final (2) conditions.
– Step 3: Rearrange the gas law to solve for the unknown (here, V2)
– Step 4: Plug your numbers into Charles’ Law and solve for V2
V1
V
= 2
T1
T2
Chapter 11
rearranges to
V1 T2
= V2
T1
(275 L)(313 K)
= 294 L
293 K
14
Gay-Lussac’s Law (P and T)
• In 1802, Joseph Gay-Lussac discovered that the pressure of a
gas is directly proportional to the temperature in Kelvin.
• Mathematically, we write:
TαP
• For a before and after situation:
P1
=
T1
P2
T2
Chapter 11
15
Gay-Lussac’s Law Problem
• A steel container of nitrous oxide at 15.0 atm is cooled from 25°C
to –40°C. What is the final volume at constant V?
– Step 1: Temperatures must be in Kelvin. Convert from °C to K if needed:
25°C + 273 = 298 K and -40°C + 273 = 233 K
– Step 2: Organize the data in a table with initial (1) and final (2) conditions.
– Step 3: Rearrange the gas law to solve for the unknown (here, P2)
– Step 4: Plug your numbers into Gay-Lussac’s Law and solve for P2
P T
P1
P
= 2 rearranges to 1 2 = P2
T1
T1
T2
(15.0 atm)(298 K)
= 11.7 atm
233 K
Chapter 11
16
Combined Gas Law
• When we introduced Boyle’s, Charles’, and GayLussac’s Laws, we assumed that one of the variables
remained constant.
• Experimentally, all three (temperature, pressure, and
volume) usually change.
• By combining all three laws, we obtain the combined gas
law:
P1V1 P2V2
=
T1
T2
Chapter 11
17
Combined Gas Law Problem
• Lets apply the combined gas law to 10.0L of carbon
dioxide gas at 300 K and 1.00 atm. If the volume and
Kelvin temperature double, what is the new pressure?
Conditions
P
V
T
Initial
1.00 atm
10.0 L
300 K
Final
P2
20.0 L
600 K
Chapter 11
18
Avogadro’s Law
• In the previous laws, the amount of gas was always constant.
• However, the amount of a gas (n) is directly proportional to the
volume of the gas, meaning that as the amount of gas increases,
so does the volume.
• Mathematically, we write:
nαV
• For a before and after situation:
V1
=
n1
V2
n2
Chapter 11
19
Avogadro’s Law Problem
• A steel container contains 2.6 mol of nitrous oxide with a volume
15.0 L. If the amount of nitrous oxide is increased to 8.4 mol, what
is the final volume at constant T and P?
– Step 1: Organize the data in a table with initial (1) and final (2) conditions.
– Step 2: Rearrange the gas law to solve for the unknown (here, V2)
– Step 3: Plug your numbers into Avogardro’s Law and solve for V2
V1
V2
V n
rearranges to 1 2 = V2
=
n1
n1
n2
(15.0 L)(8.4 mol)
= 48.5 L
2.6 mol
Chapter 11
20
Gas Laws Summary
Chapter 11
21
Molar Volume and STP
• Standard temperature and pressure (STP) are defined
as 0°C and 1 atm.
• At standard temperature and pressure,
one mole of any gas occupies 22.4 L.
• The volume occupied by one mole of
gas (22.4 L) is called the molar
volume.
1 mole Gas = 22.4 L
Chapter 11
22
Molar Volume Problem
Chapter 11
23
Molar Volume Calculation – Volume to Moles
• A sample of methane, CH4, occupies 4.50 L at STP. How many
moles of methane are present?
– Step 1: Determine what you have: 4.50 L
– Step 2: Determine what you want: ??? mol
– Step 3: Write out your plan to convert L to mol
– Step 4: Select conversion factor(s) needed for plan
Volume
4.50 L CH4 ×
Chapter 11
Molar
Volume
Moles
1 mol CH4
= 0.201 mol CH4
22.4 L CH4
24
Mole Unit Factors
• We now have three interpretations for the mole:
1 mol = 6.02 × 1023 particles
1 mol = molar mass
1 mol = 22.4 L (at STP for a gas)
• This gives us 3 unit factors to use to convert
between moles, particles, mass, and volume.
Chapter 11
25
Mole Calculation - Grams to Volume
• What is the mass of 3.36 L of ozone gas, O3, at STP?
–
–
–
–
Step 1: Determine what you have: 3.36 L
Step 2: Determine what you want: ??? g O3
Step 3: Write out your plan to convert L to grams
Step 4: Select conversion factor(s) needed for plan
Grams
3.36 L O3 ×
Chapter 11
Molar
Mass
Moles
Molar
Volume
Volume
1 mol O3
48.00 g O3
×
= 7.20 g O3
22.4 L O3
1 mol O3
26
Mole Calculation – Molecules to Volume
• How many molecules of hydrogen gas, H2, occupy 0.500 L at STP?
– Step 1: Determine what you have: 0.500 L H2
– Step 2: Determine what you want: ??? molecule H2
– Step 3: Write out your plan to convert L to molecules
– Step 4: Select conversion factor(s) needed for plan
Volume
Molar
Volume
0.500 L H2 ×
Moles
Avogadro’s
Number
Atoms
1 mol H2
6.02×1023 molecules H2
×
22.4 L H2
1 mole H2
1.34 × 1022 molecules H2
Chapter 11
27
Gas Density
• The density of a gas is much less than that of a
liquid.
• We can calculate the density of any gas at STP
easily.
• The formula for gas density at STP is:
molar mass in grams (MM)
molar volume in liters (MV)
Chapter 11
= density, g/L
28
Calculating Gas Density
• What is the density of ammonia gas, NH3, at STP?
– Step 1: Determine what you know: NH3 and the molar volume
– Step 2: Determine what you want: density of NH3
– Step 3: Determine what you need to calculate density: MM of NH3
– Step 4: Calculate what you need, then use the formula to solve for the
unknown.
• Calculate the molar mass for ammonia:
14.01 (N) + 3(1.01) (H) = 17.04 g/mol NH3
17.04 g/mol
22.4 L/mol
= 0.761 g/L
Chapter 11
29
Molar Mass of a Gas
• We can also use molar volume to calculate the
molar mass of an unknown gas.
• We can calculate the Molar Mass of any gas at
STP easily.
• The formula for Molar Mass at STP is:
Molar Mass = Density (g/L) • Molar Volume (L/mol)
MM = D • MV
Chapter 11
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Molar Mass of a Gas
• 1.96 g of an unknown gas occupies 1.00 L at STP. What
is the molar mass?
– Step 1: Determine what you have: 1.00 L and 1.96 g unknown gas
– Step 2: Determine what you want: ??? g/mol unknown gas
– Step 3: Write out your plan to convert L and grams to g/mol
– Step 4: Use the formula for Molar Mass of a gas at STP and solve.
Molar Mass = Density (g/L) x Molar Volume (L/mol)
1.96 g 22.4 L
×
= 43.9 g/mol
1.00 L 1 mole
Chapter 11
31
The Ideal Gas Law
•
The four properties used in the measurement of a gas
(Pressure, Volume, Temperature and moles) can be
combined into a single gas law:
PV = nRT
•
Here, R is the ideal gas constant and has a value of:
0.0821 atm⋅L/mol⋅K
•
Note the units of R. When working problems with the
Ideal Gas Law, your units of P, V, T and n must match
those in the constant!
Chapter 11
32
Ideal Gas Law Problem
•
How many moles of hydrogen gas occupy 0.500 L at
STP?
– Step 1: Determine what you have: 0.500 L of H2 gas at STP (so 1.00 atm and
273 K)
– Step 2: Determine what you want: ??? moles of H2 gas (nH2)
– Step 3: Determine what formula you can use to calculate nH2
– Step 4: Rearrange the formula for n and solve.
n=
PV .
RT
n=
.
(1 atm)(0.500 L)
(0.0821 atm⋅L/mol⋅K)(273K)
n = 0.0223 moles
Chapter 11
33
Gases in Chemical Reactions
• Gases are involved as reactants in
numerous chemical reactions.
• Typically, the information given for a
gas in a reaction is its Pressure (P),
volume (V) and temperature (T).
• We use this information and the Ideal
Gas Law to determine the moles of the
gas (n).
• Once we have this information, we can
proceed with the problem as we would
any other stoichiometry problem.
A (g) + X (s) → B (s) + Y (l)
Chapter 11
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Reaction with a Gas
• Hydrogen gas is formed when zinc metal reacts with
hydrochloric acid. How many liters of hydrogen gas at
STP are produced when 15.8 g of zinc reacts?
Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2 (aq)
– Step 1: Determine what you know: 15.8 g Zn reacts at STP
– Step 2: Determine what you want: ??? L of H2 are produced
– Step 3: Write out your plan to convert g Zn to L H2
– Step 4: Select conversion factor(s) and/or formulas needed for plan
Grams of Zn
Molar
Mass of Zn
Moles of Zn
Mole
Ratio
Moles of Hh
Molar
Volume
Liters of H2
Chapter 11
35
Reaction with a Gas
• Hydrogen gas is formed when zinc metal reacts with
hydrochloric acid. How many liters of hydrogen gas at a
pressure of 755 atm and 35°C are produced when 15.8 g
of zinc reacts?
Zn (s) + 2HCl (aq) → H2 (g) + ZnCl2 (aq)
– Step 1: Determine what you know: 15.8 g Zn reacts at 755 atm and 35°C
– Step 2: Determine what you want: ??? L of H2 are produced
– Step 3: Write out your plan to convert g Zn to L H2
– Step 4: Select conversion factor(s) and/or formulas needed for plan
Grams of Zn
Chapter 11
Molar
Mass of Zn
Moles of Zn
Mole
Ratio
Moles of Hh
Ideal Gas
Law
Liters of H2
36