A Quick Introduction to Gases
These notes were written to help you quickly get to the point where you
can work gas law problems on the quizzes and exams in this course.
They are not a detailed development of the topic.
The problems which will appear on exams will involve 4 "empirical" gas
laws and one additional law known as the ideal gas law.
The empirical gas laws are as follows:
Name of Gas Law

Equation 
Parameters That Must Be Constant 
Boyle's Law 
P_{1}V_{1} = P_{2}V_{2} 
Temperature, Number of moles 
Charles' Law 
V_{1}/ T_{1} = V_{2} / T_{2} 
Pressure, Number of moles 
Amonton's Law 
P_{1} / T_{1} = P_{2} / T_{2} 
Volume, Number of moles 
Combined Gas Law 
P_{1}V_{1} / T_{1} = P_{2}V_{2}
/ T_{2} 
Number of moles 
The problems which you will encounter with these laws involve having
a gas under an initial set of conditions (state 1) and then having the
gas undergo some process which changes its state of existence to another
set of conditions (state 2). All the gas parameters will be specified
except one, and the one missing can be solved for algebraically.
Generally, I ask for one of the state 2 parameters.
Boyle's Law Sample Problems
Boyle's Law Example 1
A gas was confined in a cylinder fitted with a movable piston. At
22.6 ^{o}C, the gas occupied a volume of 4.573 L under a pressure
of 1.152 atm. The gas was isothermally compressed, reducing its volume
to 2.963 L. What pressure was exerted by the comperssed gas?
Solution to Boyle's Law Example 1
We recognize that this is a Boyle's Law problem because the temperature
remains constant (we are told the process is "isothermal", which means
constant temperature). Presumably, there is also no change in the
number of moles of gas, since no amount of gas was ever mentioned in the
problem. Although the temperature is given, it is not needed, since
it does not appear in Boyle's Law.
Before the process (compression), the gas volume is 4.573 L, so this
is V_{1}, and the gas pressure is 1.152 atm, so this is P_{1}.
After the process (compression), the gas volume is 2.963 L, so this is
V_{2}. The variable left unspecified is the pressure after
the process, (P_{2}) so this is the variable we must solve for.
The equation P_{1}V_{1} = P_{2}V_{2}
can be solved for P_{2} to get
P_{2} = P_{1 }( V_{1} / V_{2 }) = 1.152
atm ( 4.573 L / 2.963 L ) = 1.778 atm <=====
ANSWER
Notice that the "new" pressure is just the "old"
pressure multiplied by a ratio of the two volumes. A pattern like
this exists in all problems involving the empirical gas laws. It
is possible to solve such problems by inspection, without actually having
to set up the algebra, once you get used to it. Notice in this case,
for example, it makes sense that the larger volume is in the numerator.
Since the gas is being compressed, the final pressure should be larger
than the initial one. Therefore, the original pressure has to be
multiplied by a number greater than one, to create a number larger than
we started with. The only other possible ratio we could write for
the two volumes is ( 2.963 L / 4.573 L ), but this ratio would
be less than one, and when multiplied by the origianl pressure, would yield
a smaller pressure than we started with. This would not make sense,
in light of the fact that the gas is being compressed, thus we could have
selected the proper ratio without ever setting up the algebra.
Boyle's Law Example 2
A gas was confined in a cylinder fitted with a movable
piston. At 19.6 ^{o}C, the gas occupied a volume of 7.336
L under a pressure of 3.196 atm. The gas was isothermally compressed
until its pressure reached 5.500 atm. What volume was occupied by
the compressed gas?
Solution to Boyle's Law Example 2
We recognize this as a Boyle's Law problem because
the temperature remains constant ("isothermal") and there is apparently
no change in the number of moles of gas, since moles (or amount of gas
in any units, for that matter), is not mentioned in the problem.
The temperature is specified in the problem, but not needed, since temperature
does not appear in Boyle's Law.
Reading through the problem, we can make the following
variable assignments:
V_{1} = 7.336 L
P_{1} = 3.196 atm
V_{2} = ?
P_{2} = 5.500 atm
The Boyle's Law equation P_{1}V_{1}
= P_{2}V_{2} can be solved for V_{2} to give
V_{2} = V_{1} ( P_{1}
/ P_{2} ) = 7.336 L ( 3.196 atm / 5.500 atm ) = 4.263 L
<===== ANSWER
Notice that the "new" volume is just the "old"
volume multiplied by a ratio of the pressures. It makes sense that
the smaller pressure is on top in the ratio. This makes the ratio
less than one, so when multiplied by the original volume, is gives us a
smaller volume than we started with. Since the gas is being compressed,
it makes sense that the final volume should be smaller than the orignial.
Boyle's Law Example 3
A gas was confined in a cylinder fitted with a movable
piston. At 21.3 ^{o}C, the gas occupied a volume of 1.994
L under a pressure of 4.357 atm. The pressure on the gas was then
reduced to 3.264 atm, and the gas was allowed to expand isothermally to
a new volume. What was the volume of the expanded gas?
Solution to Boyle's Law Example 3
We recognize this as a Boyle's Law problem because
the temperature does not change ("isothermal") and the number of moles
apparently does not change. The temperature has been given, but is
not needed in the calculation. Reading through the problem, we can
make the following variable assignments:
V_{1} = 1.994 L
P_{1} = 4.357 atm
V_{2} = ?
P_{2} = 3.264 atm
The Boyle's Law equation, P_{1}V_{1}
= P_{2}V_{2} can be solved for P_{2} to give
V_{2} = V_{1} ( P_{1}
/ P_{2} ) = 1.994 L ( 4.357atm / 3.264 atm ) = 2.662 L
<===== ANSWER
It makes sense that the larger pressure ends up
on top in the ratio. The gas is expanding, so the final volume should
be bigger than the initial one. To obtain a larger volume than we
started with, we must multiply by a number bigger than one.
Boyle's Law Example 4
A gas was confined in a cylinder fitted with a movable
piston. At 27.4 ^{o}C, the gas occupied a volume of 3.431
L under a pressure of 5.363 atm. If you would like the gas volume
to increase to 5.000 L, what pressure should you put on the gas?
Assume no change in temperature or number of moles of gas.
Solution to Boyle's Law Example 4
Lets see if we can solve this one without doing the
algegra. The unknown variable is a pressure. The pressure we
seek should be obtainable by multiplying the original pressure by a ratio
of the volumes, since the pressure change is related to the volume change.
Recall that temperature and number of moles are constant in Boyle's Law
problems. We want the gas to expand, since we are trying to go from
a volume of 3.431 L to a volume of 5.000 L. In order to let the gas
expand, the pressure on it must be reduced. Thus, the final pressure
should be smaller than the original. This means the volume ratio
should be constructed so as to be less than one  that is, the smaller
volume should be on top. Therefore the calculation is
5.363 atm ( 3.431 L / 5.000 L ) = 3.680 atm
<=====
ANSWER
So, the pressure on the gas must be reduced from
its original value to 5.363 atm down to a new, lower pressure of only 3.680
atm. At this pressure, the gas volume will be 5.000 L, as desired.
Boyle's Law Example 5
A gas was confined in a cylinder fitted with a movable
piston. At 15.6 ^{o}C, the gas occupied a volume of 3.384
L under a pressure of 4.476 atm. If you want to increase the gas
pressure to 7.603 atm (at constant temperature), to what volume should
the gas be compressed? Assume no change in the number of moles of
gas.
Solution to Boyle's Law Example 5
Lets again set up the calculation intuitively, without
doing the algebra. We want to increase the gas pressure, which means
it must be compressed into a smaller volume. The new volume can be
calculated by multiplying the original volume by a ratio of the two pressures.
Since the new volume should be smaller than the original, the pressure
ratio must be less than one. This means the smaller pressure must
be on top. The calculation is therefore
3.384 L ( 4.476 atm / 7.603 atm ) = 1.992 L
<===== ANSWER
Thus, the piston must be pushed in until the gas
volume is reduced from its original value of 3.384 L to a new value of
only 1.992 L.
Charles' Law Sample Problems
Charles' Law Example 1
A gas was confined in a cylinder fitted with a movable
piston. At 22.6 ^{o}C, the gas occupied a volume of 3.254
L under a pressure of 1.200 atm. The gas was isobarically heated
to a temperature of 100.0 ^{o}C, causing it to expand. What
was the volume of the heated gas?
Solution to Charles' Law Example 1
We recognize this as a Charles' Law problem because
the pressure does not change ("isobaric") and presumably, there is no change
in the number of moles of gas. The pressure is given in the
problem, but is not needed, because it does not appear in Charles' Law.
Reading through the problem, we can make the following variable assignments:
T_{1} = 22.6 ^{o}C = 295.75 K
(Note: Kelvin = Celsius + 273.15)
V_{1} = 3.254 L
T_{2} = 100.0 ^{o}C = 373.15 K
V_{2} = ?
The Charles' Law equation, V_{1} / T_{1}
= V_{2} / T_{2} can be solved for V_{2} to give
V_{2} = V_{1} ( T_{2}
/ T_{1} ) = 3.254 L ( 373.15 K / 295.75 K ) = 4.106 L
<===== ANSWER
Notice that a ratio pattern exists here, just
as it did in the Boyle's Law problems. The pressure and number of
moles remain constant, so changes in volume are related only to changes
in temperature. The "new" volume is found by taking the "old" volume
and multiplying it by the appropriate temperature ratio. Since the
gas is expanding, the final volume must be larger than the initial volume.
Therefore, the temperature ratio must be greater than one. This means
the larger temperature must be on top, as the algebra has shown us.
Always remember that temperature units MUST be Kelvin when doing gas calculations.
Charles' Law Example 2
A gas was confined in a cylinder fitted with a movable
piston. At 12.3 ^{o}C, the gas occupied a volume of 4.736
L under a pressure of 2.113 atm. The gas was isobarically heated
until its volume had increased to 5.000 L. What was the temperature
of the heated gas? Experss your answer in ^{o}C.
Solution to Charles' Law Example 2
We recognize this as a Charles' Law problem because
the pressure does not change ("isobaric") and presumably the number of
moles of gas does not change. The pressure is given in the problem,
but is not needed, because it does not appear in Charles' Law. Reading
through the problem, we can make the following variable assignments:
T_{1} = 12.3^{ o}C = 285.45 K
(Note: Kelvin = Celsius + 273.15)
V_{1} = 4.736 L
T_{2} = ?
V_{2} = 5.000 L
The Charles' Law equation, V_{1} / T_{1}
= V_{2} / T_{2} can be solved for T_{2} to give
T_{2} = T_{1} ( V_{2}
/ V_{1} ) = 285.45 K ( 5.000 L / 4.736 L ) = 301.36 K = 28.2 ^{o}C
<=== ANSWER
Notice that the algebra dictates that we put the
larger volume on top. This makes sense, because it creates a ratio
larger than one, so when multiplied by the original temperature, we get
a higher temperature than we started with. Notice also that even
though our starting temperature was in ^{o}C and we wanted our
final answer in ^{o}C, we still had to do the calculation in K.
We add 273.15 to the Celsius temperature to get Kelvin temperature, obtain
the final temperature in Kelvin, and then subtract 273.15 to convert the
temperature back to Celsius.
Charles' Law Example 3
A gas was confined in a cylinder fitted with a movable
piston. At 6.8^{ o}C, the gas occupied a volume of 6.225
L under a pressure of 3.219 atm. The gas was isobarically heated
to a temperature of 110.7 ^{o}C. What was the volume of the
hot gas?
Solution to Charles' Law Example 3
This time, let's solve it intuitively, instead of
going through the algebra. Since the gas is being heated, it should
expand, so the volume of the hot gas should be larger than the original
volume. This means the original volume needs to be multiplied by
a ratio that is larger than one. Since it is only the change in temperature
that is changing the volume (pressure and number of moles remain constant),
we must multiply the volume given in the problem by a temperature ratio
with the larger temperature on top. We must remember to add 273.15
to each temperature and thus enter them as Kelvin temperatures in the calculation.
6.225 L ( 383.85 K / 279.95 K ) = 8.535 L
<===== ANSWER
Thus, when heated to 110.7 ^{o}C, the
gas volume will be 8.535 L
Charles' Law Example 4
A gas was confined in a cylinder fitted with a movable
piston. At 15.4 ^{o}C, the gas occupied a volume of 2.573
L under a pressure of 1.646 atm. To what temperature should the gas
be heated to increase its volume ot 3.100 L? Assume no change in
pressure or number of moles of gas?
Solution to Charles' Law Example 4
Let's again solve the problem intuitively.
The change in temperature should be related to the change in volume, so
we should be able to get the "new" temperature by multiplying the "old"
temperature by the appropriate volume ratio. Since the gas is being
heated, the temperature we get for our answer needs to be higher than than
the temperature we started with. This means that in the volume ratio,
the larger volume needs to be on top.
288.55 K ( 3.100 L / 2.573 L ) = 347.65 K = 74.5
^{o}C
<===== ANSWER
Notice that the calculation was done in Kelvin,
even though the starting temperature was given in Celsius, and the final
answer was to be expressed in Celsius. The Kelvin answer is converted
to Celsius by subtracting 273.15.
Amonton's Law Sample Problems
Amonton's Law Example 1
A gas was contained in a rigid steel tank having
a volume of 7.528 L. At 21.4 ^{o}C, the gas exerted a pressure
of 5.173 atm. What pressure will the gas exert if it is heated to
a temperature of 89.6 ^{o}C?
Solution to Amonton's Law Example 1
We recognize this as an Amonton's Law problem because
the volume does not change and presumably, neither does the number of moles.
The problem does not explicitly say the volume does not change, but it
refers to a rigid steel tank, and we can interpret that as meaning the
volume is constant. Moles are not mentioned, so we will assume they
remain constant. The volume is given in the problem, but is not needed,
since it does not appear in Amonton's Law.
Reading through the problem, we can make the following
variable assignments:
T_{1} = 21.4 ^{o}C = 294.55 K
(Note: Kelvin = Celsius + 273.15)
P_{1} = 5.173 atm
T_{2} = 89.6 ^{o}C = 362.75 K
P_{2} = ?
The Amonton's Law equation P_{1} / T_{1}
= P_{2} / T_{2} can be solved for P_{2} to give
P_{2} = P_{1} ( T_{2}
/ T_{1} ) = 5.173 atm ( 362.75 K / 294.55 K ) = 6.371 atm
<===== ANSWER
Notice that a ratio pattern again exists.
Since the volume and number of moles are not changing, the pressure change
is related only to a temperature change. The "new" pressure can be
calculated by multiplying the "old" pressure by the appropriate temperature
ratio. Since the gas is being heated, the final pressure must be
higher than the initial pressure. This means the temperature ratio
must be larger than one, so the larger temperature must be on top.
This is exactly what the algebra led us to do. Notice that the calculation
had to be done in Kelvin temperature units.
Amonton's Law Example 2
A gas confined in a 4.473 L steel tank exerted a
pressure of 2.157 atm at 22.6 ^{o}. In a warmer environment,
the pressure inside the tank was found to be 2.306 atm. What was
the temperature in the warmer environment?
Solution to Amonton's Law Example 2
We recognize this as an Amonton's Law Problem because
the volume does not change (we can assume a steel tank has a constant volume)
and there is no mention of moles, so presumably there is no change in the
number of moles of gas. The volume is given in the problem, but is not
needed since it does not appear in Amonton's Law. Reading through
the problem, we can make the following variable assignments:
P_{1} = 2.157 atm
T_{1} = 22.6 ^{o}C = 295.75 K
(Note: Kelvin = Celsius + 273.15)
P_{2} = 2.306 atm
T_{2} = ?
The Amonton's Law equation P_{1} / T_{1}
= P_{2} / T_{2} can be solved for T_{2} to give
T_{2} = T_{1} ( P_{2}
/ P_{1} ) = 295.75 K ( 2.306 atm / 2.157 atm ) = 316.18 K = 43.0
^{o}C
<=== ANSWER
Notice that the algebra directs us to put the
larger pressure on top in the pressure ratio. This makes the ratio
larger than one, which makes sense, because if the pressure of the gas
has increased, as it has in this problem, the final temperature must be
higher than the initial temperature. As usual, gas calculations must
be done in Kelvin, but we can convert our answer to Celsius if those are
the units we want to express it in.
Amonton's Law Example 3
A sample of a gas having a volume of 3.470 L under
a pressure of 2.507 atm at 21.1 ^{o}C was isochorically heated
to a temperature of 62.8 ^{o}C. What pressure will the gas
exert under these conditions?
Solution to Amonton's Law Example 3
We recognize this as an Amonton's Law problem because
the volume and number of moles are not changing. Here, we are not
told about a steel tank, but rather, that the gas is isochorically heated.
The term isochoric refers to constant volume. No mention is made
of moles, so we shall assume the number of moles remains constant.
Let's see if we can solve it intuitively this time. If a gas is confined
to a constant volume and heated, its pressure should increase. The
"new" pressure can be obtained by multiplying the "old" pressure by the
appropriate temperature ratio. Since the "new" pressure must be larger
than the "old" pressure, the temperature ratio must be larger than one.
This means the larger temperature must be on top. Of course, the
temperatures need to be entered in Kelvin units. Therefore, the calculation
is
2.507 atm ( 335.95 K / 294.25 K ) = 2.862 atm
Amonton's Law Example 4
A steel tank at 17.3 ^{o}C contains a gas
under a pressure of 4.973 atm. For safety reasons, it is desired
that the pressure not exceed 8.000 atm. What is the maximum temperature
(expressed in ^{o}C) to which the gas in the tank can be heated?
Assume no change in the number of moles of gas during the heating process.
Solution to Amonton's Law Example 4
Since this is a steel tank the volume can be assumed
to be constant. We are also told that we should assume no change
in the number of moles. The volume of the tank is not given in this
problem, but we don't need it. Let's again solve the problem intuitively,
that is, without the algebra. The highest pressure that can be tolerated
is 8.000 atm, so we need to know at what temperature the pressure reaches
this magnitude. This temperature is clearly higher than the origianl
temperature, because pressure increases with temperature. So we must
multiply the original temperature by a pressure ratio that is larger than
one  that is, it must have the higher pressure on top. As usual,
we must carry out the calculation in Kelvin units. We can then convert
our answer to Celcius for reporting it.
290.45 K ( 8.000 atm / 4.973 atm ) = 467.24 K
= 194.1 ^{o}C <=====
ANSWER
Thus, for this tank, the temperature should not
be allowed to exceed 194.1 ^{o}C, because then the pressure would
exceed the maximum safe working pressure of 8.000 atm.
Combined Gas Law Sample Problems
Combined Gas Law Example 1
A gas was confined in a cylinder fitted with a movable
piston. At 12.8 ^{o}C, the gas occupied a volume of 8.175
L under a pressure of 1.824 atm. The gas was simultaneously heated
and compressed, so that its volume was 6.713 L and its temperature was
72.8 ^{o}C. What pressure was exerted by the hot compressed
gas?
Solution to Combined Gas Law Example 1
Notice that none of the individual laws (Boyle's,
Charles' or Amonton's) can be used to solve this problem, because pressure,
volume, and temperature of the gas are all changing. The only thing
that remains constant is the number of moles, since that was not mentioned
in the problem. Reading through the problem, we can make the following
assignments:
T_{1} = 12.8 ^{o}C = 285.95 K
(Note: Kelvin = Celsius + 273.15)
V_{1} = 8.175 L
P_{1} = 1.824 atm
T_{2} = 72.8 ^{o}C = 345.95 K
V_{2} = 6.713 L
P_{2} = ?
The combined gas law equation P_{1}V_{1}
/ T_{1} = P_{2}V_{2} / T_{2} can be solved
for P_{2} to give
P_{2} = P_{1} ( V_{1}
/ V_{2} ) ( T_{2} / T_{1} ) = 1.824 atm ( 8.175
L / 6.713 L ) ( 345.95 K / 285.95 K )
= 2.687 atm <=====
ANSWER
The ratio patterns exist here too, but there are
two ratios to consider instead of only one. Since pressure, volume,
and temperature are all changing, the change in pressure is affected both
by volume changes and temperature changes. We can analyze each of
these effects separately to deduce where the variables should be placed
in the ratios. In this problem, the gas has been compressed into
a smaller volume. The effect of reducing the gas volume is to increase
its pressure, so the volume ratio must be larger than one. Thus the
larger volume should be on top, as the algebra has directed us to do.
Looking at the temperatures, heating a gas should increase its pressure,
so the temperature ratio should also be larger than one. Again, the
algebra has directed us to put the larger temperature on top, as we would
expect.
Combined Gas Law Example 2
A gas was confined in a cylinder fitted with a movable
piston. At 18.4 ^{o}C, the gas occupied a volume of 6.312
L and exerted a pressure of 2.417 atm. The gas was simultaneously
heated and compressed, so that its pressure was 4.161 atm and its temperature
was 59.7 ^{o}C. What volume was occupied by the hot compressed
gas.
Solution to Combined Gas Law Example 2
Again, only the combined gas law can be used to solve
the problem, because volume, pressure and temperature are all changing.
Only the number of moles remains constant. Reading through the problem,
we can make the following variable assignments:
T_{1} = 18.4 ^{o}C = 291.55 K
(Note: Kelvin = Celsius + 273.15)
V_{1} = 6.312 L
P_{1} = 2.417 atm
T_{2} = 59.7 ^{o}C = 332.85 K
V_{2} = ?
P_{2} = 4.161 atm
The combined gas law equation P_{1}V_{1}
/ T_{1} = P_{2}V_{2} / T_{2} can be solved
for V_{2} to give
V_{2} = V_{1} ( P_{1}
/ P_{2} ) ( T_{2} / T_{1} ) = 6.312 L ( 2.417 atm
/ 4.161 atm ) ( 332.85 K / 291.55 K )
= 4.186 L <=====
ANSWER
The change in volume is affected both by a change
in pressure and by a change in temperature. We can consider each
of these effects separately. The pressure on the gas is being increased.
This factor alone would tend to shrink the gas, giving it a smaller volume.
Therefore, the pressure ratio must be less than one. Looking at the
calculation above, we see that that's exactly what the algebra has led
us to do. We had to put the smaller pressure on top. The gas
is also being heated. This factor alone would tend to make the gas
expand, giving it a larger volume. Therefore the temperature ratio
must be larger than one. Indeed, the algebra has led us to put the
larger temperature on top, giving a temperature ratio larger than one.
The pressure change and temperature change work against each other in this
problem, but the effect of increasing pressure "wins", because the final
volume of the gas is less than the initial volume. That is, the direction
of volume change is that predicted on the basis of the pressure change.
Combined Gas Law Example 3
A gas was confined in a cylinder fitted with a movable
piston. At 17.3 ^{o}C, the gas occupied a volume of 7.515
L under a pressure of 1.406 atm. The gas was simultaneously heated
and compressed, so that its volume was 5.221 L and its pressure was 2.515
atm. To what temperature (expressed in ^{o}C) was the gas
heated?
Solution to Combined Gas Law Example 3
Temperature, pressure and volume are all changing.
Only the number of moles of gas remains constant. Therefore, only
the combined gas law can be used to solve this problem. Reading through
the text of the problem, we can make the following variable assignments:
T_{1} = 17.3 ^{o}C = 290.45 K
(Note: Kelvin = Celsius + 273.15)
V_{1} = 7.515 L
P_{1} = 1.406 atm
T_{2} = ?
V_{2} = 5.221 L
P_{2} = 2.515 atm
The combined gas law equation, P_{1}V_{1}
/ T_{1} = P_{2}V_{2} / T_{2} can be solved
for T_{2} to give
T_{2} = T_{1} ( P_{2}
/ P_{1} ) ( V_{2} / V_{1} ) = 290.45 K ( 2.515
atm / 1.406 atm ) ( 5.221 L / 7.515 )
= 360.95 K = 87.8 ^{o}C <=====
ANSWER
The "new" temperature is obtained by multiplying
the "old" temperature by the appropriate ratios. Let's look at these
ratios and see if they make sense. The pressure of the gas increases
in this problem. Ignoring the volume change for the moment, the increasing
pressure implies increasing temperature. On this basis, the "new"
temperature must be higher than the "old" temperature, so the pressure
ratio must be larger than one. Looking at the algebraic solution,
we see that it is. Ignoring the pressure changes and looking at the
voume changes, we see that the volume is decreasing in this problem.
If we pretend pressure is not changing, then the only way to shrink the
gas is to cool it. Thus, the decreasing volume implies cooling of
the gas, or a final temperature that is lower than the initial temperature.
With this in mind, the volume ratio must be smaller than one. Looking
at the algebra, we see that it is.
Notice how we are able to isolate the effects
of each variable and deal with them one at a time. In the above example,
we temporarily pretended we were cooling the gas, even though we knew we
were in fact, heating it. In this problem, the increasing pressure
on the gas (which tends to shrink it) was overcoming the effects of temperature,
which tends to expand it. It was because of the increased pressure
on the gas that it became smaller, in spite of being heated. However,
in analyzing the requirement for the volume ratio, we ignore the pressure
change, and determine how the temperature would have to change to make
the volume get smaller IF there was no pressure change. Since temperature
would have to become smaller, we construct a volume ratio that would tend
to give us a smaller temperature than we started with. This requires
a volume ratio less than one.
Combined Gas Law Example 4
A gas was confined in a cylinder fitted with a movable
piston. At 101.4 ^{o}C, the gas occupied a volume of 9.926
L under a pressure of 2.487 atm. The gas was cooled to a temperature
of 23.8 ^{o}C, and compressed to a volume of 5.134 L. What
pressure did the gas exert under these conditions?
Solution to Gas Law Example 4
Let's try to solve this one intuitively. The
"new" pressure should be equal to the "old" pressure multiplied by the
appropriate ratios. The parameters that are causing the pressure
to change are the volume change and the temperature change, so we must
multiply by the appropriate ratios of these quantities. The volume
of the gas is getting smaller. Forcing the gas into a smaller volume
(ignoring the temperature change) should cause the pressure to increase.
Thus the volume ratio needs to be larger than one. We should put
the larger volume on top. The gas is being cooled. This fact
alone (ignoring the volume change) should reduce the pressure of the gas.
So the temperature ratio should be less than one. We should put the
smaller temperature on top. Therefore, our calculation is
2.487 atm ( 9.926 L / 5.134 L ) ( 296.95 K / 374.55
K ) = 3.812 atm <=====
ANSWER
The reduction in volume and the cooling work against
each other. The reduction in volume has a greater effect than the
cooling, because the pressure increased, as would be predicted for a compression.
The cooling offsets some of the pressure increase, however. The pressure
would have been higher if we had the same volume reduction without the
simultaneous cooling. Try it for yourself and see. Do a Boyle's
Law calculation for the isothermal compression of a gas from 9.926 L to
5.134 L, using 2.484 atm as the original pressure. If done correctly,
you should obtain 4.808 atm as the final pressure, as compared to only
3.812 atm when the gas is cooled as described in the above problem.
Ideal Gas Law Sample Problems
Ideal Gas Law Example 1
What pressure is exerted by 9.768 g of CO_{2}
in a 3.774 L vessel at 26.5 ^{o}C?
Solution to Ideal Gas Law Example 1
In this problem, unlike the ones worked so far, we
don't have a "before and after" type problem. We have a single state
of the gas, and all of the parameters have been specified but one.
Notice also, that a specific gas has been given (CO_{2}), rather
than leaving the gas unspecified, as has been done in the empirical gas
laws. When a specific gas is given, and/or you have only a single
state, this is a good indication that the problem calls for the Ideal Gas
Law.
The equation of the Ideal Gas Law is PV=nRT, where
R = 0.08205783 L atm / K mol. We can solve this equation for the
pressure to get P = nRT / V. However, we can't
use this just yet, because we don't know the number of moles of CO_{2}.
We do, however, know the number of grams, and these can be converted to
moles.
9.768 g CO_{2} ( 1 mol CO_{2}
/ 44.0098 g CO_{2} ) = 0.221950566 mol CO_{2}_{}
Now we can calculate the pressure:
P = nRT / V = (0.221950566 mol ) (0.08205783 L
atm / K mol ) ( 299.65 K ) / 3.774 L
P = 1.446 atm <=====
ANSWER
Ideal Gas Law Example 2
A steel tank contains 14.73 g of Cl_{2} gas
under a pressure of 4.825 atm at 22.7 ^{o}C. What is the
volume of the tank?
Solution to Ideal Gas Law Example 2
This is a onestate problem involving a specific
gas; the Ideal Gas Law should be used to solve it. The equation of
the Ideal Gas Law is PV=nRT. Solving this for volume gives V=
nRT / P. However, we need to know the number of moles of Cl_{2}
to solve it. We can get this from the mass of Cl_{2}.
14.73 g Cl_{2} ( 1 mol Cl_{2}
/ 70.9054 g Cl_{2} ) = 0.207741582 mol Cl_{2}_{}
Now we can calculate the volume:
V = nRT / P = ( 0.207741582 mol ) ( 0.08205783
L atm / K mol ) ( 295.85 K ) / 4.825 atm
V = 1.045 L <=====
ANSWER
Ideal Gas Law Example 3
A steel reaction vessel having a volume of 8.761
L contains 4.822 g of H_{2} gas under a pressure of 7.016 atm.
What is the temperature of the H_{2} gas?
Solution to Ideal Gas Law Example 3
We have a specified gas and only one state (no "before
and after") so we must use the Ideal Gas Law to solve it. For the
Ideal Gas Law, we must know the number of moles of gas:
4.822 g H_{2} ( 1 mol H_{2}
/ 2.01588 g H_{2} ) = 2.392007461 mol H_{2}_{}
Then, solving the Ideal Gas Law, PV=nRT for the
temperature gives
T = PV / nR = ( 7.016 atm ) (8.761 L ) / ( 2.392007461
mol ) ( 0.08205783 L atm / K mol )
T = 313.16 K = 40.0 ^{o}C
<===== ANSWER
This page was last modified Thursday April
13, 2000 at 9:14 PM CDT