CHAPTER 9: GASES
9.1
Gas Pressure
9.2
The Ideal Gas Law
9.3
Stoichiometry and Gases
9.4
Effusion and Diffusion
9.5
The Kinetic-Molecular Theory
9.6
Non-Ideal Gas Behavior
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9.1 GAS PRESSURE
Describing Gases
• Four quantities are needed to describe a sample of gas.
Pressure
Volume
Amount
(moles)
Temperature
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9.1 GAS PRESSURE
Pressure
• Pressure is the force exerted on an area of some surface.
𝑃=
𝐹
𝐴
P = pressure
F = force (N)
A = area (m2)
Force
Area
SI Unit: Pa
1 Pa = 1 N/m2
Chemistry Units: atm or bar
1 atm = 1.01325 bar
1 atm = 101325 Pa
• Gas pressure is caused by the force exerted
by gas molecules colliding with the interior
walls of a container.
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9.1 GAS PRESSURE
• Gas pressure is caused by the force exerted by gas molecules
colliding with the interior walls of a container.
Source: https://preparatorychemistry.com
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9.1 GAS PRESSURE
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9.1 GAS PRESSURE
Pressure
• Atmospheric pressure is caused by the
weight of the column of air molecules in
the atmosphere above the earth’s surface.
• Atmospheric pressure can be measured
by using a barometer.
Pressure exerted by a fluid due to gravity is
called hydrostatic pressure:
p=hρg
p → Hydrostatic pressure
h → Height of the liquid column
ρ → Density of the fluid
g → acceleration due to gravity
The height (mm) of the liquid in the tube is proportional to the
pressure exerted by the atmosphere. 1 atm = 760 mmHg
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9.1 GAS PRESSURE
Pressure
• One way to measure pressure in the laboratory is by using a
manometer.
•
A closed-end manometer can be used to measure gas trapped in a
container. The distance between the mercury level is proportional to the
pressure of the gas in the flask.
•
An open-end manometer is similar but the distance between the
mercury levels corresponds to the difference between the pressure in
the flask and the pressure in the container.
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9.2 THE IDEAL GAS LAW
𝑃 𝑎𝑛𝑑 𝑇
Pressure and Temperature: Amonton’s Law
• If a sealed flask is filled with gas, the atoms or molecules in that
gas exert pressure by colliding with the sides of the flask.
• If the temperature of the container is raised, the gas inside gets
hotter and more collisions occur causing the pressure to
increase.
Note that the volume and the amount of gas are constant through the experiment.
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9.2 THE IDEAL GAS LAW
Pressure and Temperature: Amonton’s Law
• Pressure increases with an increase in temperature for any
sample of gas where volume and amount are held constant.
From the data, it is clear that pressure is directly proportional
to temperature when volume is held constant.
P∝T or P=k T
k is a constant
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9.2 THE IDEAL GAS LAW
Pressure and Temperature: Amonton’s Law
• The direct relationship between pressure and temperature (at
constant amount and constant volume) for a gas holds at
different conditions as shown:
Condition 1
Condition 2
P1=k T 1
P1
=k
T1
P2=k T 2
P2
=k
T2
P1 P2
=
T 1 T2
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9.2 THE IDEAL GAS LAW
𝑉 𝑎𝑛𝑑 𝑇
Volume and Temperature: Charles’s Law
• Study the data and graph.
• Describe the relationship between temperature and volume.
From the data, it is clear that volume is directly proportional
to temperature when pressure is held constant.
V ∝ T or V=k T
k is a constant
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9.2 THE IDEAL GAS LAW
Volume and Temperature: Charles’s Law
• For gases, there is also direct relationship between volume and
temperature. When the temperature is raised, volume increases.
Condition 1
V 1=k T 1
V1
=k
T1
Condition 2
V 2=k T 2
V2
=k
T2
V1 V2
=
T1 T2
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9.2 THE IDEAL GAS LAW
𝑉 𝑎𝑛𝑑 𝑃
Volume and Pressure: Boyle’s Law
• An experiment concerning the
pressure-volume behavior of a
gas sample was conducted at
constant temperature.
• Study the data/graphs and
explain how P and V are related
at constant temperature.
1
k
P∝
or P=
V
V
k is a constant
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9.2 THE IDEAL GAS LAW
Volume and Temperature: Boyle’s Law
• For gases, there is an inverse relationship between
pressure and volume (at constant temperature).
Condition 1
k
P1=
V1
P1 V 1=k
Condition 2
k
P2=
V2
P2 V 2=k
P1 V 1=P2 V 2
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9.2 THE IDEAL GAS LAW
𝑉 𝑎𝑛𝑑 𝑛
Moles and Volume: Avogadro’s Law
• Avogadro developed a theory to account for the behavior of gases when
they are measured under the same conditions of temperature and pressure.
• For a confined gas, the volume (V) and number of moles (n) are directly
proportional if P and T remain constant.
• For a sample of gas, the volume is directly proportional to the amount of
gas, in moles.
V ∝n or V=k n k is a constant
Equal volumes of different gases at the same temperature and pressure
contain the same number of gaseous particles.
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9.2 THE IDEAL GAS LAW
Mass and Volume: Avogadro’s Law
Condition 1
V 1=k n1
V1
=k
n1
Condition 2
V 2=k n 2
V2
=k
n2
V1 V2
=
n1 n2
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9.2 THE IDEAL GAS LAW
The Ideal Gas Law
• Four separate laws concerning gases have been presented.
Amonton’s Law
P∝T or P=k T
Charles’s Law
V ∝ T or V=k T
Boyle’s Law
1
k
P∝
or P=
T
V
Avogadro’s Law
V ∝n or V=k n
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9.2 THE IDEAL GAS LAW
The Ideal Gas Law
• Combining the four separate laws yields the ideal gas law.
Amonton’s Law
PV =nRT
Charles’s Law
P = pressure (atm)
V = volume (L)
n = moles (mol)
T = temperature (K)
Boyle’s Law
Avogadro’s Law
𝑅=0.08206
𝐿∙ 𝑎𝑡𝑚
𝑚𝑜𝑙∙ 𝐾
Standard temperature and
pressure (STP) for gas
experiments:
T = 273.15 K (0 °C)
P = 1 atm
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9.3 STOICHIOMETRY AND GASES
Density and Molar Mass
• What is the definition of density? What are the units?
• The ideal gas law is:
𝑃𝑉 =𝑛𝑅𝑇
𝑔
• To determine the density d of a gas, the units needed are:
𝐿
• The number of grams is related to n.
• The number of liters is the volume, V.
Density:
Rearranging ideal gas equation:
Molar mass:
Relation between density and molar mass:
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9.3 STOICHIOMETRY AND GASES
Dalton’s Law of Partial Pressures
• Each individual gas in a mixture exerts the same pressure that it would exert if it were
present alone in the volume.
• Pressure exerted by each individual gas in a mixture is called its partial pressure.
• The total pressure for a mixture of gases is equal to sum of the partial pressures of
those gases.
• If the gases do not interact, each gas in the mixture obeys its own PV=nRT.
In general:
P1V1=n1RT1
P2V2=n2RT2
P3V3=n3RT3
for blue
for purple
for yellow
Ptotal = P1 + P2 + P3
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9.3 STOICHIOMETRY AND GASES
Dalton’s Law of Partial Pressures
• The volume and temperature for the gases are constant
because they are mixed together in the same container.
• The partial pressure of each gas depends upon the number of
moles (n1, n2, n3) of that gas.
• The partial pressure of each gas is related to the total pressure by the
mole fraction of that gas.
ntotal = total moles of gas
n1 = total moles of blue gas
X1 = mole fraction of blue
gas
𝑛1
𝑋 1=
𝑛𝑡𝑜𝑡𝑎𝑙
𝑃 1= 𝑋 1 × 𝑃 𝑡𝑜𝑡𝑎𝑙
𝑃 1 ∝ 𝑛1
𝑃 2 ∝ 𝑛2
𝑃 3 ∝ 𝑛3
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9.3 STOICHIOMETRY AND GASES
Chemical Stoichiometry of Gases
• In the introduction to stoichiometry, reactions were balanced,
grams were converted to moles and various calculations were
done.
• Stoichiometry for gases works exactly the same way, except
here, the number of moles might be obtained by using
calculations related to gases, usually PV=nRT.
1. Write and balance the reaction.
2. List what you know and what you want.
3. Get all units into standard gas units (L, atm, mol, K)
and look up constants or molar masses as needed.
4. Approach the calculation as you would any other
stoichiometry problem.
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9.4 EFFUSION AND DIFFUSION OF GASES
Diffusion
• The process by which molecules disperse in space in response to
difference in concentration is called diffusion.
• Diffusion ultimately results in equal concentrations of gas throughout
the system.
• The rate of diffusion is the amount of gas passing through some area
per unit time.
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9.4 EFFUSION AND DIFFUSION OF GASES
Effusion
• By contrast, effusion is the escape of gas molecules through a tiny
hole into a vacuum.
• Rates of effusion for different gases can be compared by using
Graham’s Law: The rate of effusion of a gas is inversely proportional
to the square root of the molar mass of its particles.
• For two gases A and B, the rate of their effusion and their molar mass
are related as:
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9.5 THE KINETIC MOLECULAR THEORY
Explaining Gas Laws
• The gas laws presented thus far have been derived from
experimental observations.
• As a mathematical model, PV=nRT generally describes the
behavior of most gases at or room temperature and 1-2 atm.
• Why do gases behave this way?
• Observations work on a macroscopic level.
• On a microscopic level, kinetic molecular theory is useful.
𝑛 𝑎𝑛𝑑 𝑉
𝑉 𝑎𝑛𝑑 𝑇
𝑃 𝑎𝑛𝑑 𝑇
𝑃 𝑎𝑛𝑑 𝑉
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9.5 THE KINETIC MOLECULAR THEORY
Kinetic Molecular Theory
• Particles are in continuous motion.
• The volume of each particle is negligible.
• Pressure results from collisions between particles and the
container walls.
• Particles do not exert forces on each other; collision are elastic.
• The average kinetic energy is proportional to temperature (K).
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9.5 THE KINETIC MOLECULAR THEORY
Kinetic Molecular Theory Explains the Behavior of Gases
Amonton’s Law
P and T
V constant
𝑃1
=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑇1

Boyle’s Law
P and V
Avogadro’s Law
n and V
T constant
P and T constant
𝑃 1 𝑉 1=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑉1
=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑛1
T then P
P then V
n then V
increased
frequency and force
of collisions
reduced frequency
of collisions
constant number
of collisions
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9.5 THE KINETIC MOLECULAR THEORY
Molecular Velocities and Kinetic Energy
• In a gas sample, particles have various speeds.
• Because there are an enormous number of particles in a gas
sample, the molecular speed distribution and the average
speed remain constant.
• The molecular speed distribution is known as a MaxwellBoltzmann distribution.
Maxwell-Boltzmann Distribution
The relative number of molecules in a
gas sample that possess a given speed
(the speed distribution) is shown.
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9.5 THE KINETIC MOLECULAR THEORY
Molecular Velocities and Kinetic Energy
For n number of particles with speeds u1,u2,u3,…
the root mean square speed of a particle is
defined as:
Average kinetic energy for a mole of particles,
whose molar mass is M, is given as:
Average kinetic energy for a mole of particles,
and the temperature (in Kelvin), are related as:
R= 8.314 Kgm2s-2mol-1K-1
Relation between urms and temperature →
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9.5 THE KINETIC MOLECULAR THEORY
Molecular Velocities and Kinetic Energy
• If the temperature of a gas sample increases, KE increases.
• more particles at higher speed
• distribution shifts toward higher speeds overall (to the right)
• If the temperature decreases, the average KE decreases.
• more particles at lower sped
• distribution shifts to the lower speeds overall (to the left)
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9.5 THE KINETIC MOLECULAR THEORY
Molecular Velocities and Kinetic Energy
• At a given temperature, all gases have the same average
kinetic energy.
• Gases composed of lighter particles have a speed distribution
that peaks at higher velocities.
Molecular velocity is directly related to
molecular mass. At a given temperature,
lighter particles move faster on average
than heavier particles.
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9.6 NON-IDEAL GAS BEHAVIOR
Ideal Gases: Assumptions
• Gas atoms/molecules have no volume; the particles are point
masses.
• The atoms/molecules do not experience attractions or
repulsions and only undergo elastic collisions.
Real Gases
• Are not point masses. They have a finite mass.
• They experience attraction or repulsion and undergo inelastic
collisions.
• As a consequence, the behavior of real gasses deviate from
the predictions given by the ideal gas equation.
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9.6 NON-IDEAL GAS BEHAVIOR
Non-Ideal Gases
• Non-ideal behavior of gases is measured by compressibility factor
Z.
• Compressibility factor is measured by comparing the actual volume of
1 mole of gas (its molar volume, Vm) to the molar volume of an ideal
gas at the same temperature and pressure.
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9.6 NON-IDEAL GAS BEHAVIOR
Real Gases
•
Gases deviate most from ideal behavior at high pressure and low temperature.
•
The behavior of real gases can be approximated by the van der Waals
equation. The equation corrects:
• the pressure term to account for attraction between particles.
•
the volume term to account for the real volume of gas particles (they are
not simply point masses).
Ideal gas equation →
Van der Waals equation →
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PROBLEMS
A chamber contains equal molar amounts of H₂, N₂, and O₂. If the total
chamber pressure is 2.00 atm, calculate the partial pressure of H₂
Hint: Use Dalton’s law of partial pressures and the concept of mole fraction
Solution in the next slide
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PROBLEMS
A chamber contains equal molar amounts of H₂, N₂, and O₂. If the total
chamber pressure is 2.00 atm, calculate the partial pressure of H₂
Hint: Use Dalton’s law of partial pressures
Solution:
Since the equal molar amounts of H₂, N₂, and O₂ are present, let n be the
number of moles of each gas. Then the total number of moles of glass in the
container is
n+n+n=3n
Mole fraction of H2 is n/3n = 1/3
Partial pressure of a gas = mole fraction of the gas x Total pressure.
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PROBLEMS
At what temperature does sulfur hexafluoride (SF6) have a density of 0.5550 g/L
at 0.8210 atm?
Given: Density and Pressure of SF6
Find: Temperature of SF6
Hint: What is the relation that relates Density, pressure and temperature ?
Do we have all the quantities required to find the temperature ?
Solution in the next slide
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PROBLEMS
At what temperature does sulfur hexafluoride (SF6) have a density of 0.5550 g/L at
0.8210 atm?
Pressure = 0.8210 atm
Density = 0.5550 g/L
R= 0.08206 L.atm/mol.K
Temperature
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PROBLEMS
A 12.0 L container at 313 K holds a mixture of two gases with a total pressure of
9.00 atm. If there are 2.50 mol of Gas A in the mixture, how many moles of Gas B
are present?
Hint: The container contains two gases and the total volume, pressure and
temperature of the container is given.
The number of moles of gas A is given. If we know the total number of moles of gas
in the container, then we can find the number of moles of gas B.
Solution in the next slide
39
PROBLEMS
A 12.0 L container at 313 K holds a mixture of two gases with a total pressure
of 9.00 atm. If there are 2.50 mol of Gas A in the mixture, how many moles of
Gas B are present?
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PROBLEMS
Calculate the root-mean-square velocity, in m/s, for an oxygen molecule at 35.0
°C. The universal gas constant, R= 8.314 Kg m2 s-2 mol-1 K-1
Hint: Relation between rms velocity and temperature. Is the temperature in K ?
Solution in the next slide
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PROBLEMS
Calculate the root-mean-square velocity, in m/s, for an oxygen molecule at 35.0
°C. The universal gas constant, R= 8.314 Kg m2 s-2 mol-1 K-1
TK = TC + 273.15
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PROBLEMS
When copper metal is added to nitric acid, the following reaction takes place
Cu (s) + 4 HNO₃ (aq) → Cu(NO₃)₂ (aq) + 2 H₂O (l) + 2 NO₂ (g)
Calculate the volume in liters of NO₂ gas collected over water at 25.0 °C when
2.01 g of copper is added to nitric acid and the total pressure is 726.0 mmHg. The
partial pressure of water at 25.0 °C is 23.8 mm Hg.
Step 1: Check if the equation is balanced. If not, balance the equation.
Step 2: Find out how many moles of NO2 is collected when 2.01 g of Cu is added.
Step 3: Find the pressure of NO2 gas. We have the total pressure and the partial
pressure of water. Don’t forget to convert the units from mmHg to atm. (1 atm =
760 mmHg)
Step 4: Find the temperature at which the reaction is taking place. Don’t forget to
convert the temperature units to K.
Step 5: Now that we have the number of moles, pressure and temperature of NO2
gas, use the ideal gas equation to find the volume of the gas.
Solution in the next slide
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PROBLEMS
Step 1 and 2:
Step 3 and 4:
Step 5:
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PROBLEMS
Solid sodium reacts with liquid water to form hydrogen gas according to the
equation
2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)
When 1.00 kg sodium is reacted with water at 50.0°C, the Hydrogen gas
occupies 20.0 L. What is the pressure (in atm) of hydrogen gas ?
Step 1: Check if the equation is balanced. If not, balance the equation.
Step 2: Find out how many moles of H2 is collected when 1 Kg of Na is added.
Step 3: Find the volume of the Hydrogen gas, make sure it is in liters.
Step 4: Find the temperature at which the reaction is taking place. Don’t forget
to convert the temperature units to K.
Step 5: Now that we have the number of moles, volume and temperature of H2
gas, use the ideal gas equation to find the pressure of the gas.
Solution in the next slide
45
PROBLEMS
Solid sodium reacts with liquid water to form hydrogen gas according to the
equation
2 Na(s) + 2 H₂O(l) → 2 NaOH(aq) + H₂(g)
When 1.00 kg sodium is reacted with water at 50.0°C, the Hydrogen gas
occupies 20.0 L. What is the pressure (in atm) of hydrogen gas ?
Step 2:
Step 2:
Steps 3,4 and 5:
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PROBLEMS
What volume in liters of fluorine gas is needed to form 999 L of sulfur
hexafluoride gas if the following reaction takes place at 2.00 atm and 273.15 K:
S(s) + 3F₂(g) → SF₆ (g)?
At constant temperature and pressure, volumes of gas can be related directly to
each other. The liters of fluorine can be calculated using a conversion factor
found in the balanced chemical equation.
Solution in the next slide
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PROBLEMS
What volume in liters of fluorine gas is needed to form 999 L of sulfur
hexafluoride gas if the following reaction takes place at 2.00 atm and 273.15 K:
S(s) + 3F₂(g) → SF₆ (g)?
At constant temperature and pressure, volumes of gas can be related directly to
each other. Let V1 be volume of F2 gas and V2 be volume of SF6 gas
According to Avogado’s law, under constant pressure and temperature
V=kn
We have for F2 gas: V1 = k × 3
We have for SF6 gas: V2 = k
So, the volume of Fluorine gas is : V1 = V2 × 3 = 999 L × 3 = 2997 L
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