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# Chapter 11 Gases - PowerPoint PPT Presentation

Chapter 11 Gases. 2009, Prentice Hall. Properties of Gases. Expand to completely fill their container = Compressible Take the shape of their container. Low density. Much less than solid or liquid state. Mixtures of gases are always homogeneous. The Structure of a Gas.

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### Chapter 11Gases

2009, Prentice Hall

• Expand to completely fill their container = Compressible

• Take the shape of their container.

• Low density.

• Much less than solid or liquid state.

• Mixtures of gases are always homogeneous

• Gases are composed of particles that are flying around very fast in their container(s).

• They move in straight lines until they encounter either the container wall or another particle, then they bounce off.

• If you were able to take a snapshot of the particles in a gas, you would find that there is a lot of empty space in there.

• It’s why balloons look round

• The particles of the gas (either atoms or molecules) are constantly moving & The attraction between particles is negligible.

• Therefore: When the moving particles hit another particle or the container, they do not stick, but they bounce off and continue moving in another direction.

• Like billiard balls.

• The average kinetic energy of the particles is directly proportional to the Kelvin temperature.

• As you raise the temperature of the gas, the average speed of the particles increases.

• But don’t be fooled into thinking all the particles are moving at the same speed!!

• Gas molecules are constantly in motion.

• As they move and strike a surface, they push on that surface.

• Push = force.

• If we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting.

• Pressure = force per unit area= pounds per square inch

• The pressure exerted by a gas can cause some amazing and startling effects.

Two filled balloons are connected with a long pipe. One of the balloons is plunged down into the water. Which way will the air flow? Will air flow from the lower balloon toward the top balloon; or will it flow from the top balloon to the bottom one?

9

The pressure of the air inside the straw is the same as the pressure

of the air outside

the straw—so

liquid levels are

the same on both

sides.

The pressure of the

air inside the straw is lower than the pressure

of the air outside

the straw—so

liquid is pushed

up the straw by

the outside air.

• The atmosphere exerts a pressure

• At sea level it is 14.7 psi.

• The atmosphere goes up about 370 miles, but 80% is in the first 10 miles from Earth’s surface.

• This is the same pressure that a column of water would exert if it were about 10.3 m high.

• Therefore water can not be pumped higher, in a straw for instance, than 10.3 m/33.8 ft

• Use a barometer.

• Column of mercury supported by air pressure.

• Force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury.

gravity

• The higher up in the atmosphere you go, the lower the atmospheric pressure is around you.

• At the surface, the atmospheric pressure is 14.7 psi, but at 29,028 ft it is only 4.9 psi (1/3)

• Rapid changes in atmospheric pressure may cause your ears to “pop” due to an imbalance in pressure on either side of your ear drum.

If there is a difference

in pressure across

the eardrum membrane,

the membrane will be

pushed out—what we

commonly call a

“popped eardrum.”

atm

mmHg

Example 11.1—A High-Performance Bicycle Tire Has a Pressure of 125 psi. What Is the Pressure in mmHg?

Given:

Find:

125 psi

mmHg

Solution Map:

Relationships:

1 atm = 14.7 psi, 1 atm = 760 mmHg

Solution:

Check:

Since mmHg are smaller than psi, the answer makes sense.

• Pressure of a gas is inversely proportional to its volume.

• Constant T and amount of gas.

• Graph P vs. V is curved.

• Graph P vs. 1/V is in a straight line.

• As P increases, V decreases by the same factor.

• P x V = constant.

• P1 x V1 = P2 x V2.

the volume is cut in half (as long as the

temperature and amount of gas do not change).

Gas Laws Explained— Boyle’s Law

• Boyle’s law says that the volume of a gas is inversely proportional to the pressure.

• Decreasing the volume forces the molecules into a smaller space.

• More molecules will collide with the container at any one instant, increasing the pressure.

Scuba tanks have a regulator so that the air from the tank is delivered at the same pressure as the water surrounding you.

This allows you to take in air even when the outside pressure is large.

• Since water is more dense than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm.

• At 20 m the total pressure is 3 atm.

• If your tank contained air at 1 atm of pressure, you would not be able to inhale it into your lungs.

• You can only generate enough force to overcome about 1.06 atm.

• If a diver holds her breath and rises to the surface quickly, the outside pressure drops to 1 atm.

• According to Boyle’s law, what should happen to the volume of air in the lungs?

• Since the pressure is decreasing by a factor of 3, the volume will expand by a factor of 3, causing damage to internal organs. Always Exhale When Rising!!

V1, P1, P2

V2

Example 11.2—A Cylinder with a Movable Piston Has a Volume of 6.0 L at 4.0 atm. What Is the Volume at 1.0 atm?

Given:

Find:

V1 =6.0 L, P1 = 4.0 atm, P2 = 1.0 atm

V2, L

Solution Map:

Relationships:

P1∙ V1= P2∙ V2

Solution:

Check:

Since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does.

Practice—A Balloon Is Put in a Bell Jar and the Pressure Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?(1 atm = 760 torr)

V Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?2, P1, P2

V1

Practice—A Balloon Is Put in a Bell Jar and the Pressure Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?, Continued

Given:

Find:

V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm

V1, mL

Solution Map:

Relationships:

P1∙ V1= P2∙ V2 , 1 atm = 760 torr (exactly)

Solution:

Check:

Since P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does.

Gas Laws and Temperature Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

• Gases expand when heated and contract when cooled, so there is a relationship between volume and temperature.

• Gas molecules move faster when heated, causing them to strike surfaces with more force, so there is a relationship between pressure and temperature.

• In order for the relationships to be proportional, the temperature must be measured on an absolute scale.

• When doing gas problems, always convert your temperatures to kelvins.

K = °C + 273 & °C = K - 273

°F = 1.8 °C + 32 & °C = 0.556(°F-32)

Standard Conditions Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

• Common reference points for comparing.

• Standard pressure = 1.00 atm.

• Standard temperature = 0 °C.

• 273 K.

• STP.

Volume and Temperature Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

• In a rigid container, raising the temperature increases the pressure.

• For a cylinder with a piston, the pressure outside and inside stay the same.

• To keep the pressure from rising, the piston moves out increasing the volume of the cylinder.

• As volume increases, pressure decreases.

Volume and Temperature, Continued Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

As a gas is heated, it expands.

This causes the density of the

gas to decrease.

Because the hot air in the

balloon is less dense than the

surrounding air, it rises.

Charles’s Law Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

• Volume is directly proportional to temperature.

• Constant P and amount of gas.

• Graph of V vs. T is a straight line.

• As T increases, V also increases.

• Kelvin T = Celsius T + 273.

• V = constant x T.

• If T is measured in kelvin.

We’re losing altitude. Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

Quick, Professor, give your

lecture on Charles’s law!

Absolute Zero Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

• Theoretical temperature at which a gas would have zero volume and no pressure.

• Kelvin calculated by extrapolation.

• 0 K = -273.15 °C = -459 °F

• Never attainable.

• Though we’ve gotten real close!

• All gas law problems use the Kelvin temperature scale.

Determining Absolute Zero Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

William Thomson,

the Lord of Kelvin,

extrapolated the

line graphs of

volume vs. temp-

erature to determine

the theoretical

temperature that

a gas would have

given a volume of 0.

T Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?(K) = t(°C) + 273,

V1, V2, T2

T1

Example 11.3—A Gas Has a Volume of 2.57 L at 0 °C. What Was the Temperature at 2.80 L?

Given:

Find:

V1 =2.80 L, V2 = 2.57 L, t2 = 0°C

t1, K and °C

Solution Map:

Relationships:

Solution:

Check:

Since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does.

The Combined Gas Law Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?

• Boyle’s law shows the relationship between pressure and volume.

• At constant temperature.

• Charles’s law shows the relationship between volume and absolute temperature.

• At constant pressure.

• The two laws can be combined together to give a law that predicts what happens to the volume of a sample of gas when both the pressure and temperature change.

• As long as the amount of gas stays constant.

T Is Reduced from 782 torr to 0.500 atm. If the Volume of the Balloon Is Now 2780 mL, What Was It Originally?(K) = t(°C) + 273,

P1,V1, V2, T1, T2

P2

Example 11.4—A Sample of Gas Has an Initial Volume of 158 mL at a Pressure of 735 mmHg and a Temperature of 34 °C. If the Gas Is Compressed to 108 mL and Heated to 85 °C, What Is the Final Pressure?

Given:

Find:

V1 = 158 mL, t1 = 34 °C L, P1 = 735 mmHg

V2 = 108 mL, t2 = 85 °C

P2, mmHg

Solution Map:

Relationships:

Solution:

Check:

Since T increases and V decreases we expect the pressure should increase, and it does.

Practice—A Gas Occupies 10.0 L When Its Pressure Is 3.00 atm and Temperature Is 27 °C. What Volume Will the Gas Occupy Under Standard Conditions?

Avogadro’s Law atm and Temperature Is 27 °C. What Volume Will the Gas Occupy Under Standard Conditions?

• Volume is directly proportional to the number of gas molecules.

• V = constant xn.

• Constant P and T.

• More gas molecules = larger volume.

• Count number of gas molecules by moles, n.

• Equal volumes of gases contain equal numbers of molecules.

• The gas doesn’t matter.

Avogadro’s Law, Continued atm and Temperature Is 27 °C. What Volume Will the Gas Occupy Under Standard Conditions?

mol added = atm and Temperature Is 27 °C. What Volume Will the Gas Occupy Under Standard Conditions?n2– n1,

V1, V2, n1

n2

Example 11.5—A 0.22 Mol Sample of He Has a Volume of 4.8 L. How Many Moles Must Be Added to Give 6.4 L?

Given:

Find:

V1 =4.8 L, V2 = 6.4 L, n1 = 0.22 mol

Solution Map:

Relationships:

Solution:

Check:

Since n and V are directly proportional, when the volume increases, the moles should increase, and it does.

Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy?

V What Volume Would 0.750 Moles Occupy?1, n1, n2

V2

Practice—If 1.00 Mole of a Gas Occupies 22.4 L at STP, What Volume Would 0.750 Moles Occupy?, Continued

Given:

Find:

V1 =22.4 L, n1 = 1.00 mol, n2 = 0.750 mol

V2

Solution Map:

Relationships:

Solution:

Check:

Since n and V are directly proportional, when the moles decreases, the volume should decrease, and it does.

Ideal Gas Law What Volume Would 0.750 Moles Occupy?

• By combining the gas laws, we can write a general equation.

• R is called the Gas Constant.

• The value of R depends on the units of P and V.

• We will use 0.0821 and convert P to atm and V to L.

• Use the ideal gas law when you have a gas at one condition, use the combined gas law when you have a gas whose condition is changing.

P, V, T, R What Volume Would 0.750 Moles Occupy?

n

Example 11.7—How Many Moles of Gas Are in a Basketball with Total Pressure 24.2 Psi, Volume of 3.2 L at 25 °C?

Given:

Find:

V = 3.2 L, P = 24.2 psi, t = 25 °C,

n, mol

Solution Map:

Relationships:

1 atm = 14.7 psi

T(K) = t(°C) + 273

Solution:

Check:

1 mole at STP occupies 22.4 L at STP; since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas.

Molar Mass of a Gas What Volume Would 0.750 Moles Occupy?

• One of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law.

n, m What Volume Would 0.750 Moles Occupy?

P, V, T, R

MM

n

1 atm = 760 mmHg,

T(K) = t(°C) + 273

Example 11.8—Calculate the Molar Mass of a Gas with Mass 0.311 g that Has a Volume of 0.225 L at 55 °C and 886 mmHg.

Given:

Find:

m = 0.311g, V = 0.225 L, P = 1.1658 atm, T = 328 K

Molar Mass,g/mol

m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,

Molar Mass,g/mol

Solution Map:

Relationships:

Solution:

Check:

The value 31.9 g/mol is reasonable.

Practice—What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C?

121

Mixtures of Gases Occupies 197 L at 380 torr and 127 °C?

• According to the kinetic molecular theory, the particles in a gas behave independently.

• Air is a mixture, yet we can treat it as a single gas.

• Also, we can think of each gas in the mixture as independent of the other gases.

• All gases in the mixture have the same volume and temperature.

• All gases completely occupy the container, so all gases in the mixture have the volume of the container.

Partial Pressure Occupies 197 L at 380 torr and 127 °C?

• Each gas in the mixture exerts a pressure independent of the other gases in the mixture.

• The pressure of a component gas in a mixture is called a partial pressure.

• The sum of the partial pressures of all the gases in a mixture equals the total pressure.

• Dalton’s law of partial pressures.

• Ptotal = Pgas A + Pgas B + Pgas C +...

Example 11.9—A Mixture of He, Ne, and Ar Has a Total Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

Ptot, PHe, PNe

PAr

Given:

Find:

PHe= 341 mmHg, PNe= 112 mmHg, Ptot = 558 mmHg

PAr, mmHg

Solution Map:

Relationships:

PAr = Ptot – (PHe + PNe)

Ptot= Pa + Pb + etc.

Solution:

Check:

The units are correct, the value is reasonable.

Mountain Climbing and Partial Pressure Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

• Our bodies are adapted to breathe O2 at a partial pressure of 0.21 atm.

• Sherpa, people native to the Himalaya mountains, have adapted to the much lower partial pressure of oxygen in their air.

• Partial pressures of O2—lower than 0.1 atm—leads to hypoxia.

• Unconsciousness or death.

• Climbers of Mt. Everest must carry O2 in cylinders to prevent hypoxia.

• On top of Mt. Everest:

Pair = 0.311 atm, so PO2 = 0.065 atm.

Deep Sea Divers and Partial Pressure Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

• It is also possible to have too much O2, a condition called oxygen toxicity.

• PO2 > 1.4 atm.

• Oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions.

• It is also possible to have too much N2, a condition called nitrogen narcosis.

• Also known as rapture of the deep.

• When diving deep, the pressure of the air that divers breathe increases, so the partial pressure of the oxygen increases.

• At a depth of 55 m, the partial pressure of O2 is 1.4 atm.

• Divers that go below 50 m use a mixture of He and O2 called heliox that contains a lower percentage of O2 than air.

Partial Pressure vs. Total Pressure Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

At a depth of 30 m, the total pressure of air in the divers

lungs, and the partial pressure of all the gases in the air,

Reactions Involving Gases Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

• The principles of reaction involving stoichiometry from Chapter 8 can be combined with the gas laws for reactions involving gases.

• In reactions of gases, the amount of a gas is often given as a volume.

• As we’ve seen, you must state pressure and temperature.

• The ideal gas law allows us to convert from the volume of the gas to moles; then, we can use the coefficients in the equation as a mole ratio.

P, V, T of Gas A

mole A

mole B

P, V, T of Gas B

Example 11.11—How Many Liters of O Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.2 Are Made from 294 g of KClO3 at 755 mmHg and 305 K?2 KClO3(s) → 2 KCl(s) + 3 O2(g)

g KClO3

mol KClO3

mol O2

1 atm = 760 mmHg, KClO3 = 122.5 g/mol

2 mol KClO3 : 3 mol O2

P, n, T, R

V

Given:

Find:

nO2 = 3.60 mol, P = 0.99342 atm, T = 305 K

VO2,L

mKClO3 = 294 g, P=755 mmHg, T=305 K

VO2,L

Solution Map:

Relationships:

Solution:

Practice—What Volume of O Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.2 at 0.750 atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO?2 HgO(s)  2 Hg(l) + O2(g)(MMHgO = 216.59 g/mol)

149

g HgO Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

mol HgO

mol O2

1 atm = 760 mmHg, HgO = 216.59 g/mol

2 mol HgO : 1 mol O2

P, n, T, R

V

Practice—What Volume of O2 at 0.750 atm and 313 K is Generated by the Thermolysis of 10.0 g of HgO?2 HgO(s)  2 Hg(l) + O2(g), Continued

Given:

Find:

mHgO = 10.0g, P=0.750 atm, T=313 K

VO2,L

nO2 = 0.023085 mol, P = 0.750 atm, T = 313 K

VO2,L

Solution Map:

Relationships:

Solution:

L∙atm Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

mol∙K

(1.00 atm) x V = (1.00 moles)(0.0821 )(273 K)

Calculate the Volume Occupied by 1.00 Mole of an Ideal Gas at STP.

P x V = n x R x T

• 1 mole of any gas at STP will occupy 22.4 L.

• This volume is called the molar volume and can be used as a conversion factor.

• As long as you work at STP.

1 mol  22.4 L

V = 22.4 L

Molar Volume Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

There is so much

empty space

between molecules

in the gas state that

the volume of the

gas is not effected

by the size of the

molecules (under

ideal conditions).

L H Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.2

mol H2

mol H2O

g H2O

Example 11.12—How Many Grams of H2O Form When 1.24 L H2 Reacts Completely with O2 at STP?O2(g) + 2 H2(g) → 2 H2O(g)

Given:

Find:

VH2 = 1.24 L, P = 1.00 atm, T = 273 K

massH2O,g

Solution Map:

Relationships:

H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP

2 mol H2O : 2 mol H2

Solution:

153

Practice—What Volume of O Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.2 at STP is Generated by the Thermolysis of 10.0 g of HgO?2 HgO(s)  2 Hg(l) + O2(g)(MMHgO = 216.59 g/mol)

161

g HgO Pressure of 558 MmHg. If the Partial Pressure of He Is 341 MmHg and Ne Is 112 MmHg, Determine the Partial Pressure of Ar in the Mixture.

mol HgO

mol O2

L O2

Practice—What Volume of O2 at STP is Generated by the Thermolysis of 10.0 g of HgO?2 HgO(s)  2 Hg(l) + O2(g), Continued

Given:

Find:

mHgO = 10.0 g, P = 1.00 atm, T = 273 K

VO2,L

Solution Map:

Relationships:

HgO = 216.59 g/mol, 1 mol = 22.4 L at STP

2 mol HgO : 1 mol O2

Solution:

162