Transcript CHAPTER 10 - GAS LAWS - Greater Latrobe School District

Parts of chapters 2, 10, &11
All of Chapter 12 – Gas Laws
CHEMISTRY
GREATER LATROBE HIGH SCHOOL
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TABLE OF CONTENTS
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Introduction to Gases
Boyles Law
Charles Law
Gay-Lusacc’s Law
Combined Gas Law
Ideal Gas Law
Dalton’s Law
Graham’s Law
Phase Diagrams
Surface Tension
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Click on this
picture on any
slide to return to
the table of
contents
All yellow text will take
you to additional
information
2
Introduction to Gases –
Causes of change (Ch. 11.1 and 11.2)
 Energy as Heat:
– Heat and Temperature are not the same
thing.
 Heat
 Temperature
 Enthalpy – Total energy content of the
sample.
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Introduction to Gases –
Causes of change (Ch. 11.1 and 11.2)
 Molar Heat Capacity –(Cp)– this is the
amount of heat required to raise the
temperature of 1 gram of a substance 1
degree Celsius (or 1 mole 1 Kelvin).
– Practical example: snow fall
– Calculations using Cp – These are done to
determine the amount of energy change (H) in
the process of warming or cooling matter. This
is calculated according to the following
equation:
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Introduction to Gases –
Causes of change (Ch. 11.1 and 11.2)
 H = mCp T (H can be changed for Q)
– m = mass, Cp = Specific Heat capacity (will be
given, T = Change in temperature (Final T –
Initial T)
 m can also be replaced with n for moles if necessary)
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Introduction to Gases –
Causes of change (Ch. 11.1 and 11.2)
 Problem #1: (Click for answer)
– Assuming the density of water to be 1.0 g/mL.
How much heat is lost by 4.0 L of water cooling
from 87 C to 21 C? (Heat capacity for water =
4.180 J/g* C
 Problem #2: (Click for answer)
– If 980 kJ of energy is added to 6.2L of water at
18 C, what is the final temperature of the
water?
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Introduction to Gases –
Causes of change (Ch. 11.1 and 11.2)
 When heat is added or removed from a substance
it can go through phase changes. Recall that
there are three major states (phases) of matter
(solid, liquid, and gas). A Heating Curve (in your
book on page 44) shows the relationship between
these states as the temperature changes.
– ∆Hfus – energy required to melt one mole
– ∆Hvap - energy required to vaporize one mole
– ∆Hvap > ∆Hfus
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State Changes and Phase Diagrmas
 Review of what we know at this point about solids,
liquids, and gases:
– Solids – Constant shape, constant volume
– Liquids – Variable shape, constant volume
– Gases – Variable shape, variable volume
 We now need a more detailed comparison of each
of these.
– Solids
– Liquids
– Gases
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Phase Diagrams – Vapor
Pressure and Boiling
 Vapor Pressure – the pressure of a gas on top of a liquid.
This is dependent only on the temperature.
– This is key in understanding the process of boiling. For example:
 What would happen to a an open jar of water over time?
 What would happen to a closed jar of water over time?
– Here water will evaporate until condensation = evaporation. Here vapor
pressure is a max.
 Boiling – this occurs when vapor pressure equals
atmospheric pressure.
– Substances with high vapor pressure at low temperatures are
volatile (boil / vaporize easily) i.e. perfume
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Phase Diagrams
 Phase Diagrams – Graphically represent the relationship
between the state of a substance and its pressure and
temperature.
– Processes shown phase diagram (click to go to diagram)
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(1) – melting – solid to liquid
(2) – Freezing – liquid to solid
(3) – Boiling – liquid to gas
(4) – Condensing – gas to liquid
(5) – Sublimation – solid directly to gas (dry ice)
(6) – Deposition – gas directly to solid (frost)
(7) – Triple point – Temperature and pressure at which a substance
exists as a solid, liquid, and gas at the same time.
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Characteristics of Gases
 Kinetic Molecular Theory – Describes the
behavior of gases at the molecular level
(Based on an ideal gas)
– Ideal Gas – Imaginary gas (model) that
describes the behavior of real gases at
conditions close to STP (this will be explained
later)
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Characteristics of Gases (con’t)
 5 Assumptions of the KMT:
– 1. Gases consist of large numbers of tiny particles
– 2. Particles of a gas are in constant motion and
therefore have KE
– 3. Collisions between particles of a gas and
container wall are completely elastic.
– 4. There are no forces of attraction or repulsion
between particle of a gas
– 5. The Average kinetic energy of the particles of a
gas is directly proportional to the Kelvin
Temperature of the gas.
 KE = ½ MV2
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Ideal versus Real Gases
Throughout this Unit we are going to focus on Ideal
Gases, Ideal gases differ from real gases by the following

 Real Gases
 Ideal Gases
– The volume of the particles
is negligible
– There are no attractive
forces between molecules
– Collisions between particles
are elastic
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– Particles do have volume
– As temp. decreases the
particles slow down and
attractive forces increase
– Collisions between particles
are inelastic
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Characteristics of Gases (con’t)
 Variables that describe a gas:
– Temperature – measure of the average KE of the
Particles of a gas (must be in Kelvin – T in K)
 K = C + 273
– Volume – amount of space matter occupies
 Can use L, mL, m3, or cm3
– Pressure – Force per unit area
 Measure with barometer
 Describes the forces that gases exert on the walls of a
container.
 Atmospheric Pressure = 1 atm = 760 mmHg = 760 torr = 101.3
Kpa = 29.9 inHg = 14.7 psi
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Characteristics of Gases (con’t)
 STP – Standard Temperature and Pressure
– T = 273 K
– P = 1 atm
At these conditions all gasses
occupy 22.4 L of space (standard
molar volume)
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Boyles Law
 This Gas law discovered by Robert Boyle involves
the relationship between pressure and volume at
constant temperature.
 Go to the following web site to see an illustration
of this relationship (freeze mass and
Temperature):
–
http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/frglab2.html
 This relationship is know as an inverse
relationship. (when one variable gets bigger the
other variable gets smaller)
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Boyle’s Law (con’t)
 Formula:
– PV = k (k=constant, P=pressure, V=volume)
– Since the pressure times the volume is constant this can be written
as follows: (click on the formula to go to some practice problems)
–P V = P V
–P V = P V
1
1
1
2
1
2 or
2
2
 Boyle’s Law has many practical applications. It is applied
in things such as breathing, scuba diving, submarines, and
weather balloons. We will discuss several of these.
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Charles Law
 This gas law was discovered by Joseph Louis Gay-Lussac
but is usually named for Jacques Charles Law because of
the work he did. It involves the relationship between
Volume and Temperature at constant pressure.
 Go to the following web sites to see an illustration of this
relationship (freeze mass and Pressure):
–
–
http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/frglab2.html
http://fsc.fernbank.edu/chemistry/charles.html
 This relationship is know as an direct relationship. (when
one variable gets bigger the other variable gets bigger)
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Charles Law
 Formula:
– V/T = k (k=constant, T=Temperature, V=volume)
– Since the volume divided by the temperature is constant the
following formula can be written: (click on the formula to go to some
practice problems (choose solving problems or word problems)
– V /T = V /T
– V /T = V /T
1
1
1
2
1
2 or
2
2
 Charles’s Law also has many practical applications. It is
applied in things such as effects of temperature changes
on balloons, tires, hot air balloons, etc.
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Gay-Lussac’s Law
 This Gas law discovered by Joseph Louis
Gay-Lussac involves the relationship
between temperature and Pressure at
constant volume.
 Go to the following web site to see an
illustration of this relationship (freeze mass
and volume):
– http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/frglab2.html
 This relationship is also a direct relationship.
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Gay-Lussac’s Law
 Formula:
– P/T = k (k=constant, T=Temperature, P=pressure)
– Since the volume divided by the temperature is constant the
following formula can be written: (click on the formula to go to some
practice problems (choose solving problems or word problems)
– P /T = P /T
1
1
2
2
 Gay-Lussac’s law also has many practical applications.
For example: This is why on the side of spray cans it will
tell you to keep the can stored at certain temperatures.
This is why inflatable devices such as tires or balls go flat
in the winter time.
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The Combined Gas Law
 This gas law encorporates all three of the
basic gas law (Boyle’s, Charles’s, and GayLussac’s)
– Click here to see a graphical illustration of this.
 It is often used for converting a temperature,
pressure, or volume to standard conditions
(STP)
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The Combined Gas Law
 Formula:
P1V1
P2V2
T1
T2
 Sample problem:
– 2.00 L of a gas is collected at 25.0°C and 745.0
mmHg. What is the volume at STP?
 Click for answer
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The Combined Gas Continued
 A sample of gas occupies a volume of 6.80
L at 790.0 mmHg and 300.0 K. What is the
temperature of the gas when the volume is
7.00 L and the Pressure is 2.00 atm?
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Ideal Gas Law
 The Ideal gas law encorporates all three variables that we
have been using plus it also incorporates moles into the
equation. This is the most common equation used in
working with gases.
 First written by Emil Clapeyron, and is sometimes
(although rarely called the Clapeyron equation)
 The formula is as follows:
– PV = nRT
 P = pressure, V = Volume, n = moles, R = Gas constant, T = temperature
 R = 8.314 L * kPa or 0.0821 L * atm or 62.36 mmHg * L
----------Mol * K
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----------mol * K
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--------------mol * K
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Ideal Gas Law
 Example Problems
– What is the pressure (in atm) exerted by a 0.500 mol sample of Nitrogen
gas in a 10.0 L container at 298 K?
– What mass would a sample of chlorine gas have if the pressure in a 10.0 L
tank at 27 degrees Celsius is 3.50 atm?
– Answers to Ideal Gas Law Problems
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Dalton’s Law
 This gas law was discovered by John Dalton. It
deals with the partial pressures (The pressure of
an individual gas in a mixture of gases) of a gas.
It is primarily used for situations when a gas has
been collected over water.
 Go to the following web site for an explanation of
Dalton’s Law:

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http://www.fordhamprep.com/gcurran/sho/sho/lessons/lesson74.htm
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Dalton’s Law
 Formula
– Pt = P1 + P2 + P3 + ….. + Pn (P = Pressure)
– This formula can also be written as follows:
 Patm = Pgas + PH O (P=pressure, atm=atmosphere)
2
– The second formula is the form which we will use the
most often. This is the form of Dalton’s equation that
allows us to correct for water vapor pressure (click to
see water vapor chart) when a gas is collected over
water.
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Dalton’s Law
 Example Problem:
– 2.50 L of gas is collected over water at 25 degrees C
and 795 mmHg. What is the volume of gas at STP?
– Click for answer
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Graham’s Law
 Graham’s Law is used to give the rates of effusion
and diffusion.
– Effusion – Gas escapes a container through a small pin
size hole.
– Diffusion – Gradual mixing of two gases because of
random motion of particles
 Definition: The rates of effusion and diffusion are
inversely proportional to the square roots of a
gases molar mass.
– Analogy – Big guy –vs- Small guy
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Graham’s Law

Formula:
–

Rate A Square root (MB)
-------- = ------------------Rate B Square root (MA)
Sample Problem
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Temperature
 The temperature of a substance is a
measure of the average kinetic energy of
that substances particles.
– Think of this as the intensity of energy.
– This is an intensive property – the measure of
temperature does not depend on the amount of
the sample of material.
 This is different than heat
Return
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Heat
 Heat is the measure of the total amount of
energy transferred from an object of high
temperature to one of low temperature
– Think of this as the quantity of energy
– This is an extensive property – the measure of
heat depends on the amount of the sample of
material.
– It is always High  Low
Return
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Heating Curve
(Chapter 2 – Page 44)
 The picture
to the right
represents
a heating
curve for
water.
Heating Curve for Water
120 °C
Return
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water  steam
100 °C
50°C
0 °C
What is happening to the
temperature when the
substance is changing
states?
steam
-10 °C
liquid water
ice liquid
ice
Heat added 
LecturePLUS Timberlake
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Answers to Heat Capacity
Problems
 Problem #1:
– First find mass: m = (1000 g / L)(4.0 L) = 4000 g
– H=(4000g)(4.180 J/g* C)(21-87) = 1103520 J
 Problem #2:
– First find mass: m = (1000 g / L)(6.2 L) = 6200 g
– 980 000 J = (6200g)(4.180 J/g* C)(Tf - 18 C) = 55.8 C
Return
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Barometer


The barometer The mercury barometer
measures atmospheric pressure. It works this
way. Completely fill a long glass tube with
mercury. Turn it upside down, and place the
top below the surface of more mercury in an
open basin. Rather than pouring out again, the
mercury in the tube will only fall until the
height of the column is about a meter. This is
because the pressure of the air on the
mercury in the basin is equal to the pressure
of the mercury in the column. The gap at the
top of the tube is not air but a vacuum. The
difference in height between the top of the
column and the top surface of the mercury in
the basin is a measure of the weight of the air,
which changes as the weather changes.
Weather can be predicted by this:
–
–
Low barometric pressure generally indicates
stormy weather
High barometric pressure generally indicates
good weather
Return
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Inverse Relationships Defined
 An inverse relationship
is when one of the
variables is increased
the other variable
decreases. See the
graph to the right for
an illustration of this.
Return
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Direct Relationship defined
 In a direct
relationship when
one of the variables
gets bigger the other
variable also gets
bigger. See the
graph to the right for
an illustration of this.
Return
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Water Vapor Pressure
 This is pressure in a container of gas that
results from the water evaporating slightly
as the gas is being collected over water.
The amount of evaporation increases as the
temperature increases.
Return
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Answer to Dalton’s Problem
– 2.50 L of gas is collected over water at 25 degrees C
and 795 mmHg. What is the volume of gas at STP?
 You must use the combined gas law
– 1st – Determine the pressure of the dry gas (vapor pressure
chart)
 795 mmHg = Pg + 23.8 mmHg
 Pg = 771.2 mmHg
– 2nd – Use the combined gas law to solve.
 (771.2 mmHg)(2.50L)/298K = (760 mmHg)V2/273K
 V = 2.32 L
Return
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Answer to Combined Gas
Law Problem
 The data is organized to the left.
Insert them into:
– P1V1 / T1 = P2V2 / T2
 (745.0 mmHg)(2.00L) / 298K
= ( 760.0 mmHg)(V2) / 273K
V2 = 1.79 L
Return
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Answer to Ideal Gas
Law Problem
 PV = nRT
– P(10.0L) = (0.500)(0.0821)(298K)
– P = 1.22 atm
 PV = nRT
– (3.50atm)(10.0L) = n(0.0821)(300K)
– n=1.42 moles * 70.90 g Cl2
------------------------------------- = 101
grams
1 mole of Cl2
Return
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Phase Diagrams
Return
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Combined Gas Law Graph
Trace across the
white area to see
what happens as
all three change.
Trace across a
colored area to
see what happens
when one variable
is held constant.
Return
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Water Vapor Pressure Vapor Chart
Temp (C.)
Press.
Temp (C.)
Press.
(mmHg)
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Temp (C.)
Press.
(mmHg)
(mmHg)
1
4.9
21
18.6
41
58.3
2
5.3
22
19.8
42
61.5
3
5.7
23
21.1
43
64.8
4
6.1
24
22.4
44
68.3
5
6.5
25
23.8
45
71.9
6
7.0
26
25.2
46
75.6
7
7.5
27
26.7
47
79.6
8
8.0
28
28.3
48
83.7
9
8.0
29
30.0
49
88.0
10
9.2
30
31.8
50
92.5
11
9.8
31
33.7
55
118.0
12
10.5
32
35.7
60
149.4
13
11.2
33
37.7
65
187.5
14
12.0
34
39.9
70
233.7
15
12.8
35
42.2
75
289.1
16
13.6
36
44.6
80
355.1
17
14.5
37
47.1
85
433.6
18
15.5
38
49.7
90
525.8
19
16.5
39
52.4
95
633.9
20
17.5
40
55.3
100
760.0
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Solids
 In a solid:
– The particles are packed very closely together.
– The particles are very ordered
– The particles vibrate about fixed postions.
– The particles in a solid exhibit very strong
Intermolecular Forces
Back to State Changes
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Liquids
 Particles in a liquid:
– Can slide/move past one another
– Exhibit weaker Intermolecular Forces
– Are arranged more randomly
– Exhibit:
 Surface Tension
 Capillary Action
Back To State Changes
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Gases
 Particles in a gas:
– Are very spread out (lots of empty space
between particle)
– Move rapidly in random, straight paths
– Don’t exhibit intermolecular forces
– Are very compressible
Back To State Changes
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Surface Tension
 Definition – The measure of a liquids
tendency to decrease its surface area to a
minimum.
– What does this really mean?
 Water spiders, capillary action, and water skiing are
all illustrations of surface tension.
Water Spider Picture
Back To Liquids
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Capillary Action
 Capillary action occurs when the adhessive
forces exceed the cohessive forces.
– Adhesion – liquid particles attract to a solid
surface.
– Cohesion – liquid particles attract to each other
 This explains the meniscus seen in glass
ware and how trees get water to the top of
the tree.
Back To Liquids
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