Chapter 14“The Behavior of Gases”

Section 14.1The Properties of Gases

CompressibilityGases can expand to fill its

container, unlike solids or liquidsThe reverse is also true:

They are easily compressed, or squeezed into a smaller volume

Compressibility is a measure of how much the volume of matter decreases under pressure

Compressibility This is the idea behind placing “air

bags” in automobilesIn an accident, the air compresses

more than the steering wheel or dash when you strike it

The impact forces the gas particles closer together, because there is a lot of empty space between them

Compressibility At room temperature, the distance

between particles is about 10x the diameter of the particleFig. 14.2, page 414

This empty space makes gases good insulators (example: windows, coats)

How does the volume of the particles in a gas compare to the overall volume of the gas?

Variables that describe a Gas The four variables and their common

units:1. pressure (P) in kilopascals2. volume (V) in Liters3. temperature (T) in Kelvin4. amount (n) in moles

• The amount of gas, volume, and temperature are factors that affect gas pressure.

1. Amount of GasWhen we inflate a balloon, we are

adding gas molecules. Increasing the number of gas

particles increases the number of collisionsthus, the pressure increases

If temperature is constant, then doubling the number of particles doubles the pressure

Pressure and the number of molecules are directly related

More molecules means more collisions, and…

Fewer molecules means fewer collisions.

Gases naturally move from areas of high pressure to low pressure, because there is empty space to move into – a spray can is example.

Common use? A practical application is Aerosol

(spray) cansgas moves from higher pressure to

lower pressurea propellant forces the product outwhipped cream, hair spray, paint

Fig. 14.5, page 416 Is the can really ever “empty”?

2. Volume of Gas In a smaller container, the

molecules have less room to move.

The particles hit the sides of the container more often.

As volume decreases, pressure increases. (think of a syringe)Thus, volume and pressure are

inversely related to each other

3. Temperature of Gas Raising the temperature of a gas increases

the pressure, if the volume is held constant. (Temp. and Pres. are directly related)The molecules hit the walls harder, and

more frequently! Fig. 14.7, page 417 Should you throw an aerosol can into a

fire? What could happen? When should your automobile tire pressure

be checked?

Section 14.2The Gas Laws

The Gas Laws are mathematicalThe gas laws will describe HOW

gases behave.Gas behavior can be predicted by

the theory.The amount of change can be

calculated with mathematical equations.

You need to know both of these: the theory, and the math

Robert Boyle(1627-1691)

• Boyle was born into an aristocratic Irish family

• Became interested in medicine and the new science of Galileo and studied chemistry. 

• A founder and an influential fellow of the Royal Society of London

• Wrote extensively on science, philosophy, and theology.

#1. Boyle’s Law - 1662

Pressure x Volume = a constant Equation: P1V1 = P2V2 (T = constant)

Gas pressure is inversely proportional to the volume, when temperature is held constant.

Graph of Boyle’s Law – page 418Boyle’s Law says the pressure is inverse to the volume.

Note that when the volume goes up, the pressure goes down

- Page 419

Boyle’s Law Practice Boyle’s Law Practice Ammonia gas occupies a volume of 450. Ammonia gas occupies a volume of 450.

mL at 720. mm Hg. What volume will it mL at 720. mm Hg. What volume will it occupy at standard pressure? occupy at standard pressure? VV22 = 426 = 426 mLmL

A 3.2-L sample of gas has a pressure of A 3.2-L sample of gas has a pressure of 102 kPa. If the volume is reduced to 102 kPa. If the volume is reduced to 0.65 L, what pressure will the gas 0.65 L, what pressure will the gas exert? exert? PP22 = 502 kPa = 502 kPa

Jacques Charles (1746-1823)• French Physicist• Part of a scientific

balloon flight on Dec. 1, 1783 – was one of three passengers in the second balloon ascension that carried humans

• This is how his interest in gases started

• It was a hydrogen filled balloon – good thing they were careful!

#2. Charles’s Law - 1787The volume of a fixed mass of gas is directly proportional to the Kelvin temperature, when pressure is held constant.This extrapolates to zero volume at a temperature of zero Kelvin.

VT

VT

P1

1

2

2 ( constant)

Converting Celsius to Kelvin•Gas law problems involving temperature will always require that the temperature be in Kelvin. (Remember that no degree sign is shown with the kelvin scale.)

•Reason? There will never be a zero volume, since we have never reached absolute zero.

Kelvin = C + 273 °C = Kelvin - 273and

- Page 421

Charles's Law PracticeCharles's Law Practice Helium occupies 3.8 L at -45°C. What Helium occupies 3.8 L at -45°C. What

volume will it occupy at 45°C? volume will it occupy at 45°C? VV22 = = 5.3 L5.3 L

At 27°C, fluorine occupies a volume of At 27°C, fluorine occupies a volume of 0.500 dm3. To what temperature in 0.500 dm3. To what temperature in degrees Celsius should it be lowered degrees Celsius should it be lowered to bring the volume to 200. mL? to bring the volume to 200. mL? T2 = -T2 = -153°C (120 K)153°C (120 K)

Joseph Louis Gay-Lussac (1778 – 1850) French chemist and physicist Known for his studies on the physical properties of gases. In 1804 he made balloon ascensions to study magnetic forces and to observe the composition and temperature of the air at different altitudes.

#3. Gay-Lussac’s Law - 1802•The pressure and Kelvin temperature of a gas are directly proportional, provided that the volume remains constant.

2

2

1

1

TP

TP

•How does a pressure cooker affect the time needed to cook food? (Note page 422)

•Sample Problem 14.3, page 423

Gay-Lussac’s Practice Gay-Lussac’s Practice ProblemsProblems

A gas at STP is cooled to -185°C. A gas at STP is cooled to -185°C. What pressure in kPa will it have at What pressure in kPa will it have at this temperature (volume remains this temperature (volume remains constant)? constant)? PP22 = 0.32 atm = 0.32 atm

Chlorine gas has a pressure of 1.05 Chlorine gas has a pressure of 1.05 kPa at 25°C. What pressure will it kPa at 25°C. What pressure will it exert at 75°C? exert at 75°C? PP22 = 1.23 atm = 1.23 atm

#4. The Combined Gas LawThe combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.

2

22

1

11

TVP

TVP

Sample Problem 14.4, page 424

The combined gas law contains all the other gas laws!

If the temperature remains constant...

P1 V1

T1

x = P2 V2

T2

x

Boyle’s Law

The combined gas law contains all the other gas laws!

If the pressure remains constant...

P1 V1

T1

x = P2 V2

T2

x

Charles’s Law

The combined gas law contains all the other gas laws!

If the volume remains constant...

P1 V1

T1

x = P2 V2

T2

x

Gay-Lussac’s Law

Combined Gas Law Combined Gas Law PracitcePracitce

A gas occupies 256 mL at 720 kPa and 25°C. A gas occupies 256 mL at 720 kPa and 25°C. What will its volume be at STP? What will its volume be at STP? VV22 = 220 mL = 220 mL

A gas occupies 1.5 L at 850 kPa and 15°C. At A gas occupies 1.5 L at 850 kPa and 15°C. At what pressure will this gas occupy 2.5 L at what pressure will this gas occupy 2.5 L at 30.0°C? 30.0°C? PP22 = 540 kPa = 540 kPa

A gas occupies 125 mL at 125 kPa. After being A gas occupies 125 mL at 125 kPa. After being heated to 75°C and depressurized to 100.0 kPa, heated to 75°C and depressurized to 100.0 kPa, it occupies 0.100 L. What was the it occupies 0.100 L. What was the originaloriginal temperature of the gas? temperature of the gas? TT11 = 544 K (271°C) = 544 K (271°C)

Section 14.3Ideal Gases

5. The Ideal Gas Law #1 Equation: P x V = n x R x T Pressure times Volume equals the

number of moles (n) times the Ideal Gas Constant (R) times the Temperature in Kelvin.

R = 8.31 (L x kPa) / (mol x K) The other units must match the value of

the constant, in order to cancel out. The value of R could change, if other

units of measurement are used for the other values (namely pressure changes)

We now have a new way to count moles (the amount of matter), by measuring T, P, and V. We aren’t restricted to only STP conditions:

P x V R x T

The Ideal Gas Law

n =

Ideal Gases We are going to assume the gases

behave “ideally”- in other words, they obey the Gas Laws under all conditions of temperature and pressure

An ideal gas does not really exist, but it makes the math easier and is a close approximation.

Particles have no volume? Wrong! No attractive forces? Wrong!

Ideal GasesThere are no gases for which this

is true (acting “ideal”); however,Real gases behave this way at

a) high temperature, and b) low pressure.Because at these conditions, a gas will stay a gas!

Sample Problem 14.5, page 427

#6. Ideal Gas Law 2 P x V = m x R x T M Allows LOTS of calculations, and

some new items are: m = mass, in grams M = molar mass, in g/mol

Molar mass = m R T P V

Density Density is mass divided by volume

m Vso, m M P V R T

D =

D = =

Ideal Gas Law PracticeIdeal Gas Law Practice

Write equation:                        

Substitute into equation:                                                        Solve for T2:   Recall: oC  +  273  =  KTherefore:  Temperature  =  -71oC

Gas Review Problem #1

1)  A quantity of gas has a volume of 200 dm3 at 17oC and 106.6 kPa.  To what temperature (oC) must the gas be cooled for its volume to be reduced to 150 dm3 at a pressure of 98.6 kPa?

Write given information:V1  =                                   V2  =  T1  =          T2  =  P1  =                              P2  = 

 

200 dm3 17 oC +  273   =  290 K106.6 kPa

150 dm3

_______98.6 kPa

(101.6 kPa)x(200 dm3) (98.6 kPa)x(150 dm3) 290 K T2

=

P1xV1 P2xV2

T1 T2=

T2  =  201 K

Write equation:                       Volume is constant...cancel it out from equation:               Substitute into equation:                    Solve for P2:  

Gas Review Problem #2

2)  A quantity of gas exerts a pressure of  98.6 kPa at a temperature of 22oC.  If the volume remains unchanged, what pressure will it exert at -8oC?Write given information:

V1  =                                V2  =  T1  =         T2  =  P1  =                                P2  = 

P1xV1 P2xV2

T1 T2=

P1 P2

T1 T2=

constant 22 oC+  273   =  295 K98.6 kPa

constant-8 oC+  273   =  265 K_________

98.6 kPa P2

295 K 265 K=

P2  =  88.6 kPa

(P2)(295 K) = (98.6 kPa)(265 K)(295 K) (295 K)

(98.6 kPa)(265) (295)P2 =

To solve, cross multiply and divide:

Write given information:V1  =                                       V2  =  T1  =          T2  =   P1  =                                  P2  =  R  =                    Density  =  n  =  Cl2  = 

Two approaches to solve this problem.METHOD 1:  Combined Gas Law & DensityWrite equation:                       Substitute into equation:                                           Solve for V2: Density  =  3.17 g/dm3 @ STPRecall:                       Substitute into equation:                          Solve for mass: 

P1xV1 P2xV2

T1 T2=

(98.7 kPa)x(3.34 L) (101.3 kPa)x(V2)310 K 273K

=

PVRT

=n(98.7 kPa)(3.34 dm3)

[8.314 (kPa)(dm3)/(mol)(K)](310 K)=n

Density = massvolume

3.17 g/cm3 = mass2.85 L

PV = nRT

Gas Review Problem #3

What is the mass of 3.34 dm3 sample of chlorine gas if the volumewas determined at 37oC and 98.7 kPa?  The density of chlorine gas at STP is 3.17 g/dm3.

3.34 L37 oC+  273   =  310 K98.7 kPa

8.314 kPa L / mol K___________71 g/mol

__________273  K101.3 kPa

3.17 g/dm3

V2  =  2.85 L @ STP

mass  =  9.1 g chlorine gas

2.85 L

METHOD 2:  Ideal Gas Law Write equation:                   Solve for moles:            Substitute into equation:                                              Solve for mole:  n  =  0.128 mol Cl2

Recall molar mass of diatomic chlorine is 71 g/molCalculate mass of chlorine:  x g Cl2  =  0.128 mol Cl2               =  9.1 g Cl2

 

Gas Review Problem #6Iron (II) sulfide reacts with hydrochloric acid as follows:

FeS(s)  +  2 HCl(aq)    FeCl2(aq)  +  H2S(g)

What volume of H2S, measured at 30oC and 95.1 kPa, will be producedwhen 132 g of FeS reacts?

Calculate number of moles of H2S...

        x mole H2S  =  132 g FeS                  

Write given information:P  =  n  =  R  =  T  = 

Equation:  

Substitute into Equation:                                    

Solve equation for Volume: 

132 g X L

1 mol FeS 1 mol H2S879 g FeS 1 mol FeS

= 1.50 mol H2S

95.1 kPa1.5 mole H2S8.314 L kPa/mol K 30oC +  273  =  303 K

PV  =  nRT

V  =  39.7 L

(95.1 kPa)(V) = 1.5 mol H2S 8.314 (303 K)(L)(Kpa) (mol)(K)

7)  What is the density of nitrogen gas at STP (in g/dm3 and kg/m3)?

Write given information:1 mole N2  =  28 g N2  =  22.4 dm3 @ STP

Write equation:                           Substitute into equation:                           Solve for Density:   Density  =  1.35 g/dm3

Recall:  1000 g  =  1 kg     &     1 m3  =  1000 dm3                                                                                                              

Convert m3 to dm3:

        x dm3  =  1 m3                       =  1000 dm3

 

Gas Review Problem #6

Convert:

Solve: 1.35 kg/m3

Ideal Gases don’t exist, because:

1. Molecules do take up space

2. There are attractive forces between particles

- otherwise there would be no liquids formed

Real Gases behave like Ideal Gases...

When the molecules are far apart.

The molecules do not take up as big a percentage of the space We can ignore the particle

volume. This is at low pressure

Real Gases behave like Ideal Gases…

When molecules are moving fastThis is at high temperature

Collisions are harder and faster.Molecules are not next to each

other very long.Attractive forces can’t play a role.

Section 14.4Gases: Mixtures and Movements

#7 Dalton’s Law of Partial Pressures

For a mixture of gases in a container,

PTotal = P1 + P2 + P3 + . . .

•P1 represents the “partial pressure”, or the contribution by that gas.•Dalton’s Law is particularly useful in calculating the pressure of gases collected over water.

Collecting a gas over water – one of the experiments in Chapter 14 involves this.

Connected to gas generator

If the first three containers are all put into the fourth, we can find the pressure in that container by adding up the pressure in the first 3:

2 atm + 1 atm + 3 atm = 6 atm

Sample Problem 14.6, page 434

1 2 3 4

Suppose you have a 1.00 dm3 container of oxygen gas at 202.6 kPa and a 2.00 dm3 container of nitrogen gas at 101.3 kPa.  If you transfer the oxygen to the container holding the nitrogen,

a)  what pressure would the nitrogen exert?b)  what would be the total pressure exerted by the mixture?

Write given information:

  Px Vx Vz Px,z

O2202.6 kPa

1 dm3

2 dm3

101.3 kPa

N2101.3 kPa

2 dm3

2 dm3

101.3 kPa

O2 +  N2    2

dm3 202.6 kPa

Dalton’s Law Practice Problem

Diffusion is:

Effusion: Gas escaping through a tiny hole in a container.

Both of these depend on the molar mass of the particle, which determines the speed.

Molecules moving from areas of high concentration to low concentration.Example: perfume molecules spreading across the room.

•Diffusion: describes the mixing of gases. The rate of diffusion is the rate of gas mixing.

•Molecules move from areas of high concentration to low concentration.

•Fig. 14.18, p. 435

Effusion: a gas escapes through a tiny hole in its container -Think of a nail in your car tire…

Diffusion and effusion are explained by the next gas law: Graham’s

8. Graham’s Law

The rate of effusion and diffusion is inversely proportional to the square root of the molar mass of the molecules.

Derived from: Kinetic energy = 1/2 mv2

m = the molar mass, and v = the velocity.

RateA MassB

RateB MassA

=

Sample: compare rates of effusion of Helium with Nitrogen – done on p. 436

With effusion and diffusion, the type of particle is important: Gases of lower molar mass diffuse and

effuse faster than gases of higher molar mass.

Helium effuses and diffuses faster than nitrogen – thus, helium escapes from a balloon quicker than many other gases!

Graham’s Law

Gram’s Law PracticeGram’s Law Practice At 350oC, nitrogen has At 350oC, nitrogen has a velocity of 800 m/s. a velocity of 800 m/s. Find the velocity of Find the velocity of hydrogen at the same hydrogen at the same temperature. temperature. 2993 m/s 2993 m/s