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Introductory Chemistry , 3 rd Edition Nivaldo Tro. Chapter 11 Gases. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2009, Prentice Hall. Properties of Gases. Expand to completely fill their container. Take the shape of their container. Low density.

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Introductory chemistry 3 rd edition nivaldo tro

Introductory Chemistry, 3rd EditionNivaldo Tro

Chapter 11

Gases

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA

2009, Prentice Hall


Properties of gases
Properties of Gases

  • Expand to completely fill their container.

  • Take the shape of their container.

  • Low density.

    • Much less than solid or liquid state.

  • Compressible.

  • Mixtures of gases are always homogeneous.

  • Fluid.

Tro's Introductory Chemistry, Chapter 11


The structure of a gas
The Structure of a Gas

  • 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.

Tro's Introductory Chemistry, Chapter 11


Kinetic molecular theory
Kinetic Molecular Theory

  • The particles of the gas (either atoms or molecules) are constantly moving.

  • The attraction between particles is negligible.

  • 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.

Tro's Introductory Chemistry, Chapter 11


Kinetic molecular theory of gases
Kinetic Molecular Theory of Gases

  • There is a lot of empty space between the particles in a gas.

    • Compared to the size of the particles.

  • 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!!

Tro's Introductory Chemistry, Chapter 11


Kinetic molecular theory1
Kinetic Molecular Theory

Tro's Introductory Chemistry, Chapter 11


Gas particles pushing
Gas Particles Pushing

  • 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.

Tro's Introductory Chemistry, Chapter 11


The effect of gas pressure
The Effect of Gas Pressure

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

  • Whenever there is a pressure difference, a gas will flow from area of high pressure to low pressure.

    • The bigger the difference in pressure, the stronger the flow of the gas.

  • If there is something in the gas’ path, the gas will try to push it along as the gas flows.

Tro's Introductory Chemistry, Chapter 11


Which way would air flow
Which Way Would Air Flow?

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?

Tro's Introductory Chemistry, Chapter 11

9


Is this possible at a depth of 20 m
Is This Possible at a Depth of 20 m?

Tro's Introductory Chemistry, Chapter 11

10


Soda straws and gas pressure
Soda Straws and Gas Pressure

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.

Tro's Introductory Chemistry, Chapter 11


Gas properties explained
Gas Properties Explained

  • Gases have taken the shape and volume of their container(s) because the particles don’t stick together, allowing them to move and fill the container(s) they’re in.

    • In solids and liquids, the particles are attracted to each other strongly enough so they stick together.

  • Gases are compressible and have low density because of the large amount of unoccupied space between the particles.

Tro's Introductory Chemistry, Chapter 11


Properties indefinite shape and indefinite volume
Properties—Indefinite Shape and Indefinite Volume

Because the gas

molecules have

enough kinetic

energy to overcome

attractions, they

keep moving around

and spreading out

until they fill the

container.

As a result, gases

take the shape and

the volume of the

container they

are in.

Tro's Introductory Chemistry, Chapter 11


Properties compressibility
Properties—Compressibility

Because there is a lot of unoccupied space in the structure

of a gas, the gas molecules can be squeezed closer together.

Tro's Introductory Chemistry, Chapter 11


Gas properties explained low density
Gas Properties Explained—Low Density

Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume, the result is that they have low density.

Tro's Introductory Chemistry, Chapter 11


The pressure of a gas
The Pressure of a Gas

  • Pressure is the result of the constant movement of the gas molecules and their collisions with the surfaces around them.

  • The pressure of a gas depends on several factors:

    • Number of gas particles in a given volume.

    • Volume of the container.

    • Average speed of the gas particles.

Tro's Introductory Chemistry, Chapter 11


Density and pressure
Density and Pressure

  • Pressure is the result of the constant movement of the gas molecules and their collisions with the surfaces around them.

  • When more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure.

    • Also higher density.

Tro's Introductory Chemistry, Chapter 11


Air pressure
Air Pressure

  • The atmosphere exerts a pressure on everything it contacts.

    • On average 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.

Tro's Introductory Chemistry, Chapter 11


Measuring air pressure
Measuring Air Pressure

  • 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

Tro's Introductory Chemistry, Chapter 11


Atmospheric pressure and altitude
Atmospheric Pressure and Altitude

  • 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 10,000 ft is is only 10.0 psi.

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

Tro's Introductory Chemistry, Chapter 11


Pressure imbalance in ear
Pressure Imbalance in Ear

If there is a difference

in pressure across

the eardrum membrane,

the membrane will be

pushed out—what we

commonly call a

“popped eardrum.”

Tro's Introductory Chemistry, Chapter 11


Common units of pressure
Common Units of Pressure

Tro's Introductory Chemistry, Chapter 11


Example 11 1 a high performance bicycle tire has a pressure of 125 psi what is the pressure in mmhg

psi

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.


Practice convert 45 5 psi into kpa
Practice—Convert 45.5 psi into kPa.

Tro's Introductory Chemistry, Chapter 11


Practice convert 45 5 psi into kpa continued

psi

atm

kPa

Practice—Convert 45.5 psi into kPa, Continued

Given:

Find:

45.5 psi

kPa

Concept Plan:

Relationships:

1 atm = 14.7 psi, 1 atm = 101.325 kPa

Solution:

Check:

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


Boyle s law
Boyle’s Law

  • 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.

Tro's Introductory Chemistry, Chapter 11


Boyle s experiment
Boyle’s Experiment

  • Added Hg to a J-tube with air trapped inside.

  • Used length of air column as a measure of volume.

Tro's Introductory Chemistry, Chapter 11




Boyle s experiment p x v
Boyle’s Experiment, P x V

Tro's Introductory Chemistry, Chapter 11


When you double the pressure on a gas,

the volume is cut in half (as long as the

temperature and amount of gas do not change).

Tro's Introductory Chemistry, Chapter 11


Gas laws explained boyle s law
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.

Tro's Introductory Chemistry, Chapter 11


Boyle s law and diving
Boyle’s Law and Diving

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.

Tro's Introductory Chemistry, Chapter 11


Boyle s law and diving continued
Boyle’s Law and Diving, Continued

  • 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!!

Tro's Introductory Chemistry, Chapter 11


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)

Tro's Introductory Chemistry, Chapter 11


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.


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

100°C

373 K

212°F

671 R

BP Water

0°C

273 K

32°F

459 R

MP Ice

-38.9°C

234.1 K

-38°F

421 R

BP Mercury

-183°C

90 K

-297°F

162 R

BP Oxygen

BP Helium

-269°C

4 K

-452°F

7 R

-273°C

0 K

-459 °F

0 R

Absolute

Zero

Celsius

Kelvin

Fahrenheit

Rankine


Gas laws and temperature
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)

Tro's Introductory Chemistry, Chapter 11


Standard conditions
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.

Tro's Introductory Chemistry, Chapter 11


Volume and temperature
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.

Tro's Introductory Chemistry, Chapter 11


Volume and temperature continued
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.

Tro's Introductory Chemistry, Chapter 11


Charles s law
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.

Tro's Introductory Chemistry, Chapter 11


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
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 = 0 R.

  • Never attainable.

    • Though we’ve gotten real close!

  • All gas law problems use the Kelvin temperature scale.

Tro's Introductory Chemistry, Chapter 11


Determining absolute zero
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.

Tro's Introductory Chemistry, Chapter 11


Example 11 3 a gas has a volume of 2 57 l at 0 c what was the temperature at 2 80 l

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.


Practice—The Temperature Inside a Balloon Is Raised from 25.0 °C to 250.0 °C. If the Volume of Cold Air Was 10.0 L, What Is the Volume of Hot Air?

Tro's Introductory Chemistry, Chapter 11


T(K) = 25.0 °C to 250.0 °C. If the Volume of Cold Air Was 10.0 L, What Is the Volume of Hot Air? t(°C) + 273.15,

V1, T1, T2

V2

Practice—The Temperature Inside a Balloon Is Raised from 25.0 °C to 250.0 °C. If the Volume of Cold Air Was 10.0 L, What Is the Volume of Hot Air?, Continued

Given:

Find:

V1 =10.0 L, t1 = 25.0 °C L, t2 = 250.0 °C

V2, L

Solution Map:

Relationships:

Solution:

Check:

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


The combined gas law
The Combined Gas Law 25.0 °C to 250.0 °C. If the Volume of Cold Air Was 10.0 L, What Is the Volume of Hot Air?

  • 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 25.0 °C to 250.0 °C. If the Volume of Cold Air Was 10.0 L, What Is the Volume of Hot Air? (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?

Tro's Introductory Chemistry, Chapter 11


T atm and Temperature Is 27 °C. What Volume Will the Gas Occupy Under Standard Conditions?(K) = t(°C) + 273,

V1, P1,P2, T1, T2

V2

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?, Continued

Given:

Find:

V1 = 10.0 L, t1 = 27 °C L, P1 = 3.00 atm

t2 = 0 °C, P2 = 1.00 atm

V2, L

Solution Map:

Relationships:

Solution:

Check:

When T decreases, V should decrease; when P decreases, V should increase—opposite trends make it hard to evaluate our answer.


Avogadro s law
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.

Tro's Introductory Chemistry, Chapter 11


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

Tro's Introductory Chemistry, Chapter 11


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

n2, and added moles

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

Tro's Introductory Chemistry, Chapter 11


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
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.

Tro's Introductory Chemistry, Chapter 11


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.


Practice calculate the volume occupied by 637 g of so 2 mm 64 07 at 6 08 x 10 3 mmhg and 23 c
Practice—Calculate the Volume Occupied by 637 g of SO What Volume Would 0.750 Moles Occupy?2 (MM 64.07) at 6.08 x 103 mmHg and –23 °C.

Tro's Introductory Chemistry, Chapter 11

107


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

g

V

n

Practice—Calculate the Volume Occupied by 637 g of SO2 (MM 64.07) at 6.08 x 103 mmHg and –23 °C, Continued.

Given:

Find:

mSO2 = 637 g, P = 6.08 x 103 mmHg, t = −23 °C,

V, L

Solution Map:

Relationships:

1 atm = 760 mmHg

T(K) = t(°C) + 273, 1 mol SO2 = 64.07 g

Solution:


Practice calculate the density of a gas at 775 torr and 27 c if 0 250 moles weighs 9 988 g
Practice—Calculate the Density of a Gas at 775 torr and 27 °C if 0.250 moles Weighs 9.988 g.

Tro's Introductory Chemistry, Chapter 11

109


V, m °C if 0.250 moles Weighs 9.988 g.

P, n, T, R

d

V

1 atm = 760 mmHg,

T(K) = t(°C) + 273

Practice—Calculate the Density of a Gas at 775 torr and 27 °C if 0.250 moles Weighs 9.988 g, Continued

Given:

Find:

m = 9.988g, n = 0.250 mol, P = 1.0197 atm, T = 300. K

density,g/L

m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C,

density,g/L

Solution Map:

Relationships:

Solution:

Check:

The value 1.65 g/L is reasonable.


Molar mass of a gas
Molar Mass of a Gas °C if 0.250 moles Weighs 9.988 g.

  • 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.

Tro's Introductory Chemistry, Chapter 11


n, m °C if 0.250 moles Weighs 9.988 g.

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

Tro's Introductory Chemistry, Chapter 11

121


Practice what is the molar mass of a gas if 12 0 g occupies 197 l at 380 torr and 127 c continued
Practice—What Is the Molar Mass of a Gas if 12.0 g Occupies 197 L at 380 torr and 127 °C?, Continued

n, m

P, V, T, R

MM

n

1 atm = 760 torr,

T(K) = t(°C) + 273

Given:

Find:

m = 12.0g, V = 197 L, P = 0.50 atm, T =400 K,

molar mass,g/mol

m=12.0 g, V= 197 L, P=380 torr, t=127°C,

molar mass,g/mol

Solution Map:

Relationships:

Solution:

Check:

The value 31.9 g/mol is reasonable.


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

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

  • 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 +...

Tro's Introductory Chemistry, Chapter 11


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.


Finding partial pressure
Finding 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.

  • To find the partial pressure of a gas, multiply the total pressure of the mixture by the fractional composition of the gas.

  • For example, in a gas mixture that is 80.0% He and 20.0% Ne that has a total pressure of 1.0 atm, the partial pressure of He would be:

    PHe = (0.800)(1.0 atm) = 0.80 atm

    • Fractional composition = percentage divided by 100.

Tro's Introductory Chemistry, Chapter 11


The Partial Pressure of Each Gas in a Mixture, or the Total Pressure of a Mixture, Can Be Calculated Using the Ideal Gas Law

127


Practice—Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 atm, Volume 8.7 L, Temperature 598 K, and 0.17 moles Xe.

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128


Practice—Find the Partial Pressure of Neon in a Mixture of Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

nXe, V, T, R

PXe

Ptot, PXe

PNe

Given:

Find:

Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol

PNe, atm

Solution Map:

Relationships:

Solution:

The unit is correct, the value is reasonable.

Check:


Mountain climbing and partial pressure
Mountain Climbing and Partial Pressure Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

  • 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.

Tro's Introductory Chemistry, Chapter 11


Deep sea divers and partial pressure
Deep Sea Divers and Partial Pressure Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

  • 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
Partial Pressure vs. Total Pressure Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

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,

are quadrupled!

Tro's Introductory Chemistry, Chapter 11


Collecting gases
Collecting Gases Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

  • Gases are often collected by having them displace water from a container.

  • The problem is that since water evaporates, there is also water vapor in the collected gas.

  • The partial pressure of the water vapor, called the vapor pressure, depends only on the temperature.

    • So you can use a table to find out the partial pressure of the water vapor in the gas you collect.

  • If you collect a gas sample with a total pressure of 758 mmHg at 25 °C, the partial pressure of the water vapor will be 23.8 mmHg, so the partial pressure of the dry gas will be 734 mmHg.

Tro's Introductory Chemistry, Chapter 11


Vapor pressure of water
Vapor Pressure of Water Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

Tro's Introductory Chemistry, Chapter 11


If the temperature of the water is 30 Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued., the vapor pressure of the water is 31.8 mmHg.

Zn metal reacts

with HCl(aq) to

produce H2(g).

The gas flows

through the tube

and bubbles into

the jar, where it

displaces the

water in the jar.

Because water

evaporates, some

water vapor gets

mixed in with

the H2.

If the total pressure is 760 mmHg, the partial pressure of the H2 is 760 − 31.8 mmHg = 728 mmHg.

Tro's Introductory Chemistry, Chapter 11


Practice—0.12 moles of H Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.2 Is Collected Over Water in a 10.0 L Container at 323 K. Find the Total Pressure(Vapor Pressure of Water at 50 C = 92.6 mmHg).

Tro's Introductory Chemistry, Chapter 11

136


P Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.H2, PH2O

Ptotal

nH2,V,T

PH2

1 atm = 760 mmHg

Ptotal = PA + PB,

Practice—0.12 moles of H2 Is Collected Over Water in a 10.0 L Container at 323 K. Find the Total Pressure(Vapor Pressure of Water at 50 C = 92.6 mmHg), Continued.

Given:

Find:

V = 10.0 L, nH2 = 0.12 mol, T = 323 K

Ptotal,atm

Solution Map:

Relationships:

Solution:


Reactions involving gases
Reactions Involving Gases Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

  • 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.

    • Instead of moles.

    • 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

Tro's Introductory Chemistry, Chapter 11


Example 11.11—How Many Liters of O Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.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 Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.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)

Tro's Introductory Chemistry, Chapter 11

149


g HgO Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

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:


Calculate the volume occupied by 1 00 mole of an ideal gas at stp

L∙atm Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

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

Tro's Introductory Chemistry, Chapter 11


Molar volume
Molar Volume Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

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).

Tro's Introductory Chemistry, Chapter 11


L H Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.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:

Tro's Introductory Chemistry, Chapter 11

153


Practice—What Volume of O Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.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)

Tro's Introductory Chemistry, Chapter 11

161


g HgO Ne and Xe with Total Pressure 3.9 Atm, Volume 8.7 L, Temperature 598 K, and 0.17 Moles Xe, Continued.

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:

Tro's Introductory Chemistry, Chapter 11

162


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