NAME____________________________________ PER____________ DATE DUE____________
ACTIVE LEARNING I N C HEMISTRY E DUCATION
"ALICE"
CHAPTER 4
THE GAS
LAWS
Boyle's Law
Charles' Law
Gay-Lussac's Law
The Combined Gas Law
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©1997, A.J. Girondi
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© 1997 A.J. Girondi, Ph.D.
505 Latshmere Drive
Harrisburg, PA 17109
[email protected]
Website: www.geocities.com/Athens/Oracle/2041
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©1997, A.J. Girondi
SECTION 4.1
Kinetic Theory and the Properties of Gases
We know from chapter 3 that matter normally exists in three forms - solid, liquid, and gas. While a
fourth state, plasma, does exist, it is not in our realm to be able to study this form of matter firsthand. We
can study the other three, however. This chapter will deal with gases, exclusively. The other states of
matter will be covered more extensively in other chapters.
If we attempt to define gas, we can say that it is a state of matter made up of particles that distribute
themselves uniformly throughout the space in which the gas is confined. This differs from a liquid and a
solid. A liquid takes the shape of the container that it is in. In addition, a liquid does not distribute itself
uniformly in its container. Figure 4.1 will help you to visualize this fact. A solid, of course, retains its own
shape regardless of the shape or size of the container into which it is placed.
LIQUID
SOLID
GAS
Figure 4.1 The Three Phases of Matter
Gases have three basic characteristics. They are: (1) expansibility, (2) diffusion, and (3)
compressibility. Expansibility is the ability of a gas to expand to the shape and volume of its container
whether it is a small beaker, a room, or something larger. The opposite of expansibility is compressibility.
A gas can be compressed to fill a volume many times smaller than its original volume. Diffusion is the ability
of a gas to move from a region of high concentration to a region on low concentration. If you were to open
a bottle of perfume in a room, it would evaporate into a gas, and you would soon be able to smell the odor
at all points in the room. The perfume vapor molecules have diffused through the air molecules to all parts
of the room.
The properties of gases just described provide the basis for the kinetic-molecular theory. This
theory gets its name from its close tie with the concept of kinetic energy. Any object that is in motion has
kinetic energy. The faster something is moving, the more kinetic energy it has. The kinetic-molecular
theory helps to explain why gases behave the way they do.
The main assumptions of the kinetic theory are:
1. Gases consist of molecules in continuous, random motion. The molecules collide with each other and
with the walls of the container. The pressure exerted by the gas is caused by the collisions of the gas
particles with the walls of the container.
2. Molecular "collisions" are elastic. This means that although there is a lot of "bouncing off" going on as
the molecules approach each other, there is no energy lost to friction when the gas molecules
"collide" with each other.
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©1997, A.J. Girondi
3. Gas molecules in a container move at a variety of different speeds; but, overall, an increase in
temperature increases the average speed (kinetic energy) of the molecules. This increase in the speed
of the molecules results in more frequent and more forceful collisions between the molecules and the
walls of the container. Consequently, gases exert more pressure at higher temperatures.
How does the kinetic theory explain the gas property of diffusion?
{1}___________________________
______________________________________________________________________________
______________________________________________________________________________
How does the kinetic theory explain the fact that gases exert pressure? {2}________________________
______________________________________________________________________________
______________________________________________________________________________
ACTIVITY 4.2
Measuring the Effect of Pressure on Gas Volume
The background you now have is sufficient to introduce you to the gas laws. There are several of
these. The first one we will look at was named after Robert Boyle, the scientist who first discovered that
there was a definite relationship between the pressure and the volume of a gas. Boyle's Law simply
states:
"At a constant temperature, if the pressure on a gas is increased, the
volume of the gas will decrease proportionately."
Figure 4.2 illustrates the laboratory setup for
investigating the relationship between the pressure and volume
of a gas. Your teacher has a similar setup for you. Once you have
gotten the equipment and materials, begin by setting the end of
the piston on the 30 cm3 (or cc) mark. (To set the piston where
you want it, air must be allowed to escape from the syringe. The
wire that is attached to the syringe is there for that purpose.
Place the wire into the syringe before inserting the piston. This
will allow air to escape. Then, remove the wire while holding the
piston in place.) If you gently push down on the piston and then
let up, the piston should return to its starting point. Record your
starting point in Table 4.1. Once you have done this, add 1 "unit"
of mass to the platform top (a unit of mass can be a textbook. If
you decide to use textbooks, be sure they are all the same).
Record the new volume in Table 4.1. Continue adding "units" of
mass to the plunger by 500 g or 1 book increments until you
have 5 or 6 readings. Record each new volume in the
appropriate space in Table 4.1.
Figure 4.2
What happens to the volume of the gas as you increase the mass (pressure)?
{3}_____________________________
______________________________________________________________________________
Does your result above agree with the general statement given for Boyle's law in section 4.1? Explain.
{4}___________________________________________________________________________________________________________________
______________________________________________________________________________
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©1997, A.J. Girondi
It was mentioned that temperature must remain constant for Boyle's law to be valid. At what temperature
did you conduct your experiment? ___________oC. Do you think it remained constant? ____________
Table 4.1
Effect of Pressure on Volume
Pressure Applied (books)
Volume (cm3)
1
___________
2
___________
3
___________
4
___________
5
___________
On the grid below, prepare a graph of the information in data table 4.1. Your plot should actually
be a slightly curved line. Do not try to make it absolutely straight. Remember to follow the instructions
given in chapter 2 regarding the correct way to construct a graph.
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©1997, A.J. Girondi
Your plot is slightly curved. It would be even more curved if we had used a greater range of pressures.
You may wonder why your plot is not a straight line. When two variables are both changing at constant
rates, their resulting graph will be a straight line. However, in this activity, only one variable was changing at
a constant rate. Was it pressure or volume? {5}_________________.
independent variable?
{6}__________________________.
Was this the dependent or
the
You may wonder why both variables don't
change at constant rates. Well, as pressure continues to increase in equal amounts, the amount of
decrease in volume gets smaller and smaller. In other words, as pressure changes at a constant rate,
volume changes at a rate which is not constant. Why is this? Gas atoms or molecules – like all atoms and
molecules – have electrons. These electrons are negatively charged and repel each other when they get
too close. As gas particles get closer to each other, the repulsion between them gets stronger and it's
more difficult to force them closer together. In other words, each additional "book" of pressure, has less
effect on the volume of the gas.
SECTION 4.3
Solving Problems Involving Boyle's Law
Boyle's law can be expressed mathematically as: P 1V 1 = P2V 2. This equation means that the
product of the pressure (P) and the volume (V) of a gas remains constant, provided the temperature does
not change. (If temperature were a variable, it would have to be included in the equation). So, even
though P and V can change, the product, PV, maintains a constant value. When the product of two
variables remains constant, the variables are said to be inversely proportional, meaning that as one
increases, the other decreases.
The sample problem below illustrates how Boyle's law works, mathematically. Study it carefully.
Sample problem: A quantity of air has a volume of 100. mL (V1) at 720. mm of pressure (P1). What will
the new volume (V2) be if the pressure is changed to 760. mm (P2)? Follow the steps below.
1. Write the formula P1V 1 = P2V 2.
2. Solve the equation for the unknown quantity (V2):
V2 =
P1V1
P2
3. Substitute the values into the formula including units:
V2 =
4. Do the math:
(720. mm)(100. mL)
760. mm
V2 = 94.7 mL.
Solve the following problems. Show the set-up for the problem work, and be sure to include units on all
measurements used in the problem and given in the answer. (The period appearing after some numbers
indicates that the trailing zeros are significant figures.)
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©1997, A.J. Girondi
Problem 1. A gas has a measured volume of 100. mL under a pressure of 740. mm Hg. What would the
volume be under a pressure of 780. mm Hg at constant temperature?
Problem 2. A sample of gas is confined to a 100. mL flask under a pressure of 740. mm Hg. If this same
gas were transferred to a 50.0 mL flask, what would the resulting pressure be?
Problem 3. You are given a gas that you measure under a pressure of 720. mm Hg. When the pressure
is changed to 760. mm, the volume becomes 580. mL. What was the original volume of the gas?
Problem 4. At a certain original pressure, the volume of a given amount of air is 134 mL. If the pressure
is changed to 1200. mm Hg, the volume is reduced to 45.0 mL. What was the original pressure?
Problem 5. A balloon has a volume of 800. mL when it is held at sea level where the pressure is 760.
mm Hg. Calculate the volume of the same balloon after it has floated up to where the atmospheric
pressure is only 720. mm Hg.
According to your calculations, does the balloon get larger or smaller? {7}________________________
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©1997, A.J. Girondi
ACTIVITY 4.4
Measuring the Effect of Temperature on Gas Volume
Gas volumes are also affected by temperature changes. The gas law that explains this is called
Charles' law. Charles' law states that:
"As the absolute temperature of a gas increases, its volume also
increases proportionately, when the pressure remains constant."
As we discuss Charles' law, we will explain the meaning of the words absolute and proportionately as they
are used above.
One way that we can see Charles' Law demonstrated in the lab is to use the piece of equipment
pictured in Figure 4.3. It consists of a piece of glass capillary tubing which contains a small bead of mercury
metal. One end of the tube is sealed, while the other is open. A volume of air is trapped in the tube
between the sealed end and the mercury bead. As the volume of the trapped air changes, the mercury
bead will move up or down the tube. The volume of trapped air can be calculated if you assume that the
container is a cylinder. The formula for the volume of a cylinder is V = π r 2 h , where π = 3.1416. Notice that
you will need to know the internal radius of the capillary tube as well as the height (or length) of the air
column between the sealed end of the tube and the mercury bead. Measure the internal diameter of the
tube in mm. Half of that is the internal radius, r. All measurements should be made in millimeters, which
when substituted in the formula above will give volumes in units of cubic millimeters (mm3).
h
Figure 4.3 Charles' law apparatus
Obtain a thermometer, a metric ruler, and a capillary tube with mercury bead from the materials
shelf. Measure the height of the air column between the closed end of the tube and the mercury bead in
mm at room temperature. Record this height and temperature in Table 4.2. Next, place the tube and
thermometer in a large beaker or flask of hot water at constant temperature (between 80 oC and 90 oC) so
that most of the trapped air column is immersed. When the mercury bead stops moving, read the
temperature. Without removing the tube from the hot water, measure the height of the air column in mm
again. Record the data. Finally, place the capillary tube and thermometer in an ice water bath which will be
available in the lab. The mercury bead will move down the tube. When it has stopped moving, measure
the temperature and, without removing the tube from the ice water, measure the height of the air column
in mm using a ruler. Record all data in Table 4.2. Return all materials to the proper place. Using your data,
calculate the volume of the trapped air in mm3 at each of the three temperatures.
Table 4.2
Temperature Versus Volume
Temp (oC)
Air Column Height (mm)
Air Volume (mm3)
________
_________________
_____________
________
_________________
_____________
________
_________________
_____________
internal radius of tube = ________ mm
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©1997, A.J. Girondi
Using your data in Table 4.2, plot a graph of temperature (X axis) versus volume (Y axis) on the grid
below. The scales on each axis do not necessarily have to start with zero. You may start each scale at any
convenient value. After you have plotted your three points, draw the best straight line that you can
through them. (We know from experiment, that this curve has to be a straight line, so use a ruler to draw
the best straight line you can through your points. Do not expect this line to actually intersect with each of
your points.)
Temperature is a measure of the amount of kinetic energy that particles have. The more kinetic energy
that particles have, the faster they move and the harder they "collide" with the walls of their containers.
What effect does increased temperature have on the pressure exerted by a gas?
{8}_________________
______________________________________________________________________________
The graphic representation in Figure 4.4 on the next page depicts the relationship between the
volume of gas and its temperature. Note that volume decreases as temperature decreases. Study this
carefully.
Do your data seem to follow a similar trend?_________. What happens to the volume of a gas as the
temperature goes up and down?
As you examine the graph in Figure 4.4, note that eventually a temperature is reached where the
volume of the gas appears to go to zero. Actually, it is impossible for a real gas to have a volume of zero
because no matter how close the molecules get to each other, they have volume themselves. In addition,
repulsive forces between the molecules limit how close they can get to each other. These factors are not
accounted for in the formula for Charles' law. Scientists have invented the concept of an "ideal" gas which
is a gas that obeys the gas laws, including Charles' law, at all times and under all conditions.
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©1997, A.J. Girondi
A gas with a volume of
273 mL at 0oC (273 K)
300 mL
...should occupy zero
volume at -273oC (0 K)
200 mL
coldest
possible
temperature
V
O
L
U
M
E
100 mL
-273
Temperature in oC
0
+50
Figure 4.4 Volume-Temperature Relationship
While ideal gases do not really exist, the ideal does help us to understand the gas laws. It is
possible for an "ideal" gas to have a volume of zero. Note that this would be the case at -273oC. It is
impossible for any gas including an ideal gas to have a negative volume. Therefore, at -273oC a gas would
have the smallest volume possible, and, therefore, -273oC (or 0 K) must be the coldest possible
temperature which a gas - or anything else for that matter - can have. If colder temperatures were possible,
then gases would have to have negative volumes! This temperature, -273oC, has come to be known as
absolute zero. It is the starting point for the Kelvin temperature scale. (-273oC = 0 K) Zero on the Kelvin
scale (0 K) is absolute zero. (The degree symbol is not used when expressing Kelvin temperatures.) At
absolute zero, the molecules have no more kinetic energy. Although we have been able to get very close
to this temperature in laboratories, it has never been reached.
SECTION 4.5
STP - What It Is and Why We Need It
Let's say that you have two one–liter containers. One contains gas A and the other contains gas
B. Since a gas expands to fill its container, the volume of each gas is one liter. Which container holds the
greatest amount of gas – meaning the greatest number of gas molecules?
Which of these 1.00 Liter Containers Holds More Gas?
1.00 Liter
1.00 Liter
GAS A
GAS B
28.0 oC & 700 mm Hg
26.0 oC & 662 mm Hg
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©1997, A.J. Girondi
Just because the two containers have the same volume, they do not necessarily contain the same
quantities of gas. After all, most of the volume occupied by a gas is empty space. A one–liter container
can be filled by many gas molecules or only a few. In order to determine which container holds the most
molecules, you would need to know the temperature and pressure of each gas. These factors affect how
much space exists between the molecules and, therefore, how much gas you can get into the container.
Only if both of our gases (A and B) are at the same temperature and pressure would both one–liter
volumes contain the same number of gas molecules. (This is known as Avogadro's hypothesis, and you
will learn more about it in Chapter 8.) If the two gases are not at the same temperature and pressure, then
we will have to determine what the volumes of the gases would be if they were at the same temperature
and pressure. Only then could we determine which container holds the most molecules. When this is
necessary, the conditions of temperature and pressure which have been chosen for use are 0 oC and 760
mm Hg (millimeters of mercury) pressure. These conditions are known as standard temperature and
pressure, or STP.
Converting to STP RevealsThat There is More of Gas A
0.835 Liter
0.795 Liter
GAS A
GAS B
0 oC & 760 mm Hg
Why did we develop the concept of STP?
0 oC & 760 mm Hg
{9}___________________________________________________________________
______________________________________________________________________________
The units of mm Hg (millimeters of mercury) for measuring pressure may seem somewhat mystical
to you. If you fill a test tube with water and put your thumb over it, and invert it into a shallow dish of water,
you will notice that the water will not drain out. This is because atmospheric pressure is "pushing up" on
the water column. In fact, atmospheric pressure is great enough to support a column of water over 34 feet
high! As the atmospheric pressure changes from day to day, the height of the water column which it can
support varies. Thus, we could use the height of the water column as a measure of pressure in millimeters
of water. This device would be a barometer. But, who wants a barometer that is over 34 feet high?
Mercury metal is about 13.5 times more dense than water, and can be used in place of water in a
barometer.
On an average day at sea level, the atmosphere supports a
column of mercury 760 mm high. This is more manageable,
and is why mercury barometers are found in labs and at
weather bureaus. We can, therefore, measure atmospheric
pressure by measuring the height of a column of mercury
which it can support (units of "mm Hg"). We will use mm from
now on, but "mm" really implies "mm Hg." 760 mm is also
known as 1 atmosphere Figure 4.5 (1 atm) of pressure. (760
mm Hg = 1 atm)
760 mm
Hg
Figure 4.5
Mercury Barometer
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©1997, A.J. Girondi
We can summarize standard conditions as follows:
S T P (Standard Temperature and Pressure) Means:
pressure = 1 atm or 760 mm Hg
temperature = 0oC or 273 K
When solving gas law problems that involve temperature changes, it is always necessary to use
the absolute, or Kelvin scale. Why? Well, you see the zero on the Kelvin scale really means zero. At zero
on the Kelvin scale, molecules have lost all their kinetic energy. The zeros on the Fahrenheit and Celsius
scales do not really mean zero. At those temperatures, the molecules still contain quite a bit of kinetic
energy, which is why 0 oF and 0oC are not the lowest temperatures on those scales. For this reason, only
the Kelvin scale can be used in gas law calculations. We know from Charles' law, that as the temperature of
a gas goes up, its volume goes up in proportion. In other words, if the temperature of a gas doubles, then
its volume also doubles.
Let's say that you have 1 L of a gas at 20oC. What will the volume be if you heat the gas to 40 oC?
You might think that the new volume would be 2 L. After all, doesn't the temperature double from 20oC to
40 oC? And shouldn't the volume, therefore, double from 1 L to 2 L? The answer is no, because a change
from 20 oC to 40 oC is NOT a doubling of the temperature. Remember that 0oC does not really mean 0. It
would be a doubling if 0oC really meant 0. The real zero point on the Celsius scale is -273 oC. Is 40oC twice
as far from -273 oC as is 20 oC? The answer, of course, is no! However, if we repeat the problem using
Kelvin temperatures, it will work. If you have 1 L of a gas at 20 K, and you heat it to 40 K, what will the new
volume be? It will be 2 L. 40 K is, indeed, twice as much as 20 K. Remember, you must ALWAYS use
Kelvin temperatures when solving gas law problems.
The conversion of temperatures in oC to the Kelvin scale is easy to do. Simply add 273o to the
Celsius number. Follow the examples below.
oC
+ 273 = K
25 oC + 273 = 298 K
What is the volume of an "ideal" gas at a temperature of 0 K? {10}____________________________________
Problem 6. For additional practice, convert the temperatures below to Kelvin.
a. 56oC _________
b. -34oC __________
c. 12.3oC __________
Problem 7. Convert the temperatures below to oC.
a. 123 K _________
SECTION 4.6
b. 35.6 K __________
c. 358 K ___________
Solving Problems Involving Charles' Law
Like Boyle's law, Charles' law can be represented in mathematical form. One form is the following:
V1
V
= 2
T1
T2
(When the ratios of two variables are equal, the variables are directly proportional – as one increases, the
other increases.)
By substituting into the formula, you can calculate the final volume from the change in temperature and
the original volume. Now let's look at a sample problem:
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©1997, A.J. Girondi
Sample problem: A quantity of air has a volume of 80.0 mL (V1) at a temperature of 20.0oC (T1). What
will the new volume (V 2) be if the temperature is changed to 100. oC (T2), assuming the pressure remains
constant?
Follow these steps to solve this problem:
1. You must ALWAYS use Kelvin temperatures when solving gas law problems!!! Make the changes!!
20.0 oC + 273 = 293 K; 100.oC + 273 = 373 K
2. Write out the formula, and solve for the unknown variable. In this case, that would be V2.
V2 =
V1T2
T1
3. Substitute correct values into the formula, and do the math.
V2 =
(80.0 mL)(373 K)
= 102 mL
293 K
To gain practice with Charles' law, solve the problems below. Show set-up as well as the answers.
Assume that pressure remains constant in all problems.
Problem 8. At 60.0oC, a gas has a volume of 600. mL. What is the volume of this gas at 10.0oC?
Problem 9. If 105 mL of oxygen at 25.0oC were heated until its volume expanded to 120. mL, what
would its final temperature (in oC) be?
Problem 10. A quantity of hydrogen has a volume of 103 mL at a temperature of 20.0oC. To what
temperature (oC) would this gas need to be cooled in order to reduce the volume to 92.0 mL?
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©1997, A.J. Girondi
Problem 11. If 136 mL of nitrogen at 25.0oC is cooled to 0.00oC, what will the new volume be?
Problem 12. A helium balloon has a volume of 2.50 L at a temperature of 25.0oC. What temperature
(oC) is needed to decrease the size of the balloon to 1,800. mL?
SECTION 4.7
The Effect of Pressure on the Temperature of a Gas
Gay-Lussac's law deals with temperature and pressure and has the following formula:
P1
P
= 2
T1
T2
What this formula says is that the ratio of pressure divided by temperature for a gas is constant, even
though the temperatures and pressures may change. Gay Lussac's law assumes that the volume of the
gas remains constant. The reason that this ratio remains constant can be explained using the kinetic
theory. As the temperature increases, there are more collisions of the faster-moving gas molecules with
the walls of their container, and those collisions are more forceful since the molecules have more kinetic
energy. As a result, as temperature goes up, so does the pressure being exerted by the gas. In other
words, as T increases, P increases in proportion. Therefore, the ratio of P over T remains the same. You
should use the same procedure for solving problems involving Gay-Lussac's law as you used for solving
Charles' law. Note the similarity between the equation for Gay-Lussac's Law and the equation for Charles'
Law. How do you think a graph of the P vs. T relationship would compare to a graph of the V vs. T
relationship?
{11}__________________________________________________________________________________________________________________
Problem 13. A collapsible cylinder contains a gas at 765 mm Hg pressure. As external force causes the
cylinder to collapse, the pressure reaches 988 mm Hg. The final temperature in the cylinder is 86.2oC.
What was the original temperature of the gas in the cylinder before it collapsed? (Remember, always use
Kelvin temperatures!)
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©1997, A.J. Girondi
Problem 14. If an automobile tire contains air with a pressure of 26.0 psi (pounds per square inch) at a
temperature of 20.0oC. After driving for several miles, the temperature of the air in the tire increases to
48.0 oC. Assuming volume remains constant, what is the new pressure of the air in the tire?
Problem 15. A steel cylinder contains a gas at 25.0 oC and 2.50 atm pressure. The cylinder is designed
to sustain a maximum internal pressure of 3.00 atm. How high can the temperature go before the cylinder
will explode?
Explain why a basketball seems to lose some firmness when you take it outside to shoot baskets on a cold
day.
{12}____________________________________________________________________________________________________________
______________________________________________________________________________
SECTION 4.8
Solving Combined Gas Law Problems
The volume of a gas can change with changes in both pressure and temperature. By combining
Charles' law and Boyle's law, the following mathematical formula results:
P1V1
PV
= 2 2
T1
T2
This is called the combined gas law. The sample problem below illustrates the use of the combined gas
law. Carefully study the four steps used to solve the problem. Many students like to eliminate step two
and go directly to step three. Don't do that. Complete step two before going to step three as you solve
the problems.
Sample problem: A volume of gas measured 900. mL at a temperature of 51.0oC with a pressure of
700. mm. What volume will it occupy at a temperature of 27.0oC and a pressure of 760. mm? Follow these
steps:
Step 1. Write the equation.
P1V1
PV
= 2 2
T1
T2
Step 2. Solve for the unknown variable.
V2 =
P1V1T2
T1P2
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©1997, A.J. Girondi
Step 3. Substitute numbers and units.
V2 =
Step 4. Do the math.
(700. mm)(900. mL)(300. K)
(324 K)(760 . mm )
V2 = 768 mL
Try the problems below to gain practice with the combined gas law. Show all work as well as your answer.
Problem 16. In the laboratory, a volume of gas measures 600.mL at 22oC and 735 mm of pressure.
What new volume will be established if the gas is placed under standard conditions (STP)?
Problem 17. Bromine vapor occupied a volume of 2.5 L at 575oC and 780.mm. In order to reduce the
volume to 2.3 L, the temperature is changed to 17oC. What must the pressure be changed to under
these conditions?
Problem 18. A gas has a volume of 237 mL when the pressure is 2.20 atm and the temperature is
45.0 oC. What volume will the gas occupy at 3.40 atm and 23.0oC?
Problem 19. A sample of gas if found to occupy a volume of 38.2 mL at 25.0oC and an unknown
pressure. If this same gas occupies 37.5 mL at STP, determine what its unknown original pressure was in
mm Hg.
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©1997, A.J. Girondi
Problem 20. A gas occupies a volume of 250. mL at STP. To what Celsius temperature must the
system be changed to increase the volume to 0.500 L, while the pressure is adjusted to 1.50 atm? (1.00
atm = 760. mm)
Note that the three simpler gas laws can be derived from the combined gas law as follows. The combined
law is:
P1V1
PV
= 2 2
T1
T2
If temperature is not changing (is constant), then T 1 and T2 are equal. If that's true, then T1 and T2
can be eliminated from the equation. Right? This would leave us with P 1V 1 = P2V 2 which is Boyle's Law,
the relationship between pressure and volume!
If pressure is not changing (is constant), then P1 and P2 are equal and can be eliminated from the
equation for the combined laws. This would leave us with V1 / T 1 = V2 / T 2 which is Charles' Law, the
relationship between temperature and volume!
If volume is not changing (is constant), then V1 and V 2 are equal and can be eliminated from the
equation for the combined laws. This would leave us with P1 / T1 = P2 / T2. This is Gay-Lussac's Law, the
relationship between pressure and temperature!
(So, if you know the equation for the combined laws, you can always use it to derive the other laws.)
ACTIVITY 4.9
Measuring The Effect Of Temperature On Pressure
(This activity may be done by the class as a whole, rather than by lab groups. Ask your instructor.)
One way that we can measure gas pressures in
the lab is to use a piece of equipment known as a
manometer. Figure 4.6 illustrates a manometer. To
demonstrate that temperature does have an effect on
the pressure exerted by a gas, obtain a clean, dry 125
mL flask. Note the assembly drawn in Figure 4.6.
Before inserting the stopper into the flask, open or
remove the clamp attached to the tubing. Insert the
stopper assembly into the flask so that it is tight. Handle
the flask with only your finger tips so that you do not
warm it. One tube should lead to the manometer and
the other should be open to the atmosphere. Tighten
or replace the clamp to seal the flask. At this point the
difference in levels of liquid in the manometer should be
zero since the pressure in the flask is the same as
atmospheric pressure.
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Table 4.3
Temperature vs. Pressure
Condition
Pressure of Air
Room Temp.
____0___ mm Hg
Warm Hands
________ mm Hg
©1997, A.J. Girondi
atmospheric
pressure
Ruler
clamp
rubber tubing
Glass
Tubing
difference
in levels
liquid
Figure 4.6 A Manometer
Before you actually start the investigation, predict whether the pressure in the flask will rise or fall when the
air in the flask is warmed by your hands.
Prediction: _____________________________________________________________________
You are now ready to begin. Pick up the flask and wrap your hands around it. Note any change in the
levels of liquid in the manometer. When the levels have stopped moving, determine the difference in the
levels in mm. Enter the data into Table 4.3.
The liquid in the manometer is colored water, so the difference in levels that you measured is
expressed as mm H2O. Normally, pressures are measured in mm of mercury. To convert the difference in
levels from mm H2O to mm Hg, you must now divide mm H 2O by 13.5. (Mercury is 13.5 times more dense
than water.) Do this and enter the result into Table 4.3. Comment on how your prediction matches the
result. _________________________________________________________________________
Why does the air exert more pressure when it is warmed?
{13}
_________________________________
______________________________________________________________________________
ACTIVITY 4.10
Comparing Equilibrium Vapor Pressures
(This activity may be done by the class as a whole, rather than by lab groups. Ask the instructor.)
Manometers are used to measure the vapor pressure of liquids. When a liquid in a closed
container evaporates it forms a vapor. This vapor exerts a pressure. As the liquid continues to evaporate,
the pressure increases. Some of the vapor molecules collide with the surface of the liquid in the container
and return to the liquid phase again. This is called condensation. At first, the evaporation process goes
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©1997, A.J. Girondi
faster than the condensation, so that the amount of vapor in the container increases (as does the
pressure). As the amount of vapor in the container increases, the rate of condensation will also increase.
Finally, a point is reached in which the rates of evaporation and condensation are equal. At this point, the
amount of vapor in the container will be constant, as will the pressure. (As indicated when the mercury
levels stop moving in Activity 4.9.) This condition is known as equilibrium. It is characterized by two
opposing forces going on at equal rates. When equilibrium has been reached, the pressure of the vapor
in the container is constant and is known as the equilibrium vapor pressure.
Figure 4.7 Equilibrium Vapor Pressure in a Closed System
The equilibrium vapor pressure will remain constant provided that the temperature remains
constant. If temperature changes, that will offset the equilibrium by changing evaporation and
condensation rates. Eventually, a new equilibrium will be established at the new temperature, and a new
equilibrium vapor pressure will exist. In other words, a substance has a specific equilibrium vapor pressure
at any given temperature. Different substances have different equilibrium vapor pressures.
A substance that has strong forces of attraction between its molecules is harder to vaporize and
will have a lower equilibrium vapor pressure (at a given temperature) than would a substance with weaker
forces of attraction between its molecules. We conclude that the weaker the attraction between the
molecules of a substance, the higher its equilibrium vapor pressure will be at any given temperature. We
can use the apparatus in Figure 4.6 (which you used in the previous activity) to compare the equilibrium
vapor pressures of various liquids. NOTE! All of the liquids used in this activity are flammable.
Procedure:
1. Obtain the manometer and remove the clamp from the hosing on the 125 mL Erlenmeyer flask; put
about 10 mL of acetone into the DRY flask and replace the stopper. (If the flask is not dry, rinse it with a
little acetone first. Discard the acetone used for the rinsing into the sink and run the water for a minute.)
Clamp the hose shut (as shown in Figure 4.6). Pick up the flask and swirl the liquid for about 10 seconds to
hasten the evaporation. Remove your hands from the flask. After two minutes, record the difference in
the levels of the liquid on the manometer (even if the levels are still moving). This difference is in mm of
water. To convert to mm of mercury, divide the difference by 13.5. (Mercury is 13.5 times more dense
than water.) Record this difference in levels in mm Hg in Table 4.4. Next, pick up the flask and wrap your
hands around it to warm the acetone. The bottom of the flask should press against one of your palms.
You should see the water levels moving in the manometer. After two minutes, read the difference in the
levels (in mm). Again, convert to mm Hg and record the data in Table 4.4. Part of the difference in levels
is due to the vapor pressure exerted by the acetone. Another part is due to the expansion of air which is
also present in the flask. Even though the air adds to the total pressure, this procedure will allow us to
compare the vapor pressure of acetone to that of anther substance. Discard the acetone into the sink and
run some water.
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©1997, A.J. Girondi
2. Rinse the flask with a little ethyl alcohol and discard it into the sink while running some water. Repeat
step 1 using about 10 mL of ethyl alcohol instead of acetone. Record the data in Table 4.4.
Table 4.4
Equilibrium Vapor Pressures
Liquid
Pressure of Vapor
(Room Temp)
Pressure of Vapor
(Warm Hands Temp)
Acetone
__________mm Hg
__________mm Hg
Ethyl Alcohol
__________mm Hg
__________mm Hg
According to your results, which liquid seems to have the weakest attractive forces between its
molecules? {14}_________________ Liquids which evaporate most readily have high vapor pressures.
Which liquid do you think will evaporate most readily?
{15}____________________________Why
do you think that
some liquids evaporate more easily than others?{16}____________________________________________________________
_____________________________________________________________________________________________________________________
SECTION 4.11
Review Problems To Solve
If you look back at the problems you have solved which make use of Boyle's law, Charles'
law, Gay-Lussac's law, and the combined gas law you will notice that all of these problems include what we
might call the "original" conditions and the "new" conditions. For example, you might have been given the
volume of a gas at a certain temperature and pressure, and you may have been asked to find the "new"
volume after the original temperature or pressure had changed.
There is yet another gas law, known as the ideal gas law which can be used to solve gas problems
which involve only one set of conditions. For example, what is the volume of 2.00 grams of hydrogen gas
at STP? Notice that only one set of conditions is given? You can not solve this problem using the laws
that you have learned up to this point. In order to understand the ideal gas law, you first need to learn
about the mole concept which is presented in chapters 7, 8 and 9 of ALICE. After you are familiar with the
mole concept, we will learn about this final gas law.
You will find some review problems below. Read each problem to determine which gas law you will
use to solve it. In the space to the left of each problem, indicate which law you have chosen by writing B
for Boyle's law, CH for Charles' law, G for Gay-Lussac's law, or COM for combined gas law. Then, solve
each problem showing all work.
Problem 21. _______ A basketball has a volume of 6.00 L when it is kept in a closet that is at 25.0oC.
What does the volume of the ball become when it is taken outdoors on a chilly 10.0oC day? (Assume
constant pressure.)
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©1997, A.J. Girondi
Problem 22. ________ A 10.0 L sample of water vapor at 200. oC and 1.00 atm pressure is cooled to
150. oC. The gas now occupies a volume of 7.00 L. Calculate the resulting pressure in the container.
Problem 23. ________ The volume of a gas at 1800. mm Hg pressure is 10.0 L. If the temperature is
kept constant, what will be the volume of gas at standard atmospheric pressure, 760. mm Hg?
Problem 24. ________ The pressure gauge on a cylinder of oxygen gas at a temperature of 15.0oC
reads 50.0 atm. What would the pressure gauge read if the temperature of the gas is raised to 38.0oC?
Problem 25. ________ A gas occupies a volume of 4.00 L at 470. mm and -70.0 oC. What will its
volume be at 963 mm and 122oC?
Problem 26. ________ At 755 mm Hg pressure, 1.00 cm 3 of a gas has a mass of 0.0341 grams. What
will the density of this gas be (in grams per cm3) at a pressure of 905 mm Hg pressure? (Hint: the pressure
change will affect the volume, but not the mass.)
Density = __________ g/cm3
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©1997, A.J. Girondi
ACTIVITY 4.12
A Teacher Demonstration of the Pressure–Temperature
Relationship
The relationship between pressure and temperature can also be demonstrated (with the aid of a
computer) using a syringe connected to a temperature probe. Your teacher will demonstrate. The
mechanical energy used to push the plunger of the syringe into the cylinder is transferred to the gas
molecules in the syringe where it is converted into kinetic energy. Watch the change in the temperature
of the gas in the syringe (shown on the screen of the computer display) as the gas in the syringe is
compressed.
What do you observe?
{17}__________________________________________________________
______________________________________________________________________________
SECTION 4.13
Learning Outcomes
Understanding the gas laws is essential to your mastery of chemistry. Before you move on to
chapter 5, review the learning outcomes listed below. Check the ones that you are certain you can
handle. Arrange to take the chapter quiz and move on to chapter 5.
_____1. Describe the basic principles of the kinetic-molecular theory.
_____2. Define pressure and how it is measured and expressed.
_____3. Convert between the Celsius and Kelvin temperature scales.
_____4. Explain what is meant by the absolute temperature scale.
_____5. Mathematically adjust temperature and pressure to standard conditions.
_____6. Qualitatively explain the effects of pressure and temperature changes on the volume of a gas.
_____7. Solve mathematical problems using Charles', Boyle's, Gay-Lussac's, and the combined gas laws.
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©1997, A.J. Girondi
SECTION 4.14 Answers to Questions and Problems
Questions:
{1} Constantly moving gas particles collide with air molecules causing mixing and the spread of the gas
through air; {2} Constantly moving gas particles collide with the wall of their container; {3} decreases
{4} yes - as pressure increases volume decreases; {5} pressure; {6} independent; {7} larger; {8} it
increases the pressure exerted; {9} so that volumes of gases can be compared under the same
conditions of T and P, which allows us to relate volumes of gases to actual amounts of gas; {10} zero; {11}
they would look similar; {12} when temperature decreases so does pressure; {13} molecules collide with
the walls of the container more frequently and with greater force; {14} (the one that had the highest vapor
pressure); {15} (the one that had the lowest vapor pressure); {16} liquids which evaporate more easily
have weaker attractive forces between the molecules; {17} Temperature should increase as pressure
increases.
Problems:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
94.9 mL
1480 mm Hg
612 mL
403 mm Hg
844 mL
a. 329 K; b. 239 K; c. 285.3 K
a. -150oC; b. -237.4oC; c. 85oC
510 mL
68 oC
-11 oC
125 mL
-58.0oC
5.00 oC (278 K)
28.5 psi
84.6 oC (357.6 K) - rounds to 85oC (358 K)
537 mL rounds to 540 mL
290 mm
143 mL
814 mm Hg
546 oC
CH, 5.70 L
COM, 1.28 atm
B, 23.7 L
G, 54.0 atm
COM, 3.8 L
B, 0.0409 g/cm3
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©1997, A.J. Girondi
SECTION 4.15 Student Notes
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©1997, A.J. Girondi