Kinetic Molecular Theory
and
Gas Laws Table of Contents
"Men occasionally stumble over truth, but most of them pick themselves
up and hurry off as if nothing had happened." -- Winston Churchill
Special Significant Figure Note: Textbook authors and teachers sometimes don't pay
slavish attention to significant figure rules in the gas laws unit. However, the key is to be
reasonable. For example, suppose 100 K is used in a multiplication, so you think the
answer should have only one significant figure. Wrong! 100 K is a value easily measured
in the lab (well, not a high school lab) to three or four significant figures. Chemists
students in the classroom are advised to think of this value as having three significant
figures. Other chemistry students should check with your classroom teacher on this point
and follow what you are told.
Fundamental ideas
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Basics of the Kinetic Molecular Theory
The Four Gas Law Variables: Temperature, Pressure, Volume, and Moles
Necessary conversions for doing gas law problems
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
Converting Between Celsius and Kelvin Temperature Scales
Converting Between atm, mmHg, and kPa Pressure Units
Six gas laws which hold two of the four variables constant
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Boyle's Law (PV vary, nT constant)
Charles' Law (VT vary, nP constant)
Gay-Lussac's Law (PT vary, nV constant)
Avogadro's Law (Vn vary, PT constant)
Diver's Law (Pn vary, VT constant)
The Gas Law with No Name (Tn vary, PV constant)
Additional gas laws
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Combined Gas Law (PVT vary, n constant)
Ideal Gas Law (PV = nRT)
Dalton's Law of Partial Pressures
Graham's Law
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Additional gas-related tutorials
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Molar Volume
Gas Density
Vapor Pressure
Homework

150 Gas Law Problems in a PDF file
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. Well-defined and
established laws of motion can describe this 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.
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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.
The Chemists discusses such things as Boyle's Law and other laws where volume is a
variable. That means the container involved in the experiment has a movable wall. When
the volume goes up, that wall slides out. When the volume goes down, the wall slides in.
This is VERY IMPORTANT to remember. Imagine the seal of the movable wall to be
perfect so no gas escapes.
If the volume is constant, then the container is made with thick, rigid walls that cannot
move. If the pressure increased too much, the walls would break, destroying the
experiment. However, within the limits of any experiment discussed, the walls remain
fixed and the volume stays constant. As Mr. Spock said to Bones in one of the Star Trek
movies, "Remember."
II) Temperature
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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. Your teacher may try
and trip you up on this point.
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
The molecules of gas hitting the walls of the container create gas pressure. 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 = mmHg)
3. Pascals (symbol = Pa) or, more commonly, kiloPascals (symbol = kPa)
You will find more on pressure units in another page.
Standard pressure is defined as one atm. or 760.0 mm Hg or 101.325 kPa.
Let's pause here for a second. Make sure that you nail the values for standard pressure.
You MUST have those values memorized!! Many problems will simply say "standard
pressure" and you have to already know the value. Here they are again:
Pressure Unit Standard Value
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atm
1.00
mmHg
760.0
kPa
101.325
Standard temperature and pressure is a very common phrase in chemistry, so common it
has been abbreviated to STP. What the Chemists will use as STP is actually called
standard ambient temperature and pressure (STAP), but the difference between the two is
unimportant at this stage of your chemistry training.
There is no such thing as standard volume, although you will probably learn about molar
volume in your class.
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 Converting between Celsius and Kelvin
There are not any high school level gas law problems that the Chemists are aware of that
use the Celsius temperature directly in the calculation.
If you have a Celsius temperature in the problem, you MUST change it to Kelvin, in
order to use it in your problem.
Sometimes your teacher might put a temperature in the problem, but you really don't need
to use it. Your teacher is doing what he or she is driven to do: trying the confuse poor
teenage kids who are really trying their hardest. The Chemists understands this fully for,
you see, this is what happens in his class. All teachers know this is really fun! OK, back
to work.
You can convert between Celsius and Kelvin like this: Kelvin = Celsius + 273.15. Often,
the value of 273.0 is used instead of 273.15. Check with your teacher on this point. All
examples to follow will use 273.0.
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A point before going on to some problems: very often in gas law problems it looks like
significant figure rules are being violated. The reality is that they are not. In reality, the
significant figure concept is more complex than the simple rules taught at this level.
However, having said that, be careful to watch your teacher's actions in class and ask
what the rules are in your class. Don't use this phrase" "Well, there's some guy on the
Internet who says you're wrong."
Example #1: convert 25.0°C to Kelvin.
Answer: 25.0 + 273 = 298.0
Essentially the 273 is being treated as 273.0
Everybody (except you!) in chemistry knows the true conversion value is 273.15, but the
decimal portion is usually ignored. Also, the most common type of thermometer in high
school labs is readable to the nearest tenth. So temperatures are usually written just to the
tenth place at this introductory level
The Chemists occasionally has a student who insists on using 273.15. You may be the
same. However, be careful on this. You may do a test calculation using 273.15 and the
teacher does it with 273. That difference may generate a value for your answer that
causes the teacher to deduct points. Follow the policy your classroom teacher sets down.
Example #2: convert 375 K to degrees Celsius.
Answer: 375 - 273 = 102°C
Example #3: Convert -50°C to Kelvin. That's minus 50
Answer: -50 plus 273 = 223 K
The Chemists doesn't feel any extra practice problems are needed. See your classroom
teacher if you need more.
e symbol (note: the letter is in lowercase. The others above are all caps.).
Converting between Units of Pressure: atm., mmHg and
kPa
Here is a repeat from the "Four Variables" file:
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:
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1. atmospheres (symbol = atm)
2. millimeters of mercury (symbol = mm Hg)
3. Pascals (symbol = Pa) or, more commonly, kiloPascals (symbol = kPa)
Here is a repeat from each of the problem worksheets:
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
=
Doing Pressure Conversions
I. Between atmospheres and millimeters of mercury. One atm. equals 760.0 mm Hg, so
there will be a multiplication or division based on the direction of the change.
Example #1 - Convert 0.875 atm to mmHg.
Solution - multiply the atm value by 760.0 mmHg / atm.
Notice that the atm values - one in the numerator and one in the denominator - cancel,
leaving mmHg.
Example #2 - Convert 745.0 mmHg to atm.
Solution - divide the mmHg value by 760.0 mmHg / atm
Notice that the mmHg values cancel and the atm, in the denominator of the denominator,
moves to the numerator.
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II. Between atmospheres and kilopascals. One atm equals 101.325 kPa, so there will be a
multiplication or division based on the direction of the change.
Example #3 - Convert 0.955 atm to kPa.
Solution - multiply the atm value by 101.325 kPa / atm.
Notice that the atm values - one in the numerator and one in the denominator - cancel,
leaving kPa.
Example #4 - Convert 98.35 kPa to atm.
Solution - divide the kPa value by 101.325 kPa / atm.
Notice that the kPa values cancel and the atm, in the denominator of the denominator,
moves to the numerator.
III. Between millimeters of mercury and kilopascals. 760.0 mmHg equals 101.325 kPa,
so both values will be involved. This situation is slightly unusual because most
conversions involve a one, usually in the denominator. The conversion examples above
are examples of a one being involved.
In this conversion, both 760.0 and 101.325 will be involved and the location of each
(numerator or denominator) will depend on the conversion.
Example #5 - Convert 740.0 mmHg to kPa.
Notice that the mmHg will cancel, since one is in the numerator and one is in the
denominator, leaving kPa as the unit on the answer.
Example #6 - Convert 99.25 kPa to mmHg.
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Notice that the kPa will cancel, since one is in the numerator and one is in the
denominator, leaving mmHg as the unit on the answer.
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.
In the Chemists classroom, a student will occasionally ask "What is k?"
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. The Chemist does not care what the exact numbers are, just that there
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are two numbers. 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.)
Of course, you now want to ask "Why does it have to stay at k?" The Chemist believes it
is best right now to ignore that question even though it is a perfectly valid one.
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)
Multiply the left side and divide (by 760.0 mmHg) to solve for x. Note that the units of
mmHg will cancel. x is a symbol for an unknown and, technically, does not carry units.
So do not write x L for x liters. Just keep checking to see you are using the proper
equation and you have all the right values and units. Don't put a unit on the unknown.
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)
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Hopefully you can see that Boyle's Law problems all use the same solution technique. It's
just a question of where the x is located. Two problems will arise during the gas laws unit
in your classroom:
1. How to match a given problem with what law it is, so you can solve it.
2. Watching out for questions worded in a slightly confusing manner or with
unnecessary information. Teachers like to do these sorts of things, if you have
noticed.
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.
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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?
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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?
Gas Law Problems - Boyle's Law Answers: Part One
I tried to put the answers in the form of P1V1 = P2V2. They don't have to be in that order,
except that the sub ones must be paired on one side of the equals sign and the sub twos
must be paired on the other.
Some answers at the start are provided. Work the rest out. Show units on all values, not
just the answer!! Pay attention to sig figs.
1. (40.0 mm Hg) (12.3 liters) = (60.0 mm Hg) (x); x = 8.20 L, note three significant
figures!!
2. (1.00 atm) ( 3.60 liters) = (2.50 atm) (x); x = 1.44 L
3. ( 400.0 cu. ft) (1.00 atm) = (x) (3.00 cubic foot); x = 133 atm
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4. (1.56 L) (1.00 atm) = (3.00 atm) (x); 0.520 L
5. (11.2 liters) (0.860 atm) = (x) (15.0 L); x = 0.642 atm
6. ( 745.0 mm Hg) (500.0 mL) = (760.0 mm Hg) (x)
7. (740.0 mm Hg) (350.0 mL) = (760.0 mm Hg) (x)
8. (63.0 atm) (338 L) = (1.00 atm) (x)
9. (166.0 mm Hg) (273.15 mL) = (760.0 mm Hg) (x)
10. (18.0 mm Hg) (77.0 L) = (760.0 mm Hg) (x)
11. Volume will decrease.
12. It will double in size.
13. (0.755 atm) (4.31 liters) = (1.25 atm) (x)
14. (8.00 atm) (600.0 mL) = (2.00 atm) (x)
15. (800.0 torr) (400.0 mL) = (1000.0 torr) (x)
Gas Law Problems - Boyle's Law Answers: Part Two
I tried to put the answers in the form of P1V1 = P2V2. They don't have to be in that order,
except that the sub ones must be paired on one side of the equals sign and the sub twos
must be paired on the other.
16. (6.00 atm) (4.00 L) = (2.00 atm) (x)
17. (790.5 mm Hg) (25.3 mL) = ( 0.804 atm) (x)
This is wrong!! You MUST change one of the pressures units so both are the same. I'll
change the mm Hg to atm:
(790.5 mm Hg / 760.0 mm Hg/atm) (25.3 mL) = ( 0.804 atm) (x)
18. (1.00 atm) (12.0 L) = (2.00 atm) (x)
19. (4.00 atm) (30.0 mL) = (2.00 atm) (x)
20. (760.0 mm Hg) (14.0 L) = (400.0 mm Hg) (x)
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21. (8.00 atm) (40.0 L ) = (12.0 atm) (x)
22. (2.00 atm) (200.0 L) = (600.0 kPa) (x)
This is wrong. The pressure units must be the same. I'll change the atm to kPa. You could
go the other way if you want, the answer would be the same.
(2.00 atm x 101.325 kPa/atm) (200.0 L) = (600.0 kPa) (x)
In fact, here's the problem with the kPa changed to atm:
(2.00 atm) (200.0 L) = (600.0 kPa / 101.325 kPa/atm) (x)
23. Your lungs will "explode." As you go up towards the surface, the pressure on your
body and lungs becomes less. The air in your lungs expands. What would happen is the
alveoli and small capallaries would rupture, causing massive bleeding in the lungs. You'd
die. No, your body would not swell up and burst, like a balloon.
24. V2 = (P1V1) / P2
25. B
26. (12 atmospheres) (1.5 liter) = (2.0 atmospheres) (x)
27. (645 mm Hg) (300 mL) = (one atmosphere) (x) This is wrong. I will change atm to
mm Hg. (645 mm Hg) (300 mL) = (760 mm Hg) (x)
28. (1.00 atm.) (x) = (3.00 atmospheres) (0.50 cu. ft.)
29. (50.0 torr) ( 75.0 mL) = (8.00 torr) (x)
30. (1960.0 p.s.i.) (50.0 liters) = (30.00 p.s.i.) (x)
Important point: 2000 psi - 40 psi = 1960 psi flowed out of tank. 2000
is not used in calculation.
31. (1.00 atmosphere) (196.0) = (x) (26.0 liters)
Charles' Law
Discovered by Joseph Louis Gay-Lussac (the uppermost picture to the right) in 1802. He
made reference in his paper to unpublished work done by Jacques Charles (the lower GIF
picture to the right) about 1787. Charles had found that oxygen, nitrogen, hydrogen,
carbon dioxide, and air expand to the same extent over the same 80 degree interval.
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The Chemist does admit that Charles looks like he's ready to fail a
sobriety test, but you would do well to cut the guy some slack. After
all, he did invent 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. His image is from a stamp France issued
in memory of the 100th anniversary of his death in 1950.
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 the Chemist will use.
The same year a 23-year-old Gay-Lussac discovered this law, he had occasion to walk
into a linen draper's shop in Paris and there he made a wonderous discovery. He found
the 17-year-old shop girl reading a chemistry textbook while waiting for customers.
Needless to say, he was intrigued by this and made more visits to the shop. In 1808, he
and Josephine were married and over the years, five little Gay-Lussac ankle-biters were
added to the scene.
So ladies, while the Chemists cannot guarantee that the study of chemistry will land you
the man of your dreams, you just never know what might happen through diligent study
of your chemistry.
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 span of
16
time the Chemists is referring to here is much, much shorter than that. So there.). 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, the law
does not allow this; 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.
The new volume-temperature data pair will preserve the value of k. The Chemists does
not care what the actual value of k is, only that two different volume-temperature data
pairs equal the same value and that value is called k.
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.
This graphic simply restates the above in a way HTML cannot do.
17
Notice that the right-hand equation results from cross-multiplying the first one. Some
people remember one better than the other, so both are provided.
Before going to some sample problems, let's be very clear:
EVERY TEMPERATURE USED IN A CALCULATION MUST BE
IN KELVINS, NOT DEGREES CELSIUS.
The Chemist hopes you understand this very well. Repeating it does not hurt:
DON'T YOU DARE USE CELSIUS IN A NUMERICAL
CALCULATION. USE KELVIN EVERY TIME.
Now, please don't send me e-mail asking me what I meant by that. Thanks.
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:
Remember that you have to plug into the equation in a very specific way. The
temperatures and volumes come in connected pairs and you must put them in the proper
place.
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:
18
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:
Example #4: a 2.5 liter sample of gas is at STP. When the temperature is raised to 273°C
and the pressure remains constant, what is the new volume?
Answer:
We know the gas starts at standard temperature, zero degrees Celsius. In Kelvins, that is
273 K. Now, note the ending temperature, 273°C. In Kelvins, that is 546 K.
The absolute temperature has doubled! Since Charles' Law is a direct relationship, the
volume also doubles, to 5.0 L and that is the answer.
Setting it up mathematically is left to the reader. Note also, the pressure remains constant,
so it simply drops from all consideration in the solving of tis problem. After all, it
remained constant during the entire problem.
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?
19
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
20
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?
Gas Law Problems - Charles' Law Answers: Part One
I used V1 / T1 = V2 / T2 to set up the answers.
32. (2.00 L) / 293.0 K) = (1.00 L) / (x); x = 146.5 K
33. (600.0 mL) / (293.0) = (x) / (333.0 K); x = 682 mL
34. (900.0 mL) / (300.0 K) = (x) / (405.0 K); x = 1215 mL
35. (60.0 mL) / (306.0 K) = (x) / (278.00 K)
36. (300.0) / (290.0 K) = (x) / (283.0 K)
37. (1.00 L ) / (273.0 K) = (x) / (606.0 K)
38. (6.00 L) / (300.00 K) = (x) / (423.0 K)
39. (400.0 mL) / (498.0 K) = (x) / (400.0 K)
40. (8.00 L) / (483.0 K) = (x) / (250.0 K)
41. (4.00 L) / (283.0 K) = (x) / (293.0 K)
42. (30.0 L ) / (1273 K) = (x) / (298.0 K)
21
43. (600.0 mL) / (350.0 K) = (x) / (359.0 K)
Gas Law Problems - Charles' Law Answers: Part Two
I used V1 / T1 = V2 / T2 to set up the answers.
44. 400.0 mL / 295.0 K = x / 303.0K
45. 56.05 milliliters / 315.1 K = x / 380.5 K
46. 42.3 milliliters / 371.15 K = x / 254.50 K
47. Decrease.
48. Two. Or doubled.
49. V2 = (V1 times T2) / T1
50. c. volume/temperature
51. 540.0 mL / 273.0 K = x / 373.0 K
52. 2500.0 mL / 303.0 =x / 283.0 K
53. 50.0 liters / 293.0 K = 5.00 liters / x
54. 15.0 liters / 298.0 K = 45.0 liters / x
55. 3.50 liters / 1000.0 K = x / 426.0 K
Gay-Lussac's Law
Discovered by Joseph Louis Gay-Lussac in the early 1800's. However, the Chemists's
knowledge of this is much less sure than concerning Boyle's or Charles' Law.
Gives the relationship between pressure and temperature when volume and amount are
held constant.
If the temperature of a container is increased, the pressure increases.
If the temperature of a container is decreased, the pressure decreases.
22
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, since the container has rigid walls (volume stays constant).
Gay-Lussac's 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 Gay-Lussac's Law is: P ÷ T = k
This means that the pressure-temperature fraction will always be the same value if the
volume and amount remain constant.
Let P1 and T1 be a pressure-temperature pair of data at the start of an experiment. If the
temperature is changed to a new value called T2, then the pressure will change to P2.
Keep in mind that when volume is not discussed (as in this law), it is constant. That
means a container with rigid walls.
As with the other laws, the exact value of k is unimportant in our context. It is important
to know the PT data pairs obey a constant relationship, but it is not important for us what
the exact value of the constant is. Besides which, the value of K would shift based on
what pressure units (atm, mmHg, or kPa) you were using.
We know this: P1 ÷ T1 = k
And we know this: P2 ÷ T2 = k
Since k = k, we can conclude that P1 ÷ T1 = P2 ÷ T2.
This equation of P1 ÷ T1 = P2 ÷ T2 will be very helpful in solving Gay-Lussac's Law
problems.
This graphic simply restates the above in a way HTML cannot do.
Notice the similarities to the Charles' Law graphic. This is because both laws are direct
relationships.
Make sure to convert any Celsius temperature to Kelvin before using it in your
calculation.
23
Example #1: 10.0 L of a gas is found to exert 97.0 kPa at 25.0°C. What would be the
required temperature (in Celsius) to change the pressure to standard pressure?
Answer: change 25.0°C to 298.0 K and remember that standard pressure in kPa is
101.325. Insert values into the equation (the Chemists will use the left-hand one in the
graphic above) and get:
The answer is 311.3 K, but the question asks for Celsius, so you subtract 273 to get the
final answer of 38.3°C, but then you knew that. Right?
Example #2: 5.00 L of a gas is collected at 22.0°C and 745.0 mmHg. When the
temperature is changed to standard, what is the new pressure?
Answer: convert to Kelvin and insert:
Gas Law Problems- Gay-Lussac'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
=
56. Determine the pressure change when a constant volume of gas at 1.00 atm is heated
from 20.0 °C to 30.0 °C.
57. A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at standard
temperature?
58. A gas has a pressure of 699.0 mm Hg at 40.0 °C. What is the temperature at standard
pressure?
24
59. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what
final pressure would result if the original pressure was 750.0 mm Hg?
60. If a gas in a closed container is pressurized from 15.0 atmospheres to 16.0
atmospheres and its original temperature was 25.0 °C, what would the final temperature
of the gas be?
61. A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside
an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at
3.00 atm. What is the pressure of the nitrogen after its temperature is increased?
62. A sample of gas at 3.00 x 103 mm Hg inside a steel tank is cooled from 500.0 °C to
0.00 °C. What is the final pressure of the gas in the steel tank?
63. The temperature of a sample of gas in a steel container at 30.0 kPa is increased from 100.0 °C to 1.00 x 103 °C. What is the final pressure inside the tank?
64. Calculate the final pressure inside a scuba tank after it cools from 1.00 x 103 °C to
25.0 °C. The initial pressure in the tank is 130.0 atm.
Gas Law Answers - Gay-Lussac's Law
Gay-Lussac's Law is P1 / T1 = P2 / T2
56. 1.00 atm / 293 K = x / 303 K; x = 1.03 atm.
57. 0.370 atm / 323 K = x / 273 K; x = 0.313 atm.
58. 699.0 mm Hg / 313 K = 760 mm Hg / x
59. 750.0 mm Hg / 323.0 K = x / 273.15 K
60. 15.0 atm / 298 K = 16.0 atm / x
61. 3.00 atm. / 293 K = x / 323
62. 3.00 x 103 mm Hg / 773 K = x / 273
63. 30.0 kPa / 173 K = x / 1273
64. 130.0 atm. /1273 K = x / 298 K
Avogadro's Law
25
Discovered by Amedo Avogadro, of Avogadro's Hypothesis fame. The Chemists is not
sure when, but probably sometime in the early 1800s.
Gives the relationship between volume and amount when pressure and temperature are
held constant. Remember amount is measured in moles. Also, since volume is one of the
variables, that means the container holding the gas is flexible in some way and can
expand or contract.
If the amount of gas in a container is increased, the volume increases.
If the amount of gas in a container is decreased, the volume decreases.
Why?
Suppose the amount is increased. This means there are more gas molecules and this will
increase the number of impacts on the container walls. This means the gas pressure inside
the container will increase (for an instant), becoming greater than the pressure on the
outside of the walls. This causes the walls to move outward. Since there is more wall
space the impacts will lessen and the pressure will return to its original value.
The mathematical form of Avogadro's Law is: V ÷ n = k
This means that the volume-amount fraction will always be the same value if the pressure
and temperature remain constant.
Let V1 and n1 be a volume-amount pair of data at the start of an experiment. If the
amount is changed to a new value called n2, then the volume will change to V2.
We know this: V1 ÷ n1 = k
And we know this: V2 ÷ n2 = k
Since k = k, we can conclude that V1 ÷ n1 = V2 ÷ n2.
This equation of V1 ÷ n1 = V2 ÷ n2 will be very helpful in solving Avogadro's Law
problems. Here is the Law done up in fractional form, something HTML isn't good at:
Avogadro's Law is a direct mathematical relationship.
You prove Avogadro's Law every time you blow up a balloon.
26
Example #1: 5.00 L of a gas is known to contain 0.965 mol. If the amount of gas is
increased to 1.80 mol, what new volume will result (at an unchanged temperature and
pressure)?
Answer: this time I'll use V1n2 = V2n1
(5.00 L) (1.80 mol) = (x) (0.965 mol)
There are no worksheet problems.
"Diver's" Law
The Chemists do not know the name of the discoverer. It may even be that this law has
never been "discovered" in the manner we speak of the discovery of Boyle's Law. Diver
is not a name; it refers to diving beneath the water. The deeper you go the greater the
pressure because of the larger amount of water pressing down on you.
This law gives the relationship between pressure and amount when the temperature and
volume are held constant. Remember amount is measured in moles. Also, since volume is
one of the constants, that means the container holding the gas is rigid and cannot change
in volume.
If the amount of gas in a container is increased, the pressure increases.
If the amount of gas in a container is decreased, the pressure decreases.
Why?
Suppose the amount is increased. This means there are more gas molecules and this will
increase the number of impacts on the container walls. This means the gas pressure inside
the container will increase. It will stay at this higher level because the container walls do
not move (the volume is constant).
The mathematical form of Diver's Law is: P ÷ n = k
This is a direct mathematical relationship.
Let P1 and n1 be a volume-amount pair of data at the start of an experiment. If the amount
is changed to a new value called n2, then the pressure will change to P2.
We know this: P1 ÷ n1 = k
And we know this: P2 ÷ n2 = k
27
Since k = k, we can conclude that P1 ÷ n1 = P2 ÷ n2.
This equation of P1 ÷ n1 = P2 ÷ n2 will be very helpful in solving Diver's Law problems.
There are no sample problems.
There are no worksheet problems.
The "no-name" Law
The Chemists do not know the name of the discoverer. As far as the Chemist knows, this
law does not have a name because it was never really "discovered."
Gives the relationship between amount and temperature when pressure and volume are
held constant. Remember amount is measured in moles. Also, since volume is one of the
constants, that means the container holding the gas is rigid-walled and inflexible.
If the amount of gas in a container is increased, the temperature must decrease.
If the amount of gas in a container is decreased, the temperature must increase.
Why?
Suppose the amount is increased. This means there are more gas molecules and this will
increase the number of impacts on the container walls. This means the gas pressure inside
the container will increase, since the volume is constant. However, this cannot be allowed
since pressure also remains constant. Lowering the temperature will cause the molecules
to move slower, thus lessening the impacts on the container walls and reducing the
pressure to the original value.
The mathematical form of the "no-name" Law is: nT = k
Let n1 and T1 be a amount-temperature pair of data at the start of an experiment. If the
amount is changed to a new value called n2, then the temperature will change to T2.
We know this: n1T1 = k
And we know this: n2T2 = k
Since k = k, we can conclude that n1T1 = n2T2.
This equation of n1T1 = n2T2 will be very helpful in solving "no-name" Law problems. If
you ever see one!
28
The "no-name" Law is an inverse mathematical relationship.
There are no sample problems.
There are no worksheet problems.
Combined Gas Law
To derive the Combined Gas Law, do the following:
Step 1: Write Boyle's Law:
P1V1 = P2V2
Step 2: Multiply by Charles Law:
P1V12 / T1 = P2V22 / T2
Step 3: Multiply by Gay-Lussac's Law:
P12V12 / T12 = P22V22 / T22
Step 4: Take the square root to get the combined gas law:
P1V1 / T1 = P2V2 / T2
If all six gas laws are included (the three above as well as Avogadro, Diver, and "noname"), we would get the following:
P1V1 / n1T1 = P2V2 / n2T2
However, this more complete combined gas law is rarely, if ever, discussed.
Consequently, we will ignore it in future discussions and use only the law given in step 4
above.
Example #1: This type of combined gas law problem (where everything goes to STP) is
VERY common:
2.00 L of a gas is collected at 25.0°C and 745.0 mmHg. What is the volume at STP?
STP is a common abbreviation for "standard temperature and pressure."
29
You have to recognize that five values are given in the problem and the sixth is an x.
Also, remember to change the Celsius temperatures to Kelvin.
When problems like this are solved in the Chemists classroom, I write a solution matrix,
like this:
And fill it in with data from the problem.
Here is the right-hand side filled in with the STP values:
You can be pretty sure that the term "STP" will appear in the wording of at least one test
question in your classroom. The Chemists recommends you memorize the various
standards conditions. If your teacher allows a "cheat sheet" to be used on the test, MAKE
CERTAIN those values are there. Pretty obvious, don't you think. Believe me, there are a
lot of dunderheads out there! Don't be one
Here's the solution matrix completely filled in:
Insert the values in their proper places in the combined gas law equation:
P1V1 / T1 = P2V2 / T2
and solve for x.
Example #2 - This next problem uses two gas laws in sequence. It involves using
Dalton's Law of Partial Pressures first, then use of the Combined Gas Law. The
explanation will assume you understand Dalton's Law. These two laws occurring together
in a problem is VERY COMMON.
30
1.85 L of a gas is collected over water at 98.0 kPa and 22.0°C. What is the volume of the
dry gas at STP?
The key phrase is "over water." Another phrase to look for is "wet gas." This means the
gas was collected by bubbling it into an inverted bottle filled with water which is sitting
in a water bath. The gas bubbles in and is trapped. It displaces the water which flows out
into the water bath.
The problem is that the trapped gas now has water vapor mixed in with it. This is a
consequence of the technique and cannot be avoided. However, there is a calculation
technique (Dalton's Law) that allows use to remove the effect of the water vapor and treat
the gas as "dry." For this example, we write Dalton's Law like this:
We need to know the vapor pressure of water at 22.0°C and to do this we must look it up
in a reference source. See Handbook of Chemistry and Physics: 73rd Edition (1992-93)
It is important to recognize the Ptot is the 98.0 value. Ptot is the combined pressure of the
dry gas AND the water vapor. We want the water vapor's pressure OUT.
We solve the problem for Pgas and get 95.3553 kPa. Notice that it is not rounded off. The
only rounding off done is at the FINAL answer, which this is not.
Placing all the values into the solution matrix yields this:
Solve for x in the usual manner of cross-multiplying and dividing.
Gas Law Problems- Combined Gas 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
31
=
=
1 mL
1 L
=
Most of the answers
73. A gas has a volume of 800.0 mL at minus 23.00 °C and 300.0 torr. What would the
volume of the gas be at 227.0 °C and 600.0 torr of pressure?
74. 500.0 liters of a gas are prepared at 700.0 mm Hg and 200.0 °C. The gas is placed
into a tank under high pressure. When the tank cools to 20.0 °C, the pressure of the gas is
30.0 atm. What is the volume of the gas?
75. What is the final volume of a 400.0 mL gas sample that is subjected to a temperature
change from 22.0 °C to 30.0 °C and a pressure change from 760.0 mm Hg to 360.0 mm
Hg?
76. What is the volume of gas at 2.00 atm and 200.0 K if its original volume was 300.0 L
at 0.250 atm and 400.0 K.
77. At conditions of 785.0 torr of pressure and 15.0 °C temperature, a gas occupies a
volume of 45.5 mL. What will be the volume of the same gas at 745.0 torr and 30.0 °C?
78. A gas occupies a volume of 34.2 mL at a temperature of 15.0 °C and a pressure of
800.0 torr. What will be the volume of this gas at standard conditions?
79. The volume of a gas originally at standard temperature and pressure was recorded as
488.8 mL. What volume would the same gas occupy when subjected to a pressure of
100.0 atm and temperature of minus 245.0 °C?
80. At a pressure of 780.0 mm Hg and 24.2 °C, a certain gas has a volume of 350.0 mL.
What will be the volume of this gas under STP
81. A gas sample occupies 3.25 liters at 24.5 °C and 1825 mm Hg. Determine the
temperature at which the gas will occupy 4250 mL at 1.50 atm.
82. If 10.0 liters of oxygen at STP are heated to 512 °C, what will be the new volume of
gas if the pressure is also increased to 1520.0 mm of mercury?
83. What is the volume at STP of 720.0 mL of a gas collected at 20.0 °C and 3.00 atm
pressure?
84. 2.00 liters of hydrogen, originally at 25.0 °C and 750.0 mm of mercury, are heated
until a volume of 20.0 liters and a pressure of 3.50 atmospheres is reached. What is the
new temperature?
85. A gas balloon has a volume of 106.0 liters when the temperature is 45.0 °C and the
pressure is 740.0 mm of mercury. What will its volume be at 20.0 °C and 780 .0 mm of
mercury pressure?
32
86. If the absolute temperature of a given quantity of gas is doubled and the pressure
tripled, what happens to the volume of the gas?
87. 73.0 mL of nitrogen at STP is heated to 80.0 °C and the volume increase to 4.53 L.
What is the new pressure?
88. 500.0 mL of a gas was collected at 20.0 °C and 720.0 mm Hg. What is its volume at
STP?
89. A sample of gas occupies 50.0 L at 15.0 °C and 640.0 mm Hg pressure. What is the
volume at STP?
90. A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters
to 35.0 liters by moving a large piston within a cylinder. If the original pressure was 1.00
atm, what would the final pressure be?
91. The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as the volume
of its container is increased by moving a piston from 85.0 mL to 350.0 mL. What would
the final temperature be if the original temperature was 90.0 °C?
92. If a gas is heated from 298.0 K to 398.0 K and the pressure is increased from 2.230 x
103 mm Hg to 4.560 x 103 mm Hg what final volume would result if the volume is
allowed to change from an initial volume of 60.0 liters?
Combined Gas Law which requires Dalton's Law also
IMPORTANT NOTE: A gas collected over water is always considered to be saturated
with water vapor. The vapor pressure of water varies with temperature and must be
looked up in a reference source. Handbook of Chemistry and Physics: 73rd Edition
(1992-93)
93. 690.0 mL of oxygen are collected over water at 26.0 °C and a total pressure of 725.0
mm of mercury. What is the volume of dry oxygen at 52.0 °C and 800.0 mm pressure?
94. 400.0 mL of hydrogen are collected over water at 18.0 °C and a total pressure of
740.0 mm of mercury.
a) What is the partial pressure of H2?
b) What is the partial pressure of H2O?
c) What is the volume of DRY hydrogen at STP?
95. 45.0 mL of wet argon gas is collected at 729.3 mm Hg and 25.0 °C. What would be
the volume of this dry gas at standard conditions?
96. 19.1 L of He gas is collected over water at 681.3 mm Hg and 18.5 °C. What would be
the volume of this dry gas at standard conditions?
33
97. 407 mL of H2 gas is collected over water at 785.3 mm Hg and 23.5 °C. What would
be the volume of this dry gas at standard conditions?
98. 93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0 °C. What would be
the volume of this dry gas at standard conditions?
99. 6.12 L of wet xenon gas is collected at 2.00 x 105 Pa and 80.0 °C. What would be the
volume of this dry gas at standard conditions?
100. A sample of oxygen collected over water when the atmospheric pressure was 1.002
atm and the room temperature, 25.5 °C occupied 105.8 mL. What would be the volume
of this dry gas at standard conditions?
101. 1.000 L of hydrogen gas is collected over water at 30.0 °C at a pressure of 831.8
mm Hg. Find the volume of dry hydrogen collected at STP.
102. 50.6 mL of a gas is collected over water at 18.0 °C and 755.5 mm Hg pressure.
What is the volume of dry gas at STP?
103. Write the combined gas law in equation form. Solve the combined gas law for V2.
Gas Law Problems- Combined Gas Law Answers
The form of the Combined Gas Law most often used is (P1V1) / T1 = (P2V2) / T2
Most commonly V2 is being solved for. The rearrangement looks like this:
V2 = (P1V1T2) / (T1P2)
This is what I will use below. I can't promise to get the same order, but the three in the
numerator will be there and the two in the denominator will be there. By the way, the
above equation is the answer to question #103.
I didn't do all the answers. Oh well.
73. x = [ (300 torr) (800 mL) (500 K) ] / [ (250 K) (600 torr) ]; x = 800.0 mL
Keep in mind that torr = mmHg.
74. x = [ (700/760) (500) (293) ] / [ (473) (30) ]; x = 9.51 L
Note that this problem mixes pressure units. The 700/760 fraction converts mmHg to
atm.
34
75. x = [ (760 mm Hg) (400 mL) (303 K) ] / [ (295 K) (360 mm Hg) ]; x = 867.3 mL
76. x = [ (0.25) (300) (200) ] / [ (400) (2) ]
77. x = [ (785) (45.5) (303) ] / [ (288) (745) ]
78. x = [ (800) (34.2) (273) ] / [ (288) (760) ]
79. x = [ (1) (488.8) (28) ] / [ (273) (100) ]
80. x = [ (780) (350) (273) ] / [ (297.2) (760) ]
81. x = [ (1.50 atm) (4.25 L) (297.5 K) ] / [(1825 mm Hg/760 atm mmHg¯1) (3.25 L) ]
Note that we had to change units around somewhat. Also note that a temperature was
solved for rather than the usual volume.
82. x = [ (760) (10) (785) ] / [ (273) (1520) ]
83. x = [ (3.00 atm) (720.0 mL) (273 K) ] / [ (293 K) (1.00 atm)
84. x = [ (3.50) (20) (298) ] / [ (750/760) (2.00) ]
85. x = [ (740 mmHg) (106 L) (293 K) ] / [ (318 K) (780 mmHg) ]
86. x = [ (1) (10) (2) ] / [ (1) (3) ]
87. x = [ (1.00 atm) (73.0 mL) (353 K) ] / [ (273 K) (4530 mL) ]
88. x = [ (720) (500) (273) ] / [ (293) (760) ]
89. x = [ (640) (50) (273) ] / [ (288) (760) ]
90. A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters
to 35.0 liters by moving a large piston within a cylinder. If the original pressure was 1.00
atm, what would the final pressure be?
91. The pressure of a gas is reduced from 1200.0 mm Hg to 850.0 mm Hg as moving a
piston from 85.0 mL to 350.0 mL increases the volume of its container. What would the
final temperature be if the original temperature was 90.0 °C?
92. If a gas is heated from 298.0 K to 398.0 K and the pressure is increased from 2.230 x
103 mm Hg to 4.560 x 103 mm Hg what final volume would result if the volume is
allowed to change from an initial volume of 60.0 liters?
Combined Gas Law which requires Dalton's Law also
35
IMPORTANT NOTE: A gas collected over water is always considered to be saturated
with water vapor. The vapor pressure of water varies with temperature and must be
looked up in a reference source. Handbook of Chemistry and Physics: 73rd Edition
(1992-93)
93. Pdry = 725.0 - 25.2 = 699.8 mmHg
x = [ (699.8) (690.0) (325) ] / [ (299) (800.0) ]
94.
a) Pdry = 740.0 - 15.5 = 724.5 mmHg
b) 15.5 mmHg
c) x = [ (724.5 mm Hg) (400.0 mL) (273 K) ] / [ (291 K) (760.0 mm Hg)
95. Pdry = 729.3 - 23.8 = 705.5
x = [ (705.5) (45.0) (273) ] / [ (298) (760) ]
96. The 18.5 °C requires an interpolation between 18 °C and 19 °C. This gives 16.0
mmHg.
Pdry = 681.3 - 16.0 = 665.3
x = [ (665.3 mm Hg) (19.1 L) (273 K) ] / [ (291.5 K) (760 mm Hg) ]
97. The 23.5 °C requires an interpolation between 23 °C and 24 °C. This gives 21.7
mmHg.
Pdry = 785.3 - 21.7 = 763.6
x = [ (763.6 mm Hg) (407 mL) (273 K) ] / [ (296.5 K) (760 mm Hg) ]
98. 93.0 mL of O2 gas is collected over water at 0.930 atm and 10.0 °C. What would be
the volume of this dry gas at standard conditions?
99. 6.12 L of wet xenon gas is collected at 2.00 x 105 Pa and 80.0 °C. What would be the
volume of this dry gas at standard conditions?
100. A sample of oxygen collected over water when the atmospheric pressure was 1.002
atm and the room temperature, 25.5 °C occupied 105.8 mL. What would be the volume
of this dry gas at standard conditions?
101. 1.000 L of hydrogen gas is collected over water at 30.0 °C at a pressure of 831.8
mm Hg. Find the volume of dry hydrogen collected at STP.
36
102. 50.6 mL of a gas is collected over water at 18.0 °C and 755.5 mm Hg pressure.
What is the volume of dry gas at STP?
103. V2 = (P1 V1 T2 ) / (T1 P2 )
PV = nRT: The Ideal Gas Law
The Ideal Gas Law was first written in 1834 by Emil
Clapeyron. This 13K GIF to the right is of him.
This is just one way to "derive" the Ideal Gas Law:
For a static sample of gas, we can write each of the six
gas laws as follows:
PV = k1
V / T = k2
P / T = k3
V / n = k4
P / n = k5
1 / nT = 1 / k6
Note that the last law is written in reciprocal form. The subscripts on k indicate that six
different values would be obtained.
When you multiply them all together, you get:
P3V3 / n3T3 = k1k2k3k4k5 / k6
Let the cube root of k1k2k3k4k5 / k6 be called R.
The units work out:
k1 = atm-L
k2 = L / K
k3 = atm / K
k4 = L / mol
k5 = atm / mol
1 / k6 = 1 / mol-K
Each unit occurs three times and the cube root yields L-atm / mol-K, the correct units for
R when used in a gas law context.
Resuming, we have:
37
PV / nT = R
or, more commonly:
PV = nRT
R is called the gas constant. Sometimes it is referred to as the universal gas constant. If
you wind up taking enough chemistry, you will see it showing up over and over and over.
The Numerical Value for R
R's value can be determined many ways. This is just one way:
We will assume we have 1.000 mol of a gas at STP. The volume of this amount of gas
under the conditions of STP is known to a high degree of precision. We will use the value
of 22.414 L.
By the way, 22.414 L at STP has a name. It is called molar volume. It is the volume of
ANY ideal gas at standard temperature and pressure.
Molar Volume
The molar volume is the volume occupied by one mole of ideal gas at STP. Its value is:
22.414 L mol¯1
It is actually known to several more decimal places but the number above should prove
sufficient.This value has been known for about 200 years and it is not a constant of nature
like, say, the charge on the electron. If we had picked a diffeent standard temperature,
then the molar volume would be different.
Using PV = nRT, you can calculate the value for molar volume. V is the unknown and n
= 1.00 mol. Set P and T to their standard values and use R = 0.08206.
Molar volume doesn't show up that often in problems. As a consequence, teachers
sometimes like to use molar volume on the test, in order to trip the kid up!! Let's do some
examples.
Example #1: you have 2.00 L of dry H2 at STP. How many moles is this?
Solution: we could solve this with PV = nRT, but we can shortcut since we have STP
and do this:
38
Example #2: 0.250 moles of HCl will occupy how many liters at STP?
Solution: this is just the reverse of example one:
0.250 mol x 22.414 L mol¯1
Example #3: What is molar volume at 576 K?
Solution: strictly speaking, this isn't more than a volume-temperature problem, but for
some reason putting "molar volume" in the problem messes people up.
22.414 L mol¯1 / 273 K = x / 546 K
Notice that 546 is double 273, so that just confirms the answer of 44.828 L.
Let's plug our numbers into the equation:
(1.000 atm) (22.414 L) = (1.000 mol) (R) (273.15 K)
Notice how atmospheres were used as well as the exact value for standard temperature.
Solving for R gives 0.08206 L atm / mol K, when rounded to four significant figures.
This is usually enough. Remember the value. You'll need it for problem solving.
Notice the weird unit on R: say out loud "liter atmospheres per mole Kelvin."
This is not the only value of R that can exist. It depends on which units you select. Those
of you that take more chemistry than high school level will meet up with 8.3145 Joules
per mole Kelvin, but that's for another time. The Chemists will only use the 0.08206
value in gas-related problems.
Example #1 - A sample of dry gas weighing 2.1025 grams is found to occupy 2.850 L at
22.0°C and 740.0 mmHg. How many moles of the gas are present?
Notice that the units for pressure MUST be in atm., so the 740.0 mm Hg must be
converted first.
740.0 mm Hg ÷ 760.0 mm Hg/atm = 0.9737 atm
39
However, the unrounded-off value should be used in the calculation just below.
Now, plug into the equation:
(0.9737 atm) (2.850 L) = (n) (0.08206 L atm / mol K) (295.0 K)
and solve for n.
Example #2 - Using the problem above, what is the molar mass of the gas?
This is a very common use of this law and the odds are very good you will see this type
of question on a test.
The key is to remember the units on molar mass: grams per mole.
We know from the problem statement that 2.1025 grams of the gas is involved and we
also know how many moles that is.
We know that from doing the calculation above and getting 0.1146 mol.
So all we have to do is divide the grams of gas by how many moles it is:
2.1025 g ÷ 0.1146 mol = 18.34 g/mol
Let's go over those steps for using the Ideal Gas Law to calculate the molar mass of the
gas:
1) You have to know the grams of gas involved. Usually the problem will
just give you the value, but not always. You might have to calculate it.
Gas Density
Discussing gas density is slightly more complex than discussing solid/liquid density.
Since gas volume is VERY responsive to temperature and pressure, these two factors
must be included in EVERY gas density discussion.
By the way, solid and liquid volumes are responsive to temperature and pressure, but the
response is so little that it can usually be ignored in introductory classes.
So, for gases, we speak of "standard gas density." This is the density of the gas
(expressed in grams per liter) at STP. If you discuss gas density at any other set of
conditions, you drop the word standard and specify the pressure and temperature. Also,
when you say "standard gas density," you do not need to add "at STP." STP is part of the
40
definition of the term. It does no harm to say "standard gas density at STP," it just looks a
bit silly.
You can calculate the ideal standard gas density fairly easily. Just take the mass of one
mole of the gas and divide by the molar volume. For nitrogen, we would have:
28.014 g mol¯1 / 22.414 L mol¯1 = 1.250 g / L
For water, we have:
18.015 g mol¯1 / 22.414 L mol¯1 = 0.8037 g / L
Notice I said "ideal." The behavior of "real" gases diverges from predictions based on
ideal conditions. Small gases like H2 at high temperatures approach ideal behavior almost
exactly while larger gas molecules (NH3) at low temperatures diverge the greatest
amount. These "real" gas differences are small enough to ignore right now, but in later
classes they will become important.
One place teachers like to bring gas density into play is when you calculate a molar mass
of a gas using PV = nRT.
Example Problem #1: The density of a gas is measured at 1.853 g / L at 745.5 mmHg
and 23.8 °C. What is its molar mass?
Solution #1: Convert mmHg to atm and °C to K. Use 1.000 L. Plug into PV = nRT and
solve for n. Then divide 1.853 g by n, the number of moles, for your answer.
Solution #2: This solution exploits a rearrangement of the ideal gas law. Here it is:
41
The answer: 46.008 g mol¯1
Example Problem #2: What is the molar mass of a gas which has a density of 0.00249
g/mL at 20.0 °C and 744.0 mm Hg?
Solution: Convert mmHg to atm (744.0/760.0) and °C to K (20.0 + 273.15). Use 0.001 L
(which is 1 mL converted to liters). Plug into PV = nRT and solve for n (the value of
which is calculated to be 4.069 x 10¯5 mol).
Then, divide 0.00249 g by the moles just calculated for an answer of 61.2 g/mol.
2) You are going to have to calculate the moles of gas. Use PV = nRT and
solve for n. Make sure to use L, atm and K.
3) Divide grams by moles and there's your answer.
Gas Law Problems- Ideal Gas 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
=
104. How many moles of gas are contained in 890.0 mL at 21.0 °C and 750.0 mm Hg
pressure?
105. 1.09 g of H2 is contained in a 2.00 L container at 20.0 °C. What is the pressure in
this container in mm Hg?
106. Calculate the volume 3.00 moles of a gas will occupy at 24.0 °C and 762.4 mm Hg.
107. What volume will 20.0 g of Argon occupy at STP?
108. How many moles of gas would be present in a gas trapped within a 100.0 mL vessel
at 25.0 °C at a pressure of 2.50 atmospheres?
109. How many moles of a gas would be present in a gas trapped within a 37.0 liter
vessel at 80.00 °C at a pressure of 2.50 atm?
110. If the number of moles of a gas are doubled at the same temperature and pressure,
will the volume increase or decrease?
42
111. What volume will 1.27 moles of helium gas occupy at STP?
112. At what pressure would 0.150 mole of nitrogen gas at 23.0 °C occupy 8.90 L?
113. What volume would 32.0 g of NO2 gas occupy at 3.12 atm and 18.0 °C?
114. Find the volume of 2.40 mol of gas whose temperature is 50.0 °C and whose
pressure is 2.00 atm.
115. Calculate the molecular weight of a gas if 35.44 g of the gas stored in a 7.50 L tank
exerts a pressure of 60.0 atm at a constant temperature of 35.5 °C
116. How many moles of gas are contained in a 50.0 L cylinder at a pressure of 100.0
atm and a temperature of 35.0 °C?
117. Determine the number of moles of Krypton contained in a 3.25 liter gas tank at 5.80
atm and 25.5 °C. If the gas is Oxygen instead of Krypton, will the answer be the same?
Why or why not?
118. Determine the number of grams of carbon dioxide in a 450.6 mL tank at 1.80 atm
and minus 50.5 °C. Determine the number of grams of oxygen that the same container
will contain under the same temperature and pressure.
119. Determine the volume of occupied by 2.34 grams of carbon dioxide gas at STP.
120. A sample of argon gas at STP occupies 56.2 liters. Determine the number of moles
of argon and the mass in the sample.
121. At what temperature will 0.654 moles of neon gas occupy 12.30 liters at 1.95
atmospheres?
122. A 30.6 g sample of gas occupies 22.4 L at STP. What is the molecular weight of this
gas?
123. A 40.0 g gas sample occupies 11.2 L at STP. Find the molecular weight of this gas.
124. A 12.0 g sample of gas occupies 19.2 L at STP. What is the molecular weight of this
gas?
125. 96.0 g. of a gas occupies 48.0 L at 700.0 mm Hg and 20.0 °C. What is its molecular
weight?
126. 20.83 g. of a gas occupies 4.167 L at 79.97 kPa at 30.0 °C. What is its molecular
weight?
43
127. At STP 3.00 liters of an unknown gas has a mass of 9.50 grams. Calculate its molar
mass.
128. At STP 0.250 liter of an unknown gas has a mass of 1.00 gram. Calculate its molar
mass.
129. At STP 150.0 mL of an unknown gas has a mass of 0.250 gram. Calculate its molar
mass.
130. 1.089 g of a gas occupies 4.50 L at 20.5 °C and 0.890 atm. What is its molar mass?
131. 0.190 g of a gas occupies 250.0 mL at STP. What is its molar mass? What gas is it?
Hint - calculate molar mass of the gas.
132. If 9.006 grams of a gas are enclosed in a 50.00 liter vessel at 273.15 K and 2.000
atmospheres of pressure, what is the molar mass of the gas? What gas is this?
133. What is the value of and units on R? What is R called ("A letter" is not the correct
answer!)?
134. A 50.00 liter tank at minus 15.00 °C contains 14.00 grams of helium gas and 10.00
grams of nitrogen gas.
a. Determine the moles of helium gas in the tank.
b. Determine the moles of nitrogen gas in the tank.
c. Determine the mole fraction of helium gas in the tank.
d. Determine the mole fraction of nitrogen gas in the tank.
e. Determine the partial pressure of helium gas in the tank.
f. Determine the partial pressure of nitrogen gas in the tank.
g. Determine the total pressure of the mixture in the tank.
Gas Law Problems- Ideal Gas Law Answers
104. n = PV / RT
n = [ (750.0 mmHg / 760.0 mmHg atm¯1) (0.890 L) ] / (0.08206 L atm mol¯1 K¯1) (294.0
K)
Please note the division of 750 by 760. That is done in order to convert the pressure from
mmHg to atm., because the value for R contains atm. as the pressure unit. If we used
mmHg, the pressure units would not cancel and we need to have them cancel because we
require mol. only to be in the answer.
105. P = nRT / V
44
P = [ (1.09 g / 2.02 g mol¯1) (0.08206 L atm mol¯1 K¯1) (293.0 K) ] / 2.00 L
Multiply the answer (which is in atm) by 760.0 mmHg atm¯1 to get mmHg
106. V = nRT / P
V = [ (3.00 mol) (0.08206 L atm mol¯1 K¯1) (297.0 K) ] / (762.4 mmHg / 760.0 mmHg
atm¯1
107. V = nRT / P
V = [ (20.0 g / 40.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / (1.00 atm)
108. n = PV / RT
n = [ (2.50 atm) (0.1000 L) ] / [ (0.08206 L atm mol¯1 K¯1) (298.0 K) ]
109. n = PV / RT
n = [ (2.50 atm) (37.0 L) ] / [ (0.08206 L atm mol¯1 K¯1) (353.0 K) ]
110. The volume would increase. In fact, it would double.
111. V = nRT / P
V = [ (1.27 mol) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm
or
(22.4 L / 1.00 mol) = (x / 1.27 mol)
112. P = nRT / V
P = [ (0.150 mol) (0.08206 L atm mol¯1 K¯1) (296.0 K) ] / 8.90 L
113. V = nRT / P
V = [ (32.0 g / 46.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (291.0 K) ] / 3.12 atm
114. V = nRT / P
V = [ 2.40 mol) (0.08206 L atm mol¯1 K¯1) (323.0 K) ] / 2.00 atm
115. n = PV / RT
n = [ (60.0 atm) (7.50 L) ] / [ (0.08206 L atm mol¯1 K¯1) (308.5 K) ]
45
Divide 35.44 g by the number of moles calculated to get the molecular weight.
116. n = PV / RT
n = [ (100.0 atm) (5.00 L) ] / [ (0.08206 L atm mol¯1 K¯1) (308.0 K) ]
117. n = PV / RT
n = [ (5.80 atm) (3.25 L) ] / [ (0.08206 L atm mol¯1 K¯1) (298.5 K) ]
The moles of gas would be the same if the gas was switched to oxygen. Since the
temperature and pressure would be the same, the same volume willcontain the same
number of molecules of gas, i.e. moles of gas. This is Avogadro's Hypothesis.
118. n = PV / RT
n = [ (0.4506 atm) (1.80 L) ] / [ (0.08206 L atm mol¯1 K¯1) (222.5 K) ]
This calculates the number of moles of CO2. Multiply the moles by the molecular weight
of CO2 to get the grams. Under the same set of conditions, the moles of oxygen would be
the same, so multiply the calculated moles by the molecular weight of O2 to get the
grams.
119. V = nRT / P
V = [ (2.34 g / 44.0 g mol¯1) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm
120. n = PV / RT
n = [ (1.00 atm) (56.2 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ] Multiply the moles
by the atomic weight of Ar to get the grams.
121. T = PV / nR T = [ (1.95 atm) (12.30 L) ] / [ (0.654 mol) (0.08206 L atm mol¯1 K¯1)
]
122. Since one mole of gas occupies 22.4 L at STP, the molecular weight of the gas is
30.6 g mol¯1
123. 11.2 L at STP is one-half molar volume, so there is 0.5 mol of gas present.
Therefore, the molecular weight is 80.0 g mol¯1
124. This problem, as well as the two just above can be solved with PV = nRT. You
would solve for n, the number of moles. Then you would divide the grams given by the
mole calculated.
Since it is at STP, we can also use a ratio method (see prob. 111)
46
(19.2 L / 12.0 g) = (22.4 L / x )
125. n = PV / RT
n = [ (700.0 mmHg / 760.0 mmHg atm¯1) (48.0 L) ] / [ (0.08206 L atm mol¯1 K¯1)
(293.0 K) ]
Then, divide the grams given (96.0) by the moles just calculated above. This will be the
molecular weight.
126. n = PV / RT
n = [ (79.97 kPa / 101.325 kPa atm¯1) (4.167 L) ] / [ (0.08206 L atm mol¯1 K¯1) (303.0
K) ]
Then, divide the grams given (20.83) by the moles just calculated above. This will be the
molecular weight.
127. (3.00 L / 9.50 g) = (22.4 L / x )
128. (0.250 L / 1.00 g) = (22.4 L / x )
129. (0.150 L / 0.250 g) = (22.4 L / x )
130. n = PV / RT
n = [ (0.890 atm) (4.50 L) ] / [ (0.08206 L atm mol¯1 K¯1) (293.5 K) ]
Then, divide the grams given (1.089) by the moles just calculated above. This will be the
molecular weight.
131. n = PV / RT
n = [ (1.00 atm) (0.2500 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.0 K) ]
Then, divide the grams given (0.190) by the moles just calculated above. This will be the
molecular weight.
132. If 9.006 grams of a gas are enclosed in a 50.00 liter vessel at 273.15 K and 2.000
atmospheres of pressure, what is the molar mass of the gas? What gas is this?
n = PV / RT
n = [ (2.000 atm) (50.00 L) ] / [ (0.08206 L atm mol¯1 K¯1) (273.15 K) ]
47
Then, divide the grams given (9.006) by the moles just calculated above. This will be the
molecular weight.
The answer (2.019 g mol¯1) is approximately that of hydrogen gas, H2
133. 0.08206 L atm mol¯1 K¯1
The gas constant.
134.
a. 14.0 g / 4.00 g mol¯1 = 3.50 mol
b. 49.0 g / 28.0 g mol¯1 = 1.75 mol
c. 3.50 mol / 5.25 mol
d. 1.75 mol / 5.25 mol
e. Use PV = nRT to determine the total pressure in the container.
P = nRT / V
P = [ (5.25 mol) (0.08206 L atm mol¯1 K¯1) (258.0 K) ] / 50.00 L
The total pressure times the mole fraction of He will give helium's partial pressure.
f. The total pressure times the mole fraction of N2 will give nitrogen's partial pressure.
Since it is the only other value, you could subtract helium's answer from the total.
g. Already done to answer part e.
h. V = nRT / P
V = [ (5.25 mol) (0.08206 L atm mol¯1 K¯1) (273.0 K) ] / 1.00 atm
h. Determine the volume that the mixture will occupy at STP.
Dalton's Law of Partial Pressures
This law was discovered by John Dalton in 1801.
For any pure gas (let's use helium), PV = nRT holds true. Therefore, P is directly
proportional to n if V and T remain constant. As n goes up, so would P. Or the reverse.
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Suppose you were to double the moles of helium gas present. What would happen?
Answer: the gas pressure doubles.
However, suppose the new quantity of gas added was a DIFFERENT gas. Suppose that,
instead of helium, you added neon.
What would happen to the pressure?
Answer: the pressure doubles, same as before.
Dalton's Law immediately follows from this example since each gas is causing 50% of
the pressure. Summing their two pressures gives the total pressure.
Written as an equation, it looks like this:
PHe + PNe = Ptotal
Dalton's Law of Partial Pressures: each gas in a mixture creates pressure as if the other
gases were not present. The total pressure is the sum of the pressures created by the gases
in the mixture.
Ptotal = P1 + P2 + P3 + .... + Pn
Where n is the total number of gases in the mixture.
The only necessity is that the two gases do not interact in some chemical fashion, such as
reacting with each other.
The pressure each gas exerts in mixture is called its partial pressure.
The most common use of Dalton's Law seen in high school is with water vapor.
A common method of collecting gas during an experiment is by trapping it "over water."
An inverted bottle filled with water sits in a water bath. A tube from the reaction vessel
conducts the gas into the bottle where it bubbles to the top and displaces water, which
runs out the mouth of the bottle into the water bath.
However, there is an unavoidable problem. The gas saturates with water vapor and now
the total pressure inside the bottle is the sum of two pressures - the gas itself and the
added water vapor.
WE DO NOT WANT THE WATER VAPOR PRESSURE.
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So we get rid of it by subtraction.
This means we must get the water vapor pressure from somewhere.
We get it from a table because the water vapor pressure depends only on the temperature,
NOT how big the container is or the pressure of the other gas. Usually the textbook will
have an abbreviated table with more complete tables in reference manuals like "The
Handbook of Chemistry and Physics." So finally, here is the example problem: 0.750 L
of a gas is collected over water at 23.0°C with a total pressure of 99.75 kPa. What is the
pressure of the dry gas? Look up the vapor pressure data here. Handbook of Chemistry
and Physics: 73rd Edition (1992-93)
The Chemists will leave you to work this one out!
Another common concept that crops up in a Dalton's Law context is mole fraction.
Suppose you had equal moles of two different gases in a mixture. Then the mole fraction
for each would be 0.50.
The mole fraction for each gas is simply the moles of that gas divided by the total moles
in the mixture.
Seems simple enough. How does it relate to Dalton's Law?
Answer: the mole fraction also gives the fraction of the total pressue each gas contributes.
So if the mole fraction for a gas was 0.50, then it would contribute 50% of the total
pressure. If the mole fraction of a gas was 0.15, then its partial pressure would be 0.15
times the total pressure.
The reverse is also true. If you divided the partial pressure of a gas by the total pressure,
you would get the mole fraction for that gas. (I hope you know enough by now that the
two pressures would have to be in the same units!)
By the way, mole fractions are unitless numbers. The mole (or pressure) units cancel out.
Gas Law Problems- Dalton'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)
Conversions
K = °C + 273
1 cm3 (cubic centimeter)
1 dm3 (cubic decimeter)
Standard Conditions
50
=
=
1 mL
1 L
K - Kelvin
°C - degrees Celsius
101.325 kPa = 101,325 Pa
0.00 °C = 273 K
1.00 atm = 760.0 mm Hg
=
The vapor pressure of water in kilopascals is here. Handbook of Chemistry and Physics:
73rd Edition (1992-93) Remember to convert to mmHg or atm. as needed.
65. A container holds three gases: oxygen, carbon dioxide, and helium. The partial
pressures of the three gases are 2.00 atm, 3.00 atm, and 4.00 atm, respectively. What is
the total pressure inside the container?
66. A container with two gases, helium and argon, is 30.0% by volume helium. Calculate
the partial pressure of helium and argon if the total pressure inside the container is 4.00
atm.
67. If 60.0 L of nitrogen is collected over water at 40.0 °C when the atmospheric pressure
is 760.0 mm Hg, what is the partial pressure of the nitrogen?
68. 80.0 liters of oxygen is collected over water at 50.0 °C. The atmospheric pressure in
the room is 96.00 kPa. What is the partial pressure of the oxygen?
69. A tank contains 480.0 grams of oxygen and 80.00 grams of helium at a total pressure
of 7.00 atmospheres. Calculate the following.
a) How many moles of O2 are in the tank?
b) How many moles of He are in the tank?
c) Total moles of gas in tank.
d) Mole fraction of O2.
e) Mole fraction of He.
f) Partial pressure of O2.
g) Partial pressure of He.
70. A tank contains 5.00 moles of O2, 3.00 moles of neon, 6.00 moles of H2S, and 4.00
moles of argon at a total pressure of 1620.0 mm Hg. Complete the following table
Moles
O2
Ne
H2S
Ar
Total
18.00
Mole fraction
1
Pressure fraction
1
Partial Pressure
1620.0
71. A mixture of 14.0 grams of hydrogen, 84.0 grams of nitrogen, and 2.0 moles of
oxygen are placed in a flask. When the partial pressure of the oxygen is 78.00 mm of
mercury, what is the total pressure in the flask?
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72. A flask contains 2.00 moles of nitrogen and 2.00 moles of helium. How many grams
of argon must be pumped into the flask in order to make the partial pressure of argon
twice that of helium?
Law Problems - Dalton's Law Answers
The vapor pressure of water in kilopascals is here. Handbook of Chemistry and Physics:
73rd Edition (1992-93) Remember to convert to mmHg or atm. as needed.
65. 9.00 atm.
66. PHe = 0.300 x 4.00 atm = 1.20 atm. PAr = 4.00 - 1.20
67. 760.0 mmHg minus 55.3 mmHg
68. 96.00 kPa minus 12.33 kPa
69.
a) 480.0 g O2 / 32.0 g/mol
b) 80.00 g He / 4.00 g/mol
c) 35.0 moles
d) 15.0 mol O2 / 35.0 mol
e) 20.0 mol He / 35.0 mol
f) 7.00 atm x 0.4286
g) 7.00 atm x 0.5714
Keep in mind that once one partial pressure is calculated, the other can be arrived at by
subtraction, if so desired.
70. Complete the following table
Moles
Mole fraction
Pressure fraction
Partial Pressure
O2
Ne
H2S
Ar
Total
5.00
3.00
6.00
9.00
18.00
4/18 = 0.222
1
0.222
1
5/18 = 0.278 3/18 = 0.167 6/18 = 0.333
0.278
0.167
0.333
1620 x 0.278 1620 x 0.167 1620 x 0.0.333 1620 x 0.222
1620.0
= 450.36
= 270.54
= 539.46
= 359.64
71. (14.0 g / 2.00 g/mol) + (84.0 g /28.0 g/mol) + (2.0 moles) = 12.0 moles total
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2.0 / 12.0 = 0.167 of the total pressure. 78.00 is to 0.167 as the total pressure is to one, so
468 mmHg is the answer.
72. The solution is left to the reader.
Graham's Law
Discovered by Thomas Graham [32K GIF] of Scotland in
the 1830s (I think!).
Consider samples of two different gases at the same Kelvin
temperature.
Since temperature is proportional to the kinetic energy of the
gas molecules, the kinetic energy (KE) of the two gas
samples is also the same.
In equation form, we can write: KE1 = KE2 Since KE = (1/2)
mv2, (m = mass and v = velocity) we can write the following
equation:
m1v12 = m2v22
Note that the value of one-half cancels.
The equation above can be rearranged algebraically into the following: the square root of
(m1 / m2) = v2 / v1
You may wish to assure yourself of the correctness of this rearrangement.
This last equation is the modern way of stating Graham's law.
Gas Law Problems - Graham'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
53
=
=
1 mL
1 L
=
135. If equal amounts of helium and argon are placed in a porous container and allowed
to escape, which gas will escape faster and how much faster?
136. What is the molecular weight of a gas which diffuses 1/50 as fast as hydrogen?
137. Two porous containers are filled with hydrogen and neon respectively. Under
identical conditions, 2/3 of the hydrogen escapes in 6 hours. How long will it take for half
the neon to escape?
138. If the density of hydrogen is 0.090 g/L and its rate of diffusion is 6 times that of
chlorine, what is the density of chlorine?
139. How much faster does hydrogen escape through a porous container than sulfur
dioxide?
Problems added after all the worksheets were numbered:
1) Compare the rate of diffusion of carbon dioxide (CO2) & ozone (O3) at the same
temperature.
Gas Law Problems - Graham's Law Answers
135. Set rate1 = He = x and rate2 = Ar = 1. The weight of He = 4.00 and Ar = 39.95. By
Graham's Law, x / 1 = square root (39.95 / 4.00) = 3.16 times as fast.
136. Set rate1 = other gas = 1 and rate2 = H2 = 50. The weight of H2 = 2.02 and the other
gas = x. By Graham's Law, 1 / 50 = square root (2.02 / x) = molecular weight of 5050.
137. Set rate1 = H2 = x and rate2 = Ne = 1. The weight of H2 = 2.02 and Ne = 20.18. By
Graham's Law, x / 1 = square root (20.18 / 2.02) = 3.16 times as fast.
0.67 / 3.16 = 0.211 = amount of Ne leaving in 6 hours.
Therefore 0.211 / 6 = 0.50 / x.
x = 14.2 hours
138. Set rate1 = H2 = 6 and rate2 = Cl2 = 1. The weight of H2 = 2.02 and Ne = x. By
Graham's Law, 6 / 1 = square root (x / 2.02) = 72.72 (molec. wt. of Cl2)
72.72 g / 22.414 L = 3.24 g/L. (22.414 is molar volume.)
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139. Set rate1 = H2 = x and rate2 = SO2 = 1. The weight of H2 = 2.02 and SO2 = 64.06. By
Graham's Law, x / 1 = square root (64.06 / 2.02) = 5.63 times as fast
Answers to problems after worksheet numbering was set:
1) molecular weights: CO2 = 44.0, O3 = 48.0. Since O2 is the heavier, I will set its rate = 1
and called it r2 in Graham's Law:
r1 over r2 = [square root of] molec, wt2 over [square root of] molec, wt1
Therefore: x over 1 = [square root of] (48 over 44) = [square root of] 1.0909 . . . = 1.04
CO2 diffuses 1.04 times as fast as O3
Molar Volume
The molar volume is the volume occupied by one mole of ideal gas at STP. Its value is:
22.414 L mol¯1
It is actually known to several more decimal places but the number above should prove
sufficient.This value has been known for about 200 years and it is not a constant of nature
like, say, the charge on the electron. If we had picked a different standard temperature,
then the molar volume would be different.
Using PV = nRT, you can calculate the value for molar volume. V is the unknown and n
= 1.00 mol. Set P and T to their standard values and use R = 0.08206.
Molar volume doesn't show up that often in problems. As a consequence, teachers
sometimes like to use molar volume on the test, in order to trip the kid up!! Let's do some
examples.
Example #1: you have 2.00 L of dry H2 at STP. How many moles is this?
Solution: we could solve this with PV = nRT, but we can shortcut since we have STP
and do this:
Example #2: 0.250 moles of HCl will occupy how many liters at STP?
Solution: this is just the reverse of example one:
0.250 mol x 22.414 L mol¯1
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Example #3: What is molar volume at 576 K?
Solution: strictly speaking, this isn't more than a volume-temperature problem, but for
some reason putting "molar volume" in the problem messes people up.
22.414 L mol¯1 / 273 K = x / 546 K
Notice that 546 is double 273, so that just confims the answer of 44.828 L.
Gas Density
Discussing gas density is slightly more complex than discussing solid/liquid density.
Since gas volume is VERY responsive to temperature and pressure, these two factors
must be included in EVERY gas density discussion.
By the way, solid and liquid volumes are responsive to temperature and pressure, but the
response is so little that it can usually be ignored in introductory classes.
So, for gases, we speak of "standard gas density." This is the density of the gas
(expressed in grams per liter) at STP. If you discuss gas density at any other set of
conditions, you drop the word standard and specify the pressure and temperature. Also,
when you say "standard gas density," you do not need to add "at STP." STP is part of the
definition of the term. It does no harm to say "standard gas density at STP," it just looks a
bit silly.
You can calculate the ideal standard gas density fairly easily. Just take the mass of one
mole of the gas and divide by the molar volume. For nitrogen, we would have:
28.014 g mol¯1 / 22.414 L mol¯1 = 1.250 g / L
For water, we have:
18.015 g mol¯1 / 22.414 L mol¯1 = 0.8037 g / L
Notice I said "ideal." The behavior of "real" gases diverges from predictions based on
ideal conditions. Small gases like H2 at high temperatures approach ideal behavior almost
exactly while larger gas molecules (NH3) at low temperatures diverge the greatest
amount. These "real" gas differences are small enough to ignore right now, but in later
classes they will become important.
One place teachers like to bring gas density into play is when you calculate a molar mass
of a gas using PV = nRT.
56
Example Problem #1: The density of a gas is measured at 1.853 g / L at 745.5 mmHg
and 23.8 °C. What is its molar mass?
Solution #1: Convert mmHg to atm and °C to K. Use 1.000 L. Plug into PV = nRT and
solve for n. Then divide 1.853 g by n, the number of moles, for your answer.
Solution #2: This solution exploits a rearrangement of the ideal gas law. Here it is:
The answer: 46.008 g mol¯1
Example Problem #2: What is the molar mass of a gas which has a density of 0.00249
g/mL at 20.0 °C and 744.0 mm Hg?
Solution: Convert mmHg to atm (744.0/760.0) and °C to K (20.0 + 273.15). Use 0.001 L
(which is 1 mL converted to liters). Plug into PV = nRT and solve for n (the value of
which is calculated to be 4.069 x 10¯5 mol).
Then, divide 0.00249 g by the moles just calculated for an answer of 61.2 g/mol.
Vapor Pressure
Vapor pressure is the pressure of the vapor over a liquid (and some solids) at equilibrium.
Now, what does that definition mean? I'm going to go through some explanation steps
that, hopefully, give you a correct idea of vapor pressure.
1) Imagine a closed box of several liters in size. It has rigid walls and is totally empty of
all substances.
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2) Now, I inject some liquid into the box, but the box is not full of liquid. What will
happen to the liquid?
3) That's right. Some, maybe all, of the liquid will evaporate into gas, filling the empty
space. Now if all the liquid evaporates, we just have a box of gas. That's not what we
want. So let's suppose that only some of the liquid evaporated and that there are both the
liquid state and the gas state present in the box.
The gas that is above the liquid is called its vapor and it creates a pressure called vapor
pressure.
What I mean is this: suppose you put a gas-pressure measuring device into the gas and
never touched the liquid. Would a gas pressue be recorded? The answer is YES!
However, here is a key, key point. The vapor must be in contact with the liquid at all
times. Remove the liquid and you just have a box of gas, you do not have vapor or vapor
pressure.
Let's emphasize this point!!!! For vapor pressure to exist, the vapor (gas phase) MUST be
in physical contact with the liquid (or solid) it came from. You CAN'T have vapor
pressure without two phases being present and in contact!!!!!!
How is vapor pressure created? Another way to put it - how do molecules of the liquid
become molecules of gas?
Each molecule in the liquid has energy, but not the same amount. The energy is
distributed according to the Maxwell-Boltzmann distribution. Even if you don't know
what that is, the point is that some molecules have a fairly large amount of energy
compared to the average. Those are the ones we are interested in.
We are ESPECIALLY interested if one of these high energy molecules happens to be
sitting right at the surface of the water. Now, all the molecules are in motion because of
their enery, but none have sufficient energy to break the mutual attractive force molecules
have for each other. Suppose, just suppose that our surface molecule was moving up
away from the surface AND had enough energy to break away from the attractive forces
of the molecules around it.
Where would that molecule go? It would continue to move away from the liquid surface
AND IT BECOMES A MOLECULE OF GAS. This is great because we are now making
some vapor pressure. It happens to another molecule and another and another.
But wait! The vapor pressure stops going up and winds up staying at some fixed value.
What's going on? Here's the answer, I hope you can handle it!
58
As more and more molecules LEAVE the surface, what do some start to do? That's right,
some RETURN to the surface and resume their former life as a liquid molecule. Soon the
number of molecules in the vapor phase is constant because the rate of returning equals
the rate of leaving and so the pressure stays constant.
This image is my attempt to summarize this process. I'll put it here without comment:
Here's a REAL IMPORTANT point about vapor pressure:
the vapor pressure depends ONLY on the temperature
That is true because as the temperature goes up, there are more and more molecules with
the right combination of energy and direction to break free of the liquid's surface.
Finally, you might think that, if you were to put more liquid in, the vapor pressure would
go up. That's NOT correct. Keep in mind that there are two opposing processes at work:
(1) molecules leaving the liquid's surface and (2) molecules returning to the liquid at its
surface. Process #1 depends only on temperature and process #2 depends only on how
many molecules are in the vapor phase. Adding more liquid affects neither process.
59