The Kinetic Molecular Theory
of Ideal Gases
These statements are made only for what is called an ideal gas. They cannot all be
rigorously applied (i.e. mathematically) to real gases, but can be used to explain their
observed behavior qualitatively.
1. All matter is composed of tiny, discrete particles (molecules or atoms).
2. Ideal gases consist of small particles (molecules or atoms) that are far apart in
comparison to their own size. The molecules of a gas are very small compared to the
distances between them.
3. These particles are considered to be dimensionless points which occupy zero volume.
The volume of real gas molecules is assumed to be negligible for most purposes.
This above statement is NOT TRUE. Real gas molecules do occupy volume and it does
have an impact on the behavior of the gas. This impact WILL BE IGNORED when
discussing ideal gases.
4. These particles are in rapid, random, constant straight line motion. This motion can be
described by well-defined and established laws of motion.
5. There are no attractive forces between gas molecules or between molecules and the
sides of the container with which they collide.
In a real gas, there actually is attraction between the molecules of a gas. Once again, this
attraction WILL BE IGNORED when discussing ideal gases.
6. Molecules collide with one another and the sides of the container.
7. Energy can be transferred in collisions among molecules.
8. Energy is conserved in these collisions, although one molecule may gain energy at the
expense of the other.
9. Energy is distributed among the molecules in a particular fashion known as the
Maxwell-Boltzmann Distribution.
10. At any particular instant, the molecules in a given sample of gas do not all possess the
same amount of energy. The average kinetic energy of all the molecules is proportional to
the absolute temperature.
The Four Gas Law Variables:
Temperature, Pressure, Volume, and Moles
I) Volume
All gases must be enclosed in a container that, if there are openings, can be sealed with
no leaks. The three-dimensional space enclosed by the container walls is called volume.
When the generalized variable of volume is discussed, the symbol V is used.
Volume in chemistry is usually measured in liters (symbol = L) or milliliters (symbol =
mL). A liter is also called a cubic decimeter (dm3).
Other units of volume do occur such as cubic feet (cu. ft. or ft3) or cubic centimeters (cc
or cm3). The main point to remember is: whatever units of volume are used, use them all
the way through the problem. If you must convert from one unit to another, make sure
you do it correctly.
II) Temperature
All gases have a temperature, usually measured in degrees Celsius (symbol = °C). Note
that Celsius is capitalized since this was the name of a person (Anders Celsius). When the
generalized variable of temperature is discussed, the symbol T is used.
There is another temperature scale which is very important in gas behavior. It is called
the Kelvin scale (symbol = K). Note that K does not have a degree sign and Kelvin is
capitalized because this was a person's title (Lord Kelvin, his given name was William
Thomson).
All gas law problems will be done with Kelvin temperatures. If you were to use degrees
Celsius in any of your calculations, YOU WOULD BE WRONG. You can convert
between Celsius and Kelvin like this: Kelvin = Celsius + 273.15. Often, the value of 273
is used instead of 273.15. Check with your teacher on this point. All examples to follow
will use 273. For example, 25 °C = 298 K, because 25 + 273 = 298.
Standard temperature is defined as zero degrees Celsius or 273 K.
The Kelvin temperature of a gas is directly proportional to its kinetic energy. Double the
Kelvin temperature, you double the kinetic energy.
III) Pressure
Gas pressure is created by the molecules of gas hitting the walls of the container. This
concept is very important in helping you to understand gas behavior. Keep it solidly in
mind. This idea of gas molecules hitting the wall will be used often. When the
generalized variable of pressure is discussed, the symbol P is used.
There are three different units of pressure used in chemistry. This is an unfortunate
situation, but we cannot change it. You must be able to use all three. Here they are:
1. atmospheres (symbol = atm)
2. millimeters of mercury (symbol = mm Hg)
3. Pascals (symbol = Pa) or, more commonly, kiloPascals (symbol = kPa)
Standard pressure is defined as one atm. or 760.0 mm Hg or 101.325 kPa.
Standard temperature and pressure is a very common phrase in chemistry, so common it
has been abbreviated to STP. What some will use as STP is actually called standard
ambient temperature and pressure (SATP), but the difference between the two is
unimportant at this stage of your chemistry training.
Abbreviations
atm - atmosphere
mm Hg - millimeters of mercury
(milliliter)
torr - another name for mm Hg
(liter) = 1000 mL
Pa - Pascal (kPa = kilo Pascal)
K - Kelvin
°C - degrees Celsius
101.325 kPa = 101,325 Pa
Conversions
K = °C + 273
1 cm3 (cubic centimeter)
1 dm3 (cubic decimeter)
Standard Conditions
0.00 °C = 273 K
1.00 atm = 760.0 mm Hg
=
=
1 mL
1 L
=
IV. Amount of Gas
The amount of gas present is measured in moles (symbol = mol) or in grams (symbol = g
or gm). Typically, if grams are used, you will need to convert to moles at some point.
When the generalized variable of amount in moles is discussed, the letter "n" is used as
the symbol (note: the letter is in lowercase. The others above are all caps.).
Boyle's Law
Discovered by Robert Boyle in 1662. On the continent of Europe, this law is attributed to
Edme Mariotte, therefore those counties tend to call this law by his name. Mariotte,
however, did not publish his work until 1676.
His law gives the relationship between pressure and volume if temperature and amount
are held constant.
If the volume of a container is increased, the pressure decreases.
If the volume of a container is decreased, the pressure increases.
Why?
Suppose the volume is increased. This means gas molecules have farther to go and they
will impact the container walls less often per unit time. This means the gas pressure will
be less because there are less molecule impacts per unit time.
If the volume is decreased, the gas molecules have a shorter distance to go, thus striking
the walls more often per unit time. This results in pressure being increased because there
are more molecule impacts per unit time.
The mathematical form of Boyle's Law is: PV = k
This means that the pressure-volume product will always be the same value if the
temperature and amount remain constant. This relationship was what Boyle discovered.
This is an inverse mathematical relationship. As one quantity goes up in the value, the
other goes down.
Suppose P1 and V1 are a pressure-volume pair of data at the start of an experiment. In
other words, some container of gas is created and the volume and pressure of that
container is measured. Keep in mind that the amount of gas and the temperature DOES
NOT CHANGE. When you multiply P and V together, you get a number that is called k.
We don't care what the exact value is.
Now, if the volume is changed to a new value called V2, then the pressure will
spontaneously change to P2. It will do so because the PV product must always equal k.
The PV product CANNOT just change to any old value, it MUST go to k. (If the
temperature and amount remain the same.)
So we know this: P1V1 = k
And we know that the second data pair equals the same constant: P2V2 = k
Since k = k, we can conclude that P1V1 = P2V2.
This equation of P1V1 = P2V2 will be very helpful in solving Boyle's Law problems.
Example #1: 2.00 L of a gas is at 740.0 mmHg pressure. What is its volume at standard
pressure?
Answer: this problem is solved by inserting values into P1V1 = P2V2.
(740.0 mmHg) (2.00 L) =(760.0 mmHg) (x)
Example #2: 5.00 L of a gas is at 1.08 atm. What pressure is obtained when the volume is
10.0 L?
Answer: use the same technique.
(1.08 atm) (5.00 L) =(x) (10.0 L)
Example #3: 2.50 L of a gas was at an unknown pressure. However, at standard pressure,
its volume was measured to be 8.00 L. What was the unknown pressure? Answer: notice
the units of the pressure were not specified, so any can be used. If this were a test
question, you might want to inquire of the teacher as to a possible omission of desired
units. Let's use kPa since the other two units were used above. Once again, insert into
P1V1 = P2V2 for the solution.
(x) (2.50 L) = (101.325 kPa) (8.00 L)
Gas Law Problems- Boyle's Law
Abbreviations
atm - atmosphere
mm Hg - millimeters of mercury
(milliliter)
torr - another name for mm Hg
(liter) = 1000 mL
Pa - Pascal (kPa = kilo Pascal)
K - Kelvin
°C - degrees Celsius
101.325 kPa = 101,325 Pa
Conversions
K = °C + 273
1 cm3 (cubic centimeter)
1 dm3 (cubic decimeter)
=
=
Standard Conditions
0.00 °C = 273 K
1.00 atm = 760.0 mm Hg
1 mL
1 L
=
1. A gas occupies 12.3 liters at a pressure of 40.0 mm Hg. What is the volume when the
pressure is increased to 60.0 mm Hg?
2. If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its
volume at a pressure of 2.50 atm?
3. To what pressure must a gas be compressed in order to get into a 3.00 cubic foot tank
the entire weight of a gas that occupies 400.0 cu. ft. at standard pressure?
4. A gas occupies 1.56 L at 1.00 atm. What will be the volume of this gas if the pressure
becomes 3.00 atm?
5. A gas occupies 11.2 liters at 0.860 atm. What is the pressure if the volume becomes
15.0 L?
6. 500.0 mL of a gas is collected at 745.0 mm Hg. What will the volume be at standard
pressure?
7. Convert 350.0 mL at 740.0 mm of Hg to its new volume at standard pressure.
8. Convert 338 L at 63.0 atm to its new volume at standard pressure.
9. Convert 273.15 mL at 166.0 mm of Hg to its new volume at standard pressure.
10. Convert 77.0 L at 18.0 mm of Hg to its new volume at standard pressure.
11. When the pressure on a gas increases, will the volume increase or decrease?
12. If the pressure on a gas is decreased by one-half, how large will the volume change
be?
13. A gas occupies 4.31 liters at a pressure of 0.755 atm. Determine the volume if the
pressure is increased to 1.25 atm.
14. 600.0 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at 2.00
atm?
15. 400.0 mL of a gas are under a pressure of 800.0 torr. What would the volume of the
gas be at a pressure of 1000.0 torr?
16. 4.00 L of a gas are under a pressure of 6.00 atm. What is the volume of the gas at 2.00
atm?
17. A gas occupies 25.3 mL at a pressure of 790.5 mm Hg. Determine the volume if the
pressure is reduced to 0.804 atm.
18. A sample of gas has a volume of 12.0 L and a pressure of 1.00 atm. If the pressure of
gas is increased to 2.00 atm, what is the new volume of the gas?
19. A container of oxygen has a volume of 30.0 mL and a pressure of 4.00 atm. If the
pressure of the oxygen gas is reduced to 2.00 atm and the temperature is kept constant,
what is the new volume of the oxygen gas?
20. A tank of nitrogen has a volume of 14.0 L and a pressure of 760.0 mm Hg. Find the
volume of the nitrogen when its pressure is changed to 400.0 mm Hg while the
temperature is held constant.
21. A 40.0 L tank of ammonia has a pressure of 8.00 atm. Calculate the volume of the
ammonia if its pressure is changed to 12.0 atm while its temperature remains constant.
22. Two hundred liters of helium at 2.00 atm and 28.0 °C is placed into a tank with an
internal pressure of 600.0 kPa. Find the volume of the helium after it is compressed into
the tank when the temperature of the tank remains 28.0 °C.
23. You are now wearing scuba gear and swimming under water at a depth of 66.0 ft.
You are breathing air at 3.00 atm and your lung volume is 10.0 L. Your scuba gauge
indicates that your air supply is low so, to conserve air, you make a terrible and fatal
mistake: you hold your breath while you surface. What happens to your lungs? Why?
24. Solve Boyle's Law equation for V2.
25. Boyle's Law deals what quantities?
a. pressure/temperature
b. pressure/volume
c. volume/temperature
d. volume temperature/pressure
26. A 1.5 liter flask is filled with nitrogen at a pressure of 12 atmospheres. What size
flask would be required to hold this gas at a pressure of 2.0 atmospheres?
27. 300 mL of O2 are collected at a pressure of 645 mm of mercury. What volume will
this gas have at one atmosphere pressure?
28. How many cubic feet of air at standard conditions (1.00 atm.) are required to inflate a
bicycle tire of 0.50 cu. ft. to a pressure of 3.00 atmospheres?
29. How much will the volume of 75.0 mL of neon change if the pressure is lowered
from 50.0 torr to 8.00 torr?
30. A tank of helium has a volume of 50.0 liters and is under a pressure of 2000.0 p.s.i..
This gas is allowed to flow into a blimp until the pressure in the tank drops to 40.00 p.s.i.
and the pressure in the blimp is 30.00 p.s.i.. What will be the volume of the blimp?
31. What pressure is required to compress 196.0 liters of air at 1.00 atmosphere into a
cylinder whose volume is 26.0 liters?
Charles' Law
Discovered by Joseph Louis Gay-Lussac in 1802. He made reference in his paper to
unpublished work done by Jacques Charles about 1787. Charles had found that oxygen,
nitrogen, hydrogen, carbon dioxide, and air expand to the same extent over the same 80
degree interval. He also invented the hydrogen-filled balloon and on December 1, 1783,
he ascended into the air and became possibly the first man in history to witness a double
sunset.
Gay-Lussac was no slouch in the area of ballooning. On September 16, 1804, he
ascended to an altitude of 7016 meters (just over 23,000 feet - about 4.3 miles). This
remained the world altitude record for almost 50 years and then was broken by only a few
meters.
Because of Gay-Lussac's reference to Charles' work, many people have come to call the
law by the name of Charles' Law. There are some books which call the temperaturevolume relationship by the name of Gay-Lussac's Law and there are some which call it
the Law of Charles and Gay-Lussac. Needless to say, there are some confused people out
there. Most textbooks call it Charles' Law, so that's what we will use.
This law gives the relationship between volume and temperature if pressure and amount
are held constant.
If the volume of a container is increased, the temperature increases.
If the volume of a container is decreased, the temperature decreases.
Why?
Suppose the temperature is increased. This means gas molecules will move faster and
they will impact the container walls more often. This means the gas pressure inside the
container will increase (but only for an instant. Think of a short span of time.) The greater
pressure on the inside of the container walls will push them outward, thus increasing the
volume. When this happens, the gas molecules will now have farther to go, thereby
lowering the number of impacts and dropping the pressure back to its constant value.
It is important to note that this momentary increase in pressure lasts for only a very, very
small fraction of a second. You would need a very fast, accurate pressure sensing device
to measure this momentary change.
Consider another case. Suppose the volume is suddenly increased. This will reduce the
pressure, since molecules now have farther to go to impact the walls. However, this is not
allowed by the law; the pressure must remain constant. Therefore, the temperature must
go up, in order to get the molecules to the walls faster, thereby overcoming the longer
distance and keeping the pressure constant.
Charles' Law is a direct mathematical relationship. This means there are two connected
values and when one goes up, the other also increases.
The mathematical form of Charles' Law is: V ÷ T = k
This means that the volume-temperature fraction will always be the same value if the
pressure and amount remain constant.
Let V1 and T1 be a volume-temperature pair of data at the start of an experiment. If the
volume is changed to a new value called V2, then the temperature must change to T2.
So we know this: V1 ÷ T1 = k
And we know this: V2 ÷ T2 = k
Since k = k, we can conclude that V1 ÷ T1 = V2 ÷ T2.
This equation of V1 ÷ T1 = V2 ÷ T2 will be very helpful in solving Charles' Law
problems.
Before going to some sample problems, let's be very clear:
EVERY TEMPERATURE USED IN A CALCULATION MUST BE IN KELVINS,
NOT DEGREES CELSIUS.
DON'T YOU DARE USE CELSIUS IN A NUMERICAL CALCULATION. USE
KELVIN EVERY TIME.
Example #1: A gas is collected and found to fill 2.85 L at 25.0°C. What will be its
volume at standard temperature?
Answer: convert 25.0°C to Kelvin and you get 298 K. Standard temperature is 273 K. We
plug into our equation like this:
Example #2: 4.40 L of a gas is collected at 50.0°C. What will be its volume upon cooling
to 25.0°C?
First of all, 2.20 L is the wrong answer. Sometimes a student will look at the temperature
being cut in half and reason that the volume must also be cut in half. That would be true
if the temperature was in Kelvin. However, in this problem the Celsius is cut in half, not
the Kelvin.
Answer: convert 50.0°C to 323 K and 25.0°C to 298 K. Then plug into the equation and
solve for x, like this:
Example #3: 5.00 L of a gas is collected at 100 K and then allowed to expand to 20.0 L.
What must the new temperature be in order to maintain the same pressure (as required by
Charles' Law)?
Answer:
Gas Law Problems- Charles' Law
Abbreviations
atm - atmosphere
mm Hg - millimeters of mercury
(milliliter)
torr - another name for mm Hg
(liter) = 1000 mL
Pa - Pascal (kPa = kilo Pascal)
K - Kelvin
°C - degrees Celsius
101.325 kPa = 101,325 Pa
Conversions
K = °C + 273
1 cm3 (cubic centimeter)
1 dm3 (cubic decimeter)
Standard Conditions
0.00 °C = 273 K
1.00 atm = 760.0 mm Hg
=
=
1 mL
1 L
=
32. Calculate the decrease in temperature when 2.00 L at 20.0 °C is compressed to 1.00
L.
33. 600.0 mL of air is at 20.0 °C. What is the volume at 60.0 °C?
34. A gas occupies 900.0 mL at a temperature of 27.0 °C. What is the volume at 132.0
°C?
35. What change in volume results if 60.0 mL of gas is cooled from 33.0 °C to 5.00 °C?
36. Given 300.0 mL of a gas at 17.0 °C. What is its volume at 10.0 °C?
37. A gas occupies 1.00 L at standard temperature. What is the volume at 333.0 °C?
38. At 27.00 °C a gas has a volume of 6.00 L. What will the volume be at 150.0 °C?
39. At 225.0 °C a gas has a volume of 400.0 mL. What is the volume of this gas at 127.0
°C?
40. At 210.0 °C a gas has a volume of 8.00 L. What is the volume of this gas at -23.0 °C?
41. The temperature of a 4.00 L sample of gas is changed from 10.0 °C to 20.0 °C. What
will the volume of this gas be at the new temperature if the pressure is held constant?
42. Carbon dioxide is usually formed when gasoline is burned. If 30.0 L of CO2 is
produced at a temperature of 1.00 x 103 °C and allowed to reach room temperature (25.0
°C) without any pressure changes, what is the new volume of the carbon dioxide?
43. A 600.0 mL sample of nitrogen is warmed from 77.0 °C to 86.0 °C. Find its new
volume if the pressure remains constant.
44. What volume change occurs to a 400.0 mL gas sample as the temperature increases
from 22.0 °C to 30.0 °C?
45. A gas syringe contains 56.05 milliliters of a gas at 315.1 K. Determine the volume
that the gas will occupy if the temperature is increased to 380.5 K
46. A gas syringe contains 42.3 milliliters of a gas at 98.15 °C. Determine the volume
that the gas will occupy if the temperature is decreased to -18.50 °C.
47. When the temperature of a gas decreases, does the volume increase or decrease?
48. If the Kelvin temperature of a gas is doubled, the volume of the gas will increase by
____.
49. Solve the Charles' Law equation for V2.
50. Charles' Law deals with what quantities?
a. pressure/temperature
b. pressure/volume
c. volume/temperature
d. volume/temperature/pressure
51. If 540.0 mL of nitrogen at 0.00 °C is heated to a temperature of 100.0 °C what will be
the new volume of the gas?
52. A balloon has a volume of 2500.0 mL on a day when the temperature is 30.0 °C. If
the temperature at night falls to 10.0 °C, what will be the volume of the balloon if the
pressure remains constant?
53. When 50.0 liters of oxygen at 20.0 °C is compressed to 5.00 liters, what must the new
temperature be to maintain constant pressure?
54. If 15.0 liters of neon at 25.0 °C is allowed to expand to 45.0 liters, what must the new
temperature be to maintain constant pressure?
55. 3.50 liters of a gas at 727.0 K will occupy how many liters at 153.0 K?