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Chapter 5

Gases

Chapter 5

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(5.1) Pressure
(5.2) The gas laws of Boyle, Charles, and Avogadro
(5.3) The ideal gas law
(5.4) Gas stoichiometry
(5.5) Dalton’s law of partial pressures
(5.6) The kinetic molecular theory of gases

Chapter 5

Table of Contents

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(5.7) Effusion and diffusion
(5.8) Real gases
(5.9) Characteristics of several real gases
(5.10) Chemistry in the atmosphere

Chapter 5

Table of Contents

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Properties of Gases

Uniformly fill any container
Easily compressible
Completely mix with other gases
Exert pressure on their surroundings

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Section 5.1

Pressure

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Figure 5.1 - The Collapsing Can Experiment

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Section 5.1

Pressure

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Barometer

Device used to measure atmospheric pressure

Atmospheric pressure results from the weight of the air

Variations are attributed to change in weather conditions and altitudes

Contains a glass tube filled with liquid mercury that is inverted in a dish containing mercury

Section 5.1

Pressure

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Manometer

Device used for measuring the pressure of a gas in a container

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Section 5.1

Pressure

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Units of Pressure

Based on the height of the mercury column that a gas can support
mm Hg (millimeter of mercury)

Torr and mm Hg are used interchangeably

Standard atmosphere (atm)

Section 5.1

Pressure

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Units of Pressure (Continued)

Defined as force per unit area

SI system

Newton/m² = 1 pascal (Pa)
1 standard atmosphere = 101,325 Pa
1 atmosphere ≈ 10⁵ Pa

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Section 5.1

Pressure

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Interactive Example 5.1 - Pressure Conversions

The pressure of a gas is measured as 49 torr

Represent this pressure in both atmospheres and pascals

Section 5.1

Pressure

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Interactive Example 5.1 - Solution

Section 5.1

Pressure

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Boyle’s Law

Robert Boyle studied the relationship between the pressure of trapped gas and its volume

k - Constant for a given sample of air at a specific temperature
Deduced that pressure and volume are inversely related

Ideal gas: Gas that strictly obeys Boyle’s law

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Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Plot of Boyle’s Law

A plot of P versus V shows that the volume doubles as the pressure is halved

Resulting curve is a hyperbola

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Plot of Boyle’s Law (Continued)

Boyle’s law can be rearranged to mirror the straight line equation

y = V
x = 1/P
m = k
b = 0

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Figure 5.5 - Linear Plot of Boyle’s Law

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Figure 5.6 - A Plot of PV versus P for Several Gases at Pressures below 1 atm

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.2 - Boyle’s Law I

Sulfur dioxide (SO₂), a gas that plays a central role in the formation of acid rain, is found in the exhaust of automobiles and power plants

Consider a 1.53-L sample of gaseous SO₂ at a pressure of 5.6×10³ Pa

If the pressure is changed to 1.5×10⁴ Pa at a constant temperature, what will be the new volume of the gas?

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.2 - Solution

Where are we going?

To calculate the new volume of gas

What do we know?

P₁ = 5.6×10³ Pa P₂ = 1.5×10⁴ Pa
V₁ = 1.53 L V₂ = ?

What information do we need?

Boyle’s law

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.2 - Solution (Continued 1)

How do we get there?

What is Boyle’s law (in a form useful with our knowns)?

What is V₂?

The new volume will be 0.57 L

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.2 - Solution (Continued 2)

Reality check

The new volume (0.57 L) is smaller than the original volume
As pressure increases, the volume should decrease, so our answer is reasonable

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Exercise

A particular balloon is designed by its manufacturer to be inflated to a volume of no more than 2.5 L

If the balloon is filled with 2.0 L helium at sea level, is released, and rises to an altitude at which the atmospheric pressure is only 500 mm Hg, will the balloon burst? (Assume temperature is constant)

The balloon will burst

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Example 5.3 - Boyle’s Law II

In a study to see how closely gaseous ammonia obeys Boyle’s law, several volume measurements were made at various pressures, using 1.0 mole of NH₃ gas at a temperature of 0°C

Using the results listed below, calculate the Boyle’s law constant for NH₃ at the various pressures

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Example 5.3 - Boyle’s Law II (Continued)

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Example 5.3 - Solution

To determine how closely NH₃ gas follows Boyle’s law under these conditions, we calculate the value of k (in L · atm) for each set of values

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Example 5.3 - Solution (Continued 1)

Although the deviations from true Boyle’s law behavior are quite small at these low pressures, note that the value of k changes regularly in one direction as the pressure is increased

To calculate the ideal value of k for NH₃, we can plot PV versus P, and extrapolate back to zero pressure, where, for reasons we will discuss later, a gas behaves most ideally

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Example 5.3 - Solution (Continued 2)

The value of k obtained by this extrapolation is 22.41 L· atm

Notice that this is the same value obtained from similar plots for the gases CO₂, O₂, and Ne at 0°C

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Jacques Charles

French physicist who made the first solo balloon flight
Determined that the volume of a gas at constant pressure increases linearly with the temperature of the gas

Plot of volume of gas (at constant pressure) versus temperature (in °C) gives a straight line

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Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Figure 5.8 - Plots of V versus T (°C) for Several Gases

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Charles’s Law

Volume of each gas is directly proportional to the temperature

Volume extrapolates to zero when the temperature is equal to 0 K (absolute zero)

T - Temperature in kelvins
b - Proportionality constant

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Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Critical Thinking

According to Charles’s law, the volume of a gas is directly related to its temperature in kelvins at constant pressure and number of moles

What if the volume of a gas was directly related to its temperature in degrees Celsius at constant pressure and number of moles?

What differences would you notice?

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.4 - Charles’s Law

A sample of gas at 15°C and 1 atm has a volume of 2.58 L

What volume will this gas occupy at 38°C and 1 atm?

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.4 - Solution

Where are we going?

To calculate the new volume of gas

What do we know?

T₁ = 15°C + 273 = 288 K
T₂ = 38°C + 273 = 311 K
V₁ = 2.58 L
V₂ = ?

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.4 - Solution (Continued 1)

What information do we need?

Charles’s law

How do we get there?

What is Charles’s law (in a form useful with our knowns)?

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.4 - Solution (Continued 2)

What is V₂?

Reality check

The new volume is greater than the original volume, which makes physical sense because the gas will expand as it is heated

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Avogadro’s Law

For a gas at constant pressure and temperature, the volume is directly proportional to the number of moles of gas

V - Volume of gas
n - Number of moles of gas particles
a - Proportionality constant

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Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Figure 5.10 - Representation of Avogadro’s Law

These balloons each hold 1.0 L gas at 25°C and 1 atm, and each balloon contains 0.041 mole of gas, or 2.5×10²² molecules

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.5 - Avogadro’s Law

Suppose we have a 12.2-L sample containing 0.50 mole of oxygen gas (O₂) at a pressure of 1 atm and a temperature of 25°C

If all this O₂ were converted to ozone (O₃) at the same temperature and pressure, what would be the volume of the ozone?

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.5 - Solution

Where are we going?

To calculate the volume of the ozone produced by 0.50 mole of oxygen

What do we know?

n₁ = 0.50 mol O₂
n₂ = ? mol O₃
V₁ = 12.2 L O₂
V₂ = ? L O₃

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.5 - Solution (Continued 1)

What information do we need?

Balanced equation
Moles of O₃
Avogadro’s law:

How do we get there?

How many moles of O₃ are produced by 0.50 mole of O₂?

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.5 - Solution (Continued 2)

What is the balanced equation?

What is the mole ratio between O₃ and O₂?

Now we can calculate the moles of O₃ formed

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.5 - Solution (Continued 3)

What is the volume of O₃ produced?

Avogadro’s law states that V = an, which can be rearranged to give

Since a is a constant, an alternative representation is

V₁ is the volume of n₁ moles of O₂ gas and V₂ is the volume of n₂ moles of O₃ gas

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Interactive Example 5.5 - Solution (Continued 4)

In this case we have

n₁ = 0.50 mol n₂ = 0.33 mol
V₁ = 12.2 L V₂ = ?

Solving for V₂ gives

Reality check - Note that the volume decreases, as it should, since fewer moles of gas molecules will be present after O₂ is converted to O₃

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Exercise

An 11.2-L sample of gas is determined to contain 0.50 mole of N₂

At the same temperature and pressure, how many moles of gas would there be in a 20-L sample?

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0.89 mol

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro

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Ideal Gas Law

The gas laws of Boyle, Charles, and Avogadro can be combined to give the ideal gas law

Universal gas constant: Combined proportionality constant (R)

R = 0.08206 L · atm/K · mol when pressure is expressed in atmospheres and the volume in liters

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Section 5.3

The Ideal Gas Law

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Ideal Gas Law (Continued)

Can be rearranged to:

Equation of state for a gas, where the state of the gas is its condition at a given time

Based on experimental measurements of the properties of gases
Any gas that obeys this law is said to be behaving ideally

An ideal gas is a hypothetical substance

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Section 5.3

The Ideal Gas Law

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Gas Law Problems

Types

Boyle’s law problems
Charles’s law problems
Avogadro’s law problems

Ideal gas law can be applied to any problem

Place the variables that change on one side of the equal sign and the constants on the other side

Section 5.3

The Ideal Gas Law

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Interactive Example 5.7 - Ideal Gas Law II

Suppose we have a sample of ammonia gas with a volume of 7.0 mL at a pressure of 1.68 atm

The gas is compressed to a volume of 2.7 mL at a constant temperature
Use the ideal gas law to calculate the final pressure

Section 5.3

The Ideal Gas Law

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Interactive Example 5.7 - Solution

Where are we going?

To use the ideal gas equation to determine the final pressure

What do we know?

P₁ = 1.68 atm
P₂ = ?
V₁ = 7.0 mL
V₂ = 2.7 mL

Section 5.3

The Ideal Gas Law

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Interactive Example 5.7 - Solution (Continued 1)

What information do we need?

Ideal gas law

R = 0.08206 L · atm/K · mol

How do we get there?

What are the variables that change?

P, V

Section 5.3

The Ideal Gas Law

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Interactive Example 5.7 - Solution (Continued 2)

What are the variables that remain constant?

n, R, T

Write the ideal gas law, collecting the change variables on one side of the equal sign and the variables that do not change on the other

Change

Remain

constant

Section 5.3

The Ideal Gas Law

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Interactive Example 5.7 - Solution (Continued 3)

Since n and T remain the same in this case, we can write P₁V₁ = nRT and P₂V₂ = nRT, and combining these gives

P₁ = 1.68 atm V₁ = 7.0 mL V₂ = 2.7 mL

Solving for P₂ gives

Section 5.3

The Ideal Gas Law

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Interactive Example 5.7 - Solution (Continued 4)

Reality check - The volume decreased (at constant temperature), so the pressure should increase

Note that the calculated final pressure is 4.4 atm
Most gases do not behave ideally above 1 atm

If the pressure of this gas sample was measured, the observed pressure would differ slightly from 4.4 atm

Section 5.3

The Ideal Gas Law

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Interactive Example 5.9 - Ideal Gas Law IV

A sample of diborane gas (B₂H₆), a substance that bursts into flame when exposed to air, has a pressure of 345 torr at a temperature of –15°C and a volume of 3.48 L

If conditions are changed so that the temperature is 36°C and the pressure is 468 torr, what will be the volume of the sample?

Section 5.3

The Ideal Gas Law

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Interactive Example 5.9 - Solution

Where are we going?

To use the ideal gas equation to determine the final volume

What do we know?

T₁ = 15°C + 273 = 258 K T₂ = 36°C + 273 = 309 K
V₁ = 3.48 L V₂ = ?
P₁ = 345 torr P₂ = 468 torr

Section 5.3

The Ideal Gas Law

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Interactive Example 5.9 - Solution (Continued 1)

What information do we need?

Ideal gas law

R = 0.08206 L · atm/K · mol

How do we get there?

What are the variables that change?

P, V, T

Section 5.3

The Ideal Gas Law

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Interactive Example 5.9 - Solution (Continued 2)

What are the variables that remain constant?

n, R

Write the ideal gas law, collecting the change variables on one side of the equal sign and the variables that do not change on the other

Section 5.3

The Ideal Gas Law

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Interactive Example 5.9 - Solution (Continued 3)

Solving for V₂

Section 5.3

The Ideal Gas Law

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Interactive Example 5.10 - Ideal Gas Law V

A sample containing 0.35 mole of argon gas at a temperature of 13°C and a pressure of 568 torr is heated to 56°C and a pressure of 897 torr

Calculate the change in volume that occurs

Section 5.3

The Ideal Gas Law

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Interactive Example 5.10 - Solution

Where are we going?

To use the ideal gas equation to determine the final volume

What do we know?

Section 5.3

The Ideal Gas Law

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Interactive Example 5.10 - Solution (Continued 1)

What information do we need?

Ideal gas law

R = 0.08206 L · atm/K · mol
V₁ and V₂

Section 5.3

The Ideal Gas Law

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Interactive Example 5.10 - Solution (Continued 2)

How do we get there?

What is V₁?

Section 5.3

The Ideal Gas Law

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Interactive Example 5.10 - Solution (Continued 3)

What is V₂?

Section 5.3

The Ideal Gas Law

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Interactive Example 5.10 - Solution (Continued 4)

What is the change in volume ΔV?

Conclusion

The change in volume is negative because the volume decreases

Section 5.3

The Ideal Gas Law

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Molar Volume of an Ideal Gas

Molar volume: For 1 mole of an ideal gas at 0°C and 1 atm, the volume of the gas is 22.42 L

Standard temperature and pressure (STP): Conditions 0°C and 1 atm

Section 5.4

Gas Stoichiometry

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Critical Thinking

What if STP was defined as normal room temperature (22°C) and 1 atm?

How would this affect the molar volume of an ideal gas?

Include an explanation and a number

Section 5.4

Gas Stoichiometry

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Interactive Example 5.12 - Gas Stoichiometry II

Quicklime (CaO) is produced by the thermal decomposition of calcium carbonate (CaCO₃)

Calculate the volume of CO₂ at STP produced from the decomposition of 152 g CaCO₃ by the reaction

Section 5.4

Gas Stoichiometry

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Interactive Example 5.12 - Solution

Where are we going?

To use stoichiometry to determine the volume of CO₂ produced

What do we know?

What information do we need?

Molar volume of a gas at STP is 22.42 L

Section 5.4

Gas Stoichiometry

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Interactive Example 5.12 - Solution (Continued 1)

How do we get there?

What is the balanced equation?

What are the moles of CaCO₃ (100.09 g/mol)?

Section 5.4

Gas Stoichiometry

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Interactive Example 5.12 - Solution (Continued 2)

What is the mole ratio between CO₂ and CaCO₃ in the balanced equation?

What are the moles of CO₂?

1.52 moles of CO₂, which is the same as the moles of CaCO₃ because the mole ratio is 1

Section 5.4

Gas Stoichiometry

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Interactive Example 5.12 - Solution (Continued 3)

What is the volume of CO₂ produced?

We can compute this by using the molar volume since the sample is at STP:

Thus, the decomposition of 152 g CaCO₃ produces 34.1 L CO₂ at STP

Section 5.4

Gas Stoichiometry

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Interactive Example 5.13 - Gas Stoichiometry III

A sample of methane gas having a volume of 2.80 L at 25°C and 1.65 atm was mixed with a sample of oxygen gas having a volume of 35.0 L at 31°C and 1.25 atm, and the mixture was then ignited to form carbon dioxide and water

Calculate the volume of CO₂ formed at a pressure of 2.50 atm and a temperature of 125°C

Section 5.4

Gas Stoichiometry

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Interactive Example 5.13 - Solution

Where are we going?

To determine the volume of CO₂ produced

What do we know?

Section 5.4

Gas Stoichiometry

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Interactive Example 5.13 - Solution (Continued 1)

What information do we need?

Balanced chemical equation for the reaction
Ideal gas law

R = 0.08206 L · atm/K · mol

Section 5.4

Gas Stoichiometry

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Interactive Example 5.13 - Solution (Continued 2)

How do we get there?

What is the balanced equation?

From the description of the reaction, the unbalanced equation is

This can be balanced to give

Section 5.4

Gas Stoichiometry

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Interactive Example 5.13 - Solution (Continued 3)

What is the limiting reactant?

We can determine this by using the ideal gas law to determine the moles for each reactant

Section 5.4

Gas Stoichiometry

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Interactive Example 5.13 - Solution (Continued 4)

In the balanced equation for the combustion reaction, 1 mole of CH₄ requires 2 moles of O₂
The moles of O₂ required by 0.189 mole of CH₄ can be calculated as follows:

The limiting reactant is CH₄

Section 5.4

Gas Stoichiometry

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Interactive Example 5.13 - Solution (Continued 5)

What are the moles of CO₂?

Since CH₄ is limiting, we use the moles of CH₄ to determine the moles of CO₂ produced

What is the volume of CO₂ produced?

Since the conditions stated are not STP, we must use the ideal gas law to calculate the volume

Section 5.4

Gas Stoichiometry

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Interactive Example 5.13 - Solution (Continued 6)

In this case, n = 0.189 mol, T = 125°C + 273 = 398 K, P = 2.50 atm, and R = 0.08206 L · atm/K · mol

This represents the volume of CO₂ produced under these conditions

Section 5.4

Gas Stoichiometry

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Molar Mass of a Gas

Ideal gas law is essential for the calculation of molar mass of a gas from its measured density

Substituting into the ideal gas equation gives:

Section 5.4

Gas Stoichiometry

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Molar Mass of a Gas (Continued)

Density, d, is equal to m/V

Or

Section 5.4

Gas Stoichiometry

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Interactive Example 5.14 - Gas Density/Molar Mass

The density of a gas was measured at 1.50 atm and 27°C and found to be 1.95 g/L

Calculate the molar mass of the gas

Section 5.4

Gas Stoichiometry

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Interactive Example 5.14 - Solution

Where are we going?

To determine the molar mass of the gas

What do we know?

P = 1.50 atm
T = 27°C + 273 = 300 K
d = 1.95 g/L

Section 5.4

Gas Stoichiometry

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Interactive Example 5.14 - Solution (Continued)

What information do we need?

R = 0.08206 L · atm/K · mol

How do we get there?

Reality check - These are the units expected for molar mass

Section 5.4

Gas Stoichiometry

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Law of Partial Pressures - John Dalton

For a mixture of gases in a container, the total pressure exerted is the sum of the partial pressures

Partial pressure: Pressure that a gas would exert if it were alone in a container

Represented by symbols P₁, P₂, and P₃

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Section 5.5

Dalton’s Law of Partial Pressures

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Law of Partial Pressures - John Dalton (Continued 1)

Assume that all gases behave ideally

Their partial pressures can be calculated from the ideal gas law

Total pressure of the mixture P_TOTAL

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Section 5.5

Dalton’s Law of Partial Pressures

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Law of Partial Pressures - John Dalton (Continued 2)

n_TOTAL - Sum of the number of moles of the various gases

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Section 5.5

Dalton’s Law of Partial Pressures

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Figure 5.12 - Schematic Diagram of Dalton's Law of Partial Pressures

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Section 5.5

Dalton’s Law of Partial Pressures

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Characteristics of an Ideal Gas

Since the pressure exerted by an ideal gas is unaffected by its identity:

The volume of individual gas particles must not be important
The forces among the particles must not be important

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.15 - Dalton’s Law I

Mixtures of helium and oxygen can be used in scuba diving tanks to help prevent “the bends”

For a particular dive, 46 L He at 25°C and 1.0 atm and 12 L O₂ at 25°C and 1.0 atm were pumped into a tank with a volume of 5.0 L

Calculate the partial pressure of each gas and the total pressure in the tank at 25°C

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.15 - Solution

Where are we going?

To determine the partial pressure of each gas
To determine the total pressure in the tank at 25°C

What do we know?

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.15 - Solution (Continued 1)

What information do we need?

Ideal gas law

R = 0.08206 L · atm/K · mol

How do we get there?

How many moles are present for each gas?

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.15 - Solution (Continued 2)

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.15 - Solution (Continued 3)

What is the partial pressure for each gas in the tank?

The tank containing the mixture has a volume of 5.0 L, and the temperature is 25°C
We can use these data and the ideal gas law to calculate the partial pressure of each gas

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.15 - Solution (Continued 4)

What is the total pressure of the mixture of gases in the tank?

The total pressure is the sum of the partial pressures

Section 5.5

Dalton’s Law of Partial Pressures

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Mole Fraction (χ)

Ratio of number of moles of a given component in a mixture to the total number of moles in the mixture
Example

For a given component in a mixture, χ₁ is calculated as follows:

Section 5.5

Dalton’s Law of Partial Pressures

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Mole Fraction (χ) in Terms of Pressure

Number of moles of a gas is directly proportional to the pressure of the gas, since

Representation of mole fraction in terms of pressure

Section 5.5

Dalton’s Law of Partial Pressures

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Mole Fraction (χ) in Terms of Pressure (Continued)

Mole fraction of each component in a mixture of ideal gases is directly related to its partial pressure

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.16 - Dalton’s Law II

The partial pressure of oxygen was observed to be 156 torr in air with a total atmospheric pressure of 743 torr

Calculate the mole fraction of O₂ present

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.16 - Solution

Where are we going?

To determine the mole fraction of O₂

What do we know?

P_O2 = 156 torr
P_TOTAL = 743 torr

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.16 - Solution (Continued)

How do we get there?

The mole fraction of O₂ can be calculated from the equation

Note that the mole fraction has no units

Section 5.5

Dalton’s Law of Partial Pressures

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Rearranging the Expression of Mole Fraction

This can be rearranged to give

P₁ = χ₁ + P_TOTAL

The partial pressure of a particular component of a gaseous mixture is the mole fraction of that component times the total pressure

Section 5.5

Dalton’s Law of Partial Pressures

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Vapor Pressure of Water

When the rate of escape of a gas in a container equals the rate of return:

Number of water molecules in the vapor state remain constant
Pressure of water vapor remains constant

Section 5.5

Dalton’s Law of Partial Pressures

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Figure 5.13 - The Production of Oxygen by Thermal Decomposition of Potassium Chlorate

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.18 - Gas Collection over Water

A sample of solid potassium chlorate (KClO₃) was heated in a test tube and decomposed by the following reaction:

The oxygen produced was collected by displacement of water at 22°C at a total pressure of 754 torr
The volume of the gas collected was 0.650 L, and the vapor pressure of water at 22°C is 21 torr

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.18 - Gas Collection over Water (Continued)

Calculate the partial pressure of O₂ in the gas collected and the mass of KClO₃ in the sample that was decomposed

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.18 - Solution

Where are we going?

To determine the partial pressure of O₂ in the gas collected
Calculate the mass of KClO₃ in the original sample

What do we know?

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.18 - Solution (Continued 1)

How do we get there?

What is the partial pressure of O₂?

What is the number of moles of O₂?

Now we use the ideal gas law to find the number of moles of O₂

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.18 - Solution (Continued 2)

In this case, the partial pressure of the O₂ is

To find the moles of O₂ produced, we use

V = 0.650 L

T = 22°C + 273 = 295 K

R = 0.08206 L · atm/K · mol

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.18 - Solution (Continued 3)

How many moles of KClO₃ are required to produce this amount of O₂?

What is the balanced equation?

What is the mole ratio between KClO₃ and O₂ in the balanced equation?

Section 5.5

Dalton’s Law of Partial Pressures

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Interactive Example 5.18 - Solution (Continued 4)

What are the moles of KClO₃?

What is the mass of KClO₃ (molar mass 122.6 g/mol) in the original sample?

Section 5.5

Dalton’s Law of Partial Pressures

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The Kinetic Molecular Theory (KMT)

Attempts to explain the properties of an ideal gas

Real gases do not conform to the postulates of the KMT

Based on speculations about the behavior of individual gas particles

111

Section 5.6

The Kinetic Molecular Theory of Gases

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Postulates of the Kinetic Molecular Theory

Particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero)

112

Section 5.6

The Kinetic Molecular Theory of Gases

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Postulates of the Kinetic Molecular Theory (Continued 1)

Particles are in constant motion

Collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas

Particles are assumed to exert no forces on each other

Particles are assumed neither to attract nor to repel each other

113

Section 5.6

The Kinetic Molecular Theory of Gases

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Postulates of the Kinetic Molecular Theory (Continued 2)

Average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas

114

Section 5.6

The Kinetic Molecular Theory of Gases

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Pressure and Volume (Boyle’s Law)

For a given sample of a gas at any given temperature, if the volume decreases, the pressure increases

According to KMT, decrease in volume implies that the gas particles would hit the wall more often

Increases pressure

115

Constant

Section 5.6

The Kinetic Molecular Theory of Gases

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Figure 5.16 - Effects of Decreasing the Volume of a Sample of Gas at Constant Temperature

Section 5.6

The Kinetic Molecular Theory of Gases

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Pressure and Temperature

According to the ideal gas law, for a given sample of an ideal gas at constant volume, the pressure is directly proportional to the temperature

According to KMT, as the temperature increases, the speed of the gas particles increases, and they hit the wall with greater force resulting in increased pressure

Constant

Section 5.6

The Kinetic Molecular Theory of Gases

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Figure 5.16 - Effects of Increasing the Temperature of a Sample of Gas at Constant Volume

Section 5.6

The Kinetic Molecular Theory of Gases

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Critical Thinking

You have learned the postulates of the KMT

What if we could not assume the third postulate to be true?

How would this affect the measured pressure of a gas?

Section 5.6

The Kinetic Molecular Theory of Gases

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Volume and Temperature (Charles’s Law)

According to the ideal gas law, for a sample of a gas at constant pressure, the volume of the gas is directly proportional to the temperature in Kelvins

According to the KMT, at higher temperature, the speed of gas molecules increases

Hit the walls with more force

Constant

Section 5.6

The Kinetic Molecular Theory of Gases

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Volume and Temperature (Charles’s Law) (Continued)

Pressure can be kept constant only by increasing the volume of the container

Section 5.6

The Kinetic Molecular Theory of Gases

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Volume and Number of Moles (Avogadro’s Law)

According to the ideal gas law, the volume of a gas at constant temperature and pressure directly depends on the number of gas particles present

According to the KMT, increase in number of gas particles at the same temperature would cause the pressure to increase if the volume were held constant

Constant

Section 5.6

The Kinetic Molecular Theory of Gases

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Volume and Number of Moles (Avogadro’s Law) (Continued)

Pressure can return to its original value if volume is increased

Section 5.6

The Kinetic Molecular Theory of Gases

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Mixture of Gases (Dalton’s Law)

Total pressure exerted by a mixture of gases is the sum of the pressures of the individual gases
The KMT assumes that:

All gas particles are independent of one another
Volume of individual particles are not important

Section 5.6

The Kinetic Molecular Theory of Gases

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Deriving the Ideal Gas Law

Pressure can be expressed differently by applying the definitions of velocity, momentum, force, and pressure to the collection of particles in an ideal gas

Section 5.6

The Kinetic Molecular Theory of Gases

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Deriving the Ideal Gas Law (Continued 1)

Here,

P - Pressure of the gas
n - Number of moles of gas
N_A - Avogadro’s number
m - Mass of each particle
- Average of the square of the velocities of the particles
V - Volume of the container
- Average kinetic energy of a gas particle

Section 5.6

The Kinetic Molecular Theory of Gases

127 of 169

Deriving the Ideal Gas Law (Continued 2)

Average kinetic energy for a mole of gas particles can be ascertained by multiplying the average kinetic energy of an individual particle by N_A

The expression for pressure can be rewritten as:

Section 5.6

The Kinetic Molecular Theory of Gases

128 of 169

Deriving the Ideal Gas Law (Continued 3)

Since (KE)_avg ∝ T:

The agreement between the ideal gas law and the postulates of the KMT prove the validity of the KMT model

Section 5.6

The Kinetic Molecular Theory of Gases

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The Meaning of Temperature

According to the KMT, the Kelvin temperature indicates the average kinetic energy of gas particles

This relationship between temperature and average kinetic energy can be obtained when the following equations are combined:

Section 5.6

The Kinetic Molecular Theory of Gases

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The Meaning of Temperature (Continued)

This shows that the Kelvin temperature is an index of the random motions of particles of a gas

As temperature increases, the motion of the particles becomes greater

Section 5.6

The Kinetic Molecular Theory of Gases

131 of 169

Root Mean Square Velocity

Refers to the square root of
Symbolized by u_rms

The expression for u_rms can also be attained from the following equations:

131

Section 5.6

The Kinetic Molecular Theory of Gases

132 of 169

Root Mean Square Velocity (Continued 1)

Combination of the equations gives

Taking square root on both sides, we get:

m - Mass (kg) of a single gas particle
N_A - Number of particles in a mole

132

Section 5.6

The Kinetic Molecular Theory of Gases

133 of 169

Root Mean Square Velocity (Continued 2)

Substituting M for N_Am in the equation for u_rms, we obtain:

M - Mass of a mole of gas particles (kg)
R = 8.3145 J/K·mol

J = joule = kg^.m²/s²

133

Section 5.6

The Kinetic Molecular Theory of Gases

134 of 169

Interactive Example 5.19 - Root Mean Square Velocity

Calculate the root mean square velocity for the atoms in a sample of helium gas at 25°C

Section 5.6

The Kinetic Molecular Theory of Gases

135 of 169

Interactive Example 5.19 - Solution

Where are we going?

To determine the root mean square velocity for the atoms of He

What do we know?

T = 25°C + 273 = 298 K
R = 8.3145 J/K · mol

What information do we need?

Root mean square velocity is

Section 5.6

The Kinetic Molecular Theory of Gases

136 of 169

Interactive Example 5.19 - Solution (Continued 1)

How do we get there?

What is the mass of a mole of He in kilograms?

What is the root mean square velocity for the atoms of He?

Section 5.6

The Kinetic Molecular Theory of Gases

137 of 169

Interactive Example 5.19 - Solution (Continued 2)

Since the units of J are kg · m²/s², this expression gives:

Reality check - The resulting units are appropriate for velocity

Section 5.6

The Kinetic Molecular Theory of Gases

138 of 169

Range of Velocities in Gas Particles

Real gases have large number of collisions between particles

Path of a gas particle is erratic

Mean free path

Average distance travelled by a particle between collisions in a gas sample
1×10^–7 m for O₂ at STP

Section 5.6

The Kinetic Molecular Theory of Gases

139 of 169

Effect of Collisions among Gas Particles

When particles collide and exchange kinetic energy, a large range of velocities is produced

This plot depicts the relative number of O₂ molecules that have a given velocity at STP

Section 5.6

The Kinetic Molecular Theory of Gases

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Effect of Temperature on Velocity Distribution

As the temperature increases, the range of velocities becomes larger

Peak of the curve reflects the most probable velocity

Section 5.6

The Kinetic Molecular Theory of Gases

141 of 169

Effusion and Diffusion - An Introduction

Diffusion: Describes the mixing of gases

Rate of diffusion is the rate of mixing of gases

Effusion: Describes the passage of a gas through a tiny orifice into an evacuated chamber

Rate of effusion measures the speed at which the gas is transferred into the chamber

Inversely proportional to the square root of the mass of the gas particles

141

Section 5.7

Effusion and Diffusion

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Figure 5.22 - The Effusion of a Gas into an Evacuated Chamber

142

Section 5.7

Effusion and Diffusion

143 of 169

Graham’s Law of Effusion

Relative rates of effusion of two gases at the same T and P are given by the inverse ratio of the square roots of the masses of the gas particles

M₁ and M₂ - Molar masses of the gases

143

Section 5.7

Effusion and Diffusion

144 of 169

Interactive Example 5.20 - Effusion Rates

Calculate the ratio of the effusion rates of hydrogen gas (H₂) and uranium hexafluoride (UF₆), a gas used in the enrichment process to produce fuel for nuclear reactors

Relative molecular speed distribution of H₂ and UF₆

Section 5.7

Effusion and Diffusion

145 of 169

Interactive Example 5.20 - Solution

First we need to compute the molar masses

Molar mass of H₂ = 2.016 g/mol, and molar mass of UF₆ = 352.02 g/mol

Using Graham’s law, we have:

The effusion rate of the very light H₂ molecules is about 13 times that of the massive UF₆ molecules

Section 5.7

Effusion and Diffusion

146 of 169

Prediction of the Relative Effusion Rates of Gases by the KMT

The kinetic molecular model fits the experimental results for the effusion of gases
Prediction for two gases at the same pressure and temperature

Section 5.7

Effusion and Diffusion

147 of 169

Figure 5.24 - Demonstration of the Relative Diffusion Rates of NH₃ and HCl Molecules

Section 5.7

Effusion and Diffusion

148 of 169

Ideal Gas Behavior

Exhibited by real gases under certain conditions of:

Low pressure
High temperature

148

Section 5.8

Real Gases

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Figure 5.25 - Plots of PV/nRT versus P for Several Gases (200 K)

149

Section 5.8

Real Gases

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Figure 5.26 - Plots of PV/nRT versus P for Nitrogen Gas at Three Temperatures

150

Section 5.8

Real Gases

151 of 169

van der Waals Equation

Actual volume available to a given gas molecule can be calculated as follows:

V - Volume of the container
nb - Correction factor for the volume of the molecules

n - Number of moles of gas
b - Empirical constant

Section 5.8

Real Gases

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van der Waals Equation (Continued 1)

The ideal gas equation can be modified as follows:

When gas particles come close together, attractive forces occur

Cause the particles to hit the walls very slightly and less often

Section 5.8

Real Gases

153 of 169

van der Waals Equation (Continued 2)

Size of the correction factor depends on the concentration of gas molecules

Higher the concentration, the more likely that the particles will attract each other

a - Proportionality constant whose value can be determined by observing the actual behavior of the gas

Section 5.8

Real Gases

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van der Waals Equation (Continued 3)

Values of a are b vary until the best fit for the observed pressure is obtained

Corrected volume

V_ideal

Section 5.8

Real Gases

155 of 169

Table 5.3 - Values of the van der Waals Constants for some Common Gases

Section 5.8

Real Gases

156 of 169

Critical Thinking

You have learned that no gases behave perfectly ideally, but under conditions of high temperature and low pressure (high volume), gases behave more ideally

What if all gases always behaved perfectly ideally?

How would the world be different?

Section 5.8

Real Gases

157 of 169

Behavior of Real Gases - Conclusions

For a real gas, the actual observed pressure is lower than the pressure expected for an ideal gas

Caused due to intermolecular attractions that occur in real gases, which increase in the following order:

H₂ < N₂ < CH₄ < CO₂

157

Section 5.9

Characteristics of Several Real Gases

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The Atmosphere

Surrounds the earth's surface
Contains essential gases such as N₂, O₂, H₂O, and CO₂
Composition is not constant due to gravitational effects

Heavy molecules stay closer to the earth’s surface
Light molecules migrate to higher altitudes

158

Section 5.10

Chemistry in the Atmosphere

159 of 169

Figure 5.30 - The Variation of Temperature with Altitude

159

Section 5.10

Chemistry in the Atmosphere

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Chemistry in the Atmosphere - An Introduction

Higher levels of atmosphere

Chemistry is affected by high-energy radiation and particles from the sun and other sources in space
Ozone - Prevents ultraviolet radiation from reaching the earth

Troposphere - Layer that is closest to the earth’s surface

Chemistry is highly affected by human activities

Section 5.10

Chemistry in the Atmosphere

161 of 169

Air Pollution

Sources

Transportation
Production of electricity

Combustion of petroleum produces CO, CO₂, NO, and NO₂

When the mixture is trapped close to the ground in stagnant air, reactions occur to produce chemicals that harm living systems

161

Section 5.10

Chemistry in the Atmosphere

162 of 169

Figure 5.31 - Concentration for Some Smog Components versus Time of Day

162

Section 5.10

Chemistry in the Atmosphere

163 of 169

Pollution Due to Transportation

Caused by the presence of nitrogen oxides in the air

Leads to the production of photochemical smog

163

Section 5.10

Chemistry in the Atmosphere

164 of 169

Pollution Due to the Production of Electricity

Caused due to the presence of sulfur in coal, which, when burned, produces SO₂

After further oxidation, SO₂ is converted to SO₃ in air

Section 5.10

Chemistry in the Atmosphere

165 of 169

Pollution due to the Production of Electricity (Continued)

SO₃ can combine with water droplets to form sulfuric acid

Sulfuric acid is corrosive to living beings and building materials

Can lead to acid rains

Section 5.10

Chemistry in the Atmosphere

166 of 169

Sulfur Dioxide Pollution

Complicated due to the energy crisis

Lower petroleum supplies would mean a shift to the usage of high-sulfur coal

High-sulfur coal can be used without harming the air quality by removing the SO₂ from the exhaust gas by a system called the scrubber

Involves the decomposition of CaCO₃ to lime and carbon dioxide

Section 5.10

Chemistry in the Atmosphere

167 of 169

Sulfur Dioxide Pollution (Continued)

Lime combines with the sulfur dioxide to form calcium sulfite

An aqueous suspension of lime is injected into the exhaust gases to produce a slurry

Helps remove calcium sulfite and any remaining unreacted SO₂

Section 5.10

Chemistry in the Atmosphere

168 of 169

Figure 5.33 - A Schematic Diagram of a Scrubber

168

Section 5.10

Chemistry in the Atmosphere

169 of 169

Drawbacks of Scrubbing

Complicated and expensive
Consumes huge amounts of energy
Calcium sulfite that is produced is buried in landfills

No use has been found for calcium sulfite

Section 5.10

Chemistry in the Atmosphere