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Chemistry: A Molecular Approach , 1 st Ed. Nivaldo Tro. Chapter 5 Gases. Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA. 2008, Prentice Hall. Air Pressure & Shallow Wells. water for many homes is supplied by a well less than 30 ft. deep with a pump at the surface

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chapter 5 gases

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Chapter 5Gases

Roy Kennedy

Massachusetts Bay Community College

Wellesley Hills, MA

2008, Prentice Hall

air pressure shallow wells
Air Pressure & Shallow Wells
  • water for many homes is supplied by a well less than 30 ft. deep with a pump at the surface
  • the pump removes air from the pipe, decreasing the air pressure in the pipe
  • the outside air pressure then pushes the water up the pipe
  • the maximum height the water will rise is related to the amount of pressure the air exerts

Tro, Chemistry: A Molecular Approach

atmospheric pressure
Atmospheric Pressure
  • pressure is the force exerted over an area
  • on average, the air exerts the same pressure that a column of water 10.3 m high would exert
    • 14.7 lbs./in2
    • so if our pump could get a perfect vacuum, the maximum height the column could rise is 10.3 m

Tro, Chemistry: A Molecular Approach

gases pushing
Gases Pushing
  • gas molecules are constantly in motion
  • as they move and strike a surface, they push on that surface
    • push = force
  • if we could measure the total amount of force exerted by gas molecules hitting the entire surface at any one instant, we would know the pressure the gas is exerting
    • pressure = force per unit area

Tro, Chemistry: A Molecular Approach

the effect of gas pressure
The Effect of Gas Pressure
  • the pressure exerted by a gas can cause some amazing and startling effects
  • whenever there is a pressure difference, a gas will flow from area of high pressure to low pressure
    • the bigger the difference in pressure, the stronger the flow of the gas
  • if there is something in the gas’s path, the gas will try to push it along as the gas flows

Tro, Chemistry: A Molecular Approach

atmospheric pressure effects
Atmospheric Pressure Effects
  • differences in air pressure result in weather and wind patterns
  • the higher up in the atmosphere you climb, the lower the atmospheric pressure is around you
    • at the surface the atmospheric pressure is 14.7 psi, but at 10,000 ft it is only 10.0 psi
  • rapid changes in atmospheric pressure may cause your ears to “pop” due to an imbalance in pressure on either side of your ear drum

Tro, Chemistry: A Molecular Approach

pressure imbalance in ear
Pressure Imbalance in Ear

If there is a difference

in pressure across

the eardrum membrane,

the membrane will be

pushed out – what we

commonly call a

“popped eardrum.”

Tro, Chemistry: A Molecular Approach

the pressure of a gas
The Pressure of a Gas
  • result of the constant movement of the gas molecules and their collisions with the surfaces around them
  • the pressure of a gas depends on several factors
    • number of gas particles in a given volume
    • volume of the container
    • average speed of the gas particles

Tro, Chemistry: A Molecular Approach

measuring air pressure

gravity

Measuring Air Pressure
  • use abarometer
  • column of mercury supported by air pressure
  • force of the air on the surface of the mercury balanced by the pull of gravity on the column of mercury

Tro, Chemistry: A Molecular Approach

common units of pressure
Common Units of Pressure

Tro, Chemistry: A Molecular Approach

example 5 1 a high performance bicycle tire has a pressure of 132 psi what is the pressure in mmhg

psi

atm

mmHg

Example 5.1 – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg?

Given:

Find:

132 psi

mmHg

Concept Plan:

Relationships:

1 atm = 14.7 psi, 1 atm = 760 mmHg

Solution:

Check:

since mmHg are smaller than psi, the answer makes sense

manometers
Manometers
  • the pressure of a gas trapped in a container can be measured with an instrument called a manometer
  • manometers are U-shaped tubes, partially filled with a liquid, connected to the gas sample on one side and open to the air on the other
  • a competition is established between the pressure of the atmosphere and the gas
  • the difference in the liquid levels is a measure of the difference in pressure between the gas and the atmosphere

Tro, Chemistry: A Molecular Approach

manometer
Manometer

for this sample, the gas has a larger pressure than the atmosphere, so

Tro, Chemistry: A Molecular Approach

boyle s law
Boyle’s Law
  • pressure of a gas is inversely proportional to its volume
    • constant T and amount of gas
    • graph P vs V is curve
    • graph P vs 1/V is straight line
  • as P increases, V decreases by the same factor
  • P x V = constant
  • P1 x V1 = P2 x V2

Tro, Chemistry: A Molecular Approach

boyle s experiment
Boyle’s Experiment
  • added Hg to a J-tube with air trapped inside
  • used length of air column as a measure of volume

Tro, Chemistry: A Molecular Approach

boyle s experiment p x v
Boyle’s Experiment, P x V

Tro, Chemistry: A Molecular Approach

slide19

When you double the pressure on a gas,

the volume is cut in half (as long as the

temperature and amount of gas do not change)

Tro, Chemistry: A Molecular Approach

boyle s law and diving
Boyle’s Law and Diving

if your tank contained air at 1 atm pressure you would not be able to inhale it into your lungs

  • since water is denser than air, for each 10 m you dive below the surface, the pressure on your lungs increases 1 atm
    • at 20 m the total pressure is 3 atm

Tro, Chemistry: A Molecular Approach

slide21

V1, P1, P2

V2

Example 5.2 – A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?

Given:

Find:

V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm

V2, L

Concept Plan:

Relationships:

P1∙ V1 = P2∙ V2

Solution:

Check:

since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does

slide22

Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally?

Tro, Chemistry: A Molecular Approach

slide23

V1, P1, P2

V2

A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally?

Given:

Find:

V2 =2780 mL, P1 = 762 torr, P2 = 0.500 atm

V1, mL

Concept Plan:

Relationships:

P1∙ V1 = P2∙ V2 , 1 atm = 760 torr (exactly)

Solution:

Check:

since P and V are inversely proportional, when the pressure decreases ~2x, the volume should increase ~2x, and it does

charles law
Charles’ Law
  • volume is directly proportional to temperature
    • constant P and amount of gas
    • graph of V vs T is straight line
  • as T increases, V also increases
  • Kelvin T = Celsius T + 273
  • V = constant x T
    • if T measured in Kelvin

Tro, Chemistry: A Molecular Approach

charles law a molecular view
Charles’ Law – A Molecular View
  • the pressure of gas inside and outside the balloon are the same
  • at high temperatures, the gas molecules are moving faster, so they hit the sides of the balloon harder – causing the volume to become larger
  • the pressure of gas inside and outside the balloon are the same
  • at low temperatures, the gas molecules are not moving as fast, so they don’t hit the sides of the balloon as hard – therefore the volume is small

Tro, Chemistry: A Molecular Approach

slide26

The data fall on a straight line.

If the lines are extrapolated back to a volume of “0,” they all show the same temperature, -273.15°C, called absolute zero

example 5 3 a gas has a volume of 2 57 l at 0 00 c what was the temperature at 2 80 l

T(K) = t(°C) + 273.15,

V1, V2, T2

T1

Example 5.3 – A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L?

Given:

Find:

V1 =2.57 L, V2 = 2.80 L, t2 = 0.00°C

t1, K and °C

Concept Plan:

Relationships:

Solution:

Check:

since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does

slide28

Practice – The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air?

Tro, Chemistry: A Molecular Approach

slide29

T(K) = t(°C) + 273.15,

V1, T1, T2

V2

The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air?

Given:

Find:

V1 =10.0 L, t1 = 25.0°C L, t2 = 250.0°C

V2, L

Concept Plan:

Relationships:

Solution:

Check:

since T and V are directly proportional, when the temperature increases, the volume should increase, and it does

avogadro s law
Avogadro’s Law
  • volume directly proportional to the number of gas molecules
    • V = constant x n
    • constant P and T
    • more gas molecules = larger volume
  • count number of gas molecules by moles
  • equal volumes of gases contain equal numbers of molecules
    • the gas doesn’t matter

Tro, Chemistry: A Molecular Approach

slide31

mol added = n2 – n1,

V1, V2, n1

n2

Example 5.4 – A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?

Given:

Find:

V1 =4.65 L, V2 = 6.48 L, n1 = 0.225 mol

n2, and added moles

Concept Plan:

Relationships:

Solution:

Check:

since n and V are directly proportional, when the volume increases, the moles should increase, and it does

ideal gas law

By combing the gas laws we can write a general equation

  • R is called the gas constant
  • the value of R depends on the units of P and V
    • we will use 0.08206 and convert P to atm and V to L
    • the other gas laws are found in the ideal gas law if
    • two variables are kept constant
    • allows us to find one of the variables if we know the other 3
Ideal Gas Law

Tro, Chemistry: A Molecular Approach

slide33

1 atm = 14.7 psi

T(K) = t(°C) + 273.15

P, V, T, R

n

Example 5.6 – How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C?

Given:

Find:

V = 3.24 L, P = 24.3 psi, t = 25 °C,

n, mol

Concept Plan:

Relationships:

Solution:

Check:

1 mole at STP occupies 22.4 L, since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas

standard conditions
Standard Conditions
  • since the volume of a gas varies with pressure and temperature, chemists have agreed on a set of conditions to report our measurements so that comparison is easy – we call these standard conditions
    • STP
  • standard pressure = 1 atm
  • standard temperature = 273 K
    • 0°C

Tro, Chemistry: A Molecular Approach

slide35
Practice – A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions?

Tro, Chemistry: A Molecular Approach

a gas occupies 10 0 l at 44 1 psi and 27 c what volume will it occupy at standard conditions

1 atm = 14.7 psi

T(K) = t(°C) + 273.15

P1, V1, T1, R

n

P2, n, T2, R

V2

A gas occupies 10.0 L at 44.1 psi and 27°C. What volume will it occupy at standard conditions?

Given:

Find:

V1 = 10.0 L, P1 = 44.1 psi, t1 = 27 °C, P2 = 1.00 atm, t2 = 0°C

V2, L

Concept Plan:

Relationships:

Solution:

Check:

1 mole at STP occupies 22.4 L, since there is more than 1 mole, we expect more than 22.4 L of gas

molar volume
Molar Volume
  • solving the ideal gas equation for the volume of 1 mol of gas at STP gives 22.4 L
    • 6.022 x 1023 molecules of gas
    • notice: the gas is immaterial
  • we call the volume of 1 mole of gas at STP the molar volume
    • it is important to recognize that one mole of different gases have different masses, even though they have the same volume

Tro, Chemistry: A Molecular Approach

molar volume1
Molar Volume

Tro, Chemistry: A Molecular Approach

density at standard conditions
Density at Standard Conditions
  • density is the ratio of mass-to-volume
  • density of a gas is generally given in g/L
  • the mass of 1 mole = molar mass
  • the volume of 1 mole at STP = 22.4 L

Tro, Chemistry: A Molecular Approach

gas density
Gas Density
  • density is directly proportional to molar mass

Tro, Chemistry: A Molecular Approach

example 5 7 calculate the density of n 2 at 125 c and 755 mmhg

1 atm = 760 mmHg, MM = 28.01 g

T(K) = t(°C) + 273.15

P, MM, T, R

d

Example 5.7 – Calculate the density of N2at 125°C and 755 mmHg

Given:

Find:

P = 755 mmHg, t = 125 °C,

dN2,g/L

Concept Plan:

Relationships:

Solution:

since the density of N2 is 1.25 g/L at STP, we expect the density to be lower when the temperature is raised, and it is

Check:

molar mass of a gas
Molar Mass of a Gas
  • one of the methods chemists use to determine the molar mass of an unknown substance is to heat a weighed sample until it becomes a gas, measure the temperature, pressure, and volume, and use the ideal gas law

Tro, Chemistry: A Molecular Approach

slide43

n, m

P, V, T, R

MM

n

1 atm = 760 mmHg,

T(K) = t(°C) + 273.15

Example 5.8 – Calculate the molar mass of a gas with mass 0.311 g that has a volume of 0.225 L at 55°C and 886 mmHg

Given:

Find:

m=0.311g, V=0.225 L, P=886 mmHg, t=55°C,

molar mass,g/mol

m=0.311g, V=0.225 L, P=1.1658 atm, T=328 K,

molar mass,g/mol

Concept Plan:

Relationships:

Solution:

Check:

the value 31.9 g/mol is reasonable

practice calculate the density of a gas at 775 torr and 27 c if 0 250 moles weighs 9 988 g
Practice - Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g

Tro, Chemistry: A Molecular Approach

calculate the density of a gas at 775 torr and 27 c if 0 250 moles weighs 9 988 g

V, m

P, n, T, R

d

V

1 atm = 760 mmHg,

T(K) = t(°C) + 273.15

Calculate the density of a gas at 775 torr and 27°C if 0.250 moles weighs 9.988 g

Given:

Find:

m=9.988g, n=0.250 mol, P=775 mmHg, t=27°C,

density,g/L

m=9.988g, n=0.250 mol, P=1.0197 atm, T=300. K

density,g/L

Concept Plan:

Relationships:

Solution:

Check:

the value 1.65 g/L is reasonable

mixtures of gases
Mixtures of Gases
  • when gases are mixed together, their molecules behave independent of each other
    • all the gases in the mixture have the same volume
      • all completely fill the container  each gas’s volume = the volume of the container
    • all gases in the mixture are at the same temperature
      • therefore they have the same average kinetic energy
  • therefore, in certain applications, the mixture can be thought of as one gas
    • even though air is a mixture, we can measure the pressure, volume, and temperature of air as if it were a pure substance
    • we can calculate the total moles of molecules in an air sample, knowing P, V, and T, even though they are different molecules

Tro, Chemistry: A Molecular Approach

partial pressure
Partial Pressure
  • the pressure of a single gas in a mixture of gases is called its partial pressure
  • we can calculate the partial pressure of a gas if
    • we know what fraction of the mixture it composes and the total pressure
    • or, we know the number of moles of the gas in a container of known volume and temperature
  • the sum of the partial pressures of all the gases in the mixture equals the total pressure
    • Dalton’s Law of Partial Pressures
    • because the gases behave independently

Tro, Chemistry: A Molecular Approach

composition of dry air
Composition of Dry Air

Tro, Chemistry: A Molecular Approach

the partial pressure of each gas in a mixture can be calculated using the ideal gas law
The partial pressure of each gas in a mixture can be calculated using the ideal gas law

Tro, Chemistry: A Molecular Approach

example 5 9 determine the mass of ar in the mixture

Ptot, PHe, PNe

PAr

PAr, V, T

nAr

mAr

Ptot = Pa + Pb + etc.,

1 atm = 760 mmHg, MMAr = 39.95 g/mol

Example 5.9 – Determine the mass of Ar in the mixture

PHe=341 mmHg, PNe=112 mmHg, Ptot = 662 mmHg,

V = 1.00 L, T=298 K

massAr,g

PAr = 0.275 atm, V = 1.00 L, T=298 K

massAr,g

Given:

Find:

Concept Plan:

Relationships:

PAr = Ptot – (PHe + PNe)

Solution:

Check:

the units are correct, the value is reasonable

slide51
Practice – Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe.

Tro, Chemistry: A Molecular Approach

slide52

nXe, V, T, R

PXe

Ptot, PXe

PNe

Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe

Given:

Find:

Ptot = 3.9 atm, V = 8.7 L, T = 598 K, Xe = 0.17 mol

PNe, atm

Concept Plan:

Relationships:

Solution:

the unit is correct, the value is reasonable

Check:

mole fraction
Mole Fraction

the fraction of the total pressure that a single gas contributes is equal to the fraction of the total number of moles that a single gas contributes

the ratio of the moles of a single component to the total number of moles in the mixture is called the mole fraction, c

for gases, = volume % / 100%

the partial pressure of a gas is equal to the mole fraction of that gas times the total pressure

Tro, Chemistry: A Molecular Approach

mountain climbing partial pressure
Mountain Climbing & Partial Pressure
  • our bodies are adapted to breathe O2 at a partial pressure of 0.21 atm
    • Sherpa, people native to the Himalaya mountains, are adapted to the much lower partial pressure of oxygen in their air
  • partial pressures of O2 lower than 0.1 atm will lead to hypoxia
    • unconsciousness or death
  • climbers of Mt Everest carry O2 in cylinders to prevent hypoxia
    • on top of Mt Everest, Pair = 0.311 atm, so PO2 = 0.065 atm

Tro, Chemistry: A Molecular Approach

deep sea divers partial pressure
Deep Sea Divers & Partial Pressure
  • its also possible to have too much O2, a condition called oxygen toxicity
    • PO2 > 1.4 atm
    • oxygen toxicity can lead to muscle spasms, tunnel vision, and convulsions
  • its also possible to have too much N2, a condition called nitrogen narcosis
    • also known as Rapture of the Deep
  • when diving deep, the pressure of the air divers breathe increases – so the partial pressure of the oxygen increases
    • at a depth of 55 m the partial pressure of O2 is 1.4 atm
    • divers that go below 50 m use a mixture of He and O2 called heliox that contains a lower percentage of O2 than air

Tro, Chemistry: A Molecular Approach

partial pressure diving
Partial Pressure & Diving

Tro, Chemistry: A Molecular Approach

slide57

mgas

ngas

cgas

ntot, V, T, R

Ptot

cgas, Ptotal

Pgas

MMHe = 4.00 g/mol

MMO2 = 32.00 g/mol

Ex 5.10 – Find the mole fractions and partial pressures in a 12.5 L tank with 24.2 g He and 4.32 g O2 at 298 K

Given:

Find:

mHe = 24.2 g, mO2 = 43.2 g V = 12.5 L, T = 298 K

cHe, cO2, PHe, atm, PO2, atm, Ptotal, atm

nHe = 6.05 mol, nO2 = 0.135 mol V = 12.5 L, T = 298 K

cHe=0.97817, cO2=0.021827, PHe, atm, PO2, atm, Ptotal, atm

Concept Plan:

Relationships:

Solution:

collecting gases
Collecting Gases
  • gases are often collected by having them displace water from a container
  • the problem is that since water evaporates, there is also water vapor in the collected gas
  • the partial pressure of the water vapor, called the vapor pressure, depends only on the temperature
    • so you can use a table to find out the partial pressure of the water vapor in the gas you collect
  • if you collect a gas sample with a total pressure of 758.2 mmHg* at 25°C, the partial pressure of the water vapor will be 23.78 mmHg – so the partial pressure of the dry gas will be 734.4 mmHg
    • Table 5.4*

Tro, Chemistry: A Molecular Approach

vapor pressure of water
Vapor Pressure of Water

Tro, Chemistry: A Molecular Approach

collecting gas by water displacement
Collecting Gas by Water Displacement

Tro, Chemistry: A Molecular Approach

slide61

Ptot, PH2O

PO2

PO2,V,T

nO2

gO2

1 atm = 760 mmHg,

Ptotal = PA + PB, O2 = 32.00 g/mol

Ex 5.11 – 1.02 L of O2 collected over water at 293 K with a total pressure of 755.2 mmHg. Find mass O2.

Given:

Find:

V=1.02 L, P=755.2 mmHg, T=293 K

mass O2,g

V=1.02 L, PO2=737.65 mmHg, T=293 K

mass O2,g

Concept Plan:

Relationships:

Solution:

slide62
Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

Tro, Chemistry: A Molecular Approach

0 12 moles of h 2 is collected over water in a 10 0 l container at 323 k find the total pressure

PH2, PH2O

Ptotal

nH2,V,T

PH2

1 atm = 760 mmHg

Ptotal = PA + PB,

0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

Given:

Find:

V=10.0 L, nH2=0.12 mol, T=323 K

Ptotal,atm

Concept Plan:

Relationships:

Solution:

reactions involving gases
Reactions Involving Gases
  • the principles of reaction stoichiometry from Chapter 4 can be combined with the gas laws for reactions involving gases
  • in reactions of gases, the amount of a gas is often given as a volume
    • instead of moles
    • as we’ve seen, must state pressure and temperature
  • the ideal gas law allows us to convert from the volume of the gas to moles; then we can use the coefficients in the equation as a mole ratio
  • when gases are at STP, use 1 mol = 22.4 L

P, V, T of Gas A

mole A

mole B

P, V, T of Gas B

Tro, Chemistry: A Molecular Approach

slide65

g CH3OH

mol CH3OH

mol H2

1 atm = 760 mmHg, CH3OH = 32.04 g/mol

1 mol CH3OH : 2 mol H2

P, n, T, R

V

Ex 5.12 – What volume of H2 is needed to make 35.7 g of CH3OH at 738 mmHg and 355 K?CO(g) + 2 H2(g) → CH3OH(g)

Given:

Find:

mCH3OH = 37.5g, P=738 mmHg, T=355 K

VH2,L

nH2 = 2.2284 mol, P=0.97105 atm, T=355 K

VH2,L

Concept Plan:

Relationships:

Solution:

slide66

L H2

mol H2

mol H2O

g H2O

Ex 5.13 – How many grams of H2O form when 1.24 L H2 reacts completely with O2 at STP?O2(g) + 2 H2(g) → 2 H2O(g)

Given:

Find:

VH2 = 1.24 L, P=1.00 atm, T=273 K

massH2O,g

Concept Plan:

Relationships:

H2O = 18.02 g/mol, 1 mol = 22.4 L @ STP

2 mol H2O : 2 mol H2

Solution:

slide67

Practice – What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO?2 HgO(s)  2 Hg(l) + O2(g)(MMHgO = 216.59 g/mol)

Tro, Chemistry: A Molecular Approach

slide68

g HgO

mol HgO

mol O2

1 atm = 760 mmHg, HgO = 216.59 g/mol

2 mol HgO : 1 mol O2

P, n, T, R

V

What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO?2 HgO(s)  2 Hg(l) + O2(g)

Given:

Find:

mHgO = 10.0g, P=0.750 atm, T=313 K

VO2,L

nO2 = 0.023085 mol, P=0.750 atm, T=313 K

VO2,L

Concept Plan:

Relationships:

Solution:

properties of gases
Properties of Gases
  • expand to completely fill their container
  • take the shape of their container
  • low density
    • much less than solid or liquid state
  • compressible
  • mixtures of gases are always homogeneous
  • fluid

Tro, Chemistry: A Molecular Approach

kinetic molecular theory
Kinetic Molecular Theory
  • the particles of the gas (either atoms or molecules) are constantly moving
  • the attraction between particles is negligible
  • when the moving particles hit another particle or the container, they do not stick; but they bounce off and continue moving in another direction
    • like billiard balls

Tro, Chemistry: A Molecular Approach

kinetic molecular theory1
Kinetic Molecular Theory
  • there is a lot of empty space between the particles
    • compared to the size of the particles
  • the average kinetic energy of the particles is directly proportional to the Kelvin temperature
    • as you raise the temperature of the gas, the average speed of the particles increases
      • but don’t be fooled into thinking all the particles are moving at the same speed!!

Tro, Chemistry: A Molecular Approach

gas properties explained indefinite shape and indefinite volume
Gas Properties Explained – Indefinite Shape and Indefinite Volume

Because the gas

molecules have

enough kinetic

energy to overcome

attractions, they

keep moving around

and spreading out

until they fill the

container.

As a result, gases

take the shape and

the volume of the

container they

are in.

Tro, Chemistry: A Molecular Approach

gas properties explained compressibility
Gas Properties Explained - Compressibility

Because there is a lot of unoccupied space in the structure

of a gas, the gas molecules can be squeezed closer together

Tro, Chemistry: A Molecular Approach

gas properties explained low density
Gas Properties Explained – Low Density

Because there is a lot of unoccupied space in the structure of a gas, gases do not have a lot of mass in a given volume, the result is they have low density

Tro, Chemistry: A Molecular Approach

density pressure
Density & Pressure
  • result of the constant movement of the gas molecules and their collisions with the surfaces around them
  • when more molecules are added, more molecules hit the container at any one instant, resulting in higher pressure
    • also higher density

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gas laws explained boyle s law
Gas Laws Explained - Boyle’s Law
  • Boyle’s Law says that the volume of a gas is inversely proportional to the pressure
  • decreasing the volume forces the molecules into a smaller space
  • more molecules will collide with the container at any one instant, increasing the pressure

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gas laws explained charles s law
Gas Laws Explained - Charles’s Law
  • Charles’s Law says that the volume of a gas is directly proportional to the absolute temperature
  • increasing the temperature increases their average speed, causing them to hit the wall harder and more frequently
    • on average
  • in order to keep the pressure constant, the volume must then increase

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gas laws explained avogadro s law
Gas Laws ExplainedAvogadro’s Law
  • Avogadro’s Law says that the volume of a gas is directly proportional to the number of gas molecules
  • increasing the number of gas molecules causes more of them to hit the wall at the same time
  • in order to keep the pressure constant, the volume must then increase

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gas laws explained dalton s law of partial pressures
Gas Laws Explained – Dalton’s Law of Partial Pressures
  • Dalton’s Law says that the total pressure of a mixture of gases is the sum of the partial pressures
  • kinetic-molecular theory says that the gas molecules are negligibly small and don’t interact
  • therefore the molecules behave independent of each other, each gas contributing its own collisions to the container with the same average kinetic energy
  • since the average kinetic energy is the same, the total pressure of the collisions is the same

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dalton s law pressure
Dalton’s Law & Pressure
  • since the gas molecules are not sticking together, each gas molecule contributes its own force to the total force on the side

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deriving the ideal gas law from kinetic molecular theory
Deriving the Ideal Gas Law from Kinetic-Molecular Theory
  • pressure = Forcetotal/Area
  • Ftotal = F1 collisionx number of collisions
    • in a particular time interval
    • F1 collision = mass x 2(velocity)/time interval
    • no. of collisions is proportional to the number of particles within the distance (velocity x time interval) from the wall
    • Ftotalαmass∙velocity2x Areaxno. molecules/Volume
  • Pressure α mv2 x n/V
  • Temperature α mv2
  • P α T∙n/V,  PV=nRT

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calculating gas pressure
Calculating Gas Pressure

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molecular velocities
Molecular Velocities
  • all the gas molecules in a sample can travel at different speeds
  • however, the distribution of speeds follows a pattern called a Boltzman distribution
  • we talk about the “average velocity” of the molecules, but there are different ways to take this kind of average
  • the method of choice for our average velocity is called the root-mean-square method, where the rms average velocity, urms, is the square root of the average of the sum of the squares of all the molecule velocities

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boltzman distribution
Boltzman Distribution

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kinetic energy and molecular velocities
Kinetic Energy and Molecular Velocities
  • average kinetic energy of the gas molecules depends on the average mass and velocity
    • KE = ½mv2
  • gases in the same container have the same temperature, the same average kinetic energy
  • if they have different masses, the only way for them to have the same kinetic energy is to have different average velocities
    • lighter particles will have a faster average velocity than more massive particles

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molecular speed vs molar mass
Molecular Speed vs. Molar Mass
  • in order to have the same average kinetic energy, heavier molecules must have a slower average speed

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temperature and molecular velocities
Temperature and Molecular Velocities

_

  • KEavg = ½NAmu2
    • NA is Avogadro’s number
  • KEavg = 1.5RT
    • R is the gas constant in energy units, 8.314 J/mol∙K
      • 1 J = 1 kg∙m2/s2
  • equating and solving we get:
    • NA∙mass = molar mass in kg/mol
  • as temperature increases, the average velocity increases

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temperature vs molecular speed
Temperature vs. Molecular Speed
  • as the absolute temperature increases, the average velocity increases
    • the distribution function “spreads out,” resulting in more molecules with faster speeds

Tro, Chemistry: A Molecular Approach

ex 5 14 calculate the rms velocity of o 2 at 25 c

T(K) = t(°C) + 273.15, O2 = 32.00 g/mol

MM, T

urms

Ex 5.14 – Calculate the rms velocity of O2 at 25°C

Given:

Find:

O2, t = 25°C

urms

Concept Plan:

Relationships:

Solution:

mean free path
Mean Free Path
  • molecules in a gas travel in straight lines until they collide with another molecule or the container
  • the average distance a molecule travels between collisions is called the mean free path
  • mean free path decreases as the pressure increases

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diffusion and effusion
Diffusion and Effusion
  • the process of a collection of molecules spreading out from high concentration to low concentration is called diffusion
  • the process by which a collection of molecules escapes through a small hole into a vacuum is called effusion
  • both the rates of diffusion and effusion of a gas are related to its rms average velocity
  • for gases at the same temperature, this means that the rate of gas movement is inversely proportional to the square root of the molar mass

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effusion
Effusion

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graham s law of effusion
Graham’s Law of Effusion
  • for two different gases at the same temperature, the ratio of their rates of effusion is given by the following equation:

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ex 5 15 calculate the molar mass of a gas that effuses at a rate 0 462 times n 2

N2 = 28.01 g/mol

rateA/rateB, MMN2

MMunknown

Ex 5.15 – Calculate the molar mass of a gas that effuses at a rate 0.462 times N2

Given:

Find:

MM, g/mol

Concept Plan:

Relationships:

Solution:

ideal vs real gases
Ideal vs. Real Gases
  • Real gases often do not behave like ideal gases at high pressure or low temperature
  • Ideal gas laws assume
    • no attractions between gas molecules
    • gas molecules do not take up space
    • based on the kinetic-molecular theory
  • at low temperatures and high pressures these assumptions are not valid
the effect of molecular volume
The Effect of Molecular Volume
  • at high pressure, the amount of space occupied by the molecules is a significant amount of the total volume
  • the molecular volume makes the real volume larger than the ideal gas law would predict
  • van der Waals modified the ideal gas equation to account for the molecular volume
    • b is called a van der Waals constant and is different for every gas because their molecules are different sizes

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real gas behavior
Real Gas Behavior
  • because real molecules take up space, the molar volume of a real gas is larger than predicted by the ideal gas law at high pressures

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the effect of intermolecular attractions
The Effect of Intermolecular Attractions
  • at low temperature, the attractions between the molecules is significant
  • the intermolecular attractions makes the real pressure less than the ideal gas law would predict
  • van der Waals modified the ideal gas equation to account for the intermolecular attractions
    • a is called a van der Waals constant and is different for every gas because their molecules are different sizes

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real gas behavior1
Real Gas Behavior
  • because real molecules attract each other, the molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures

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van der waals equation
Van der Waals’ Equation
  • combining the equations to account for molecular volume and intermolecular attractions we get the following equation
    • used for real gases
    • a and b are called van der Waal constants and are different for each gas

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real gases
Real Gases
  • a plot of PV/RT vs. P for 1 mole of a gas shows the difference between real and ideal gases
  • it reveals a curve that shows the PV/RT ratio for a real gas is generally lower than ideality for “low” pressures – meaning the most important factor is the intermolecular attractions
  • it reveals a curve that shows the PV/RT ratio for a real gas is generally higher than ideality for “high” pressures – meaning the most important factor is the molecular volume

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pv rt plots
PV/RT Plots

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structure of the atmosphere
Structure of the Atmosphere
  • the atmosphere shows several layers, each with its own characteristics
  • the troposphere is the layer closest to the earth’s surface
    • circular mixing due to thermal currents – weather
  • the stratosphere is the next layer up
    • less air mixing
  • the boundary between the troposphere and stratosphere is called the tropopause
  • the ozone layeris located in the stratosphere

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air pollution
Air Pollution
  • air pollution is materials added to the atmosphere that would not be present in the air without, or are increased by, man’s activities
    • though many of the “pollutant” gases have natural sources as well
  • pollution added to the troposphere has a direct effect on human health and the materials we use because we come in contact with it
    • and the air mixing in the troposphere means that we all get a smell of it!
  • pollution added to the stratosphere may have indirect effects on human health caused by depletion of ozone
    • and the lack of mixing and weather in the stratosphere means that pollutants last longer before “washing” out

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pollutant gases so x
Pollutant Gases, SOx
  • SO2 and SO3, oxides of sulfur, come from coal combustion in power plants and metal refining
    • as well as volcanoes
  • lung and eye irritants
  • major contributor to acid rain

2 SO2 + O2 + 2 H2O  2 H2SO4

SO3 + H2O  H2SO4

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pollutant gases no x
Pollutant Gases, NOx
  • NO and NO2, oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plants
    • as well as lightning storms
  • NO2 causes the brown haze seen in some cities
  • lung and eye irritants
  • strong oxidizers
  • major contributor to acid rain

4 NO + 3 O2 + 2 H2O  4 HNO3

4 NO2 + O2 + 2 H2O  4 HNO3

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pollutant gases co
Pollutant Gases, CO
  • CO comes from incomplete burning of fossil fuels in cars, trucks, and power plants
  • adheres to hemoglobin in your red blood cells, depleting your ability to acquire O2
  • at high levels can cause sensory impairment, stupor, unconsciousness, or death

Tro, Chemistry: A Molecular Approach

pollutant gases o 3
Pollutant Gases, O3
  • ozone pollution comes from other pollutant gases reacting in the presence of sunlight
    • as well as lightning storms
    • known as photochemical smog and ground-level ozone
  • O3 is present in the brown haze seen in some cities
  • lung and eye irritants
  • strong oxidizer

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major pollutant levels
Major Pollutant Levels
  • government regulation has resulted in a decrease in the emission levels for most major pollutants

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stratospheric ozone
Stratospheric Ozone
  • ozone occurs naturally in the stratosphere
  • stratospheric ozone protects the surface of the earth from over-exposure to UV light from the sun

O3(g) + UV light  O2(g) + O(g)

  • normally the reverse reaction occurs quickly, but the energy is not UV light

O2(g) + O(g)  O3(g)

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ozone depletion
Ozone Depletion
  • chlorofluorocarbons became popular as aerosol propellants and refrigerants in the 1960s
  • CFCs pass through the tropopause into the stratosphere
  • there CFCs can be decomposed by UV light, releasing Cl atoms

CF2Cl2 + UV light  CF2Cl + Cl

  • Cl atoms catalyze O3 decomposition and removes O atoms so that O3 cannot be regenerated
    • NO2 also catalyzes O3 destruction

Cl + O3 ClO + O2

O3 + UV light  O2 + O

ClO + O  O2 + Cl

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ozone holes
Ozone Holes
  • satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions

Tro, Chemistry: A Molecular Approach