The Gaseous State of Matter Chapter 12

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The Gaseous State of Matter Chapter 12. Hein and Arena . Version 1.1. Chapter Outline. 12.2 The Kinetic MolecularTheory. 12.9 Combined Gas Laws. 12.10 Dalton’s Law of Partial Pressures. 12.3 Measurement of Pressure of Gases. 12.11 Avogadro’s Law.

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### The Gaseous State of MatterChapter 12

Hein and Arena

Version 1.1

Chapter Outline

12.2 The KineticMolecularTheory

12.9Combined Gas Laws

12.10Dalton’s Law of PartialPressures

12.3 Measurement of Pressure ofGases

12.4 Dependence of Pressure on Number of Molecules and Temperature

12.12Mole-Mass-VolumeRelationships of Gases

12.13Density of Gases

12.5 Boyle’s Law

12.14 Ideal Gas Equation

12.6Charles’ Law

12.15 Gas Stoichiometry

12.7Gay Lussac’s Law

12.8Standard Temperature andPressure

12.16 Real Gases

The Kinetic-Molecular Theory
• KMT is based on the motions of gas particles.
• A gas that behaves exactly as outlined by KMT is known as an ideal gas.
• While no ideal gases are found in nature, real gases can approximate ideal gas behavior under certain conditions of temperature and pressure.
Principle Assumptions of the KMT
• Gases consist of tiny subatomic particles.
• The distance between particles is large compared with the size of the particles themselves.
• Gas particles have no attraction for one another.
Principle Assumptions of the KMT
• Gas particles move in straight lines in all directions, colliding frequently with one another and with the walls of the container.
• No energy is lost by the collision of a gas particle with another gas particle or with the walls of the container. All collisions are perfectly elastic.
Principle Assumptions of the KMT
• The average kinetic energy for particles is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin temperature.

mH

= 2

vH

4

2

2

=

mO

= 32

1

vO

2

2

Kinetic Energy
• All gases have the same average kinetic energy at the same temperature.
• As a result lighter molecules move faster than heavier molecules.

Stopcock closed No diffusion occurs

Stopcock open Diffusion occurs

Diffusion

The ability of two or more gases to mix spontaneously until they form a uniform mixture.

Effusion

A process by which gas molecules pass through a very small orifice from a container at higher pressure to one at lower pressure.

Graham’s Law of Effusion

The rates of effusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their densities, or molar masses.

The pressure resulting from the collisions of gas molecules with the walls of the balloon keeps the balloon inflated.
The pressure exerted by a gas depends on the
• Number of gas molecules present.
• Temperature of the gas.
• Volume in which the gas is confined.

Mercury Barometer

The barometer is used to measure atmospheric pressure.

Average Composition of Dry Air

Gas

Volume Percent

Gas

Volume Percent

N2

78.08%

He

0.0005%

O2

20.95%

CH4

0.0002%

Ar

0.93%

Kr

0.0001%

CO2

0.033%

Xe, H2, and N2O

Trace

Ne

0.0018%

• Pressure is produced by gas molecules colliding with the walls of a container.
• At a specific temperature and volume, the number of collisions depends on the number of gas molecules present.
• For an ideal gas the number of collisions is directly proportional to the number of gas molecules present.

The pressure exerted by a gas is directly proportional to the number of molecules present.

V = 22.4 LT = OoC

Dependence of Pressure on Temperature
• The pressure of a gas in a fixed volume increases with increasing temperature.
• When the pressure of a gas increases, its kinetic energy increases.
• The increased kinetic energy of the gas results in more frequent and energetic collisions of the molecules with the walls of the container.

Increased pressure is due to more frequent and more energetic collisions of the gas molecules with the walls of the container at the higher temperature.

Lower T

Lower P

Higher T

Higher P

The pressure of a gas in a fixed volume increases with increasing temperature.

Boyle’s Law

At constant temperature (T), the volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P).

An 8.00 L sample of N2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

Method A. Conversion Factors

Step 1. Determine whether volume is being increased or decreased.

Initial volume = 8.00 L Final volume = 3.00 L

volume decreases  pressure increases

An 8.00 L sample of N2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

Step 2. Multiply the original pressure by a ratio of volumes that will result in an increase in pressure.

new pressure = original pressure x ratio of volumes

P1 = 500 torr

V1 = 8.00 L

P2 = ?

V2 = 3.00 L

An 8.00 L sample of N2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

Method B. Algebraic Equation

Step 1. Organize the given information:

An 8.00 L sample of N2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

Step 2. Write and solve the equation for the unknown.

An 8.00 L sample of N2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

Step 3. Put the given information into the equation and calculate.

Charles’ Law

• If a given volume of any gas at 0oC is cooled by 1oC the volume of the gas decreases by .
• If a given volume of any gas at 0oC is cooled by 20oC the volume of the gas decreases by .
Absolute Zero on the Kelvin Scale

If a given volume of any gas at 0oC is cooled by 273oC the volume of the gas decreases by .

Absolute Zero on the Kelvin Scale
• -273oC (more precisely –273.15oC) is the zero point on the Kelvin scale. It is the temperature at which an ideal gas would have 0 volume.
Charles’ Law

At constant pressure the volume of a fixed mass of gas is directly proportional to the absolute temperature.

Effect of temperature on the volume of a gas. Pressure is constant at 1 atm. When temperature increases at constant pressure, the volume of the gas increases.

A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC?

Method A. Conversion Factors

Step 1. Change oC to K:

oC + 273 = K

75oC + 273 = 348 K

250oC + 273 = 523 K

A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC?

Step 2: Multiply the original volume by a ratio of Kelvin temperatures that will result in an increase in volume:

V1 = 255 mL

T1 = 75oC = 348 K

V2 = ?

T2 = 250oC = 523 K

A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC?

Method B. Algebraic Equation

Step 1. Organize the information (remember to make units the same):

A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC?

Step 2. Write and solve the equation for the unknown:

V1 = 255 mL

T1 = 75oC = 348 K

V2 = ?

T2 = 250oC = 523 K

A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC?

Step 3. Put the given information into the equation and calculate:

Gay-Lussac’s Law

The pressure of a fixed mass of gas, at constant volume, is directly proportional to the Kelvin temperature.

At a temperature of 40oC an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen?

Method A. Conversion Factors

Step 1. Change oC to K:

oC + 273 = K

40oC + 273 = 313 K

100oC + 273 = 373 K

Determine whether temperature is beingincreased or decreased.

temperature increases  pressure increases

At a temperature of 40oC an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen?

Step 2: Multiply the original pressure by a ratio of Kelvin temperatures that will result in an increase in pressure:

At a temperature of 40oC an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen?

A temperature ratio greater than 1 will increase the pressure

P1 = 21.5 atm

T1 = 40oC = 313 K

P2 = ?

T2 = 100oC = 373 K

At a temperature of 40oC an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen?

Method B. Algebraic Equation

Step 1. Organize the information (remember to make units the same):

At a temperature of 40oC an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen?

Step 2. Write and solve the equation for the unknown:

P1 = 21.5 atm

T1 = 40oC = 313 K

P2 = ?

T2 = 100oC = 373 K

At a temperature of 40oC an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen?

Step 3. Put the given information into the equation and calculate:

STP

Standard Conditions

Standard Temperature and Pressure

Standard Temperature and Pressure

Selected common reference points of temperature and pressure.

273.15 K or 0.00oC

1 atm or 760 torr or 760 mm Hg

Combined Gas Laws

• A combination of Boyle’s and Charles’ Law.
• Used when pressure and temperature change at the same time.
• Solve the equation for any one of the 6 variables

ratio of pressures

ratio of temperatures

initial volume

calculate final volume

final volume =

ratio of pressures

ratio of temperatures

initial volume

increases or decreases volume

final volume =

ratio of pressures

ratio of temperatures

initial volume

increases or decreases volume

final volume =

A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

Step 1. Organize the given information, putting temperature in Kelvins:

oC + 273 = K

0oC + 273 = 273 K

-15oC + 273 = 258 K

A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

Step 1. Organize the given information, putting temperature in Kelvins:

P1 = 760 torr P2 = 950 torr

V1 = 465 mL V2 = ?

T1 = 273 K T2 = 258 K

A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

Method A Conversion Factors

Step 2. Set up ratios of T and P

A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

Step 3. Multiply the original volumes by the ratios:

P1 = 760 torr P2 = 950 torr

V1 = 465 mL V2 = ?

T1 = 273 K T2 = 258 K

A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

Method B Algebraic Equation

Step 2. Write and solve the equation for

the unknown V2.

A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

Step 2 Put the given information into the equation and calculate.

Dalton’s Law ofPartial Pressures

Each gas in a mixture exerts a pressure that is independent of the other gases present.

The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture.

Ptotal = Pa + Pb + Pc + Pd + ….

A container contains He at a pressure of 0.50 atm, Ne at a pressure of 0.60 atm, and Ar at a pressure of 1.30 atm. What is the total pressure in the container?

Ptotal = PHe + PNe+ PAr

Ptotal = 0.5 atm + 0.6 atm + 1.30 atm = 2.40 atm

Collecting a Gas Sample Over Water
• The pressure in the collection container is equal to the atmospheric pressure.
• The pressure of the gas collected plus the pressure of water vapor at the collection temperature is equal to the atmospheric pressure.

A sample of O2 was collected in a bottle over water at a temperature of 25oC when the atmospheric pressure was 760 torr. The vapor pressure of water at 25oC is 23.8 torr.

N2

+

3 H2

2 NH3

1 volume

2 volumes

3 volumes

+

Gay Lussac’s Law of Combining Volumes

When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

Gay Lussac’s Law of Combining Volumes

When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

There are 2 molecules of hydrogen chloride.

hydrogen + chlorine  hydrogen chloride

1 volume 1 volume 2 volumes

1 molecule 1 molecule 2 molecules

1 mol 1 mol 2 mol

Each molecule of hydrogen chloride contains at least 1 atom of hydrogen and 1 atom of chlorine.

hydrogen + chlorine → hydrogen chloride

1 volume 1 volume 2 volumes

1 molecule 1 molecule 2 molecules

1 mol 1 mol 2 mol

Each molecule of hydrogen and each molecule of chlorine must contain at least 2 atoms.

H2 + Cl2→ 2 HCl

Mole-Mass-Volume Relationships

• Volume of one mole of any gas at STP = 22.4 L.
• 22.4 L at STP is known as the molar volume of any gas.
Density of Gases

grams

liters

Density of Gases

depends on T and P

1 mole of any gas occupies 22.4 L at STP

nT

V a

P

Ideal Gas Equation

liters

nT

V a

P

moles

nT

V a

P

Kelvin

nT

V a

P

nT

Ideal Gas Constant

V a

P

A balloon filled with 5.00 moles of helium gas is at a temperature of 25oC. The atmospheric pressure is 750. torr. What is the balloon’s volume?

Step 1. Organize the given information.

Convert temperature to kelvins.

K = oC + 273

K = 25oC + 273 = 298K

Convert pressure to atmospheres.

A balloon filled with 5.00 moles of helium gas is at a temperature of 25oC. The atmospheric pressure is 750. torr. What is the balloon’s volume?

Step 2. Write and solve the ideal gas equation for the unknown.

Step 3. Substitute the given data into the equation and calculate.

Calculate the molar mass of an unknown gas, if 0.020 g occupies 250 mL at a temperature of 305 K and a pressure of 0.045 atm.

V = 250 mL = 0.250 L

g = 0.020 g

P = 0.045 atm

T = 305 K

Gas Stoichiometry

• All calculations are done at STP.
• Gases are assumed to behave as ideal gases.
• A gas not at STP is converted to STP.
Definition

Stoichiometry: The area of chemistry that deals with the quantitative relationships among reactants and products in a chemical reaction.

Gas Stoichiometry

Primary conversions involved in stoichiometry.

Mole-Volume Calculations

Mass-Volume Calculations

What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?
• Step 1 Write the balanced equation

2 KClO3 2 KCl + 3 O2

• Step 2 The starting amount is 0.500 mol KClO3. The conversion is

moles KClO3 moles O2 liters O2

What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?

2 KClO3 2KCl + 3 O2

• Step 3. Calculate the moles of O2, using the mole-ratio method.
• Step 4. Convert moles of O2 to liters of O2
What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?

The problem can also be solved in one continuous calculation.

2 KClO3 2KCl + 3 O2

What volume of hydrogen, collected at 30.0oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g)

Step 1 Calculate moles of H2.

grams Al  moles Al  moles H2

What volume of hydrogen, collected at 30.0oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

2 Al(s) + 6 HCl(aq)  2AlCl3(aq) + 3 H2(g)

Step 2 Calculate liters of H2.

• Convert oC to K: 30.oC + 273 = 303 K
• Convert torr to atm:
What volume of hydrogen, collected at 30.0oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?
• Solve the ideal gas equation for V

PV = nRT

For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as mole-mole relationships.

2 x 22.4 L

22.4 L

22.4 L

STP

STP

STP

H2(g) + Cl2(g) 2HCl(g)

2 mol HCl

1 mol H2

1 mol Cl2

1 volume

1 volume

2 volumes

Y volume

Y volume

2Y volumes

What volume of nitrogen will react with 600. mL of hydrogen to form ammonia? What volume of ammonia will be formed?

N2(g) + 3H2(g) 2NH3(g)

Real Gases

Ideal Gas
• An ideal gas obeys the gas laws.
• The volume the molecules of an ideal gas occupy is negligible compared to the volume of the gas. This is true at all temperatures and pressures.
• The intermolecular attractions between the molecules of an ideal gas are negligible at all temperatures and pressures.
Real Gases
• Deviations from the gas laws occur at high pressures and low temperatures.
• At high pressures the volumes of the real gas molecules are not negligible compared to the volume of the gas
• At low temperatures the kinetic energy of the gas molecules cannot completely overcome the intermolecular attractive forces between the molecules.
Key Concepts

12.2 The KineticMolecularTheory

12.9 Combined Gas Laws

12.10 Dalton’s Law of PartialPressures

12.3 Measurement of Pressure ofGases

12.4 Dependence of Pressure on Number of Molecules and Temperature

12.12 Mole-Mass-VolumeRelationships of Gases

12.13 Density of Gases

12.5 Boyle’s Law

12.14 Ideal Gas Equation

12.6 Charles’ Law

12.15 Gas Stoichiometry

12.7 Gay Lussac’s Law

12.8 Standard Temperature andPressure

12.16 Real Gases