Download Chapter 14: Gases

Chapter 13: Gases
Nature of gases
• Assumptions of Kinetic-Molecular theory are based on
four factors:
1)
2)
3)
4)
Number of particles present
Temperature
Pressure
Volume
• When one variable changes, it affects the other three
Boyle’s Law
• Boyle’s Law: volume of a given amount of gas held at
constant temperature varies inversely with pressure.
• Increase volume = decrease pressure (less collisions)
• Decrease volume = increase pressure (more collisions)
Boyle’s Law
P1V1 = P2V2
initial
*** You MUST memorize this equation!!!
final
Using Boyle’s Law
A sample of helium gas in a balloon is compressed from 4.0
L to 2.5 L at a constant temperature. If the pressure of the gas
in the 4.0-L volume is 210 kPa, what will the pressure be at
2.5 L?
Using Boyle’s Law
A sample of helium gas in a balloon is compressed from 4.0
L to 2.5 L at a constant temperature. If the pressure of the gas
in the 4.0-L volume is 210 kPa, what will the pressure be at
2.5 L?
What equation do we use?
Using Boyle’s Law
A sample of helium gas in a balloon is compressed from 4.0
L to 2.5 L at a constant temperature. If the pressure of the gas
in the 4.0-L volume is 210 kPa, what will the pressure be at
2.5 L?
What equation do we use?
P1V1 = P2V2
Using Boyle’s Law
A sample of helium gas in a balloon is compressed from 4.0
L to 2.5 L at a constant temperature. If the pressure of the gas
in the 4.0-L volume is 210 kPa, what will the pressure be at
2.5 L?
Known:
Unknown:
Using Boyle’s Law
A sample of helium gas in a balloon is compressed from 4.0
L to 2.5 L at a constant temperature. If the pressure of the gas
in the 4.0-L volume is 210 kPa, what will the pressure be at
2.5 L?
Known:
V1 = 4.0 L
V2 = 2.5 L
P1 = 210 kPa
Unknown: P2 = ?
Using Boyle’s Law
A sample of helium gas in a balloon is compressed from 4.0
L to 2.5 L at a constant temperature. If the pressure of the gas
in the 4.0-L volume is 210 kPa, what will the pressure be at
2.5 L?
P1V1 = P2V2
Using Boyle’s Law
A sample of helium gas in a balloon is compressed from 4.0
L to 2.5 L at a constant temperature. If the pressure of the gas
in the 4.0-L volume is 210 kPa, what will the pressure be at
2.5 L?
P1V1 = P2V2
(210 kPa) (4.0 L) = (P2) (2.5 L)
Using Boyle’s Law
A sample of helium gas in a balloon is compressed from 4.0
L to 2.5 L at a constant temperature. If the pressure of the gas
in the 4.0-L volume is 210 kPa, what will the pressure be at
2.5 L?
P1V1 = P2V2
(210 kPa) (4.0 L) = (P2)
(2.5 L)
Using Boyle’s Law
A sample of helium gas in a balloon is compressed from 4.0
L to 2.5 L at a constant temperature. If the pressure of the gas
in the 4.0-L volume is 210 kPa, what will the pressure be at
2.5 L?
P1V1 = P2V2
(210 kPa) (4.0 L) = (P2)
(2.5 L)
340 kPa = (P2)
Practice Problem: Boyle’s Law
Example: The pressure of a sample of helium in a 1.00-L
container is 0.988 atm. What is the new pressure if the
sample is placed in a 2.00-L container?
Practice Problem: Boyle’s Law
Example: The pressure of a sample of helium in a 1.00-L
container is 0.988 atm. What is the new pressure if the
sample is placed in a 2.00-L container?
P1V1 = P2V2
Charles’s Law
• Charles’s Law: volume of a given mass of gas is directly
proportional to its kelvin temperature at constant pressure.
• Increase temperature = Increase volume (faster particles)
• Decrease temperature = Decrease volume (slower particles)
Charles’s Law
V1
V2
=
T1
T2
initial
*** You MUST memorize this equation!!!
final
Using Charles’s Law
A gas sample at 40.0 °C occupies a volume of 2.32 L. If the
temperature is raised to 75.0 °C, what will the volume be,
assuming the pressure remains constant?
Using Charles’s Law
A gas sample at 40.0 °C occupies a volume of 2.32 L. If the
temperature is raised to 75.0 °C, what will the volume be,
assuming the pressure remains constant?
What equation do we use?
V1
V2
=
T1
T2
Using Charles’s Law
A gas sample at 40.0 °C occupies a volume of 2.32 L. If the
temperature is raised to 75.0 °C, what will the volume be,
assuming the pressure remains constant?
Known:
Unknown:
Using Charles’s Law
A gas sample at 40.0 °C occupies a volume of 2.32 L. If the
temperature is raised to 75.0 °C, what will the volume be,
assuming the pressure remains constant?
Known:
T1 = 40.0 °C
V1 = 2.32 L
T2 = 75.0 °C
Unknown: V2 = ?
Using Charles’s Law
A gas sample at 40.0 °C occupies a volume of 2.32 L. If the
temperature is raised to 75.0 °C, what will the volume be,
assuming the pressure remains constant?
V1
V2
=
T1
T2
Using Charles’s Law
A gas sample at 40.0 °C occupies a volume of 2.32 L. If the
temperature is raised to 75.0 °C, what will the volume be,
assuming the pressure remains constant?
2.32 L
V2
=
40.0 °C
75.0 °C
Using Charles’s Law
A gas sample at 40.0 °C occupies a volume of 2.32 L. If the
temperature is raised to 75.0 °C, what will the volume be,
assuming the pressure remains constant?
2.32 L
75.0 °C x
=
40.0 °C
V2
Using Charles’s Law
A gas sample at 40.0 °C occupies a volume of 2.32 L. If the
temperature is raised to 75.0 °C, what will the volume be,
assuming the pressure remains constant?
2.58 L =
V2
Practice Problem: Charles’s Law
The celsius temperature of a 3.00-L sample of gas is lowered
from 80.0 °C to 30.0 °C. What will be the resulting volume
of this gas?
Gay-Lussac’s Law
• Gay-Lussac’s Law: pressure of a given mass of gas varies
directly with kelvin temperature when the volume remains
constant.
• Increase temperature = Increase pressure
• Decrease temperature = Decrease pressure
Gay-Lussac’s Law
P1
P2
=
T1
T2
*** You MUST memorize this equation!!!
Using Gay-Lussac’s Law
The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the
temperature rises to 60.0 °C, what will be the pressure in the
tank?
Using Gay-Lussac’s Law
The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the
temperature rises to 60.0 °C, what will be the pressure in the
tank?
What equation do we use?
Using Gay-Lussac’s Law
The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the
temperature rises to 60.0 °C, what will be the pressure in the
tank?
P1
P2
=
T1
T2
Using Gay-Lussac’s Law
The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the
temperature rises to 60.0 °C, what will be the pressure in the
tank?
Known:
Unknown:
Using Gay-Lussac’s Law
The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the
temperature rises to 60.0 °C, what will be the pressure in the
tank?
Known:
T1 = 22.0 °C
P1 = 3.20
T2 = 60.0 °C
Unknown: P2 = ?
Using Gay-Lussac’s Law
The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the
temperature rises to 60.0 °C, what will be the pressure in the
tank?
P1
P2
=
T1
T2
Using Gay-Lussac’s Law
The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the
temperature rises to 60.0 °C, what will be the pressure in the
tank?
3.20 atm
P2
=
22.0 °C
60.0 °C
Using Gay-Lussac’s Law
The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the
temperature rises to 60.0 °C, what will be the pressure in the
tank?
60.0 °C X
3.20 atm
= P2
22.0 °C
Using Gay-Lussac’s Law
The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the
temperature rises to 60.0 °C, what will be the pressure in the
tank?
3.61 atm = P2
Overview of Gas Laws
Gas Law
Temperature
Boyle’s
Charles’s
yes
Gay-Lussac’s
yes
Pressure
Volume
yes
yes
yes
yes
14.2 Combined Gas Law
• We can combine Boyle’s Law, Charles’s Law and GayLussac’s Law into one law (“Combined Gas Law”).
• States the relationship between pressure, volume, and
temperature of a fixed amount of gas.
P1V1 P2V2
=
T1
T2
Converting to Kelvin (K)
• For the rest of the chapter, we NEED to convert
temperature to Kelvin (K) first, BEFORE we use the
combined gas law.
• To convert to Kelvin temperature (K), use the following
conversion:
TK = 273 + TC
Converting to Kelvin (K)
Convert the following temperatures to Kelvin.
1) -25.0 °C
2) 0 °C
3) 23 °C
4) 80.0 °C
Solving for a Variable
Rearrange the Combined Gas Law to isolate the appropriate
variable.
1) Solve for P1
2) Solve for V1
3) Solve for T1
4) Solve for T2
5) Solve for V2
P1V1 P2V2
=
T1
T2
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
What equation do we use?
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
What equation do we use?
P1V1 P2V2
=
T1
T2
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
FIRST, what do we do to the temperature values?
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
FIRST, what do we do to the temperature values?
CONVERT TO KELVIN!
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
Convert to Kelvin:
T1 = 30.0 °C + 273 = 303 K
T2 = 80.0 °C + 273 = 353 K
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
Known Values:
Unknown:
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
Known Values:
T1 = (30.0 °C + 273) 303 K
P1 = 110 kPa
V1 = 2.00 L
T2 = (80.0 °C + 273) 353 K
P2 = 440 kPa
Unknown:
V1 = ?
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
T1 = 303 K
P1 = 110 kPa
V1 = 2.00 L
T2 = 353 K
P2 = 440 kPa
V2 = ?
P1V1 P2V2
=
T1
T2
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
**Plug & Chug
(110 kPa)(2.00L)
303K
=
(440 kPa)(V2)
353 K
Using the Combined Gas Law
A gas at 110 kPa and 30.0 °C fills a flexible container with
an initial volume of 2.00 L. If the temperature is raised to
80.0 °C and the pressure increased to 440 kPa, what is the
new volume?
**Plug & Chug
(110 kPa)(2.00L)
303K
=
(V2) =
(440 kPa)(V2)
353 K
Practice Problem: Combined Gas Law
Example: At 0.00 °C and 1.00 atm pressure, a sample of gas
occupies 30.0 mL. If the temperature is increased to 30.0 °C
and the entire gas sample is transferred to a 20.0 mL
container, what will be the gas pressure inside the
container?
Practice Problem: Combined Gas Law
Example: At 0.00 °C and 1.00 atm pressure, a sample of gas
occupies 30.0 mL. If the temperature is increased to 30.0 °C
and the entire gas sample is transferred to a 20.0 mL
container, what will be the gas pressure inside the
container?
What is the first step?
What is the second step?
Practice Problem: Combined Gas Law
Example: At 0.00 °C and 1.00 atm pressure, a sample of gas
occupies 30.0 mL. If the temperature is increased to 30.0 °C
and the entire gas sample is transferred to a 20.0 mL
container, what will be the gas pressure inside the
container?
What is the first step?
* CONVERT TO KELVIN
What is the second step?
*Determine Known and Unknown Variables
Practice Problem: Combined Gas Law
Example: At 0.00 °C and 1.00 atm pressure, a sample of gas
occupies 30.0 mL. If the temperature is increased to 30.0 °C
and the entire gas sample is transferred to a 20.0 mL
container, what will be the gas pressure inside the
container?
Avogadro’s Principle
• The size of large krypton atoms and small helium atoms
have no influence on the volume occupied by a fixed
number of particles.
• Avogadro’s Principle: states that equal volumes of
gases at the same temperature and pressure contain equal
number of particles.
Molar Volume
• Molar volume for a gas is the volume that one mole
occupies at 0.00 °C (273 K) and 1.00 atm.
• STP: standard temperature and pressure;
0.00 °C (273 K) and 1.00 atm
• Avogadro showed that 1 mol gas occupies 22.4 L at STP.
Molar Volume Conversion Factor
• You can use the following conversion factor to find the
number of moles, the mass, and even the number of
particles in a gas sample.
conversion factor:
22.4 L
1 mol
(Oh NO! Conversion Factors!)
OR
1 mol
22.4 L
Molar Volume
How do you find the number of particles in a sample of gas
that has a volume of 3.72 L at STP?
Molar Volume
How do you find the number of particles in a sample of gas
that has a volume of 3.72 L at STP?
Step 1: Calculate the number of moles of gas in 3.72 L.
Molar Volume
How do you find the number of particles in a sample of gas
that has a volume of 3.72 L at STP?
Step 1: Calculate the number of moles of gas in 3.72 L.
3.72 L x
Molar Volume
How do you find the number of particles in a sample of gas
that has a volume of 3.72 L at STP?
Step 1: Calculate the number of moles of gas in 3.72 L.
3.72 L x
1 mol = 0.166 mol
22.4 L
Molar Volume
How do you find the number of particles in a sample of gas
that has a volume of 3.72 L at STP?
Step 2: Convert moles to particles using Avogadro’s
number (1 mol = 6.022 x 1023 particles)
Molar Volume
How do you find the number of particles in a sample of gas
that has a volume of 3.72 L at STP?
Step 2: Convert moles to particles using Avogadro’s
number (1 mol = 6.022 x 1023 particles)
0.166 mol x 6.022x1023 particles = 9.99x1022 particles
1 mol
Avogadro’s Principle
Calculate the volume that 0.861 mol of gas at standard
temperature and pressure (STP) will occupy.
Avogadro’s Principle
Calculate the volume that 0.861 mol of gas at standard
temperature and pressure (STP) will occupy.
Hint: Use molar volume conversion factor to calculate the
unknown volume.
0.861 mol x
Avogadro’s Principle
Calculate the volume that 0.861 mol of gas at standard
temperature and pressure (STP) will occupy.
Hint: Use molar volume conversion factor to calculate the
unknown volume.
0.861 mol x 22.4 L = 19.7 L
1 mol
Practice Problem:Avogadro’s Principle
Example: How many moles of nitrogen gas will be
contained in a 2.00 L flask at STP?
Avogadro’s Principle using Mass
Calculate the volume that 2000 g of methane gas (CH4) will
occupy at STP.
Avogadro’s Principle using Mass
Calculate the volume that 2000 g of methane gas (CH4) will
occupy at STP.
STEP 1: Convert mass to moles (using molar mass)
Avogadro’s Principle using Mass
Calculate the volume that 2000 g of methane gas (CH4) will
occupy at STP.
STEP 1: Convert mass to moles (using molar mass)
2000 g CH4 x 1 mol
= 125 mol CH4
16.05 g
Avogadro’s Principle using Mass
Calculate the volume that 2000 g of methane gas (CH4) will
occupy at STP.
STEP 2: Convert moles to volume (since conditions are
already at STP, use 22.4 L/mol)
125 mol CH4 x 22.4 L
1 mol
= 2800 L
Practice Problems
How many grams of carbon dioxide gas are in a 1.0 L
balloon at STP?
Practice Problems
How many grams of carbon dioxide gas are in a 1.0 L
balloon at STP?
STEP 1: Convert volume to moles (since conditions are at
STP, use 22.4 L/mol)
1.0 L CO2 x 1 mol =
22.4 L
Practice Problems
How many grams of carbon dioxide gas are in a 1.0 L
balloon at STP?
STEP 2: Convert moles to mass (use molar mass CO2)
mol CO2 x 44.01 g
1 mol
=
14.3 The Ideal Gas Law
Combine the laws of Avogadro, Boyle, Charles, and GayLussac into one equation (“The Ideal Gas Law”) to
describe the relationship between:
-
Pressure
Volume
Temperature
Number of moles
The Ideal Gas Law
The Ideal Gas Law:
PV = nRT
P = pressure (atm)
V = volume (L)
n = number of moles (mol)
R = gas constant
T = temperature (K)
The Ideal Gas Constant, R
The Gas Constant, R, depends on the unit of pressure
Units of R
Numerical
value of R
Units of P
Units of V
Units of T
Units of n
𝐿 · 𝒂𝒕𝒎
𝑚𝑜𝑙 · 𝐾
0.0821
atm
L
K
mol
𝐿 · 𝒌𝑷𝒂
𝑚𝑜𝑙 · 𝐾
8.314
kPa
L
K
mol
𝐿 · 𝒎𝒎 𝑯𝒈
𝑚𝑜𝑙 · 𝐾
62.4
Mm Hg
L
K
mol
Real vs. Ideal
Ideal Gases:
- Particles take up no space and have no intermolecular
attractive forces.
- Follow all gas laws under all conditions of T & P.
Real Gases:
- No gases are ideal.
- Gas particles have volume (due to size and shape).
- Subject to intermolecular forces
* Most gases behave like “ideal” gases, except at high
pressure and low temps
Applying Ideal Gas Law
Calculate the number of moles if gas contained in a 3.0 L
vessel at 300 K with a pressure of 1.50 atm.
Applying Ideal Gas Law
Calculate the number of moles if gas contained in a 3.0 L
vessel at 300 K with a pressure of 1.50 atm.
What equation do we use?
Applying Ideal Gas Law
Calculate the number of moles if gas contained in a 3.0 L
vessel at 300 K with a pressure of 1.50 atm.
What equation do we use?
Ideal Gas Law!
PV = nRT
Applying Ideal Gas Law
Calculate the number of moles of gas contained in a 3.0 L
vessel at 300 K with a pressure of 1.50 atm.
Known:
Unknown:
Applying Ideal Gas Law
Calculate the number of moles of gas contained in a 3.0 L
vessel at 300 K with a pressure of 1.50 atm.
Known:
V = 3.0 L
T = 300 K
P = 1.50 atm
R=?
Unknown: n = ?
Applying Ideal Gas Law
Calculate the number of moles of gas contained in a 3.0 L
vessel at 300 K with a pressure of 1.50 atm.
Known:
V = 3.0 L
T = 300 K
P = 1.50 atm
𝐿 · 𝑎𝑡𝑚
R = 0.0821
𝑚𝑜𝑙 ·𝐾
Unknown: n = ?
Applying Ideal Gas Law
Calculate the number of moles if gas contained in a 3.0 L
vessel at 300 K with a pressure of 1.50 atm.
PV = nRT
(1.50 atm)(3.0 L)
=
(1.5 atm)(3.0 L)
(0.0821
𝐿 ·𝑎𝑡𝑚
)(300
𝑚𝑜𝑙 ·𝐾
𝐿 ·𝑎𝑡𝑚
n(0.0821
)(300 K)
𝑚𝑜𝑙 ·𝐾
=n
K)
n = 0.18 mol
Practice Problem
If the pressure exerted by a gas at 25 °C in a volume of
0.044 L is 3.81 atm, how many moles of gas are present?
Applying the Ideal Gas Law
The Ideal Gas Law can calculate:
1) Moles
The Ideal Gas Law can also be used to calculate:
2) Molar mass (if mass of sample is known)
3) Density (if mass of sample is known)
Ideal Gas Law & Molar Mass
Relationship between moles and molar mass:
moles (n) =
PV = nRT
𝑚𝑎𝑠𝑠 (𝑚)
𝑀𝑜𝑙𝑎𝑟 𝑀𝑎𝑠𝑠 (𝑀)
changes to
rearrange to solve for M,
;
𝑚
(n = )
𝑀
𝑚𝑅𝑇
PV =
𝑀
𝑚𝑅𝑇
M =
𝑃𝑉
Ideal Gas Law & Density
Relationship between moles and density:
𝑚𝑎𝑠𝑠 (𝑚)
Density (D) =
;
𝑉𝑜𝑙𝑢𝑚𝑒 (𝐿)
𝑚𝑅𝑇
M=
𝑃𝑉
(D =
𝑚
𝑉
changes to
𝐷𝑅𝑇
M =
𝑃
Rearrange to solve for D,
𝑀𝑃
D =
𝑅𝑇
)
Ideal Gas Law & Molar Mass
What is the molar mass of a pure gas that has a density of
1.40 g/L at STP?
Ideal Gas Law & Molar Mass
What is the molar mass of a pure gas that has a density of
1.40 g/L at STP?
What equation do we use?
Hint: given a density, and need to calculate molar mass…
Ideal Gas Law & Molar Mass
What is the molar mass of a pure gas that has a density of
1.40 g/L at STP?
What equation do we use?
M =
𝐷𝑅𝑇
𝑃
Ideal Gas Law & Molar Mass
What is the molar mass of a pure gas that has a density of
1.40 g/L at STP?
M =
Known:
Unknown:
𝐷𝑅𝑇
𝑃
Ideal Gas Law & Molar Mass
What is the molar mass of a pure gas that has a density of
1.40 g/L at STP?
M =
Known:
D: 1.40 g/L
T: ?
P: ?
𝐿 ·𝑎𝑡𝑚
R: 0.0821
𝑚𝑜𝑙 ·𝐾
Unknown: M: ?
𝐷𝑅𝑇
𝑃
Ideal Gas Law & Molar Mass
What is the molar mass of a pure gas that has a density of
1.40 g/L at STP?
M =
Known:
D: 1.40 g/L
T: 273 K
P: 1 atm
𝐿 ·𝑎𝑡𝑚
R: 0.0821
𝑚𝑜𝑙 ·𝐾
Unknown: M: ?
𝐷𝑅𝑇
𝑃
Ideal Gas Law & Molar Mass
What is the molar mass of a pure gas that has a density of
1.40 g/L at STP?
M =
𝐷𝑅𝑇
𝑃
𝑔
M =
𝐿 ·𝑎𝑡𝑚
(1.40 𝐿 )(0.0821𝑚𝑜𝑙 ·𝐾)(273 𝐾)
(1𝑎𝑡𝑚)
M = 31.4 g/mol
Ideal Gas Law & Molar Mass
How many grams of gas are present in a sample that has a
molar mass of 70.0 g/mol and occupies a 2.00-L container at
117 kPa, and 35.1 ° C?
Ideal Gas Law & Molar Mass
How many grams of gas are present in a sample that has a
molar mass of 70.0 g/mol and occupies a 2.00-L container at
117 kPa, and 35.1 ° C?
What equation do we use?
Ideal Gas Law & Molar Mass
How many grams of gas are present in a sample that has a
molar mass of 70.0 g/mol and occupies a 2.00-L container at
117 kPa, and 35.1 ° C?
What equation do we use?
PV =
𝑚𝑅𝑇
𝑀
Ideal Gas Law & Molar Mass
How many grams of gas are present in a sample that has a
molar mass of 70.0 g/mol and occupies a 2.00-L container at
117 kPa, and 35.1 ° C?
PV =
Known:
Unknown:
𝑚𝑅𝑇
𝑀
Ideal Gas Law & Molar Mass
How many grams of gas are present in a sample that has a
molar mass of 70.0 g/mol and occupies a 2.00-L container at
117 kPa, and 35.1 ° C?
PV =
Known:
𝑚𝑅𝑇
𝑀
M: 70.0 g/mol
V: 2.00 L
P: 117 kPa
T: (35.1 °C + 273) = 308.1 K
𝐿 ·𝑘𝑃𝑎
R: 8.314
𝑚𝑜𝑙 ·𝐾
Unknown: m: ?
Ideal Gas Law & Molar Mass
How many grams of gas are present in a sample that has a
molar mass of 70.0 g/mol and occupies a 2.00-L container at
117 kPa, and 35.1 ° C?
m=
𝑃𝑉𝑀
𝑅𝑇
14.4 Gas Stoichiometry
Gas Laws we learned can be applied to calculate the
stoichiometry of reactions w/ gases (either reactants or
products)
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
Gas Stoichiometry
Coefficients in balanced equation tell us…
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
Gas Stoichiometry
Coefficients in balanced equation tell us…
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
1) Number of moles
2) Number of representative particles, and …
Gas Stoichiometry
Coefficients in balanced equation tell us…
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
1) Number of moles
2) Number of representative particles
3) Number of Liters!
Gas Stoichiometry
How many moles of each gas do we have?
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
___ mol of C4H10(g)
___ mol of O2(g)
___ mol of CO2(g)
___ mol of H2O(g)
Gas Stoichiometry
How many liters of each gas do we have?
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
___ L of C4H10(g)
___ L of O2(g)
___ L of CO2(g)
___ L of H2O(g)
Gas Stoichiometry
Moles and Liters in a balanced chemical equation
2C4H10(g)
2 mol
2L
+ 13O2(g)
13 mol
13 L
→ 8CO2(g)
8 mol
8L
+ 10H2O(g)
10 mol
10 L
Gas Stoichiometry with Volume
Write some conversion factors relating the Liters of gases in
the following chemical equation.
2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(g)
Gas Stoichiometry with Volume
What volume of methane (CH4) is needed to produce 26 L
of water?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Gas Stoichiometry with Volume
What volume of methane (CH4) is needed to produce 26 L
of water?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Step 1: Write the given!!!
Gas Stoichiometry with Volume
What volume of methane (CH4) is needed to produce 26 L
of water?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Step 1: Write the given!!!
26 L H2O
Gas Stoichiometry with Volume
What volume of methane (CH4) is needed to produce 26 L
of water?
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Step 2: Write the conversion factor comparing
L of water to L of methane, using volume relationship from
coefficients in balanced chemical equation.
26 L H2O x
Practice Problem: Gas Stoichiometry
What volume of oxygen gas is needed for the complete
combustion of 4.00 L of propane gas (C3H8)? Assume
constant temperature and pressure.
Practice Problem: Gas Stoichiometry
What volume of oxygen gas is needed for the complete
combustion of 4.00 L of propane gas (C3H8)? Assume
constant temperature and pressure.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Practice Problem: Gas Stoichiometry
What volume of oxygen gas is needed for the complete
combustion of 4.00 L of propane gas (C3H8)? Assume
constant temperature and pressure.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Known:
Unknown:
Practice Problem: Gas Stoichiometry
What volume of oxygen gas is needed for the complete
combustion of 4.00 L of propane gas (C3H8)? Assume
constant temperature and pressure.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
Known:
4.00 L propane
Unknown: ? L of oxygen
Practice Problem: Gas Stoichiometry
What volume of oxygen gas is needed for the complete
combustion of 4.00 L of propane gas (C3H8)? Assume
constant temperature and pressure.
C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g)
4.00 L C3H8 x
Hint: find Liter to Liter ratio between C3H8 and O2
Practice Problem: Gas Stoichiometry
Determine the volume of hydrogen gas needed to react
completely with 5.00 L of oxygen to form water.
Calculations w/ Volume & Mass
We can do stoichiometric calculations involving both gas
volumes & masses if we know the following information:
1) Balanced chemical equation
2) At least one mass or volume value for a reactant or
product
3) The conditions the gas volume was measured (T & P)
*Then use the Ideal Gas Law w/ mole or volume ratios
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Whoa… where do we even begin?!?!
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Analyze the problem…need to convert volume (L) N2 to
mass (g) NH3
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Analyze the problem…need to convert volume (L) N2 to
mass (g) NH3
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Step 1: convert L of N2 → L of NH3 (Hint: use coefficients
to convert volume to volume)
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Step 1: convert L of N2 → L of NH3 (Hint: use coefficients
to convert volume to volume)
5.00 L N2
x
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Step 1: convert L of N2 → L of NH3 (Hint: use coefficients
to convert volume to volume)
5.00 L N2
x
2 L NH3 = 10.0 L NH3
1 L N2
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Step 2: Convert 10.0 L of NH3 → mol HN3 (Hint: use PV =
nRT, to solve for mol)
𝑃𝑉
n=
𝑅𝑇
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Step 2: Convert 10.0 L of NH3 → mol HN3 (Hint: use PV =
nRT, to solve for mol)
V: 10.0 L
P: 3.00 atm
T: 298 K
𝐿·𝑎𝑡𝑚
R: 0.0821
𝑚𝑜𝑙·𝐾
n=
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Step 2: Convert 10.0 L of NH3 → mol HN3 (Hint: use PV =
nRT, to solve for mol)
n = 1.23 mol NH3
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Step 3: Convert mol 1.23 mol NH3 → mass NH3 (Hint: Use
Molar Mass)
Volume-Mass Problems
If 5.00 L of nitrogen reacts completely by the following
reaction at a constant pressure and temperature of 3.00 atm
and 298 K, how many grams of ammonia are produced?
N2(g) + 3H2(g) → 2NH3(g)
Step 3: Convert mol 1.23 mol NH3 → mass NH3 (Hint: Use
Molar Mass)
1.23 mol NH3 x 17.04 g NH3 = 21.0 g NH3
1 mol NH3
Practice Problem: Volume-Mass
Use the reaction below to calculate the mass of solid
ammonium nitrate that must be used to obtain 0.100 L of
dinitrogen oxide gas at STP.
NH4NO3(s) → N2O(g) + 2H2O(g)