Unit 10: Gases

Unit 10: GasesUnit 10: Gases

Chapter 14

Chemistry 1L

Cypress Creek High School

Table of ContentsTable of Contents• Chapter 14: Gases

– 14.4: Gas Stoichiometry– 14.1: KMT– 14.2: The Combined Gas Laws– 14.3: The Ideal Gas Law

The MoleThe Mole

14.414.4 Gas StoichiometryGas Stoichiometry

Avogadro’s Principle Avogadro’s Principle • Don’t forget that Mole Ratios indicate the molar

relationship between two chemicals in an equation

• In the early 1900s, Avogadro proposed that equal volumes of gases at the same conditions contain the same number of particles. – Avogadro’s principle states that one

mole (6.02 x 1023 particles) of any gas at STP occupies a volume of 22.4 L.

– Avogadro’s principle allows you to interrelate mass, moles, pressure, volume, and temperature for any sample of gas.


Gas StoichiometryGas Stoichiometry

• 1 mole = 22.4 L of a gas at STP– All stoichiometric calculations will be done at STP

2 22 . ..


STPSTP• Standard temperature and pressure (STP)

has been designated as:– Temperature at 273 K– Pressure at 1 atm

–1 mole = 22.4 L• Used to compare gases

• Creates ideal conditions for describing behavior of gases.


Stoichiometric CalculationsStoichiometric Calculations• There are three basic gas stoichiometric

calculations:– mole-to-volume conversions– volume-to-volume conversions– mass-to-volume conversions.

• All stoichiometric calculations begin with a balanced equation and mole ratios.


Mole-to-Volume ConversionsMole-to-Volume Conversions• A zeppelin combusts H2 and O2 to form water. There are

25 moles of hydrogen gas in a zeppelin. How many liters of water vapor does it produce at STP?

• Given: 25 moles X L

H2 (g) + O2 (g) H2O (g)• Equation: 1 mol 1*(22.4L)

X L H2O = 25 mol H2

1*22.4 L H2O 1 mol H2

= 25 L H2O


Volume-to-Volume ConversionsVolume-to-Volume Conversions• When you are grilling steaks, how many liters of oxygen

are required to burn 1.5 liters of propane in the reaction: C3H8 + 5O2 3CO2 + 4H2O?

• Given: 1.5 L x L

C3H8 + 5O2 3CO2 + 4H2O• Eq.: 1*(22.4L) 5*(22.4L)

X L O2 = 1.5 L C3H8

5* 22.4L O2 1*22.4L C3H8

= 7.5 L O2


• The following reaction shows the production of ammonia. How many L of nitrogen are required to produce 85 grams of ammonia?

• Given: X L 85 g

• Equation: 1*(22.4L) 2*(17g)

X L N2 = 85g NH3

1*22.4L N2 2*(17 g NH3)

= 56 L NH3

Mass-to-Volume ConversionsMass-to-Volume Conversions



Gas Stoichiometry PracticeGas Stoichiometry Practice• Ammonium sulfate can be prepared by a

reaction between ammonia gas and sulfuric acid as follows.

• What volume of NH3 gas, measured at 78°C and a pressure of 1.66 atm, will be needed to produce 5.00 x 103 g of (NH4)2SO4?

Gas CharacteristicsGas Characteristics• Gases have no definite shape or volume

• Gases diffuse rapidly

• Gases have low density

• Gases are compressible/expandable

• Gases exert pressure on their containers


Kinetic Molecular TheoryKinetic Molecular Theory• KMT explains the behavior of all matter

(solids, liquids, and gases) at a particle level - kinetic means ‘motion’

• As related to gases, there are several basic principles of kinetic molecular theory (KMT):

1) Gas particles are in constant, random motion

2) Gas particles do not attract or repel each other

3) Gas particles have elastic collisions, meaning they do not lose kinetic energy when they collide

4) Gas particles’ kinetic energy depends on their temperature

14.114.1 KMTKMT

Physical Properties: TemperaturePhysical Properties: Temperature• Temperature is a measure of

the average kinetic energy of particles in a system– Different from heat -

amount of energy in a system

• Temperature is measured in units of:– Fahrenheit (oF)– Celsius (oC)– Kelvin (K)

• Temperature is measured by a thermometer

14.114.1 KMTKMT

How does temperature change? As a result of its change, what does it effect?

Physical Properties: TemperaturePhysical Properties: Temperature• When working with gases, we never use Celsius

– only Kelvin!• Converting Celsius to Kelvin

– K = oC + 273– Ex: Room temperature is about 22oC. In Kelvin,

this would be 296 K.

• Absolute Zero (0 Kelvin or -273oC) is the temperature at which all particle motion ceases– Absolute zero can never be achieved artificially,

though it is possible to reach temperatures close to it through the use of cryocoolers.

14.114.1 KMTKMT

Physical Properties: VolumePhysical Properties: Volume• Volume is the space matter

occupies• Gases always occupy the

volume of their container– Volume of gas is measured in

units of liters (L) or milliliters (mL) 1 L = 1000 Ml

• Gas volume can be expanded or compressed due to changes in…– Temperature– Pressure– Amount of particles– (mass or moles)

• Describe the similarities and differences between the balloons. What accounts for their differences?

14.114.1 KMTKMT

Physical Properties: PressurePhysical Properties: Pressure• Pressure is the force over a given area

– If someone stepped on your foot, which shoe would you prefer they wore?

• Pressure is measured in units of:– Atmospheres (atm)– Pascals (Newtons/m2)– psi (pounds per square inch)– mmHg (mm of Mercury)

• Pressure is measured by:– Barometer– Manometer

14.114.1 KMTKMT

Physical Properties: PressurePhysical Properties: Pressure• Gas pressure can be

altered due to changes in…– Volume– Temperature

– Amount of particles (mass or moles)

• The more often gas particles collide with the walls of their container, the greater the pressure.

Describe the similarities and differences between the two basketballs. What accounts for their differences?

14.114.1 KMTKMT

Click box to view movie clip.

Dalton’s LawDalton’s Law• The total pressure of a gas mixture is the

sum of the partial pressures of each individual gas

• Air is a mixture!

I’m John Dalton

14.114.1 KMTKMT

Dalton’s LawDalton’s Law

• Ex: The pressure on a tank of air with…+ 20.9 atm oxygen+ 78.1 atm nitrogen+ 0.97 atm argon+ 1.28 atm water vapor+ 0.05 atm carbon dioxide

= 101.3 atm

Ptotal = P1 + P2 + P3…

14.114.1 KMTKMT

Gas LawsGas Laws• The gas laws apply to ideal gases,

which are described by the kinetictheory in the following five statements. – Gas particles do not attract or repel each other. – Gas particles are much smaller than the spaces

between them. – Gas particles are in constant, random motion. – No kinetic energy is lost when gas particles collide with

each other or with the walls of their container. – All gases have the same kinetic energy at a given

temperature.

• The following laws explain the relationships between temperature, volume, and pressure:

14.214.2 The Gas LawsThe Gas Laws

Boyle’s LawBoyle’s Law• Explains the effect pressure has on

volume• Temperature stays constant• Inverse relationship

– As pressure increases, volume decreases PV

– As pressure decreases, volume increases PV

I’m Robert Boyle


V

P

Boyle’s LawBoyle’s Law

• KMT connection: the less space particles have to move, the more forces they exert on each other


Boyle’s LawBoyle’s Law

• Practice:– If the pressure is

tripled, what happens to the volume?

– If the pressure is halved, what happens to the volume?

• Example:– Squeezing syringe



Charles’ LawCharles’ Law• Explains the effect temperature has

on volume• Pressure stays constant• Direct relationship

– As temperature increases, volume increases TV

– As temperature decreases, volume decreases TV

I’m JacquesCharles


V

T

Charles’ LawCharles’ Law

• KMT connection: the more avg. kinetic energy particles have, the greater the distance between particles


Charles’ LawCharles’ Law• Practice:

– If the temperature is quadrupled, what happens to the volume?

– If the temperature is decreased by 1/3, what happens to the volume?

• Example:– Hot air balloon



Gay-Lussac’s LawGay-Lussac’s Law• Explains the effect temperature has on pressure• Volume stays constant• Direct relationship

– As temperature increases, pressure increases TP

• At higher temperatures, the particles in a gas have greater kinetic energy.

– As temperature decreases, pressure decreases TP

P PI’m Joseph LouisGay-Lussac


Gay-Lussac’s LawGay-Lussac’s Law

• KMT connection: the more avg. kinetic energy particles have, the more forces they exert on each other


Gay-Lussac’s LawGay-Lussac’s Law• Practice:

– If the temperature is doubled, what happens to the pressure?

– If the temperature is decreased by 1/4, what happens to the pressure?

• Example:– Pressure cooker


Combined Gas LawCombined Gas Law• The gas laws may be

integrated into a single equation called the combined gas law

• Where…– P = pressure in atm– V = volume in L– T = temperature in K– “1” means initial– “2” means final

• Steps to solving– Assign variables– Convert oC to K (if necessary)– “Drop” constants (see example)– Solve problem


Combined Gas LawCombined Gas Law• Example: In the fall, at a temperature of 32oC, you fill your

tires to a pressure of 2.18 atm. A cold front blows through, with temperatures dropping to 5oC, and your tires become flat. Knowing that the volume of your tires has not changed, what is the new pressure of the tires?– P1 = 2.18 atm– V1 = constant– T1 = 32oC + 273 = 305 K– P2 = ? atm– V2 = constant– T2 = 5oC + 273 = 278 K

P1V1 = P2V2 substitute 2.18 atm = ? atm

T1 T2 305 K 278 K P2 = 1.987 atm

What law best illustrates what happened to the tires in this problem?


• A sample of nitrogen monoxide has a volume of 72.6 mL at a temperature of 16°C and a pressure of 104.1 kPa. – What volume will the sample occupy at 24°C and 99.3

kPa?

Applying the Combined Gas LawApplying the Combined Gas Law


Ideal Gas LawIdeal Gas Law• Ideal gases are theoretical

– Real gases behave like ideal gases at STP

• The ideal gas law relates pressure, temperature, volume, and number of moles

• The equation includes universal gas constant R, which “corrects” conditions to STP

• Where…– P = pressure in atm– V = volume in L– n = number of particles in moles– R = universal gas constant– T = temperature in K 0.0821 L · atm

mol · K

14.314.3 The Ideal Gas LawThe Ideal Gas Law

Ideal Gas LawIdeal Gas Law• Example: Tyler is scuba diving along

a coral reef. His 10 liter air tankcontains 2 moles of oxygen gas at20oC. What is the pressure of hisoxygen tank?– P = ? atm– V = 10 L– n = 2 moles– R = 0.0821 L · atm

mol · K– T = 20oC + 273 = 293 K

• PV = nRT? atm * 10 L = 2 mol * 0.0821 L · atm * 293 K

mol · KP = 4.811 atm


• What pressure in atmospheres will 18.6 mol of methane exert when it is compressed in a 12.00-L tank at a temperature of 45°C?

• Determine the molar mass of an unknown gas if a sample has a mass of 0.290 g and occupies a volume of 148 mL at 13°C and a pressure of 107.0 atm.


Applying the Ideal Gas LawApplying the Ideal Gas Law

End of Unit 10End of Unit 10

Be Prepared for Unit 10 Test on Feb 25th.

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Unit 10: Gases