| 
>>
View the other Key Equations and Concepts in this
chapter
Gas Laws
>> Parts of this equation/concept include:
The combined gas law is used when the conditions on a gas (pressure,
volume, or temperature) are changed. It will be appropriate to use
this whenever you have two values for any given variable. It works
for any gas. The key to using this type of gas law is to assign
the variable appropriately, paying careful attention to which variables
are grouped together and the units for the variable. For pressure
(P) and volume (V) any unit can be used, provided
it is the same unit for both sets of conditions. If the units
are not the same, change one. For temperature (T), the unit
must be Kelvin. This is true for all gas laws.
The combined gas law is
(Equation 8.8)
where the subscript 1 refers to one set of conditions and the subscript
2 refers to the other. It doesn't matter whether a set of conditions
is designated 1 or 2, but the entire set must be kept together!
If a variable remains constant, it is canceled out of the equation.
Notice that if temperature is constant and you remove the Ts
from the combined gas law, you have Boyle's law, the relationship
between pressure and volume at a constant temperature. At constant
pressure, the combined gas law becomes Charles' law.
The combined gas law assumes a constant number of moles. In other
words, no gas is added or allowed to escape. If the number of moles
is changing, it can be included in the combined gas law as follows
One possible set of conditions is STP, standard temperature and
pressure. At STP, T = 273.15 K and P = 1 atm.
Since STP is a defined condition, these values are exact.
>> Example 1
If the pressure on a 100.0-mL gas is changed from 1.0 atm to
680 torr, what will the volume be?
Solution:
Use the combined gas law PV/T = PV/T to start, because
the question talks about changing a parameter. The question also
refers to pressure and volume, but not to temperature (or moles).
The assumption you are supposed to make is that temperature and
moles are constant. If a value is constant, it cancels from the
equation, so the equation to use is PV = PV.
Most gas laws are easily solved by assigning variables, taking
each number as it appears in the problem,
V1 = 100.0 mL (You know it is V because
mL is a unit of volume, and the subscript is 1 because you saw
it first.)
P1 = 1.0 atm (atm is a pressure unit. The
sentence also tells you it is talking about pressure here. The
subscript is 1 because it goes with the volume labeled with
subscript 1. Another way of looking at the subscripts is that
these are the initial pressure and volume values.)
P2 = 680 torr (Torr is a pressure unit. It
is your second one and it goes with the final volume, not with
the starting volume of 100.0 mL, so the subscript is 2.)
V2 = ? (This is the value you are solving
for. It goes with the second pressure, so its subscript is 2.)
Before the values can go into the equation, the units must be
correct. For the combined gas law, this means temperature must
be in units of Kelvin (not relevant for this problem), P1
and P2 must have the same pressure units, and
V1 and V2 must have the same
volume units.
In this example, that means changing the units of either
P1 or P2. Changing atm to
torr
| 1.0 atm |
 |
|
= |
7.6 x 102 torr (only two significant figures!) |
Using the values in the equation
(760 torr)(100.0 mL) = (680 torr)V
111.7 mL = V
The units are mL because that was the unit of volume used on
the other V. However, since P1 has only
two significant figures, the answer has only two significant figures,
so the final answer is
1.1 x 102 mL = V
If you had changed torr of P2 to atm
| 680 torr |
|
|
= |
0.895 atm |
(1.0 atm)(100.0 mL) = (0.895 atm)V
111.7 mL = V (same answer!)
You can do a quick check on your answer by recalling the general
gas law that pressure and volume are inversely proportional. Since
the pressure decreased, the volume should increase. Since the
final volume is greater than the initial volume, there is some
assurance that the math was done correctly.
>> Example 2
At what temperature will 0.860 L of gas at 9 °C expand to
1.00 L at constant pressure?
Solution:
The temperature is changing, pressure is constant, so the combined
gas law without pressure is used.
Assigning variables, the 0.890 L = V and the 9 °C
goes with that volume, so
V1 = 0.890 L
T1 = 9 °C = 282 K (Temperature must always
be in Kelvin units!)
V2 = 1.00 L
T2 = ?
The volume units are the same for both sets of conditions, so
the numbers can be used as they stand. Putting the values into
the equation,
3.156 x 10–3T2 = 1.00
T2 = 317 K (temperature units are always Kelvin)
The answer has three significant figures because each number
used in the equation had three significant figures. Although the
temperature value has only one significant figure in Celsius,
it is converted to Kelvin by adding. The significant figure rule
for addition counts decimal places, not significant figures. Therefore
the temperature value in Kelvin really does have three significant
figures. It is the value with three significant figures that is
multiplied and divided in the gas law.
If you choose to report the final answer in Celsius,
317 – 273.15 = 44 °C (The two significant figures
are correct, because subtraction also counts significant figures
by the decimal place.)
Check the answer by recalling that volume and temperature are
directly proportional. Since volume increased, temperature should
also increase.
>> Example 3
What is the final pressure when the conditions on 20.8 mL of
neon gas at STP are changed to 0.800 L and 23 °C?
Solution:
Initial conditions are V = 20.8 mL and STP. Final conditions
are V = 0.800 L and T = 23 °C.
STP stands for standard temperature and pressure. These conditions
are defined as T = 273.15 K and P = 1 atm. Because
the conditions are defined, the values for temperature and pressure
have infinite significant figures.
Because two conditions are changing, the entire combined gas
law is used.
Assigning variables,
V1 = 20.8 mL
T1 = 273.15 K
P1 = 1 atm
V2 = 0.800 L
T2 = 23 °C
P2 = ?
The units for volume must be the same, so either mL must be changed
to L or L to mL. In addition, since temperature must be in Kelvin,
the units for T2 must be changed. Since pressure
units must be the same, using the value for pressure given, the
answer will have units of atm. Making the changes,
V1 = 20.8 mL
T1 = 273.15 K
P1 = 1 atm
| V2 |
= |
0.800 L |
|
|
= |
800 mL |
T2 = 23 °C + 273.15 = 296 K
P2 = ?
Using the values in the combined gas law,
| (1 atm)(20.8 mL) |
|
| 273.15 K |
|
= |
| P2800 mL |
|
| 296 K |
|
| (1)(20.8)(296) |
|
| (273.15)(800) |
|
= |
P2 |
| 0.0282 atm |
= |
P2 |
The volume measurements and T2 all have three
significant figures, so the final answer has three significant
figures.
Checking your answer: The volume increases, so the prediction
is that pressure will decrease. Temperature increased. Since temperature
is proportional to pressure, the pressure should increase. Since
one change predicts one direction and the other change predicts
a different direction, the actual change depends on which effect
is more important. In this case, you just have to do the math
and see. The direction cannot be predicted without it.
>> back
to the Top of the Page
| B. Ideal Gas Law, Equation 8.9 |
The ideal gas law is used when there is one set of conditions and
you are missing one of the four (pressure, P, volume, V,
temperature, T, or moles, n) variables. Like the combined
gas law, a good strategy is to assign the numbers given in the problem
to variables and solve for what is missing.
The ideal gas law is PV = nRT (equation 8.9), where
R is the gas constant. The value of the gas constant depends
on the units used for the four variables. It is traditional to learn
the value of R as 0.08206 L•atm/mol•K. If you
learn the value of R with the units, you know which ones
are appropriate to use in the equation. Some teachers will require
you to memorize a value for R; some will give that information.
The examples shown here will always use R = 0.08206 L•atm/mol•K,
since it is the most commonly used value. Note that this value has
four significant figures. A value given in the text, 0.082058 L•atm/mol•K,
has five significant figures. Either one is appropriate.
>> Example 4
What is the pressure of 0.363 g of N2 in 750.0 mL
at 24 °C?
Solution:
Assigning variables,
n = 0.363 g (It isn't in moles, but it does reflect
amount of substance, so it goes with this variable.)
V = 750.0 mL
T = 24 °C
P = ?
Only one of each of the variables or conditions is not changing.
Consequently, the ideal gas law is used. That means a value for
R is needed. If R = 0.08206 L•atm/mol•K,
the units of each variable must be changed to reflect the units
of R.
| n |
= |
0.363 g N2 |
|
|
= |
0.012955 mol (Only three digits are significant.) |
Note: Gas laws do not depend on the identity of the
gas. This means the law is the same, regardless of the type
of gas. However, gram-to-mole conversions do require knowing
the identity to determine molecular weight.
| V |
= |
750.0 mL |
|
|
= |
0.7500 L |
T = 24 °C + 273.15 = 297 K (Significant figures
depend on decimal places.)
P = ? atm
Using these values in the gas law
P(0.7500 L) = (0.012955 mol)(0.08206 L•atm/mol•K)(297
K)
P = 0.421 atm
The temperature and the moles have three significant figures;
R and V have four. Therefore the answer has three
significant figures.
>> Example 5
What is the molar mass of 0.7314 g of gas in 100.0 mL at STP?
Solution:
Since grams are given in the problem, all that remains is to
find the number of moles, which is conveniently one of the variables
in the gas laws. Since conditions are not changing, that law must
be the ideal gas law.
PV = nRT
P = 1 atm, as defined by STP
| V |
= |
100.0 mL |
|
|
= |
0.1000 L, so we can use the normal value of R |
n = ? moles
R = 0.08206 L•atm/mol•K
T = 273.15 K, including to two decimal places, since
STP is a defined value and using K as you must for all gas laws.
(1 atm)(0.1000 L) = n(0.08206 L•atm/mol•K)(273.15
K)
4.461 x 10–3 mol = n
| molar mass |
= |
|
= |
| 0.7314 g |
|
| 4.461 x 10–3 mol |
|
= |
163.9 g/mol |
Four significant figures, since grams, R, and V
have four. T and P have infinite significant figures.
Notice the units on molar mass. They do not cancel, so g/mol is
appropriate. (Remember, m is the symbol for meter or molality,
not mole.)
>> Example 6
What is the density of F2 gas at 513 torr and 33 °C?
Solution:
Density = grams/volume. There is one set of conditions, so the
ideal gas law might be relevant. PV = nRT. Pressure
and temperature are provided, volume is part of what you want
the answer to be, and grams is amount and therefore is related
to moles.
n = grams/molar mass. Using MM for molar mass and substituting
that into the ideal gas law gives
PV = (g/MM)RT
Solving for g/V gives
Assigning variables, and changing units so that R = 0.08206
L•atm/mol•K can be used
| P |
= |
513 torr |
|
|
= |
0.675 atm |
MM = 19.00(2) = 38.00 g/mol
T = 33 °C + 273.15 = 306 K
Using the variables in the derived equation is
| (0.675 atm)(38.00 g/mol) |
|
| (0.08206 L•atm/mol•K)(306 K) |
|
= |
density |
1.02 g/L = density
Both pressure and temperature limit the number of significant
figures to three.
>> Example 7
What is the molar mass of a gas with a density of 1.39 g/L at
STP?
Solution:
There are two ways to solve this problem. One method is to use
the equation derived in Example 6 and solve for molar mass.
P = 1 atm, T = 273.15 K
| (1)MM |
|
| (0.08206 L•atm/mol•K)(273.15) |
|
= |
1.39 g/L |
MM = 31.2 g/mol
Method 2 would be to recall that the volume of 1 mole of gas
at STP = 22.4 L. By assuming 1 mole of gas at STP, density = molar
mass/22.4 L.
31.1 g/mol = MM
This second method will work only at STP, the first at any temperature
and pressure.
>> Example 8
When calcium carbonate is strongly heated, it will produce calcium
oxide and carbon dioxide gas. If 0.5707 g calcium carbonate completely
reacts, what volume of carbon dioxide will be produced at 99.40
°C and 731.8 torr?
Solution:
This equation involves a reaction. It gives information about
one substance and asks about a different one. This means that
stoichiometric relationships will have to be used.
The question states that the reaction is
CaCO3  CaO + CO2
Writing the equation requires that you recall the naming rules
from Chapter 4. The equation must be balanced to be useful. Fortunately,
it is balanced without having to add any stoichiometric coefficients.
The conditions refer to the gas, which is carbon dioxide. The
gas laws assume that all variables refer to the same substance.
Only one set of conditions is given, so the ideal gas law is used.
Assigning variables and converting units so that the value R = 0.08206 L•atm/mol•K
can be used:
| P |
= |
731.8 torr |
|
|
= |
0.9629 atm |
The atm/torr relationship is exact, so four significant figures
is correct.
V = ? L
n = moles of CO2 (from stochiometry, next
step)
T = 99.40 °C + 273.15 = 372.55 K
The data has two decimal places, so the Kelvin value goes to
two decimal places too.
Moles of CO2 are not given directly but can be determined
from the stoichiometric relationships
| 0.5707 g CaCO3 |
|
|
|
|
= |
0.005702 mol |
Using these values in the ideal gas law
PV = nRT
(0.9629 atm)V = (0.005702 mol)( 0.08206 L•atm/mol•K)(372.55
K)
V = 0.1810 L
Four significant figures, since P, n, and R
have four. Temperature has five significant figures.
>> back
to the Top of the Page
| C. Dalton's Law of Partial Pressures, Equation 8.12 |
Dalton's law is used for mixtures of gases. Different components
of a mixture are measured with either the moles of that gas or the
pressure of the gas. Because gases fill all the available space,
each component occupies the entire volume. The various components
of the mixture will all exist at the same temperature.
If you consider pressure as the force at which the gas molecules
hit the side of container, the pressure (force) due to the individual
components can be separated. The pressure due to one component of
a mixture is called its partial pressure.
Dalton's law states that the total pressure is the sum of the partial
pressures.
>> Example 9
What is the total pressure of a mixture of oxygen, nitrogen,
and helium if their partial pressures are PO2 = 0.057 atm,
PN2 = 0.535 atm, and PHe = 592 torr?
Solution:
The total pressure is the sum of the partial pressures, so in this case
PT = PN2 + PO2
+ PHe (Equation 8.12)
The only thing to be careful of is that the units of pressure
are the same for every pressure involved in the equation. In this
example, the pressure of nitrogen and oxygen are in atmospheres
and the pressure of helium is in torr. Since it is easier to change
one pressure than two
| PHe |
= |
592 torr |
|
|
= |
0.779 atm |
PT = 0.535 atm + 0.057 atm + 0.779 atm =
1.371 atm
Since each of the values has three decimal places, the answer
should have three decimal places.
>> Mole Fractions
Another variation of Dalton's law is PA = XAPT,
where XA is the mole fraction of gas A.
>> Example 10
For a mixture of 0.258 g CO2, 0.935 g Ne, and 0.682
g O2, what is the mole fraction of each component?
Solution:
Since the mole fraction is the ratio of moles of each substance,
each substance must be converted to moles. Remember that molar
mass applies only to the pure substances, not to mixtures.
| 0.258 g CO2 |
|
|
= |
0.00586 mol |
| 0.935 g Ne |
|
|
= |
0.0463 mol |
| 0.682 g O2 |
|
|
= |
0.0213 mol |
total moles = 0.00586 + 0.0463 + 0.0213 = 0.0735 mol (Significant
figures for addition count decimal places.)
| XCO2 |
= |
|
= |
|
= |
0.0797 |
(Mole fraction does not have units. Moles cancel.) |
Note that the sum of the mole fractions must equal 1, so
0.0797 + 0.630 + 0.290 = 1.00
Alternately, the last mole fraction can be calculated by subtracting
from 1
XO2 = 1 – 0.0797 – 0.630 = 0.2903
= 0.290
>> Example 11
If a mixture of 5.08 g N2 and 6.29 g F2
has a total pressure of 881 torr, what is the partial pressure
of each gas?
Solution:
The partial pressure is determined from the total pressure by
PA = XAPT.
In this example PT = 881 torr. The mole fraction
of each gas is also needed.
| 5.08 g N2 |
|
|
= |
0.181 mol N2 |
| 6.29 g F2 |
|
|
= |
0.166 mol F2 |
total moles = 0.181 + 0.166 = 0.347 mol total
| partial pressure of nitrogen |
= |
XN2PT |
= |
|
|
881 torr |
= |
460 torr |
(three significant figures) |
There are two ways to calculate the partial pressure of fluorine.
Since the total pressure is the sum of the partial pressures
| PT |
= |
PN2 + PF2 |
| 881 |
= |
460 + PF2 |
| 421 |
= |
PF2 |
Or the same formula as was used for nitrogen can be used.
| PF2 |
= |
XF2PT |
= |
|
|
881 |
= |
421 torr |
Notice that you may use any units for pressure, provided the
units are the same for the partial pressure and the total pressure.
>> Vapor Pressures
A vapor is a gas formed by molecules escaping from a liquid. When
both the gas and the liquid are present in a closed system, the
partial pressure of the vapor will reach a constant value. This
value depends only on the identity of the vapor and the temperature
of the system. From the gas laws it can be shown the pressure is
proportional to temperature, which is one reason for the increase.
However, a temperature increase will also allow more molecules to
escape from the liquid. The resulting increase in moles will also
cause an increase in pressure. Because the ability of the molecules
to escape depends on the identity of the liquid (vapor), the gas
laws cannot be used to predict vapor pressure. Vapor pressures are
generally determined by consulting a table of experimental values.
Because many gases are collected over water, the gas collected
is actually a mixture of water vapor and the gas collected. Therefore
the vapor pressure of water is of particular interest and values
at several temperatures are listed in Table 8.2.
Dalton's law of partial pressures allows the partial pressure of
water to be separated from the partial pressure of the gas collected.
The gas laws can be used to relate pressure of one gas to moles
of the same gas.
>> Example 12
How many moles of oxygen were produced when 54.9 mL were collected
over water at 746 torr and 30 °C?
Solution:
Since the question gives only one set of conditions, the ideal
gas law is used.
PV = nRT
Since the question asks about moles of oxygen, the partial pressure
of oxygen must be used for the variable P.
Since the oxygen was collected "over water," the gas has both
oxygen and water vapor. Dalton's law of partial pressures can
be used to separate the two.
By consulting Table 8.2, the vapor pressure of water at 30 °C
is 31.8 mmHg = 31.8 torr = PH2O. According to
the problem, the total pressure is 746 torr. Dalton's law says
| PT |
= |
PO2 + PH2O |
| 746 torr |
= |
PO2 + 31.8 torr |
| 714 torr |
= |
PO2 |
Using R = 0.08206 L•atm/mol•K, the other variables
are
| V |
= |
54.9 mL |
|
|
= |
0.0549 L |
T = 30 °C + 273.15 = 303 K
For n to represent moles of oxygen, the pressure must
be the partial pressure of oxygen
| P |
= |
714 torr |
|
|
= |
0.940 atm |
Using these values in the ideal gas law results in
PV = nRT
(0.949 atm)(0.0549 L) = n(0.08206 L•atm/mol•K)(303
K)
0.00208 mol = n = moles oxygen
>> Example 12
Lithium metal was added to water and the resulting hydrogen gas
collected. If 86.0 mL of hydrogen gas was collected over the water
at 25 °C and 1.04 atm, how many grams of lithium reacted?
Li + H2O LiOH + H2
Solution:
Hydrogen gas was collected and the question was asked about lithium.
Different substances are related by balanced chemical reactions,
so balance the reaction
2 Li + 2 H2O 2 LiOH + H2
Chemical reactions relate substance in moles. Since you are given
information about hydrogen, the moles of hydrogen need to be determined.
The information about hydrogen is volume, temperature, and pressure,
so the ideal gas law can be used to get information about moles.
To use the ideal gas law to obtain moles of hydrogen, the pressure
must be partial pressure of hydrogen. Since the gas was "collected
over water," the pressure given must be the pressure of both hydrogen
and water vapor.
PT = PH2 + PH2O
The vapor pressure of water at 25 °C is 23.8 mmHg, according
to Table 8.2. The total pressure is 1.04 atm, according to the
problem. To use Dalton's law, the pressure units must be the same.
Since we need to wind up in units of atm for the ideal gas law,
converting mmHg to atm is probably a good strategy, but it would
work equally well to convert atm to mmHg (or torr).
| 23.8 mmHg |
|
|
= |
0.0313 atm |
= |
PH2O |
PH2 = PT – PH2O
= 1.04 – 0.0313 = 1.01 atm
Assigning values to the other variables and converting to the
appropriate units for
R = 0.08206 L•atm/mol•K
T = 25 °C + 273.15 = 298 K
| V |
= |
86.0 mL |
|
|
= |
0.0860 L |
PH2V = nH2RT
(1.01 atm)(0.0860 L) = nH2(0.08206 L•atm/mol•K)(298
K)
0.00355 mol = nH2
The stochiometric method from Chapter
4 is then used to convert to grams of lithium.
- Balance the equation.
2 Li + 2 H2O
2 LiOH + H2
- Convert given unit to moles.
0.00355 mol H2 is already in units of moles
- Convert moles of substance given to moles of substance desired.
| 0.00355 mol H2 |
|
|
- Convert moles to unit desired.
| 0.00355 mol H2 |
|
|
|
|
- Calculate and round to appropriate significant figures.
| 0.00355 mol H2 |
|
|
|
|
= |
0.0493 g Li |
>> back
to the Top of the Page
| D. Henry's Law, Equation 8.14 |
Solubility, amount of gas dissolved in a liquid, is related to
pressure by Henry's law. The proportionality constant, kH,
depends on the identity of the gas and solvent and the temperature.
A list of constants for water at 20 °C is in Table 8.3.
C = kHP (Equation
8.14)
>> Example 14
How many grams of helium can dissolve in 50.00 mL water at 20
°C and 770 torr?
Solution:
There are two values for Henry's constant, depending on what
unit of pressure is used. Since the pressure units are torr, the
value 5.1 x 10–7 mol/kg•mmHg is appropriate.
Recall that 1 mmHg is 1 torr. (The units are also useful for remembering
the order of pressure and concentration, since the units must
cancel.)
C = (5.1 x 10–7 mol/ kg•mmHg)(770
mmHg)
C = 3.9 x 10–4 mol/kg
Since this is concentration, the kg must refer to the amount
of solvent, using the density of water as 1.00 g/mL.
| 50.00 mL |
|
|
|
|
= |
0.0500 kg |
To get the amount of helium,
|
|
|
0.0500 kg |
= |
1.96 x 10–5 mol He |
|
|
= |
7.9 x 10–5 g |
>> Example 15
What pressure is required to dissolve 0.0045 g O2
in 1.00 L water at 20 °C?
Solution:
So little gas dissolves in water that it doesn't matter whether
it is liters of solvent or liters of solution. The concentration
is
| 0.0045 g |
|
|
= |
|
= |
1.4 x 10–4 mol/L |
Henry's constant for oxygen is 1.3 x 10–3 mol/L•atm, so
1.4 x 10–4 = (1.3 x 10–3) P
0.11 atm = P
>> back
to the Top of the Page
| E. Graham's Law, Equation 8.18 |
The rate of effusion is related to molar mass. The equation compares
two gases and assumes that gases are at the same temperature.
|
|
= |
( |
| molar mass y |
|
| molar mass x |
|
)1/2 |
(Equation 8.18)
>> Example 16
If hydrogen (H2) effuses at a rate of 1.5 m/s, at
what rate does carbon dioxide effuse?
Solution:
The molar mass of hydrogen is 2.02 g/mol. The molar mass of CO2
is 44.01 g/mol. If you assign the lighter (or faster) molecule
as x, you won't have to work with fractions.
1.5 = 4.67rate
0.32 m/s = rate
>> Example 17
What is the molar mass of a molecule if it effuses at one-eighth
the rate of helium?
Solution:
The molar mass of helium is 4.00 g/mol.
The problem says rate y = 1/8 rate x, so
| 64 |
= |
|
| 256 g/mol |
= |
molar mass y |
>> back
to the Top of the Page
| F. Van der Waals Law, Equation 8.19 |
The van der Waals law is applied to real gases, when more exact
numbers than those obtained from the ideal gas law are required.
(The ideal gas law will give an approximate value.) The values a
and b depend on the identity of the gas and can be obtained
from Table 8.4. The value a takes into account the attraction
of molecules to each other and the value b reflects the amount of
space taken up by the molecule.
|
|
|
(V – nb) |
= |
nRT |
(Equation 8.19)
>> Example 18
What is the pressure of 3.45 g of nitrogen in a 10.00-mL container
if it acts as a real gas at –15 °C.
Solution:
Since nitrogen is behaving as a real gas, the van der Waals equation
is used. If 0.082058 L•atm/mol•K is used for R,
the other variables are
P = unknown (what we're looking for)
| n |
= |
3.45 g |
|
|
= |
0.123 mol |
| V |
= |
10.00 mL |
|
|
= |
0.01000 L |
T = –15 °C + 273.15 = 258 K
a = 1.39
b = 0.0391
| P + (0.123)2(1.39) |
|
| (0.01)2 |
|
|
(0.01 – (0.123)(0.0392)) |
= |
(0.123)(0.082058)(258) |
(P + 210.3)(0.00518) = 2.60
P + 210.3 = 502.7
P = 292 atm
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the other Key Equations and Concepts in this chapter
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