Chapter 10 Gases Teacher Note The solutions to many of the calculations are worked out in a packet...

Chapter 10

Gases

Barometers and Standard Atmospheric Pressure


• Standard atmospheric pressure defined as the pressure sufficient to support a mercury column of 760mm (units of mmHg, or torr).



• Another unit was introduced to simplify things, the atmosphere (1 atm = 760 mmHg).



• Another unit was introduced to simplify things, the atmosphere (1 atm = 760 mmHg).

• 1 atm = 760 mmHg = 760 torr = 101.325 kPa (page 262).

STP standard temperature and pressure

Standard temperature 0°C or 273 K

Standard pressure 1 atm (or equivalent)

– Pressure varies inversely with volume

– Volume varies inversely with pressure

“The volume of a sample of gas is inversely proportional to its pressure, if temperature

remains constant.”

Boyle’s Law

Boyle’s Law: Pressure & Volume (Figure 10.6 (a) page 263)

P V

P V

Boyle’s Law: Pressure-Volume Relationships

A sample of air occupies 73.3 mL at 98.7 atm and 0 ºC. What volume will the air occupy at 4.02 atm and 0 ºC?

1800 mL

Boyle’s Law: Pressure-Volume RelationshipsA sample of helium occupies 535 mL at 988 mmHg and 25 °C. If the sample is transferred to a 1.05-L flask at 25 °C, what will be the gas pressure in the flask?

503 mm Hg

• Effects of temperature on a gas

• Volume varies directly with Temperature

“The volume of a quantity of gas, held at constant pressure, varies

directly with the Kelvin temperature.”

Charles’s Law

a

Charles Law: Volume and Temperature (Figure 10.8 Page 266)

Charles’ Law and Absolute Zero

•Extrapolation to zero volume gives a temperature of

-273°C or 0 K

A sample of oxygen gas occupies a volume of 2.10 L at 25 °C. What volume will this sample occupy at 150 °C? (Assume no change in pressure.)

Charles’s Law: Temperature-Volume Relationships

2.98 L

A sample of oxygen gas occupies a volume of 2.10 L at 25 °C. At what Celsius temperature will the volume of oxygen occupy 0.750 L? (Assume no change in pressure.)

Charles’s Law: Temperature-Volume Relationships

-167°C

Pressure vs. Temperature

• Pressure varies directly with Temperature

• If the temperature of a fixed volume of gas doubles its pressure doubles.


• The pressure exerted by a gas is directly related to the Kelvin temperature.

• V is constant.


Example

A gas has a pressure of 645 torr at 128°C. What is the

temperature in Celsius if the pressure increases to 1.50 atm?

Pi = 645 torr Pf = 1.50 atm 760 torr = 1140 torr

1 atm

Ti = 128°C + 273

= 401 K Tf = ?K

Solution

T2 = 401 K x 1140 torr = 709K645 torr

709K - 273 = 436°C

Combined Gas Law Problem

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

Combined Gas Law Problem

A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?

x 3.20 atm x 90.0 mL 0.800 atm 180.0 mL

604 K - 273 = 331 °C

302 K = 604 K

Combined Gas Law• A 10.0 cm3 volume of gas measured 75.6 kPa

and 60.0C is to be corrected to correspond to the volume it would occupy at STP.

6.12 cm3

Gay-Lussac’s Law of combining volumes: at a given temperature and pressure, the volumes of gases which react are ratios of small whole numbers.

Gay-Lussac’s Law

How many liters of steam can be formed from 8.60L of oxygen gas?

17.2 L

How many liters of hydrogen gas will react with 1L of nitrogen gas to

form ammonia gas?

3L H2

A

How many mL of hydrogen are needed to How many mL of hydrogen are needed to produce 13.98 mL of ammonia?produce 13.98 mL of ammonia?

A

20.97 ml NH3

Avogadro’s Law: Equal volumes of gases at the same temperature and pressure contain the same number of particles.

The molar volume of a gas at STP = 22.4L

22.4 L

Ideal Gas• An ideal gas is defined as one for which both the

volume of molecules and forces of attraction between the molecules are so small that they have no effect on the behavior of the gas.

Ideal Gas Equation

PV=nRT

R values

a

I•R values for atm and kPa on Page 272 in book.

A•

•

Calculate the volume occupied by 0.845 mol of nitrogen gas at a pressure of 1.37 atm and a

temperature of 315 K.

15.9 L

Find the pressure in millimeters of mercury of a 0.154 g sample of helium gas at 32°C and contained

in a 648 mL container.

1130 mm Hg

An experiment shows that a 113 mL gas sample has a mass of 0.171 g at a pressure of

721 mm Hg and a temperature of 32°C. What is the molar mass (molecular weight) of the gas?

40.0 g/mol

Can the ideal “gas” equation be used to determine the molar mass of a liquid?

Homework

• Do the lab summary for “The Molecular Mass of a Volatile Liquid”. It is due ____.

• Attempt the pre-lab for “The Molecular Mass of a Volatile Liquid”. It is due ____.

Problem: A volatile liquid is placed in a flask whose volume is 590.0 mland allowed to boil until all of the liquid is gone, and only vapor fills the flask at a temperature of 100.0 oC and 736 mm Hg pressure. If the mass of the flask before and after the experiment was 148.375g and 149.457 g,what is the molar mass of the liquid?

57.9 g/mol

What is the density of methane gas (natural What is the density of methane gas (natural gas), CHgas), CH44, at 125, at 125ooC and 3.50 atm?C and 3.50 atm?

1.71 g/L

Calculate the density in g/L of O2 gas at STP.

1.43 g/L

Dalton’s Law of Partial Pressure • The total pressure in a container is the sum of the partial

pressures of all the gases in the container.• In a gaseous mixture, a gas’s partial pressure is the one

the gas would exert if it were by itself in the container.

• Ptotal = P1 + P2 + P3

• Ptotal = 100 KPa + 250 KPa + 200 KPa = 550 KPa

A B

Total = 6.0 atm

P V Vmixture P

A 2.0 atm 1.0 L 2.0 atm

B 4.0 atm 1.0 L 4.0 atm

Two 1.0 L containers, A and B, contain gases with 2.0 atm and 4.0 atm, respectively. Both gases are forced into Container B. Find the total pressure of the gas mixture in B.

1.0 L

Dalton’s Law Problem• Air contains oxygen, nitrogen, carbon dioxide,

and trace amounts of other gases. What is the partial pressure of oxygen at standard conditions if the partial pressure of nitrogen, carbon dioxide, and other gases are 79.1 KPa, 0.04 KPa, and 0.94 KPa respectively?

• Ptotal = PO2 + PN2 + PCO2 + POther gases

• 101.3 KPa = PO2 + 79.1 KPa + 0.04 KPa + 0.94KPa

• PO2 = 101.3 KPa – (79.1 KPa + 0.04 KPa + 0.94KPa)

• PO2 = 21.2 KPa

A B Z

Two 1.0 L containers, A and B, contain gases with 2.0 atm and 4.0 atm, respectively. Both gases are forced into Container Z (vol. 2.0 L). Find the total pressure of mixture in Z.

A B Z

Total = 3.0 atm

Two 1.0 L containers, A and B, contain gases with 2.0 atm and 4.0 atm, respectively. Both gases are forced into Container Z (vol. 2.0 L). Find the total pressure of mixture in Z.

PX VX VZ PX,Z

A 2.0 atm 1.0 L2.0 L

1.0 atm

B 4.0 atm 1.0 L 2.0 atm

A B ZC

Find total pressure of the gas mixture in Container Z.

1.3 L 2.6 L 3.8 L 2.3 L3.2 atm 1.4 atm 2.7 atm X atm

A B ZC

Total = 7.9 atm

Find total pressure of the gas mixture in Container Z.

1.3 L 2.6 L 3.8 L 2.3 L3.2 atm 1.4 atm 2.7 atm X atm

PX VX VZ PX,Z

A 3.2 atm 1.3 L

2.3 L

1.8 atm

B 1.4 atm 2.6 L 1.6 atm

C 2.7 atm 3.8 L 4.5 atm

Dalton’s Law

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 422

Dalton’s Partial Pressures

Zumdahl, Zumdahl, DeCoste, World of Chemistry 2002, page 421

Dalton’s Law of Partial Pressures

The mole ratio in a mixture of gases determines each gas’s partial pressure.

Total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4 kPa.

Find partial pressure of each gas

Dalton’s Law of Partial Pressures

kPa 41.7 kPa 97.4 gas mol 7He mol 3

P He

kPa 55.7 kPa 97.4 gas mol 7Ne mol 4

P Ne

Total pressure of mixture (3.0 mol He and 4.0 mol Ne) is 97.4 kPa.

Find partial pressure of each gas.

80.0 g each of He, Ne, and Ar are in a container. The total pressure is 780 mm Hg. Find each gas’s partial pressure.

80.0 g each of He, Ne, and Ar are in a container. The total pressure is 780 mm Hg. Find each gas’s partial pressure.

He mol 20 g 4

mol 1 He g 80

Ne mol 4 g 20

mol 1 Ne g 80

Armol 2 g 40

mol 1 Arg 80

Hg mm 60 P Hg, mm 120 P Hg, mm 600 P ArNeHe

Total:26 mol gas

PHe = 20/26

of total

PNe = 4/26

of total

PAr = 2/26

of total

Example: A student generates oxygen gas and collects it over water. If the volume of the gas is 245 mL and the barometric pressure is 758.0 torr at 25oC, what is the volume of the “dry” oxygen gas at STP? (Pwater = 23.8 torr at 25oC)

PO2 = PT - Pwater = 758.0 torr - 23.8 torr = 734.2 torr

Find the molar mass of an unknown gas if a 0.16 g sample of the gas is collected over water and equalized to a pressure of 781.7 torr and a volume of 90.0 mL at a temperature of 28°C .

Find the molar mass of an unknown gas if a 0.16 g sample of the gas is collected over water and equalized to a pressure of 781.7 torr and a volume of 90.0 mL at a temperature of 28°C .

44 g/mol

Homework

• Do the AP sample problem (1999 Test question #5) in notebook. It will be included as part of your homework.

• Don’t forget the pre-lab and lab summary for “The Molecular Mass of a Volatile Liquid”.

Gas Diffusion and Effusion

Graham's Law: governs the rate of effusion and diffusion of gas molecules.

“Stink” or “Die”

a

The Root Mean Square Speed

Fig. 10.17 Page 285

NETNET MOVEMENT

To use Graham’s Law, both gases must be at same temperature.

diffusiondiffusion: particle movement from

high to low concentration

effusioneffusion: diffusion of gas particles

through an opening

For gases, rates of diffusion & effusion obey Graham’s law: more massive = slow; less massive = fast

Gas Diffusion and Effusion

Graham's Law: governs the rate of effusion and diffusion of gas molecules.

Rate of diffusion/effusion is inversely proportional to its molar mass.

Rate of diffusion/effusion is inversely proportional to its molar mass.

A of massmolar B of massmolar

B of RateA of Rate

Determine the relative rate of diffusion for krypton and bromine.

1.381

Kr diffuses 1.381 times faster than Br2.

Kr

Br

Br

Kr

m

m

v

v2

2

A

B

B

A

m

m

v

v

g/mol83.80

g/mol159.80

Graham’s Law

The lightest gas is “Gas A” and the heavier gas is “Gas B”.Relative rate means find the ratio “vA/vB”.

Kr83.80

36

Br79.904

35

A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions?

A

B

B

A

m

m

v

v

2

2

2

2

H

O

O

H

m

m

v

v

g/mol 2.02

g/mol32.00

m/s 12.3

vH 2

Graham’s Law

3.980m/s 12.3

vH 2

m/s49.0 vH 2

O15.9994

8

H1.00794

1

An unknown gas diffuses 4.0 times faster than O2. Find its molar mass.

Am

g/mol32.00 16

A

B

B

A

m

m

v

v

A

O

O

A

m

m

v

v2

2

Am

g/mol32.00 4.0

16

g/mol32.00 mA

2

Am

g/mol32.00 4.0

g/mol2.0

Graham’s Law

The lightest gas is “Gas A” and the heavier gas is “Gas B”. The ratio “vA/vB” is 4.0.

Square both sides to get rid of the square

root sign.

O15.9994

8

H22.0

1

Theory developed to explain gas behavior.

• Theory of moving molecules.

• Assumptions:

– Gases consist of a large number of molecules in constant random motion.

– Volume of individual molecules negligible compared to volume of container.

– Intermolecular forces (forces between gas molecules) negligible.

– Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature.

– Average kinetic energy of molecules is proportional to temperature.

Kinetic Molecular TheoryKinetic Molecular Theory

Kinetic molecular theory gives us an understanding of pressure and temperature on the molecular level.

• Pressure of a gas results from the number of collisions per unit time on the walls of container.

• Magnitude of pressure given by how often and how hard the molecules strike.

• Gas molecules have an average kinetic energy.

• Each molecule has a different energy.


There is a spread of individual energies of gas molecules in any sample of gas.


As the temperature increases, the average kinetic energy of the gas molecules increases

As kinetic energy increases, the velocity of the gas molecules increases.

• Root mean square speed, u, is the speed of a gas molecules having the certain average kinetic energy.

• Average kinetic energy, , is related to root mean square speed, u:

221mu


a

As kinetic energy increases, the velocity of the gas molecules increases.

• Root mean square speed, u, is the speed of a gas molecules having the certain average kinetic energy.

• Average kinetic energy, , is related to root mean square speed, u:

221mu


a

As the volume of a container of gas increases at constant temperature, the gas molecules have to travel further to hit the walls of the container. There are fewer collisions by the gas molecules with the walls of the container. Therefore, pressure decreases.

If temperature increases at constant volume, the average kinetic energy of the gas molecules increases. Therefore, there are more collisions with the container walls and the pressure increases.

How does this theory explain Boyles Law?

If temperature increases at constant volume, the average kinetic energy of the gas molecules increases and they speed up. Therefore, there are more frequent and more forceful collisions with the container walls by the gas molecules and the pressure increases.

How does this theory explain Charles Law?

Ideal Gases vs. Real Gases Ideal Gases vs. Real Gases

• An ideal gas is an “imaginary gas” made up of particles with negligible particle volume and negligible attractive forces.

Ideal Gases vs. Real Gases Ideal Gases vs. Real Gases

• In a “Real Gas” the molecules of a gas do have volume and the molecules do attract each other.

• Therefore anything that makes gas particles more likely to stick together or stay close to one another make them behave less ideally.

• As the volume becomes smaller, the molecules get closer together, and a greater fraction of the occupied space is actually taken up by gas molecules.

• Therefore, the higher the pressure, the less the gas resembles an ideal gas.

Real Gases: Deviations from Ideal Behavior

As the pressure on a gas increases, the molecules are forced into a smaller volume.

• The smaller the distance between gas molecules, the more likely attractive forces will develop between the molecules.

• As temperature increases, the gas molecules move faster and are further apart.

• Also, higher temperatures mean more energy available to break intermolecular forces.

• Therefore, the higher the temperature, the more ideal the gas.

Real Gases: Deviations from Ideal Behavior

• A real gas typically exhibits behavior closest to “ideal gas” behavior at low pressures and high temperatures.

Real Gases and Ideal Behavior

We add two terms to the ideal gas equation one to correct for volume of molecules and the other to correct for intermolecular attractions

The correction terms generate the van der Waals equation:

where a and b are empirical constants.

Real Gases: The van der Waals equation

2

2

V

an

nbV

nRTP

nRTnbVV

anP

2

2

a corrects for the effect of molecular attractions (van der Waals forces), and b corrects for the molecular volume

We add two terms to the ideal gas equation one to correct for volume of molecules and the other to correct for intermolecular attractions

The correction terms generate the van der Waals equation:

• You will not be required to solve this equation but you should You will not be required to solve this equation but you should know its form and which variables need to be corrected.know its form and which variables need to be corrected.

Real Gases: The van der Waals equation

nRTnbVV

anP

2

2

a corrects for the effect of molecular attractions (van der Waals forces), and b corrects for the molecular volume

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Chapter 10 Gases Teacher Note The solutions to many of the calculations are worked out in a packet...

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