4 Avogadro’s Law—How Volume and Moles Are Related
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AVOGADRO’S LAW—HOW VOLUME AND MOLES ARE RELATED
203
As the number of moles of a gas increases, its volume increases as well. Thus, dividing the
volume by the number of moles is a constant (k). The value of k is the same regardless of the
identity of the gas.
When the number of moles increases...
∝
V
n
...the volume increases.
V
n
=
k
Thus, if the pressure and temperature of a system are held constant, increasing the number of
moles increases the volume of a gas.
Increase the
number of moles.
V=1 L
V=2 L
Volume increases.
n = 1 mol
n = 2 mol
Since dividing the volume of a gas by the number of moles is a constant, knowing the volume and
number of moles initially (V1 and n1) means we can calculate a new volume or number of moles
(V2 and n2) when one of these quantities is changed.
V1
V2
=
n1
initial conditions
n2
new conditions
To solve a problem of this sort, we follow the same three steps listed in the How To outlined in
Section 7.3A, using Avogadro’s law in step [2].
SAMPLE PROBLEM 7.6
The lungs of an average male hold 0.25 mol of air in a volume of 5.8 L. How many moles of air
do the lungs of an average female hold if the volume is 4.6 L?
ANALYSIS
This question deals with volume and number of moles, so Avogadro’s law is used to determine
a final number of moles when three quantities are known—the initial volume and number of
moles (V1 and n1), and the final volume (V2).
SOLUTION
[1]
Identify the known quantities and the desired quantity.
V1 = 5.8 L
V2 = 4.6 L
n1 = 0.25 mol
known quantities
[2]
n2 = ?
desired quantity
Write the equation and rearrange it to isolate the desired quantity, n2, on one side.
• Use Avogadro’s law. To solve for n2, we must invert the numerator and denominator on
both sides of the equation, and then multiply by V2.
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204
GASES, LIQUIDS, AND SOLIDS
V1 V2
=
n1
n2
n1
V1
Switch V and n
on both sides.
n1 V2
V1
[3]
n2
=
=
Solve for n2 by multiplying both sides by V2.
V2
n2
Solve the problem.
• Substitute the three known quantities into the equation and solve for n2.
n2
=
n1V2
=
V1
(0.25 mol)(4.6 L)
(5.8 L)
=
0.20 mol
Liters cancel. Answer
PROBLEM 7.17
A balloon that contains 0.30 mol of helium in a volume of 6.4 L develops a leak so that its
volume decreases to 3.85 L at constant temperature and pressure. How many moles of helium
does the balloon now contain?
PROBLEM 7.18
A sample of nitrogen gas contains 5.0 mol in a volume of 3.5 L. Calculate the new volume
of the container if the pressure and temperature are kept constant but the number of moles of
nitrogen is changed to each of the following values: (a) 2.5 mol; (b) 3.65 mol; (c) 21.5 mol.
Avogadro’s law allows us to compare the amounts of any two gases by comparing their volumes.
Often amounts of gas are compared at a set of standard conditions of temperature and pressure, abbreviated as STP.
• STP conditions are: 1 atm (760 mm Hg) for pressure
273 K (0 °C) for temperature
• At STP, one mole of any gas has the same volume, 22.4 L, called the standard molar
volume.
Under STP conditions, one mole of nitrogen gas and one mole of helium gas each contain
6.02 × 1023 molecules of gas and occupy a volume of 22.4 L at 0 °C and 1 atm pressure. Since
the molar masses of nitrogen and helium are different (28.0 g for N2 compared to 4.0 g for
He), one mole of each substance has a different mass.
same volume
same number of particles
1 mol N2
1 mol He
22.4 L
6.02 × 1023 particles
28.0 g
22.4 L
6.02 × 1023 particles
4.0 g
The standard molar volume can be used to set up conversion factors that relate the volume and
number of moles of a gas at STP, as shown in the following stepwise procedure.
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AVOGADRO’S LAW—HOW VOLUME AND MOLES ARE RELATED
HOW TO
EXAMPLE
Step [1]
Step [2]
205
Convert Moles of Gas to Volume at STP
How many moles are contained in 2.0 L of N2 at standard temperature and pressure?
Identify the known quantities and the desired quantity.
2.0 L of N2
? moles of N2
original quantity
desired quantity
Write out the conversion factors.
• Set up conversion factors that relate the number of moles of a gas to volume at STP. Choose the conversion
factor that places the unwanted unit, liters, in the denominator so that the units cancel.
22.4 L
1 mol
Step [3]
1 mol
22.4 L
or
Choose this conversion
factor to cancel L.
Solve the problem.
• Multiply the original quantity by the conversion factor to obtain the desired quantity.
2.0 L
×
1 mol
22.4 L
=
0.089 mol of N2
Answer
Liters cancel.
By using the molar mass of a gas, we can determine the volume of a gas from a given number of
grams, as shown in Sample Problem 7.7.
SAMPLE PROBLEM 7.7
Burning 1 mol of propane in a gas grill adds 132.0 g of carbon dioxide (CO2) to the
atmosphere. What volume of CO2 does this correspond to at STP?
ANALYSIS
To solve this problem, we must convert the number of grams of CO2 to moles using the molar
mass. The number of moles of CO2 can then be converted to its volume using a mole–volume
conversion factor (1 mol/22.4 L).
SOLUTION
[1]
[2]
Identify the known quantities and the desired quantity.
132.0 g CO2
? L CO2
known quantity
desired quantity
Convert the number of grams of CO2 to the number of moles of CO2 using the molar mass.
molar mass
conversion factor
132.0 g CO2
×
1 mol CO2
44.0 g CO2
=
3.00 mol CO2
Grams cancel.
[3]
Convert the number of moles of CO2 to the volume of CO2 using a mole–volume
conversion factor.
mole–volume
conversion factor
3.00 mol CO2
×
22.4 L
Moles cancel.
smi26573_ch07.indd 205
1 mol
=
67.2 L CO2
Answer
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206
GASES, LIQUIDS, AND SOLIDS
PROBLEM 7.19
How many liters does each of the following quantities of O2 occupy at STP: (a) 4.5 mol;
(b) 0.35 mol; (c) 18.0 g?
PROBLEM 7.20
How many moles are contained in the following volumes of air at STP: (a) 1.5 L; (b) 8.5 L;
(c) 25 mL?
7.5 THE IDEAL GAS LAW
All four properties of gases—pressure, volume, temperature, and number of moles—can be combined into a single equation called the ideal gas law. The product of pressure and volume divided
by the product of moles and Kelvin temperature is a constant, called the universal gas constant
and symbolized by R.
PV
nT
=
R
universal gas constant
More often the equation is rearranged and written in the following way:
PV = nRT
For atm:
R
=
0.0821
For mm Hg:
R
=
62.4
Ideal gas law
L atm
mol K
L mm Hg
mol K
The value of the universal gas constant R depends on its units. The two most common values of R
are given using atmospheres or mm Hg for pressure, liters for volume, and kelvins for temperature.
Be careful to use the correct value of R for the pressure units in the problem you are solving.
The ideal gas law can be used to find any value—P, V, n, or T—as long as three of the quantities are
known. Solving a problem using the ideal gas law is shown in the stepwise How To procedure and
in Sample Problem 7.8. Although the ideal gas law gives exact answers only for a perfectly “ideal”
gas, it gives a good approximation for most real gases, such as the oxygen and carbon dioxide in
breathing, as well (Figure 7.6).
HOW TO
Carry Out Calculations with the Ideal Gas Law
EXAMPLE How many moles of gas are contained in a typical human breath that takes in 0.50 L of air at 1.0 atm pressure
and 37 °C?
Step [1]
Identify the known quantities and the desired quantity.
P = 1.0 atm
V = 0.50 L
T = 37 °C
known quantities
Step [2]
n = ? mol
desired quantity
Convert all values to proper units and choose the value of R that contains these units.
• Convert °C to K. K = °C + 273 = 37 °C + 273 = 310. K
• Use the value of R in atm since the pressure is given in atm; that is, R = 0.0821 L · atm/mol · K.
Step [3]
Write the equation and rearrange it to isolate the desired quantity on one side.
• Use the ideal gas law and solve for n by dividing both sides by RT.
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PV
=
nRT
PV
RT
=
n
Solve for n by dividing both sides by RT.
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THE IDEAL GAS LAW
207
How To, continued. . .
Step [4]
Solve the problem.
• Substitute the known quantities into the equation and solve for n.
n
▼
FIGURE 7.6
=
PV
RT
=
(1.0 atm)(0.50 L)
(
0.0821
L · atm
mol · K
)
=
0.0196 rounded to 0.020 mol
(310. K)
Answer
Focus on the Human Body: The Lungs
average lung
capacity—4–6 L
trachea
average tidal
volume—0.5 L
right lung
with its
three lobes
left lung
with its
two lobes
heart
pulmonary
artery
pulmonary vein
• Humans have two lungs that contain
a vast system of air passages,
allowing gases to be exchanged
between the atmosphere and with
the bloodstream. The lungs contain
about 1,500 miles of airways that
have a total surface area about the
size of a tennis court.
• Lungs are in a sense “overbuilt,” in
that their total air volume is large
compared to the tidal volume, the
amount of air taken in or expelled
with each breath. This large reserve
explains why people can smoke for
years without noticing any significant
change in normal breathing.
• In individuals with asthma, small
airways are constricted and
inflamed, making it difficult to breathe.
alveolus
section of
alveoli
cut open
Blood in pulmonary arteries gives
up waste CO2 to the lungs so that
it can be expelled to the air.
Blood in pulmonary veins picks up
O2 in the lungs so that it can be
pumped by the heart to the body.
SAMPLE PROBLEM 7.8
ANALYSIS
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If a person exhales 25.0 g of CO2 in an hour, what volume does this amount occupy at 1.00 atm
and 37 °C?
Use the ideal gas law to calculate V, since P and T are known and n can be determined by using
the molar mass of CO2 (44.0 g/mol).
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208
GASES, LIQUIDS, AND SOLIDS
SOLUTION
[1]
Identify the known quantities and the desired quantity.
P = 1.00 atm
T = 37 °C
25.0 g CO2
V=?L
known quantities
[2]
desired quantity
Convert all values to proper units and choose the value of R that contains these units.
• Convert °C to K. K = °C + 273 = 37 °C + 273 = 310. K
• Use the value of R with atm since the pressure is given in atm; that is,
R = 0.0821 L · atm/mol · K.
• Convert the number of grams of CO2 to the number of moles of CO2 using the molar
mass (44.0 g/mol).
molar mass
conversion factor
25.0 g CO2
×
1 mol CO2
44.0 g CO2
=
0.568 mol CO2
Grams cancel.
[3]
Write the equation and rearrange it to isolate the desired quantity, V, on one side.
• Use the ideal gas law and solve for V by dividing both sides by P.
[4]
PV
=
nRT
V
=
nRT
P
Solve for V by dividing both sides by P.
Solve the problem.
• Substitute the three known quantities into the equation and solve for V.
V
=
nRT
P
(
(0.568 mol) 0.0821
=
L · atm
mol · K
)
(310. K)
1.0 atm
=
14.5 L
Answer
PROBLEM 7.21
How many moles of oxygen (O2) are contained in a 5.0-L cylinder that has a pressure of
175 atm and a temperature of 20. °C?
PROBLEM 7.22
Determine the pressure of N2 under each of the following conditions.
a. 0.45 mol at 25 °C in 10.0 L
PROBLEM 7.23
b. 10.0 g at 20. °C in 5.0 L
Determine the volume of 8.50 g of He gas at 25 °C and 750 mm Hg.
7.6 DALTON’S LAW AND PARTIAL PRESSURES
Since the partial pressure of O2
is low at very high altitudes, most
mountain climbers use supplemental
O2 tanks above about 24,000 ft.
smi26573_ch07.indd 208
Since gas particles are very far apart compared to the size of an individual particle, gas particles
behave independently. As a result, the identity of the components of a gas mixture does not
matter, and a mixture of gases behaves like a pure gas. Each component of a gas mixture is said
to exert a pressure called its partial pressure. Dalton’s law describes the relationship between
the partial pressures of the components and the total pressure of a gas mixture.
• Dalton’s law: The total pressure (Ptotal) of a gas mixture is the sum of the partial
pressures of its component gases.
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DALTON’S LAW AND PARTIAL PRESSURES
209
Thus, if a mixture has three gases (A, B, and C) with partial pressures PA, PB, and PC, respectively, the total pressure of the system (Ptotal) is the sum of the three partial pressures. The partial
pressure of a component of a mixture is the same pressure that the gas would exert if it were a
pure gas.
Ptotal
total pressure
SAMPLE PROBLEM 7.9
ANALYSIS
SOLUTION
=
PA
+
PB
+
PC
partial pressures of A, B, and C
A sample of exhaled air from the lungs contains four gases with the following partial pressures:
N2 (563 mm Hg), O2 (118 mm Hg), CO2 (30 mm Hg), and H2O (50 mm Hg). What is the total
pressure of the sample?
Using Dalton’s law, the total pressure is the sum of the partial pressures.
Adding up the four partial pressures gives the total:
563 + 118 + 30 + 50 = 761 mm Hg (total pressure)
PROBLEM 7.24
CO2 was added to a cylinder containing 2.5 atm of O2 to give a total pressure of 4.0 atm of gas.
What is the partial pressure of O2 and CO2 in the final mixture?
We can also calculate the partial pressure of each gas in a mixture if two quantities are known—
[1] the total pressure and [2] the percent of each component—as shown in Sample Problem 7.10.
SAMPLE PROBLEM 7.10
ANALYSIS
Air is a mixture of 21% O2, 78% N2, and 1% argon by volume. What is the partial pressure of
each gas at sea level, where the total pressure is 760 mm Hg?
Convert each percent to a decimal by moving the decimal point two places to the left. Multiply
each decimal by the total pressure to obtain the partial pressure for each component.
SOLUTION
Fraction O2: 21% = 0.21
Fraction N2: 78% = 0.78
Fraction Ar: 1% = 0.01
PROBLEM 7.25
HEALTH NOTE
Partial pressure
0.21 × 760 mm Hg = 160 mm Hg (O2)
0.78 × 760 mm Hg = 590 mm Hg (N2)
0.01 × 760 mm Hg =
8 mm Hg (Ar)
758 rounded to 760 mm Hg
A sample of natural gas at 750 mm Hg contains 85% methane, 10% ethane, and 5% propane.
What are the partial pressures of each gas in this mixture?
The composition of the atmosphere does not change with location, even though the total atmospheric pressure decreases with increasing altitude. At high altitudes, therefore, the partial pressure of oxygen is much lower than it is at sea level, making breathing difficult. This is why mountain climbers use supplemental oxygen at altitudes above 8,000 meters.
In contrast, a hyperbaric chamber is a device that maintains air pressure two to three times higher
than normal. Hyperbaric chambers have many uses. At this higher pressure the partial pressure of
O2 is higher. For burn patients, the higher pressure of O2 increases the amount of O2 in the blood,
where it can be used by the body for reactions that fight infections.
The high pressures of a hyperbaric
chamber can be used to treat
patients fighting infections and scuba
divers suffering from the bends.
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When a scuba diver surfaces too quickly, the N2 dissolved in the blood can form microscopic
bubbles that cause pain in joints and can occlude small blood vessels, causing organ injury. This
condition, called the bends, is treated by placing a diver in a hyperbaric chamber, where the
elevated pressure decreases the size of the N2 bubbles, which are then eliminated as N2 gas from
the lungs as the pressure is slowly decreased.
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