Unit 6 Gases and Gas Laws

1
Unit 6Gases and Gas Laws
2
Gases in the Atmosphere

• The atmosphere of Earth is a layer of gases
  surrounding the planet that is retained by
  Earth's gravity.
• By volume, dry airis 78 nitrogen, 21 oxygen,
  0.9 argon, 0.04 CO2, and small amounts of
  other gases.

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Air Pollution

• Human activity has polluted the air with other
  gases
• Sulfur Oxides (SO2 SO3) produced from coal
  burning. Contribute to acid rain.
• Nitrogen Oxides (NO NO2) produced by burning
  fossil fuels. Contribute to acid rain.
• Carbon Monoxide (CO) emitted by motor
  vehicles.
• Ground-level Ozone (O3) produced when products
  offossil fuel combustion reactin the presence
  of sunlight.

4
The Ozone Layer

• O3 in the troposphere (ground-level ozone) is a
  pollutant, but O3 in the stratosphere is a
  necessary part of our atmosphere.
• Stratospheric O3 protectsus by absorbing UV
  light.
• CFCs destroy stratospheric O3, and have been
  banned in the US.
• Ozone Good up high,bad nearby.

5
Atmospheric Pressure

• Atmospheric pressure is the force per unit area
  exerted on a surface by the weight of the gases
  that make up the atmosphere above it.

Force
Pressure
Area
6
Measuring Pressure

• A common unit of pressure is millimeters of
  mercury (mm Hg).
• 1 mm Hg is also called 1 torr in honor of
  Evangelista Torricelli whoinvented the barometer
  (used tomeasure atmospheric pressure).
• The average atmospheric pressure at sea level at
  0C is 760 mm Hg, so one atmosphere (atm) of
  pressure is 760 mm Hg.

7
Measuring Pressure (continued)

• The pressure of a gas sample in the laboratory is
  often measured with a manometer.
• the difference in the liquid levels is a measure
  of the difference in pressure between the gas and
  the atmosphere.

For this sample, the gas has a larger pressure
than the atmosphere.
8
Measuring Pressure (continued)

• Pressure can also be measured in pascals (Pa) 1
  Pa 1 N/m2.
• One pascal is verysmall, so usually kilopascals
  (kPa) are used instead.
• One atm is equal to 101.3 kPa.

1 atm 760 mm Hg (Torr) 101.3 kPa
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Units of Pressure
10
Converting PressureSample Problem

• The average atmospheric pressure in Denver, CO is
  0.830 atm. Express this pressure in
• a. millimeters of mercury (mm Hg)
• b. kilopascals (kPa)

mm Hg
760
x
0.830 atm
631 mm Hg

atm
1
kPa
101.3
x
0.830 atm
84.1 kPa

atm
1
11
Daltons Law of Partial Pressures

• Daltons law of partial pressures - the total
  pressure of a gas mixture is the sum of the
  partial pressures of the component gases.

PT P1 P2 P3
12
Daltons Law of Partial PressuresSample Problem

• A container holds a mixture of gases A, B C.
  Gas A has a pressure of 0.5 atm, Gas B has a
  pressure of 0.7 atm, and Gas C has a pressure of
  1.2 atm.
• What is the total pressure of this system?
• b. What is the total pressure in mm Hg?

PT P1 P2 P3
PT
0.5 atm
0.7 atm
1.2 atm

2.4 atm
mm Hg
760
x
2.4 atm
1800 mm Hg

atm
1
13
The Kinetic-Molecular Theory A Model for Gases

• Matter is composed of particles which are
  constantly moving.
• The average kinetic energyof a particle is
  proportionalto its Kelvin temperature.
• The size of a particle is negligibly small.
• Collisions are completelyelastic energy may
  beexchanged, but not lost(like billiard balls.)

14
Ideal Gases

• The kinetic-molecular theory assumes
• no attractions between gas molecules
• gas molecules do not take up space
• An Ideal Gas is a hypothetical gas that perfectly
  fits the assumptions of the kinetic-molecular
  theory.
• Many gases behave nearlyideally if pressure is
  not veryhigh and temperature is not very low.

15
Properties of Gases Fluidity

• Gas particles glide easily past one another.
  Because liquids and gases flow, they are both
  referred to as fluids.

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Properties of Gases Expansion

• Since theres no significantattraction between
  gasmolecules, they keep moving around and
  spreading out until they fill their container.
• As a result, gases take the shape and the volume
  of the container they are in.

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Properties of Gases Low Density

• Gas particles are very far apart.There is a lot
  of unoccupied space in the structure of a gas.
• Since gases do not have a lot of mass in a given
  volume, they have a very low density
• The density of a gas is about 1/1000 the density
  of the same substance as a liquid or solid.

18
Properties of Gases Compressibility

• Because there is a lot of unoccupied space in the
  structure of a gas, the gas molecules can easily
  be squeezed closer together.

19
Diffusion and Effusion

• Diffusion is the gradual mixing of two or more
  gases due to their spontaneous, random motion.
• Effusion is the process whereby the molecules
  of a gas confined in a container randomly pass
  through a tiny opening in the container.

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Rate of Diffusion

• Light molecules move faster than heavy ones.
• The greater the molar mass of a gas,the slower
  it will diffuse and/or effuse.

21
Grahams Law of Effusion

• Grahams Law of Effusion states that the rate of
  effusion is inversely proportional to the square
  root of the molar mass of the gas.
• The ratio of effusion rates of two different
  gases is given by the following equation

22
Grahams Law of EffusionSample Problem

• Calculate the molar mass of a gas that effuses at
  a rate 0.462 times N2 .
• Solution

28.0 g/mol
(0.462)2
MMunknown
28.0 g/mol
28.0 g/mol


131 g/mol

MMunknown
(0.462)2
(0.213)
23
Gases and Pressure

• Gas pressure is caused by collisions of the gas
  molecules with each other and with the walls of
  their container.
• The greater the number of collisions, the higher
  the pressure will be.

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Pressure Volume Relationship

• When the volume of a gas is decreased, more
  collisions will occur.
• Pressure is caused by collisions.
• Therefore, pressure will increase.

25
Boyles Law

• Boyles Law The volume of a fixed mass of gas
  varies inversely with the pressure at a constant
  temperature.
• P1 and V1 representinitial conditions, andP2
  and V2 representanother set of conditions.

P1V1 P2V2
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Boyles LawSample Problem

• A sample of oxygen gas has a volume of 150.0 mL
  when its pressure is 0.947 atm. What will the
  volume of the gas be at a pressure of 0.987 atm
  if the temperature remains constant?
• Solution

P1V1 P2V2
(0.947 atm)
(150.0 mL)

(0.987 atm)
V2
(0.947 atm)
(150.0 mL)


V2
144 mL
(0.987 atm)
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Volume Temperature Relationship

• the pressure of gas inside and outside the
  balloon are the same.
• at low temperatures, the gas molecules dont move
  as much therefore the volume is small.
• at high temperatures, the gas molecules move more
  causing the volume to become larger.

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Charless Law

• Charless Law The volume of a fixed mass of gas
  at constant pressure varies directly with the
  Kelvin temperature.

V1
V2

T1
T2
V1 and T1 represent initial conditions,
and V2 and T2 represent another set of
conditions.
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The Kelvin Temperature Scale

• Absolute zero The theoretical lowest possible
  temperature where all molecular motion stops.
• The Kelvin temperature scale starts at absolute
  zero (-273oC.)
• This gives the followingrelationship between the
  two temperature scales

K oC 273
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Charless LawSample Problem

• A sample of neon gas occupies a volume of 752 mL
  at 25C. What volume will the gas occupy at 50C
  if the pressure remains constant?
• Solution

K oC 273
T1

25 273
298
T2

50 273
323
752 mL
V2

298 K
323 K
752 mL
V2

x

323 K
815 mL
298 K
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Pressure Temperature Relationship

• Increasing temperature means increasing kinetic
  energy of the particles.
• The energy and frequency of collisions depend on
  the average kinetic energy of the molecules.
• Therefore, if volume is kept constant, the
  pressure of a gas increases with increasing
  temperature.

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Gay-Lussacs Law

• Gay-Lussacs Law The pressure of a fixed mass
  of gas varies directly with the Kelvin
  temperature.
• P1 and T1 representinitial conditions.P2 and T2
  representanother set of conditions.

P1
P2

T1
T2
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Gay-Lussacs LawSample Problem

• The gas in a container is at a pressure of 3.00
  atm at 25C. What would the gas pressure in the
  container be at 52C?
• Solution

K oC 273
T1

25 273
298
T2

52 273
325
3.00 atm
P2

298 K
325 K
3.00 atm
P2

x

325 K
3.27 atm
298 K
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The Combined Gas Law

• The combined gas law is written as follows
• Each of the other simple gas laws can be obtained
  from the combined gas law when the proper
  variable is kept constant.

P1
V1
P2
V2

T1
T2
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The Combined Gas LawSample Problem

• A helium-filled balloon has a volume of 50.0 L at
  25C and 1.08 atm. What volume will it have at
  0.855 atm and 10.0C?
• Solution

K oC 273
T1

25 273
298
T2

10 273
283
(1.08 atm)
(0.855 atm)
(50.0 L)
V2

298 K
283 K
(1.08 atm)
(50.0 L)
(283 K)
V2


60.0 L
(298 K)
(0.855 atm)
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Avogadros Law

• In 1811, Amedeo Avogadro discovered that the
  volume of a gas is proportional to the number of
  molecules (or number of moles.)
• Avogadros Law - equal volumes of gases at the
  same temperature and pressure contain equal
  numbers of molecules, or

V1
V2

n1
n2
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The Ideal Gas Law

• All of the gas laws you have learned so far can
  be combined into a single equation, the ideal
  gas law
• R represents the ideal gas constant which has a
  value of 0.0821 (Latm)/(molK).

PV nRT
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The Ideal Gas LawSample Problem

• What is the pressure in atmospheres exerted by a
  0.500 mol sample of nitrogen gas in a 10.0 L
  container at 298 K?
• Solution

PV nRT
P
(10.0 L)

(298 K)
(0.500 mol)
(0.0821 Latm/molK)
(0.500 mol)
(298 K)
(0.0821 Latm/molK)
1.22 atm


P
(10.0 L)
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Molar Mass of a Gas

• The ideal gas law can beused in combination
  withmass measurements to calculate the molar
  mass of an unknown gas.
• Molar mass is calculated by dividing the mass (in
  grams) by the amount of gas (in moles.)

g
Molar Mass

mol
40
Molar Mass of a GasSample Problem

• Calculate the molar mass of a gas with mass 0.311
  g that has a volume of 0.225 L at 55C and 886
  mmHg.
• Solution

1 atm

886 mmHg
P
1.17 atm

PV nRT
760. mmHg
T
55 273 328 K
(1.17 atm)
(0.225 L)
PV

0.00978 mol

n

RT
(328 K)
(0.0821 Latm/molK)
0.311 g
grams
MM
31.8 g/mol



mole
0.00978 mol
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Standard Molar Volume

• Standard Temperature and Pressure (STP) is 0oC
  (273 K) and 1 atm.
• The Standard Molar Volume of a gas is the
  volume occupied by one mole of a gas at STP.
  It has been found to be 22.4 L.

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Molar Volume Conversion Factor

• Standard Molar Volume can be used as a conversion
  factor to convert from the number of moles of a
  gas at STP to volume (L), or vice versa.

43
Molar Volume ConversionSample Problem

• a. What quantity of gas, in moles, is contained
  in 5.00 L at STP?
• b. What volume does 0.768 moles of a gas occupy
  at STP?

mol
1
x
5.00 L
0.223 mol

L
22.4
L
22.4
x
0.768 mol
17.2 L

mol
1
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The Mole Map Revisited

• As you recall, you can convert between number of
  particles, mass (g), and volume (L) by going
  through moles.

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Stoichiometry Revisited

• Remember that you can use mole ratios and volume
  ratios (gases only) as conversion factors
• 2CO(g) O2(g) ?
  2CO2(g)
• 2 molecules 1 molecule 2 molecules
• 2 mole 1 mole 2 mol
• 2 volumes 1 volume 2 volumes
• Example What volume of O2 is needed to react
  completely with 0.626 L of CO at the same
  temperature pressure conditions to form CO2?

L O2
1
x
0.626 L CO
0.313 L O2

L CO
2
46
Gas StoichiometrySample Problem

• What volume (in L) of H2 at 355 K and 738 mmHg is
  required to synthesize 35.7 g of methanol, given
• CO(g) 2H2(g) ? CH3OH(g)
• Solution
• First, use stoichiometry to solve for moles of
  H2
• Then, use the ideal gas law to find the volume of
  H2

mol CH3OH
1
mol H2
2
35.7 g CH3OH
2.23 mol H2

mol CH3OH
1
g CH3OH
32.0
(355 K)
(2.23mol)
(0.0821 Latm/molK)
nRT
66.9 L H2


V

P
(0.971 atm)