CHAPTER 11Gases

10. 1 Kinetic Molecular Theory State the kinetic-molecular theory of matter, and

describe how it explains certain properties of matter.

List the five assumptions of the kinetic-molecular theory of gases.

Define the terms ideal gas and real gas.

Describe each of the following characteristic properties of gases: expansion, density, fluidity, compressibility, diffusion, and effusion.

Describe the conditions under which a real gas deviates from “ideal” behavior.

What is the Kinetic Molecular Theory?

Break it down: Kinetic: movement Molecular: particles Theory: tested ideas

Tested ideas about the movement of particles!

Describes microscopic properties of molecules and their interactions, leading to observerable macroscopic properties pressure, volume, temperature

KMT of Gases Ideal gas: hypothetical gas that satisfies all 5 ideas

of KMT pressure is not too high temperature is not too low

1. Gases consist of large numbers of tiny particles that are far apart relative to their size.

Most of the volume is empty space

2. Collisions between gas particles and between particles and container walls are elastic collisions.

elastic collision when there is no net loss of total kinetic energy

KMT cont.3. Gas particles are in continuous, rapid, random

motion. 4. There are no forces of attraction between gas

particles.5. The temperature of a gas depends on the

average kinetic energy of the particles of the gas.

The kinetic energy of any moving object is given by the following equation:

KE m 212

Effusion Because gases have motion, they can

travel. Effusion: process by which gas particles

pass through a tiny opening

How does effusion & diffusion relate to KMT? Gases at the same temperature have the

same KE so… Heavier gases travel (effuse/diffuse) slower Lighter gases travel faster

Gas Behavior KMT applies only to ideal gasses.

Which parts are not true for real gases?

Phase Diagramsshows the relationship between the

physical state of a substance and the temperature and pressure of the substance.

Phase Diagram for CO2

Phase Diagram for H2O

• A phase diagram has three lines:• One line shows the liquid-gas equilibrium.• Another line shows the liquid-solid equilibrium• A third line shows the solid-gas equilibrium.

• triple point, the temperature and pressure at which all three states of a substance coexist at equilibrium.

• Used to identify boiling point/freezing point at any pressure.

critical point: temperature above which the gas cannot be made to liquefy, no matter how great the applied pressure.

above the critical point the substance becomes s a supercritical fluid

11.1 Gases and Pressures Define pressure, give units of pressure,

and describe how pressure is measured.

State the standard conditions of temperature and pressure and convert units of pressure.

Use Dalton’s law of partial pressures to calculate partial pressures and total pressures.

Pressure (P) Volume (V)Temperature (T) Mols (n)

What causes pressure?• collisions of the gas molecules with each other and

with surfaces with which they come into contact.

• depends on volume (mL or L), temperature (oF, oC, K), and the number of molecules present (mol, mmol).

4 Variables of Gases

Equation for PressurePressure (P): the force per unit area on a

surface.

Pressure = Force Area

What are two ways you could increase the pressure on an object? More force on a given area, the greater the pressure.

decrease the area the force acts on, the greater the pressure.

Relationship Between Pressure, Force and Area

Measuring Pressure barometer: device used to measure

atmospheric pressure Pressure of

atmosphere supports a column of Hg about 760 mm above surface of mercury in dish

Can change depending on weather & elevation

Measuring Pressure

Units for Measuring Pressure

mm Hg : millimeters of mercuryA pressure of 1 mm Hg is also called 1 torr in honor of

Torricelli for his invention of the barometer. atm : atmosphere of pressure kPa : kiloPascal

Others…psi : pounds per square inchBar torr

1 atm = 101.3 kPa = 760 mmHg = 760 torr

STP STP : Standard Temperature & Pressure

1.0 atm (or any of units of equal value) 0 oC

Used by scientists to compare volumes of gases

Pressure ConversionsThe average atmospheric pressure in Denver, Colorado is 0.830 atm. Express this pressure in:

a. millimeters of mercury (mm Hg) andb. kilopascals (kPa)

Given: atmospheric pressure = 0.830 atm

Unknown: a. pressure in mm Hg b. pressure in kPa

Pressure Conversions Answers

A)

B)

760 mm Hgatm mm Hg; atm mm Hg

atm

101.325 kPaatm kPa; atm kPa

atm

760 mm Hg0.830 atm

a631

tmmm Hg

101.325 kPa0.830 atm

at84.1

m kPa

Dalton’s Law of Partial Pressures

The pressure of each gas in a mixture is called the partial pressure.

John Dalton discovered that the pressure exerted by each gas in a mixture is independent of that exerted by other gases present.

Dalton’s law of partial pressures: the total pressure of a gas mixture is the sum of the partial pressures of each gas.

Dalton’s Law of Partial Pressures

Dalton derived the following equation:

PT = P1 + P2 + P3 + …

Total Pressure = sum of pressures of each individual gas

Dalton’s Law of Partial Pressures

Gases Collected by Water Displacement

Water molecules at the liquid surface evaporate and mix with the gas molecules. Water vapor, like other gases, exerts a pressure known as vapor pressure.

Gases produced in the laboratory are often collected over water. The gas produced by the reaction displaces the water in the reaction bottle.

Particle Model for a Gas Collected Over Water

Gases Collected by Water Displacement (ctd)

Step 1: Raise bottle until water level inside matches the water level outside. (Ptot = Patm)

Step 2: Dalton’s Law of Partial Pressures states:

Patm = Pgas + PH2O

To get Patm, record atmospheric pressure.

Step 3: look up the value of PH2O at the temperature of the experiment in a table, you can then calculate Pgas.

Dalton’s Law of Partial Pressures Sample Problem

KClO3 decomposes and the oxygen gas was collected by water displacement. The barometric pressure and the temperature during the experiment were 731.0 torr and 20.0°C. respectively. What was the partial pressure of the oxygen collected?

Given:PT = Patm = 731.0 torrPH2O = 17.5 torr (vapor pressure of water at 20.0°C, from table A-8 in your book)Patm = PO2 + PH2O

Unknown: PO2 in torr

Dalton’s Law Sample Problem Solution

Solution:Patm = PO2 + PH2O

PO2 = Patm - PH2O

substitute the given values of Patm and into the equation: PO2 =731.0 torr – 17.5 torr = 713.5 torr

Mole Fractions (X)

Mole fraction of a gas(XA) = Moles of gas A (nA)

Total number of moles of a gas(ntot)

mole fraction: ratio of the number of moles of one component of a mixture to the total number of moles

Calculating Partial Pressure

PA = XA PT

Partial pressures can be determined from mole fractions using the following equation:

11.2 The Gas LawsUse the kinetic-molecular theory to explain the relationships between gas volume, temperature and pressure.

Use Boyle’s law to calculate volume-pressure changes at constant temperature.

Use Charles’s law to calculate volume-temperature changes at constant pressure.

Use Gay-Lussac’s law to calculate pressure-temperature changes at constant volume.

Use the combined gas law to calculate volume-temperature-pressure changes.

Boyle’s Law If you increase the pressure on a gas in a

flexible container, what happens to the volume? If you decrease the pressure, what happens the

volume?

Pressure and volume are ________ related.

P1V1 = P2V2

Variables: pressure & volumeConstant: temperature, amount of gas

Boyle’s Law

Boyle’s Law Video

Boyle’s Law ProblemA sample of oxygen gas has a volume of

150.0 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?

P1 = 0.947 atm P2 = 0.987 atmV1 = 150.0 mL V2 = ?

Boyle’s Law Problem Solution

1 12

2

PVVP 2

2(0.947 atm)(150.0 mL O )0.987 at

144 mL Om

Charles’ Law

If you increase the temperature of gas, what will happen to the volume?

If you decrease the temperature of a gas, what will happen to the volume?

Volume and temperature are ______ related.

Variables: volume & temperature Constant: pressure & amount of gas

1 2

1 2

V VT T

Charles’ Law

Charles’ Law Video

Temperature in Charles Law To Convert to Kelvin K = 273 + °C. absolute zero: when all motion stops

O K = -273 oC

Charles’ Law ProblemA sample of neon gas occupies a volume of

752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?

Temperature must be in KELVIN!!!V1 = 752 mL V2 = ?T1 = 25°C T2 = 50°C

Charles’ Law Sample Problem Solution

1 22

1

VTVT

(752 mL Ne)(323 K) 298 K

815 mL Ne1 22

1

VTVT

Gay-Lussac’s Law If you increase the temperature of a gas

what will happen to the pressure? If you decrease the temperature of gas

what will happen to the pressure? Pressure and temperature are _____ related.

Variables: pressure & temperature Constant: volume & amount of gas

1 2

1 2

P PT T

Gay-Lussac’s Law

GL Law Video

Gay-Lussac’s Law ProblemThe gas in a container is at a pressure of

3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?Temperature must also be in KELVIN!!!

P1 = 3.00 atm P2 = ?T1 = 25°C T2 = 52°C

Gay-Lussac’s Law Problem Solution

P2 = P1T2 = (3.00 atm) (325 K) = 3.27 atm T1 298 K

1 22

1

PTPT

The Combined Gas LawConstant: amount of gas

combined gas law: used when pressure, temperature, and volume change within a system

1 1 2 2

1 2

PV PVT T

NOTE: P & V are directly related to T, while P is inversely related to V

Combined Gas Law Problem

A helium-filled balloon has a volume of 50.0 L at 25.0°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C?

Temperature must be in KELVIN!!

P1 = 1.08 atm P2 = 0.855 atmV1 = 50.0 L V2 = ?T1 = 25.0°C T2 = 10.0°C

Combined Gas Law Problem Solution

1 1 22

2 1

PVTVPT

(1.08 atm)(50.0 L He)(283 K) (0.855 atm)(298 K)

60.0 L He1 1 22

2 1

PVTVPT

End of Material for Quiz #1

11.3 Gas Volumes and the Ideal Gas Law

State Avogadro’s law and explain its significance.

Define standard molar volume of a gas and use it to calculate gas masses and volumes.

State the ideal gas law.

Using the ideal gas law, calculate pressure, volume, temperature, or amount of gas when the other three quantities are known.

Avogadro’s Law If you increase the amount of moles, what

happens to the volume? If you decrease the amount of moles what

happens to the volume?

Amount of moles & volume are ____ related.

Variables: volume , moles Constants: pressure, temperature

V1 = V2 n1 n2

Because of Avogadro’s law equal volumes of gases at constant temperature and pressure contain equal numbers of molecules.

Avogadro determined one mole of any gas (regardless of mass differences) will expand to the same volume every time

standard molar volume of a gas:22.41410 L (rounded to 22.4 L)

Molar Volume Video

Deriving the Ideal Gas LawReview: Write down the combined gas law; where

do you think “n” fits in?

If both sides must equal each other, we can set one side equal to a constant. We’ll call this constant “R.”

The Ideal Gas Law Equation

PV = nRT ideal gas law: relates all variables –

pressure, volume, moles, temperature

Deriving the Ideal Gas Law Constant

R: ideal gas constantIts value depends on the units chosen for

pressure, volume, and temperature in the rest of the equation.

What are the standard conditions for an ideal gas?

P = n = V = T =

Plug in values into the equation and calculate. What is the constant that you get?

Usually rounded to 0.0821 (Latm/molK)

Numerical Values of The Gas Constant “R”

ALWAYS MATCH UP YOUR UNITS!!!!

Gas Stoichiometry Avogadro’s law can be applied in calculating the

stoichiometry of reactions involving gases. The coefficients in chemical equations of gas

reactions reflect not only mole ratios, but also volume ratios (assuming conditions remain the same).Discovered by Dalton, while exploring why water was a ratio

of 2H to 1O example2H2(g) + O2(g) → 2H2O(g)

2 molecules 1 molecule 2 molecules2 mole 1 mole 2 mol2 volumes 1 volume 2 volumes

Gas Stoichiometry ProblemNumber 1 on Practice Sheet What volume of nitrogen at STP would

be required to react with 0.100 mol of hydrogen to produce ammonia?

N2 + 3 H2 2 NH3

Gas Stoichiometry Problem Solution

0.100 mol H2 x 1 mol N2 x 22.4 L N2

3 mol H2 1 mol N2

= 0.747 L N2

Ideal Gas Law Sample Problem A sample of carbon dioxide with a mass

of 0.250 g was placed in a 350. mL container at 400 K. What is the pressure exerted by the gas?

P = ?V = 350. mL = 0.350 Ln = 0.250 g = ? molT = 400 K

Ideal Gas Law Problem Solution

nRTPV

P = nRT = .00568 mol (.0821 Latm/molK) 400 K V .350 L

= 0.533 atm

Gas Stoich and Ideal Gas Law

Number 2 on Practice Sheet What volume of nitrogen at 215OC and

715 mmHg would be required to react with 0.100 mol of hydrogen to produce ammonia?

N2 + 3 H2 2 NH3

Note: This system is NOT at STP!!

Gas Stoichiometry Problem Solution

0.100 mol H2 x 1 mol N2 = 0.0333 mol N2

3 mol H2

P = 715 mmHgV = ?n = 0.0333 mol N2

R = 62.4 LmmHg/molKT = 25OC + 273 = 488 K

11.4 Diffusion and Effusion Describe the process of diffusion.

State Graham’s law of effusion.

State the relationship between the average molecular velocities of two gases and their molar masses.

Diffusion and EffusionREMEMBER:

EFFUSION: process when the molecules of a gas confined in a container randomly pass through a tiny opening in the container

DIFFUSION: the gradual mixing of two or more gases due to their spontaneous, random motion

Graham’s Law of Effusion

Graham’s Law Of EffusionGraham’s law of effusion:

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

B

A

MAB M

rate of effusion of rate of effusion of

Sample Problem What is the rate of effusion of hydrogen if oxygen has a velocity of 175 m/s at the same temperature and pressure.

Substitute the given values into the equation:

Hydrogen rate of effusion is …

Graham’s Law of Effusion, continued

32.00 g/molrate of effusion of 32.00 g/mol 3.98rate of effusion of 2.02 g/mol2.02 g/mol

B

A

MAB M

Graham’s Law- Visual Problem