Gas Behavior Chapter 12 The Beginning The first gas studied was air. The first gas studied was air. The studies were very important to understanding

Gas BehaviorGas Behavior

Chapter 12

The BeginningThe Beginning

The first gas studied was air.The first gas studied was air.

The studies were very important to The studies were very important to understanding gas behavior because:understanding gas behavior because:– Air is a mixture of gases.Air is a mixture of gases.– It still behaved as one gas.It still behaved as one gas.– So…all gases, under similar conditions, behave So…all gases, under similar conditions, behave

similarlysimilarly

Kinetic-Molecular Theory of Kinetic-Molecular Theory of GasesGases

Assumes gas particles are in constant Assumes gas particles are in constant motionmotion

Used to explain gas behaviorUsed to explain gas behavior

Based on five assumptionsBased on five assumptions

Kinetic-Molecular Theory Kinetic-Molecular Theory of of GasesGases

Assumption #1Assumption #1– Gases consist of large numbers of tiny particles Gases consist of large numbers of tiny particles

whose sizes are negligible in comparison to whose sizes are negligible in comparison to their distance from each other.their distance from each other.

– Translation: Gas molecules are very small and Translation: Gas molecules are very small and far apart, giving the gas a low density and the far apart, giving the gas a low density and the property of compressibilityproperty of compressibility

Kinetic-Molecular Theory Kinetic-Molecular Theory of of GasesGases

Assumption #2Assumption #2– Collisions between gas particles and collisions Collisions between gas particles and collisions

between gas particles and the walls of the between gas particles and the walls of the container are elastic.container are elastic.

– Translation: No kinetic energy is lost in the Translation: No kinetic energy is lost in the collisions; it is transferredcollisions; it is transferred

Kinetic-Molecular Theory ofKinetic-Molecular Theory of GasesGases

Assumption #3Assumption #3– Particles of gas are continually, rapidly, and Particles of gas are continually, rapidly, and

randomly moving, thereby possessing kinetic randomly moving, thereby possessing kinetic energyenergy..

– Translation: Gas particles are not attracted to Translation: Gas particles are not attracted to each other because their kinetic energy is too each other because their kinetic energy is too strong; gas particles never stop; they are strong; gas particles never stop; they are always movingalways moving


Assumption #4Assumption #4– Gas particles have no forces of attraction or Gas particles have no forces of attraction or

repulsion between them.repulsion between them.

– Translation: Gas particles will not “stick” Translation: Gas particles will not “stick” together when they collide; they bounce off together when they collide; they bounce off each othereach other


Assumption #5Assumption #5– The temperature of the gas affects the average The temperature of the gas affects the average

kinetic energy of gas particles.kinetic energy of gas particles.

– Translation: Temperature is a measure of Translation: Temperature is a measure of average kinetic energy; higher temp.=higher KEaverage kinetic energy; higher temp.=higher KE

Kinetic-Molecular Theory of Kinetic-Molecular Theory of GasesGases

The KMT only applies to “ideal gases”—The KMT only applies to “ideal gases”—theoretical gasestheoretical gases

Other gases are called “real gases”Other gases are called “real gases”– Real gases behave almost ideally when the Real gases behave almost ideally when the

pressure is not too high and/or the temperature pressure is not too high and/or the temperature is not too lowis not too low

Kinetic Energy (KE)Kinetic Energy (KE)

The energy of motionThe energy of motion KE = ½ mvKE = ½ mv22

m = massm = mass v = speedv = speed

Try ThisTry This

A 68 kg track runner is running at 10 m/s A 68 kg track runner is running at 10 m/s and a 136 kg football player is running at 5 and a 136 kg football player is running at 5 m/s. Which has more kinetic energy?m/s. Which has more kinetic energy?

KE = ½ (68 kg)(10 m/s)KE = ½ (68 kg)(10 m/s)22 = 3400 J = 3400 J KE = ½ (136)(5 m/s)KE = ½ (136)(5 m/s)22 = 1700 J = 1700 J Track runnerTrack runner

VolumeVolume

Volume is measured in:Volume is measured in:– Liters (L)Liters (L)– Milliliters (mL)Milliliters (mL)– Cubic centimeters (cmCubic centimeters (cm33 or cc) or cc)

PressurePressure

Pressure is measured in:Pressure is measured in:– Atmospheres (atm)Atmospheres (atm)– Millimeters of mercury (mm Hg)Millimeters of mercury (mm Hg)– Kilopascals (kPa)Kilopascals (kPa)– Torricelli (torr)Torricelli (torr)– Pounds per square inch (psi)Pounds per square inch (psi)

TemperatureTemperature

Temperature is measured in:Temperature is measured in:– Celsius (°C)Celsius (°C)– Kelvin (K)Kelvin (K)– Fahrenheit (°F)Fahrenheit (°F)

Standard Temperature and Standard Temperature and PressurePressure

0°C or 273 K0°C or 273 K 1 atm or 760 mm Hg or 760 torr or 101.3 kPa1 atm or 760 mm Hg or 760 torr or 101.3 kPa

The Gas LawsThe Gas Laws

Developed to help create relationships Developed to help create relationships between volume, pressure, temperature, between volume, pressure, temperature, and amount of a gas. and amount of a gas.

First person to collect and analyze a gas—First person to collect and analyze a gas—Robert BoyleRobert Boyle


Boyle’s Law: Boyle’s Law: volume of a fixed mass of gas volume of a fixed mass of gas varies inversely with the pressure at varies inversely with the pressure at constant temperatureconstant temperature

Translation: When pressure (P) increases, Translation: When pressure (P) increases, volume (V) decreases. When P decreases, volume (V) decreases. When P decreases, V increasesV increases

Mathematically: Mathematically: PP11VV11==PP22VV22

Important!Important!

Units must be the same thing. For instance Units must be the same thing. For instance if Pif P11 is in mm of Hg then P is in mm of Hg then P22 must also be must also be

measured in mm of Hg. measured in mm of Hg.

If VIf V11 is measured in mL then V is measured in mL then V22 must also be must also be

measured in mL. measured in mL.


Boyle’s Law Problem:Boyle’s Law Problem:– A gas at a pressure of 608 mm Hg is held in a A gas at a pressure of 608 mm Hg is held in a

container with a volume of 545 L. If the volume container with a volume of 545 L. If the volume of the container is increased to 1065 L, and of the container is increased to 1065 L, and temperature is held constant, what is the new temperature is held constant, what is the new pressure of the gas?pressure of the gas?


Boyle’s Law Solution:Boyle’s Law Solution:– Have: PHave: P11=608 mm Hg=608 mm Hg

VV11=545 L=545 L

VV22=1065 L=1065 L

– Want: PWant: P22

– Use Boyle’s Law Equation Use Boyle’s Law Equation PP11VV11==PP22VV22


Boyle’s Law SolutionBoyle’s Law Solution PP11VV11 = P = P22VV22

(608 mm Hg)(545 L) = (P(608 mm Hg)(545 L) = (P22)(1065 L))(1065 L)

311 mm Hg311 mm Hg

Try this!Try this!

A high- altitude balloon contains 30.0 L of A high- altitude balloon contains 30.0 L of helium gas at 103 kPa. What is the volume helium gas at 103 kPa. What is the volume when the balloon rises to an altitude where when the balloon rises to an altitude where the pressure is only 25.0 kPa? (Assume that the pressure is only 25.0 kPa? (Assume that the temperature remains constant) the temperature remains constant)

AnswerAnswer

PP11VV11 = P = P22VV22

(103 kPa)(30.0 L) = (25.0 kPa)(V(103 kPa)(30.0 L) = (25.0 kPa)(V22))

124 L 124 L

Have you ever had the experience of buying a helium filled balloon and then taking Have you ever had the experience of buying a helium filled balloon and then taking it outside on a cold day? If you have you noticed that the balloon shrunk and it outside on a cold day? If you have you noticed that the balloon shrunk and

looked like there was not enough helium put in it. However if you ever put a helium looked like there was not enough helium put in it. However if you ever put a helium balloon in your car on a HOT day you will return to find that the balloon hasballoon in your car on a HOT day you will return to find that the balloon has

exploded. exploded.

Why do these things happen? Why do these things happen?


Charles’ Law—relationship between gas Charles’ Law—relationship between gas temperature and volumetemperature and volume– the volume of a fixed mass of gas at constant pressure the volume of a fixed mass of gas at constant pressure

varies directly with the Kelvin temperaturevaries directly with the Kelvin temperature

– Translation: when V increases, T increases, and when Translation: when V increases, T increases, and when V decreases, T decreasesV decreases, T decreases

– Mathematically: Mathematically: 1 2

1 2

V V

T T=

Important! Important!

The Kinetic Theory of Gases states that the The Kinetic Theory of Gases states that the kinetic energy of a gas is proportional to its kinetic energy of a gas is proportional to its temperature and using a Celsius scale would temperature and using a Celsius scale would cause the kinetic energy of a gas to be cause the kinetic energy of a gas to be negative…which is impossible! negative…which is impossible!

Change the temperature to Kelvin!!!!Change the temperature to Kelvin!!!!

Remember…all temperatures must Remember…all temperatures must be in Kelvinbe in Kelvin

K = ºC + 273K = ºC + 273


Charles’ Law ProblemCharles’ Law Problem– A sample of neon gas occupies a volume of 752 A sample of neon gas occupies a volume of 752

mL at 25ºC. What volume will the gas occupy mL at 25ºC. What volume will the gas occupy at 50ºC if the pressure remains constant?at 50ºC if the pressure remains constant?

– Remember…the gas laws will be applied; all Remember…the gas laws will be applied; all temperatures must be in Kelvin!temperatures must be in Kelvin!


Charles’ Law SolutionCharles’ Law Solution– Have: VHave: V11=752 mL=752 mL

TT11=25ºC + 273=298 K=25ºC + 273=298 K

TT22=50ºC + 273=323 K=50ºC + 273=323 K

– Want: VWant: V22 in mL in mL

– Use:Use:

– Solve: Solve: 752 mL 752 mL == V V22

298 K 323 K298 K 323 K– 815 mL815 mL

1 2

1 2

V V

T T=

Try This! Try This!

A balloon inflated in a room at 24°C has a A balloon inflated in a room at 24°C has a volume of 4.00 L. The balloon is then heated volume of 4.00 L. The balloon is then heated to a temperature of 58°C. What is the new to a temperature of 58°C. What is the new volume if the pressure remains constant? volume if the pressure remains constant?

AnswerAnswer

VV11 == V V22

TT11 T T22

TT11: 24°C + 273 = 297 K: 24°C + 273 = 297 K

TT22: 58°C + 273 = 331 K : 58°C + 273 = 331 K

4.00 L 4.00 L = = V V22

297 K 331 K297 K 331 K 4.46 L4.46 L

The Gas Laws The Gas Laws

Gay-Lussac’s Law—relationship between gas Gay-Lussac’s Law—relationship between gas pressure and temperaturepressure and temperature– pressure of a fixed mass of gas at a constant pressure of a fixed mass of gas at a constant

volume varies directly with the Kelvin volume varies directly with the Kelvin temperaturetemperature

– Translation: When pressure increases, temp. Translation: When pressure increases, temp. increases; when pressure decreases, temp. increases; when pressure decreases, temp. decreasesdecreases

– MathematicallyMathematically:: 1 2

1 2

P P

T T=


Gay-Lussac’s Law Problem:Gay-Lussac’s Law Problem:– The gas in an aerosol can is at a pressure of The gas in an aerosol can is at a pressure of

3.00 atm at 25ºC. Directions on the can warn 3.00 atm at 25ºC. Directions on the can warn the user not to keep the can in a place where the user not to keep the can in a place where the temperature exceeds 52ºC. What would the the temperature exceeds 52ºC. What would the gas pressure in the can be at 52ºC?gas pressure in the can be at 52ºC?

– Remember to convert temperatures to Kelvin!!Remember to convert temperatures to Kelvin!!

AnswerAnswer

PP11 == P P22

TT11 T T22

TT11: 25°C + 273 = 298 K: 25°C + 273 = 298 K

TT22: 52°C + 273 = 325 K : 52°C + 273 = 325 K

3.00 atm 3.00 atm = = P P22

298 K 325 K298 K 325 K 3.27 atm3.27 atm

Try This!Try This!

A gas has a pressure of 6.58 kPa at 539 K. What A gas has a pressure of 6.58 kPa at 539 K. What will be the pressure at 211 K if the volume does will be the pressure at 211 K if the volume does not change? not change?

AnswerAnswer

PP11 == P P22

TT11 T T22

6.58 kPa 6.58 kPa = = P P22

539 K 211 K539 K 211 K 2.58 kPa2.58 kPa


Combined Gas Law: combination of all three Combined Gas Law: combination of all three lawslaws

Mathematically:Mathematically:

1 1 2 2

1 2

PV PV

T T=

The volume of a gas -filled balloon is The volume of a gas -filled balloon is 30.0 L at 40°C and 153 kPa pressure. 30.0 L at 40°C and 153 kPa pressure. What volume will the balloon have at What volume will the balloon have at

standard temperature and pressure?standard temperature and pressure?

1 1 2 2

1 2

PV PV

T T=

Standard Pressure = 760 torr = 1 atm = 101.3kPa

Standard Temp. = 273 K

AnswerAnswer

PP11VV11 == P P22VV22

TT11 T T22

TT11: 40°C + 273 = 313 K: 40°C + 273 = 313 K

TT22: 0°C + 273 = 273 K : 0°C + 273 = 273 K

(153 kPa)(30.0 L) (153 kPa)(30.0 L) = = (101.3 kPa)(V (101.3 kPa)(V22) )

313 K 273 K313 K 273 K 39.5 L39.5 L


Dalton’s Law of Partial Pressures: Dalton’s Law of Partial Pressures: the total the total pressure of a mixture of gases is the sum of pressure of a mixture of gases is the sum of the pressures of each individual gasthe pressures of each individual gas

Translation: the sum of the parts equals the Translation: the sum of the parts equals the wholewhole

Mathematically: Mathematically: PPTT = P = P11 + P + P22 + P + P33 +… +…

Air contains oxygen, nitrogen , carbon Air contains oxygen, nitrogen , carbon dioxide and trace amounts of other gases. dioxide and trace amounts of other gases. What is the partial pressure of oxygen (PWhat is the partial pressure of oxygen (PO2O2) )

at 101.3 kPa of total pressure if the partial at 101.3 kPa of total pressure if the partial pressure of nitrogen,carbon dioxide and pressure of nitrogen,carbon dioxide and other gases are 79.10 kPa, 0.040 kPa and other gases are 79.10 kPa, 0.040 kPa and 0.94 kPa respectively. 0.94 kPa respectively.

AnswerAnswer

PPTT == PP11 + P + P22 + P + P33 ++……

101.3 kPa = 79.10 kPa + 0.040 kPa + 0.94 101.3 kPa = 79.10 kPa + 0.040 kPa + 0.94 kPa + PkPa + PO2O2

PPO2O2 = 21.2 kPa = 21.2 kPa

Ideal Gas LawIdeal Gas Law Gases behave differently under different Gases behave differently under different

circumstances (each gas has a different molar circumstances (each gas has a different molar mass)mass)

– Use term “ideal gas” to describe gas behavior under all Use term “ideal gas” to describe gas behavior under all circumstancescircumstances

– No such thing as ideal gas…they are “real gases”No such thing as ideal gas…they are “real gases”

– In reality gases can be liquefied and sometimes In reality gases can be liquefied and sometimes solidified by cooling and by applying pressure solidified by cooling and by applying pressure

whereas whereas ideal gases cannot be. So real gases do ideal gases cannot be. So real gases do not behave not behave like ideal gases under high pressures and like ideal gases under high pressures and at low at low temperatures. temperatures.

Ideal Gas LawIdeal Gas Law

PV = nRTPV = nRT P is pressure may be labeled kPa, atm, mm P is pressure may be labeled kPa, atm, mm

Hg, or torrHg, or torr V is volume must be labeled LV is volume must be labeled L n is molesn is moles

Ideal Gas Law ContinuedIdeal Gas Law Continued R is a constant whose value is determined R is a constant whose value is determined

by P.by P.– If P is labeled kPa If P is labeled kPa R = 8.314 R = 8.314– If P is labeled atm If P is labeled atm R = 0.0821 R = 0.0821– If P is labeled mm Hg or torr If P is labeled mm Hg or torr R = 62.4 R = 62.4

T is temperature must be labeled KT is temperature must be labeled K

Try This!Try This!

You fill a rigid steel cylinder that has a You fill a rigid steel cylinder that has a volume of 20.0 L with nitrogen gas to a final volume of 20.0 L with nitrogen gas to a final pressure of 150049 torrs at 28°C. How pressure of 150049 torrs at 28°C. How many moles of nitrogen gas does the many moles of nitrogen gas does the cylinder contain? cylinder contain?

Convert Temperature to Kelvin!Convert Temperature to Kelvin!

AnswerAnswer

PV = nRTPV = nRT 28°C + 273 = 301 K28°C + 273 = 301 K (150049 torr)(20.0 L) = n (62.4)(301 K)(150049 torr)(20.0 L) = n (62.4)(301 K) 160. moles of N160. moles of N22

Try This! Try This!

What volume is occupied by 5.03 g of What volume is occupied by 5.03 g of hydrogen gas at 28°C and a pressure of 2.0 hydrogen gas at 28°C and a pressure of 2.0 atm? atm?

Hint: Convert grams to moles and °C to K! Hint: Convert grams to moles and °C to K!

AnswerAnswer

PV = nRTPV = nRT 5.03 g ÷ 2.0 g/mol = 2.515 mol H5.03 g ÷ 2.0 g/mol = 2.515 mol H22

28°C + 273 = 301 K28°C + 273 = 301 K (2 atm)(V) = (2.515 mol)(0.0821)(301 K)(2 atm)(V) = (2.515 mol)(0.0821)(301 K) 31 L31 L

Ideal Gas LawIdeal Gas Law Finding the molar massFinding the molar mass

MM == mRT mRT

PVPV M = molar mass and m = gramsM = molar mass and m = grams

What is the molar mass of a gas if 372 ml What is the molar mass of a gas if 372 ml have a mass of 0.920 grams at 100ºC and have a mass of 0.920 grams at 100ºC and 108 kPa of pressure?108 kPa of pressure?

AnswerAnswerMM == mRT mRT

PVPV What is the molar mass of a gas if 372 ml have a mass What is the molar mass of a gas if 372 ml have a mass

of 0.920 grams at 100ºC and 108 kPa of pressure?of 0.920 grams at 100ºC and 108 kPa of pressure? 372 mL ÷ 1000 = 0.372 L372 mL ÷ 1000 = 0.372 L 100°C + 273 = 373 K100°C + 273 = 373 K M = M = (0.920 g)(8.314)(373 K)(0.920 g)(8.314)(373 K) (108 kPa)(0.372 L)(108 kPa)(0.372 L) 71.0 g/mol71.0 g/mol

Try This!Try This!

A container holds 2240 L of methane gas A container holds 2240 L of methane gas (CH(CH44) at a pressure of 1.50 kPa and a ) at a pressure of 1.50 kPa and a

temperature of 42°C. How many grams of temperature of 42°C. How many grams of CHCH44 does this container hold? does this container hold?

AnswerAnswerMM == mRT mRT

PVPV 42°C + 273 = 315 K42°C + 273 = 315 K CHCH44 = 12.0 + 4.0 = 16.0 g/mol = 12.0 + 4.0 = 16.0 g/mol 16.0 g.mol = 16.0 g.mol = (x g)(8.314)(315 K) (x g)(8.314)(315 K) (1.50 kPa)(2240L)(1.50 kPa)(2240L) 20.5 g20.5 g

Diffusion and EffusionDiffusion and Effusion

Diffusion is the gradual mixing of gases due to the Diffusion is the gradual mixing of gases due to the random, spontaneous motion of the gas particlesrandom, spontaneous motion of the gas particles

Effusion is the process by which gas molecules Effusion is the process by which gas molecules trapped in a container randomly pass through tiny trapped in a container randomly pass through tiny openings in the container openings in the container

What are everyday examples of diffusion or What are everyday examples of diffusion or effusion?effusion?

ExamplesExamples

Diffusion: perfume spreading, smelling Diffusion: perfume spreading, smelling cooking food, and smell something burningcooking food, and smell something burning

Effusion: tire puncture and a pin hole in a Effusion: tire puncture and a pin hole in a balloonballoon


Rates of diffusion/effusion depends on Rates of diffusion/effusion depends on the velocity of the moleculesthe velocity of the molecules– Velocity depends on temperature and Velocity depends on temperature and

massmass Would hot or cold particles move faster?Would hot or cold particles move faster? Would heavy particles move slower or faster?Would heavy particles move slower or faster?


Graham’s Law—relationship between rate of Graham’s Law—relationship between rate of effusion (diffusion) and molar masseffusion (diffusion) and molar mass– the rate of effusion of gases at the same temperature the rate of effusion of gases at the same temperature

and pressure are inversely proportional to the square and pressure are inversely proportional to the square root of the molar massroot of the molar mass

– Mathematically:Mathematically:B

A

Mrate of effusion of A

rate of effusion of B M=

Graham noticed that gases of lower molar mass effuse Graham noticed that gases of lower molar mass effuse

faster than gases of higher molar massfaster than gases of higher molar mass.. Nitrogen effuses at Nitrogen effuses at

535 m/s. How much 535 m/s. How much faster will helium gas faster will helium gas effuse?effuse?

535 m/s535 m/s == √4√4

x m/s √28x m/s √28 1415 m/s ÷ 535 m/s1415 m/s ÷ 535 m/s 2.6 times faster2.6 times faster

B

A

Mrate of effusion of A

rate of effusion of B M=

Documents

Gas Behavior Chapter 12 The Beginning The first gas studied was air. The first gas studied was air. The studies were very important to understanding