Gas Laws. Gas Pressure Pressure is defined as force per unit area Gas pressure is defined as collisions with the walls of a container Internal collisions.

Gas Laws

Gas Pressure

• Pressure is defined as force per unit area

• Gas pressure is defined as collisions with the walls of a container

• Internal collisions between molecules don't count

• As molecules move faster more collisions, hence greater pressure

• Measured using a barometer

• Pressure can also be changed by changing the area struck by the particles

Dalton’s Law of Partial Pressure

When Dalton studied the properties of gases, he found that each gas in a

mixture exerts pressure independently of the other gases present

Dalton’s Law of Partial Pressures

• States that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture

Dalton’s Law of Partial Pressures

• the pressure contributed by a single gas is called its partial pressure

– Partial pressure depends on the # of moles of gas, size of the container, and temperature of the mixture

– At a given temperature and pressure, the partial pressure of 1 mole of any gas is the same

Dalton’s Law of Partial Pressures

Ptotal = P1 + P2 + P3 + …Pn

• Ptotal represents the total pressure of the mixture of gases, P1, P2, P3…represents the partial pressures

Example

• What is the partial pressure of hydrogen gas in a mixture of hydrogen and helium if the total pressure is 600 mmHg and the partial pressure of helium is 439 mmHg?

Ptotal = P1 + P2 + P3 + …Pn

600 mmHg = 439 mmHg + PHydrogen

PHydrogen = 161 mmHg

Practice Problems

Practice 1

Find the total pressure for a mixture that contains four gases with partial pressures of 5.00 kPa, 4.56 kPa, 3.02 kPa, and 1.20

kPa.

13.78 kPa

Practice 2

Find the partial pressure of carbon dioxide in a gas mixture with a total pressure of 30.4

kPa if the partial pressures of the other two gases in the mixture are 16.5 kPa, and

3.7 kPa.

10.2 kPa

Boyle’s Law

Robert Boyle

• Performed experiments to study the relationship between pressure and the volume of a gas

• Displayed that under constant temperature, doubling of the pressure would result in the volume being cut in half and vice versa

– Discovered that pressure and volume are inversely proportional

Boyle’s Law

• States that volume of a given amount of gas held at constant temperature varies inversely with the pressure

P1V1 = P2V2

– P1 & V1 represent the initial set of conditions for the gas and P2 & V2 represent the new set of conditions for the gas

Boyle’s Law

• If you know any 3 of the 4 values for a gas at constant temperature, you can solve for the 4th by rearranging the equation

• Example

P2 = P1V1

V2

which can be done for any of the values in the equation

Example

The volume of a gas at 99.0 kPa is 300.0 mL. If the pressure is increased to 188 kPa, what will be the

new volume?

P1V1 = P2V2

V2 = P1V1 = (99 kPa)(300 mL)P2 188 kPa

158 mL

Practice Problems

Practice 1

• The pressure of a sample of helium in a 1.00-L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00-L container?

0.494 atm

Practice 2

• Air trapped in a cylinder fitted with a piston occupies 145.7 mL at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston?

110mL or 0.110 L

Homework

Dalton’s Law of Partial Pressures and Boyle’s

Law Handout

Review Questions

Problem 10

An ion with 8 protons, 8 neutrons, and 10 electrons is represented by

– Ne

– O+2

– O-2

– O

Problem 11

• What is the mass number of an atom which contains 28 protons, 28 electrons, and 34 neutrons?

– 28

– 56

– 62

– 90

Problem 12

The chart shows isotopes of some common elements. In what two

properties do the isotopes of carbon differ?

– Atomic number and number of neutrons

– Atomic mass and number of

neutrons

– Atomic mass and number of

protons

– Number of protons and number

of neutrons

Day 2

Charles’s Law

Jacques Charles

• French physicist that studied the relationship between volume and temperature

• Observed that volume rose as temperature did when the pressure was held constant

Charles’s Law and KMT

• As stated in the KMT, at higher temperatures, gas particles move faster, striking each other and the walls of their container more frequently and with greater force

• If pressure is constant, volume must increase so the particles have farther to travel before striking the walls, which decrease the frequency at which the particles strike the walls of the container

Charles’s Law• States that the volume of a given mass of gas is directly

proportional to its Kelvin temperature at constant pressure

V1 = V2

T1 T2

– V1 and T1 represent initial conditions and V2 and T2 represent the final conditions of the gas

• TEMPERATURE MUST BE IN KELVINS IN ORDER TO USE CHARLES’S LAW

TK = 273 + TC

TK is the temperature in Kelvin and TC is the Celsius temperature

Isolating a Variable

Charles’s LawV1 = V2

T1 T2

• Cross MultiplyV1T2 = V2T1

• Isolate the variable by dividing on both sides

V1T2 = V2T1

T2 T2

V1 = V2T1

T2

ExampleA gas at 89˚C occupies a volume of 0.67L. At what Celsius

temperature will the volume increase to 1.12L?

TK= 273 + 89 = 362K

V1 = V2

T1 T2

T2 = T1V2

V1

T2 = (362 )(1.12) = 605 K .67

TC = 605 – 273 = 332 ˚C

Practice Problems

Practice 1

The Celsius temperature of a 3.00-L sample of gas is lowered from 80˚C to 30˚C. What

will be the resulting volume of this gas?

2.58L

Practice 2

What is the volume of the air in a balloon that occupies 0.620L at 25˚C if the

temperature is lowered to 0˚C?

0.56L

Gay Lussac’s Law

Joseph Gay-Lussac

• Studied the relationship between temperature and pressure of a contained gas at a fixed volume

• Discovered that temperature in Kelvin and pressure are directly proportional

Gay-Lussac’s Law

• States that the pressure of a given gas varies directly with the Kelvin temperature when the volume remains constant

P1 = P2

T1 T2

• P1 and T1 are the initial conditions and P2 and T2 are the final conditions of the gas

ExampleA gas in a sealed container has a pressure of 125kPa at a temperature of

30°C. If the pressure in the container is increased to 201 kPa, what is the new temperature?

TK = 273 + 30 = 303K

P1 = P2

T1 T2

T2 = T1P2

P1

T2 = (303)(201) = 487K125

T2= 487-273 = 214°C

Practice Problems

Practice 1

The pressure in an automobile tire is 1.88atm at 25°C. What will be the

pressure if the temperature warms up to 37°C?

1.96atm

Practice 2

If a gas sample has a pressure of 30.7 kPa at 0°C, what would the temperature have to be to lower the pressure to 28.4 kPa?

-21°C

Homework

• Charles’s Law

• Gay-Lussac’s Law

Review Questions

Problem 13

What is the chemical formula for a compound formed from calcium ions (Ca2+) and chloride

ions (Cl–1)?

– a. CaCl

– b. Ca2Cl

– c. CaCl2

– d. Ca2Cl2

Problem 14

What is the correct formula for aluminum oxide?

– A AlO

– B Al3O2

– C AlO2

– D Al2O3

Day 3

Combined Gas Law

Combined Gas Law

• Combination of Boyle’s, Charles’s, & Gay-Lussac’s Laws

• States the relationship among pressure, volume and temperature of a fixed amount of gas

– Pressure is inversely proportional to volume

– Pressure is directly proportional to temperature and temperature is directly proportional to volume

Combined Gas Law

P1V1 = P2V2

T1 T2

PTV NOT MTV

Isolating a Variable

COMBINED GAS LAWP1V1 = P2V2

T1 T2

• Cross MultiplyP1V1T2 = P2V2T1

• Isolate the variable by dividing on both sides

P1V1T2 = P2V2T1

V1T2 V1T2

P1 = P2V2T1

V1T2

Example

A helium-filled balloon at sea level has a volume of 2.1L at 0.998 atm and 36°C. If it is released and rises to an elevation at which the pressure is 0.900 atm and the

temperature is 28°C, what will be the new volume of the balloon?

2.27L

Practice Problems

Practice Problem 1

At 0°C and 1 atm pressure, a sample of gas occupies 30mL. If the temperature is increased to 30°C and the entire gas

sample is transferred to a 20mL container, what will be the gas pressure inside the

container?

1.66atm

Practice Problem 2

A sample of gas of unknown pressure occupies 0.766L at a temperature of 298K.

The same sample of gas is then tested under known conditions and has a pressure of 32.6 kPa and occupies 0.644L at 303K. What was the original pressure of the gas?

26.96 kPa

Ideal Gas Law

Ideal Gas Law

• Describes the physical behavior of an ideal gas in terms of the pressure, volume, temperature and number of moles of gas present

PV = nRT

-where n is the number of moles of gas and R is the ideal gas constant

Ideal Gas Constant

Pg. 435 Table 14-1

Example

If the pressure exerted by a gas at 25°C in a volume of 0.044L is 3.81 atm, how many moles of gas are

present?

PV = nRT

n = PV RT

n = (3.81atm)(0.044L) (.0821)(298K)

n = .0069 or 6.9 x 10-3 mol

Practice Problems

Practice Problem 1

Determine the Celsius temperature of 2.49 moles of gas contained in a 1.00L vessel

at a pressure of 143 kPa.

-266°C

Practice Problem 2

Calculate the volume that a 0.323 mol sample of a gas will occupy at 265K and a

pressure of 0.900 atm.

7.81L

Homework

Gas Laws Handout

Review Questions

Problem 15

What is the Stock system name for FeO?

– A Iron oxide

– B Iron (II) oxide

– C Iron (III) oxide

– D Iron oxygen

Problem 16

What is the correct name for the compound NH4Cl?

– A nitrogen chloride

– B nitrogen chlorate

– C ammonium chloride

– D ammonium chlorate

Problem 17

What is the formula for nitrogen trifluoride?

– A NiF3

– B NF3

– C N3F

– D Ni3F

Day 4

Test Content• Kinetic Molecular Theory (KMT)

– 5 postulates

• Gas Laws

– Boyles’, Dalton’s, Charles’s, Gay-Lussac’s, Combined, Ideal

• Molar Volume

• Volume Stoichiometry

– Liters Liters

– Grams Liters

Molar Volume

Avogadro’s Principle

States equal volumes of gases at the same temperature and pressure

contain equal numbers of particles

Molar Volume

• Volume that one mole of gas occupies at STP

– Represented by the constant 22.414 L/mol

• the volume of one mole of gas at STP.

– Standard Temperature and Pressure (STP)

• atmospheric pressure of 1 and 0°C or 273K

The Molar Volume Conversion Factor

• Molar Volume Conversion Factor

– 22.414L/mol

• Can be used to calculate the volume that a gas occupies at STP conditions given any of the following:

– moles of a sample of gas

– grams of a sample of gas

– atoms, molecules, or particles of a sample of gas

Calculations

Once you have the number of MOLES of a sample of gas, you can use the formula below to calculate the molar volume of a

gas at STP

Molar volume (V) = number of moles of gas x 22.414L

1 mol

Example- Given Moles

• 0.250 mole of HCl gas will occupy how many liters at STP?

0.250 moles 22.414L

1 mol

= 5.604 liters of HCl

Given Mass

• Example: What volume is occupied by 100 grams of chlorine gas (Cl2) at STP?

100 g 1 mol Cl2 22.414L 70.906 g 1 mol

= 31.611 liters Cl2

Given Volume

• If given the volume of a gas at STP, you can calculate the moles of gas by dividing the molar volume by 22.414L/mol.

• Example: How many moles are equivalent to 2L of dry H2 at STP?

2 L H2 1 mol H2

22.414L

= .089 moles of H2

Practice

What volume is occupied by 8 moles of any gas at STP?

8 moles 22.414L

1 mol

= 179.312 liters

What volume is occupied by 3.49 moles of gas at STP?

3.49 moles 22.414L

1 mol

= 78.224 liters

What volume is occupied by 27.48 moles of nitrogen gas at STP?

27.48 moles 22.414L

1 mol

= 615. 937 liters nitrogen gas

How many liters will 94.5 grams of O2 occupy at STP?

94.5 grams O2 1 mol 22.414L

32 g 1 mol

= 66.12 liters O2

How many liters will 38.57 grams of CO2 occupy at STP?

38.57 grams CO2 1 mole 22.414L

44 g 1 mol

= 19. 657 liters CO2

BONUS

What is the mass of 5 liters of O2 gas?

5 liters of O2 1 mol 32 grams 22.414 L 1 mol

= 7.136 grams O2

Volume Stoichiometry

Volume Stoichiometry

• Liters Grams

Given 1 mol molar molar mass 22.414L ratio 1 mol

• Grams Liters

Given 1 mol molar 22.414L molar mass ratio 1 mol

• Liters Liters

Given 1 mole molar 22.414L22.414L ratio 1 mol

12 CO2 (g) + 11 H2O (l) C12H22O11 (s) + 12 O2 (g)

• a.) How many liters of sucrose (C12H22O11) are produced from 25.0 grams of carbon dioxide (CO2) at STP?

• b.) How many liters of carbon dioxide (CO2) are necessary to produce 2.0L of oxygen (O2)? (4 pts)

Consider the following equation: P4 (s) + 5O2 (g) ---> P4O10 (g)

• a.) If 3.54 liters of phosphorus (P4 ) is ignited in a flask containing oxygen (O2) at STP, how many grams of O2 are needed?

• b.) How many liters of tetraphosphorus decaoxide (P4O10) are produced from 3.0 L of oxygen (O2)?

Test Cheat Sheet

• You are allowed to fill 1 3x5 index card front and back with as much as you can to use on the test tomorrow

• Suggestions

– Volume stoichiometry problems in notes

– Example of each gas law

– PTV

Review Questions

Problem 18

A physical change occurs when

– A a peach spoils

– B a bracelet turns your wrist green

– C a copper bowl tarnishes

– D a glue gun melts a glue stick

Problem 19

What substance has a melting point of -94°C and a boiling point of 65°C?

– a. Ethanol

– b. Chlorine

– c. Hexane

– d. Methanol

Problem 20

22.4 liters of a gas has a mass of 36.5 grams. What is the identity of the gas?

– a. Chlorine

– b. Hydrogen chloride

– c. Nitrogen

– d. Hydrogen