Download Chapter 10: Gases

10
Gases
Unit Outline
10.1
10.2
10.3
10.4
10.5
Properties of Gases
Historical Gas Laws
The Combined and Ideal Gas Laws
Partial Pressure and Gas Law Stoichiometry
Kinetic Molecular Theory
In This Unit…
AlbertSmirnov/iStockphoto.com
Matter exists in three main physical states under conditions we encounter in everyday life: gaseous, liquid, and solid. Of these, the most fluid
and easily changed is the gaseous state. Gases differ significantly from
liquids and solids in that both liquids and solids are condensed states
with molecules packed close to one another, whereas gases have molecules spaced far apart. This unit examines the bulk properties of gases
and the molecular scale interpretation of those properties.
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10.1 Properties of Gases
10.1a Overview of Properties of Gases
Gases are one of the three major states of matter. The physical properties of gases can be
manipulated and measured more easily than those of solids or liquids. Because of this, the
mathematical relationships between different gas properties were among the first quantitative aspects of chemistry to be studied.
In general, gases differ from liquids and solids more than they differ from each other.
Solid
Liquid
Gas
Density
High
High
Low
Compressible
No
No
Yes
Fluid
No
Yes
Yes
The most striking property of gases is the simple relationship between the pressure,
volume, and temperature of a gas and how a change in one of these properties affects the
other properties (Interactive Figure 10.1.1).
Interactive Figure 10.1.1
Charles D. Winters
Explore the properties of gases.
As the gaseous water vapor inside this can condenses to a liquid,
the pressure inside the can drops and the can is crushed by the
greater external pressure.
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These same simple relationships do not exist for solids or liquids. The major properties
of gases are given in Table 10.1.1.
Table 10.1.1 Properties of Gases and Their Units
Property
Common Unit
Other Units
Symbol
Mass
grams, g
kg (SI unit), mg
—
Amount
mole, mol (SI unit)
—
n
Volume
liters, L
mL, m (SI unit)
V
Pressure
atmosphere, atm
Pa (SI unit), kPa, bar, mm Hg, psi
P
Temperature
kelvin, K (SI unit)
ºC, ºF
T
3
10.1b Pressure
Gases exert pressure on surfaces, measured as a force exerted on a given area of surface.
force
area
A confined gas exerts pressure on the interior walls of the container holding it and the gases
in our atmosphere exert pressure on every surface with which they come in contact.
Gas pressure is commonly measured using a barometer. The first barometers consisted
of a long, narrow tube that was sealed at one end, filled with liquid mercury, and then inverted
into a pool of mercury (Figure 10.1.2a). The gases in the atmosphere push down on the
mercury in the pool and balance the weight of the mercury column in the tube. The higher
the atmospheric pressure, the higher the column of mercury in the tube. The height of the
mercury column, when measured in millimeters, gives the atmospheric pressure in units of
millimeters of mercury (mm Hg). The torr is a unit of pressure (1 torr 5 1 mm Hg) named
in honor of the inventor of the barometer, Evangelista Torricelli (1608–1647).
Pressure of a gas sample in the laboratory is measured with a manometer, which is
shown in Figure 10.1.2b. In this case, mercury is added to a U-shaped tube. One end of the
tube is connected to the gas sample under study. The other is open to the atmosphere. If
the pressure of the gas sample is equal to atmospheric pressure, the height of the mercury
is the same on both sides of the tube. In Figure 10.1.2b, the pressure of the gas is greater
than atmospheric pressure by “h” mm Hg.
Gas pressure is expressed in different units. The SI unit for pressure is the pascal, Pa,
which is equal to the force in newtons exerted on 1 square meter (1 Pa 5 1 N/m2). The
pressure 5
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Patm
Vacuum
Atmospheric
pressure
Height of
mercury
column (mm)
h
Gas
(a)
(b)
© 2013 Cengage Learning
Mercury, Hg
Figure 10.1.2 Measuring pressure using (a) a barometer
and (b) a manometer
English pressure unit, pounds per square inch (psi), is a measure of how many pounds of
force a gas exerts on 1 square inch of a surface. Atmospheric pressure at sea level is approximately 14.7 psi. This means that an 8½″ 3 11″ piece of paper has a total force on it of
more than 1370 pounds. Commonly used pressure units are given in Table 10.1.2. Early
pressure units were based on pressure measurements at sea level, where on average
the mercury column has a height of 760 mm. This measurement was used to define the
standard atmosphere (1 atm 5 760 mm Hg). Modern scientific studies generally use gas
pressure units of atm, kPa, bar (1 atm 5 1.013 bar), and mm Hg.
Table 10.1.2 Common Units of Gas Pressure
1 atm
5 1.013 bar
(bar)
5 101.3 kPa
(kilopascal)
5 760 mm Hg
(millimeters of mercury)
5 760 torr
(torr)
5 14.7 psi
(pounds per square inch)
Example Problem 10.1.1 Convert between pressure units.
A gas sample has a pressure of 722 mm Hg. What is this pressure in atmospheres?
Solution:
You are asked to convert between pressure units.
You are given a pressure value.
Use the relationship 1 atm 5 760 mm Hg to convert between these pressure units.
722 mm Hg 3
1 atm
5 0.950 atm
760 mm Hg
Is your answer reasonable? The pressure, 722 mm Hg, is less than 760 mm Hg, which is
the equivalent of 1 atm. Therefore, the pressure expressed in units of atmospheres should be
less than 1 atm.
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Video Solution
Tutored Practice
Problem 10.1.1
Section 10.1 Mastery
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10.2 Historical Gas Laws
Interactive Figure 10.2.1
Explore Boyle’s law.
10.2a Boyle’s Law: P 3 V 5 kB
500
Volume, V (mL)
400
1
or P 3 V 5 kB
volume ~
pressure
Because the product of gas pressure and volume is a constant (when temperature and
amount of gas are held constant), it is possible to calculate the new pressure or volume of
a gas sample when one of the properties is changed.
P1V15 P2V2
At low pressure, volume is large.
300
200
At high pressure,
volume is small.
100
0
(10.1)
1.0
2.0
3.0
Pressure, P (atm)
4.0
5.0
© 2013 Cengage Learning
Boyle’s law states that the pressure and volume of a gas sample are inversely related
when the amount of gas and temperature are held constant. For example, consider a syringe that is filled with a sample of a gas and attached to a pressure gauge and a thermostat (used to keep the system at a constant temperature). When the plunger is depressed
(decreasing the volume of the gas sample), the pressure of the gas increases (Interactive
Figure 10.2.1).
When the pressure on the syringe is low, the gas sample has a large volume. When the
pressure is high, the gas is compressed and the volume is smaller. The relationship between
pressure and volume is therefore an inverse one:
Volume changes upon applying pressure to a gas-filled
syringe. Temperature is constant and no gas escapes
from the syringe.
The subscripts “1” and “2” in Equation 10.1 indicate the different experimental conditions
before and after pressure or volume is changed.
Example Problem 10.2.1 Use Boyle’s law to calculate volume.
A sample of gas has a volume of 458 mL at a pressure of 0.970 atm. The gas is compressed
and now has a pressure of 3.20 atm. Predict if the new volume is greater or less than the
initial volume, and calculate the new volume. Assume temperature is constant and no gas
escaped from the container.
Solution:
You are asked to calculate the volume of a gas sample when only the pressure is changed.
You are given the original volume and pressure and the new pressure of the gas sample.
According to Boyle’s law, pressure and volume are inversely related when the temperature
and amount of gas are held constant. In this case, the pressure increases from 0.970 atm to
3.20 atm, so the new volume should decrease. It will be less than the original volume.
c
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b
Example Problem 10.2.1 (continued)
To calculate the new volume, first make a table of the known and unknown pressure and
volume data. In this case, the initial volume and pressure and the new pressure are known
and the new volume must be calculated.
P1 5 0.970 atm
P2 5 3.20 atm
V1 5 458 mL
V2 5 ?
Rearrange Boyle’s law to solve for V2 and calculate the new volume of the gas sample.
P1V15 P2V2
Video Solution
10.970 atm2 1458 mL2
PV
V2 5 1 1 5
5 139 mL
P2
3.20 atm
Tutored Practice
Problem 10.2.1
The pressure units (atm) cancel, leaving volume in units of milliliters.
Is your answer reasonable? The final volume is less than the initial volume, as we
predicted.
Interactive Figure 10.2.2
Explore Charles’s law.
10.2b Charles’s Law: V 5 kC 3 T
50
At high temperature, volume is large.
40
Gas volume (L)
Charles’s law states that the temperature and volume of a gas sample are directly related
when the pressure and the amount of gas are held constant. For example, heating the air
in a hot air balloon causes it to expand, filling the balloon. Consider a sample of gas held
in a syringe attached to a pressure gauge and a temperature control unit (Interactive
Figure 10.2.2).
When the pressure and amount of gas are held constant, decreasing the temperature of
the gas sample decreases the volume of the gas. The two properties are directly related.
30
20
Hydrogen (H2)
At low temperature,
volume is small.
V 5 kC 3 T
Because the ratio of gas volume and temperature (in kelvin units) is a constant (when
pressure and amount of gas are held constant), it is possible to calculate the new volume
or temperature of a gas sample when one of the properties is changed.
V1
V
5 2
T1
T2
(10.2)
The subscripts “1” and “2” in Equation 10.2 indicate the different experimental conditions
before and after volume or temperature is changed.
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–300
–200
–100
0
100
Temperature (°C)
200
300
© 2013 Cengage Learning
Oxygen (O2)
10
Volume changes upon changing the temperature of a
gas sample. The pressure is held constant, and no gas
escapes the syringe.
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As shown in Interactive Figure 10.2.2, extending a volume–temperature plot for any
gas to the point at which the gas volume is equal to zero shows that this occurs at a temperature of 2273.15 ºC. This temperature is known as absolute zero, or 0 K.
Example Problem 10.2.2 Use Charles’s law to calculate volume.
A sample of gas has a volume of 2.48 L at a temperature of 58.0 ºC. The gas sample is cooled to a
temperature of 25.00 ºC (assume pressure and amount of gas are held constant). Predict whether
the new volume is greater or less than the original volume, and calculate the new volume.
Solution:
You are asked to calculate the volume of a gas sample when only the temperature is changed.
You are given the original volume and temperature and the new temperature of the gas sample.
According to Charles’s law, temperature and volume are directly related when the pressure
and amount of gas are held constant. In this case, the temperature decreases from 58.00 ºC to
25.00 ºC, so the volume should also decrease. It will be less than the original volume.
To calculate the new volume, first make a table of the known and unknown volume and temperature data. In this case, the initial volume and temperature and the new temperature are
known and the new volume must be calculated. Note that all temperature data must be in
kelvin temperature units.
V1 5 2.48 L
V2 5 ?
T1 5 58.00 ºC 1 273.15 5 331.15 K
T2 5 25.00 ºC 1 273.15 5 268.15 K
Rearrange Charles’s law to solve for V2, and calculate the new volume of the gas sample.
V1
V
5 2
T1
T2
V2 5
12.48 L2 1268.15 K2
V1T2
5
5 2.01 L
T1
331.15 K
The temperature units (K) cancel, leaving volume in units of liters.
Is your answer reasonable? The final volume is less than the initial volume, as we predicted.
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Video Solution
Tutored Practice
Problem 10.2.2
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10.2c Avogadro’s Law: V 5 kA 3 n
3
Avogadro’s law can also be used to calculate the new volume or amount of a gas sample
when one of the properties is changed.
V1
V
5 2
(10.3)
n1
n2
The subscripts “1” and “2” in Equation 10.3 indicate the different experimental conditions
before and after volume or the amount of gas is changed.
Avogadro’s law is independent of the identity of the gas, as shown in a plot of volume
versus amount of gas (Interactive Figure 10.2.4a). This means that a 1-mol sample of Xe
2
1.5
1
With no gas in the syringe,
the volume is zero.
0
0.1
0.2
Amount of gas (mol)
0.3
© 2013 Cengage Learning
V 5 kA 3 n
With a large amount of
N2 in the syringe, the
volume is over 2 L.
2.5
Gas volume (L)
Avogadro’s hypothesis states that equal volumes of gases have the same number of particles
when they are at the same temperature and pressure. One aspect of the hypothesis is called
Avogadro’s law, which states that the volume and amount (in moles) of a gas are directly
related when pressure and temperature are held constant. Consider an experiment where
the volume of gas in a syringe is measured as a function of the amount of gas in the syringe
(at constant pressure and temperature) (Figure 10.2.3).
As the amount of gas in the syringe is increased (at constant temperature and pressure), the volume of the gas increases.
Figure 10.2.3 Plot of volume of samples of N2
gas with differing amounts of N2 present
Interactive Figure 10.2.4
2
1
0
(a)
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0.25 0.50 0.75 1.0
Amount of gas (mol)
1.25
3
Ar (39.9 g/mol)
2
1
0
Xe (131.1 g/mol)
100
Mass (mg)
200
© 2013 Cengage Learning
3
N2 (28.0 g/mol)
The slope of the line
is the same for
N2, Ar and Xe
Gas volume (L)
Gas volume (L)
Explore Avogadro’s law.
(b)
(a) Avogadro’s law and (b) volume versus mass plots for N2, Ar, and Xe
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has the same volume as a 1-mol sample of N2 (at the same temperature and pressure),
even though the Xe sample has a mass more than 4.5 times as great. The same is not true
for gas samples with equal mass, as shown in Interactive Figure 10.2.4b.
Example Problem 10.2.3 Use Avogadro’s law to calculate volume.
A sample of gas contains 2.4 mol of SO2 and 1.2 mol O2 and occupies a volume of 17.9 L. The
following reaction takes place: 2 SO2(g) 1 O2(g) S 2 SO3(g)
Calculate the volume of the sample after the reaction takes place (assume temperature and
pressure are constant).
Solution:
You are asked to calculate the volume of a gas sample when the number of moles of gas
changes.
You are given the original volume and amount of the gases in the sample and the balanced
equation for the reaction that takes place.
Make a table of the known and unknown volume and temperature data. In this case, the initial
volume and amount of reactants and the amounts of product are known and the new volume
must be calculated.
V1 5 17.9 L
V2 5 ?
n1 5 3.6 mol SO2 and O2
n2 5 2.4 mol SO3
The reactants are present in a 2:1 stoichiometric ratio, so they are consumed completely upon
reaction to form SO3.
2.4 mol SO2
2 mol SO2
5
1.2 mol O2
1 mol O2
Use the balanced equation to calculate the amount of SO3 produced.
2.4 mol SO2 3
2 mol SO3
5 2.4 mol SO3
2 mol SO2
Rearrange Avogadro’s law to solve for V2, and calculate the volume of the gas sample after the
reaction is complete.
V1
V
5 2
n1
n2
117.9 L2 12.4 mol2
Vn
V2 5 1 2 5
5 12 L
n1
3.6 mol
Is your answer reasonable? The new volume is smaller than the initial volume, which
makes sense because the amount of gas decreased as a result of the chemical reaction.
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Video Solution
Tutored Practice
Problem 10.2.3
Section 10.2 Mastery
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10.3 The Combined and Ideal Gas Laws
10.3a The Combined Gas Law
We can rewrite the three historical gas laws, solving for volume:
Boyle’s Law
V 5 kB 3
1
P
Charles’s Law
Avogadro’s Law
V 5 kC 3 T
V 5 kA 3 n
These three equations can be combined into a single equation that relates volume,
pressure, temperature, and the amount of any gas.
V 5 constant 3
nT
P
or
PV
5 constant
nT
Because the ratio involving pressure, volume, amount, and temperature of a gas is a
constant, it can be used in the form of the combined gas law to calculate the new pressure,
volume, amount, or temperature of a gas when one or more of these properties is changed.
P1V1
P2V2
5
n1T1
n2T2
(10.4)
The combined gas law is most often used to calculate the new pressure, volume, or
temperature of a gas sample when two of these properties are changed. Under typical conditions, the amount of the gas is held constant (n1 5 n2).
Example Problem 10.3.1 Use the combined gas law to calculate pressure.
A 2.68-L sample of gas has a pressure of 1.22 atm and a temperature of 29 ºC. The sample is
compressed to a volume of 1.41 L and cooled to 217 ºC. Calculate the new pressure of the
gas, assuming that no gas escaped during the experiment.
Solution:
You are asked to calculate the pressure of a gas when only its volume and temperature are
changed.
You are given the original volume, pressure, and temperature of the gas and the new volume
and temperature of the gas sample.
Make a table of the known and unknown pressure, volume, and temperature data. Note that
temperature must be converted to kelvin temperature units and that the amount of gas (n)
does not change.
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b
Example Problem 10.3.1 (continued)
P1 5 1.22 atm
V1 5 2.68 L
T1 5 29 ºC 1 273 5 302 K
n1 5 n2
P2 5 ?
V2 5 1.41 L
T2 5 217 ºC 1 273 5 256 K
Rearrange the combined gas law to solve for P2, and calculate the new pressure of the gas sample.
P1V1
PV
5 2 2
n1T1
n2T2
Video Solution
11.22 atm2 12.68 L2 1256 K2
PVT
P2 5 1 1 2 5
5 1.97 atm
11.41 L2 1302 K2
V2T1
Tutored Practice
Problem 10.3.1
10.3b The Ideal Gas Law
The ideal gas law incorporates the three historical gas laws and a constant, which is
given the symbol R and called the universal gas constant or the ideal gas constant
(R 5 0.082057 L·atm/K·mol).
PV 5 nRT
(10.5)
One property of an ideal gas is that it follows the ideal gas law; that is, its variables (P,
V, n, and T) vary according to the ideal gas law. The universal gas constant is independent
of the identity of the gas. Note that the units of R (L·atm/K·mol) control the units of P, V,
T, and n in any equation that includes this constant.
When three of the four properties of a gas sample are known, the ideal gas law can be
used to calculate the unknown property.
Example Problem 10.3.2 Use the ideal gas law to calculate an unknown property.
A sample of O2 gas has a volume of 255 mL, has a pressure of 742 mm Hg, and is at a temperature of 19.6 ºC. Calculate the amount of O2 in the gas sample.
Solution:
You are asked to calculate the amount of gas in a sample.
You are given the pressure, volume, and temperature of the gas sample.
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b
Example Problem 10.3.2 (continued)
The ideal gas law contains a constant (R 5 0.082057 L·atm/K·mol), so all properties must have
units that match those in the constant.
1 atm
5 0.976 atm
P 5 742 mm Hg 3
760 mm Hg
V 5 255 mL 3
1L
5 0.255 L
1000 mL
T 5 (19.6 1 273.15) K 5 292.8 K
n5?
Rearrange the ideal gas law to solve for n, and calculate the amount of oxygen in the sample.
PV 5 nRT
Video Solution
10.976 atm2 10.255 L2
PV
nO2 5
5
5 0.0104 mol
10.082057 L # atm /K # mol2 1292.8 K2
RT
Tutored Practice
Problem 10.3.2
10.3c The Ideal Gas Law, Molar Mass, and Density
If a compound exists in gaseous form at some temperature, it is very easy (and relatively
inexpensive) to determine its molar mass from simple laboratory experiments. Molar
mass can be calculated from pressure, temperature, and density measurements and the
use of the ideal gas law. First, we use the definition of molar mass to incorporate the mass
of a gas sample and the molar mass of a gas into the ideal gas law.
m 1mass, in g2
molar mass 1 M2 5
n 1amount, in mol2
PV 5 a
m
b RT
M
Next, we rearrange this equation to derive a relationship between molar mass, density,
temperature, and pressure of a gas.
M5
mRT
m RT
5a b
PV
V P
M5
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dRT
P
(10.6)
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In Equation 10.6, M 5 molar mass (g/mol) and d 5 gas density (g/L). This form of the ideal
gas law can be used to calculate the molar mass or density of a gas, as shown in the following examples.
Example Problem 10.3.3 Use the ideal gas law to calculate molar mass.
A 4.07-g sample of an unknown gas has a volume of 876 mL and a pressure of 737 mm Hg at
30.4 ºC. Calculate the molar mass of this compound.
Solution:
You are asked to calculate the molar mass of a compound.
You are given the mass, volume, pressure, and temperature of the gas sample.
There are two ways to solve this problem: using Equation 10.6 or using the ideal gas law in its
original form. Note that both of these equations contain a constant (R 5 0.082057 L·atm/K·mol),
so all properties must have units that match those in the constant.
Method 1:
P 5 737 mm Hg 3
1 atm
5 0.970 atm
760 mm Hg
T 5 (30.4 1 273.15) K 5 303.6 K
V 5 876 mL 3
d5
1L
5 0.876 L
1000 mL
m
4.07 g
5
5 4.65 g /L
V
0.876 L
Use Equation 10.6 to calculate the molar mass of the unknown gas.
M5
Method 2:
14.65 g /L2 10.082057 L # atm /K # mol2 1303.6 K2
dRT
5
5 119 g /mol
P
0.970 atm
Rearrange the ideal gas law to calculate the amount (n) of gas present.
n5
10.970 atm2 10.876 L2
PV
5
5 0.0341 mol
10.082057 L # atm /K # mol2 1303.6 K2
RT
Use the mass of gas and the amount to calculate molar mass.
M5
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m
4.07 g
5
5 119 g /mol
n
0.0341 mol
Video Solution
Tutored Practice
Problem 10.3.3
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Example Problem 10.3.4 Use the ideal gas law to calculate density.
Calculate the density of oxygen gas at 788 mm Hg and 22.5 ºC.
Solution:
You are asked to calculate the density of a gas at a given pressure and temperature.
You are given the identity of the gas and its pressure and temperature.
Equation 10.6 contains a constant (R 5 0.082057 L·atm/K·mol), so all properties must have
units that match those in the constant.
1 atm
P 5 788 mm Hg 3
5 1.04 atm
760 mm Hg
T 5 (22.5 1 273.15) K 5 295.7 K
M(O2) 5 32.00 g/mol
Solve Equation 10.6 for density, and use it to calculate the density of oxygen under these
conditions.
dRT
M5
P
d5
11.04 atm2 132.00 g /mol2
PM
5
5 1.37 g /L
1
RT
0.082057 L # atm /K # mol2 1295.7 K2
Gas Densities at STP
Gas densities are often reported under a set of standard temperature and pressure
conditions (STP) of 1.00 atm and 273.15 K (0 ºC). Do not confuse STP with standard state
conditions, which typically use a temperature of 25 ºC. Under STP conditions, one mole of
an ideal gas has a standard molar volume of 22.4 L.
11 mol2 10.082057 L # atm /K # mol2 1273.15 K2
nRT
V5
5
5 22.4 L
P
1.00 atm
The relationship between gas density and molar mass is shown in Table 10.3.1, which
contains gas densities at STP for some common and industrially important gases. Notice
that some flammable gases such as propane and butane are denser than either O2 or N2.
This means that these gases sink in air and will collect near the ground, so places where
these gases might leak (such as in a house or apartment) will have a gas detector mounted
near the floor. Carbon monoxide (CO) has a density very similar to that of O2 and N2, so it
mixes well with these gases. For this reason, CO detectors can be placed at any height on
a wall.
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Video Solution
Tutored Practice
Problem 10.3.4
Table 10.3.1 Gas Densities at STP
Gas
Molar Mass
(g/mol)
Density at
STP (g/L)
H2
2.02
0.0892
He
4.00
0.178
N2
28.00
1.25
CO
28.01
1.25
O2
32.00
1.42
CO2
44.01
1.96
Propane (C3H8)
44.09
1.97
Butane (C4H10)
58.12
2.59
351.99
15.69
UF6
Section 10.3 Mastery
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10.4 Partial Pressure and Gas Law Stoichiometry
10.4a Introduction to Dalton’s Law of Partial Pressures
Our atmosphere is a mixture of many gases, and this mixture changes composition constantly. For example, every time you breathe in and out, you make small changes to the
amount of oxygen, carbon dioxide, and water in the air around you. Dalton’s law of partial
pressures states that the pressure of a gas mixture is equal to the sum of the pressures due
to the individual gases of the sample, called partial pressures. For a mixture containing the
gases A, B, and C, for example,
Ptotal 5 PA 1 PB 1 PC
(10.7)
where Ptotal is the pressure of the mixture and PX is the partial pressure of gas X in the
mixture. Each gas in a mixture behaves as an ideal gas and as if it alone occupies the
container. This means that although individual partial pressures may differ, all gases in
the mixture have the same volume (equal to the container volume) and temperature.
Example Problem 10.4.1 Calculate pressure using Dalton’s law of partial pressures.
A gas mixture is made up of O2 (0.136 g), CO2 (0.230 g), and Xe (1.35 g). The mixture has a
volume of 1.82 L at 22.0 ºC. Calculate the partial pressure of each gas in the mixture and the
total pressure of the gas mixture.
Solution:
You are asked to calculate the partial pressure of each gas in a mixture of gases and the total
pressure of the gas mixture.
You are given the identity and mass of each gas in the sample and the volume and temperature of the gas mixture.
The partial pressure of each gas is calculated from the ideal gas equation.
nRT
PO2 5
5
V
a0.136 g 3
1 mol O2
b 10.082057 L # atm /K # mol2 122.0 1 273.15 K2
32.00 g
5 0.0566 atm
1.82 L
nRT
5
PCO2 5
V
a0.230 g 3
1 mol CO2
b 10.082057 L # atm /K # mol2 122.0 1 273.15 K2
44.01 g
5 0.0695 atm
1.82 L
c
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b
Example Problem 10.4.1 (continued)
nRT
PXe 5
5
V
a1.35 g 3
1 mol Xe
b 10.082057 L # atm /K # mol2 122.0 1 273.15 K2
131.3 g
5 0.137 atm
1.82 L
Notice that the three gases have the same volume and temperature but different pressures.
Video Solution
The total pressure is the sum of the partial pressures for the gases in the mixture.
Tutored Practice
Problem 10.4.1
Ptotal 5 PO2 1 PCO2 1 PXe 5 0.0566 atm 1 0.0695 atm 1 0.137 atm 5 0.263 atm
Collecting a Gas by Water Displacement
A common laboratory experiment involves collecting the gas generated during a chemical
reaction by water displacement (Interactive Figure 10.4.1). Because water can exist in gaseous
form (as water vapor), the gas that is collected is a mixture of both the gas formed during the
chemical reaction and water vapor.
According to Dalton’s law of partial pressures, the pressure of the collected gas mixture
is equal to the partial pressure of the gas formed during the chemical reaction plus the
partial pressure of the water vapor (the vapor pressure of the water).
Ptotal 5 Pgas 1 PH2O
Interactive Figure 10.4.1
Apply Dalton’s law of partial pressures.
Water vapor pressure varies with temperature. Table 10.4.1 shows vapor pressure values
for moderate temperatures; a more complete table is found in the reference tools.
Example Problem 10.4.2 Calculate the amount of gas produced when collected by
water displacement.
Aluminum reacts with strong acids such as HCl to form hydrogen gas.
2 Al(s) 1 6 HCl(aq) S 3 H2(g) 1 2 AlCl3(aq)
In one experiment, a sample of Al reacts with excess HCl and the gas produced is collected by
water displacement. The gas sample has a temperature of 22.0 ºC, a volume of 27.58 mL, and a
pressure of 738 mm Hg. Calculate the amount of hydrogen gas produced in the reaction.
Charles D. Winters
Solution:
You are asked to calculate the amount of gas produced in a chemical reaction when it is
collected over water at a given pressure and temperature.
You are given the volume, pressure, and temperature of the gas produced in the reaction.
c
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b
Example Problem 10.4.2 (continued)
The gas collected is a mixture containing water vapor, which has a vapor pressure of
19.83 mm Hg at 22.0 ºC (Table 10.4.1). Subtract the water vapor pressure from the total
pressure of the mixture to calculate the partial pressure of the hydrogen gas in the mixture.
PH2 5 Ptotal 2 PH2O 5 739 mm Hg 2 19.83 mm Hg 5 718 mm Hg
Use the ideal gas law to calculate the amount of gas produced in the reaction.
P 5 718 mm Hg 3
V 5 27.58 mL 3
1 atm
5 0.945 atm
760 mm Hg
1L
5 0.02758 L
1000 mL
T 5 (22.0 1 273.15) K 5 295.2 K
Video Solution
10.945 atm2 10.02758 L2
PV
nH2 5
5
5 1.08 3 1023 mol
10.082057 L # atm /K # mol2 1295.2 K2
RT
Tutored Practice
Problem 10.4.2
10.4b Partial Pressure and Mole Fractions of Gases
Within a gas mixture, the total pressure is the sum of the partial pressures of each of the
component gases. The ideal gas law shows that pressure and amount (in moles) of any gas
are directly related.
RT
P5n
V
Therefore, the degree to which any one gas contributes to the total pressure is directly related to the amount (in moles) of that gas present in a mixture. In other words,
the greater the amount of a gas in a mixture, the greater its partial pressure and the
greater amount its partial pressure contributes to the total pressure. Quantitatively, this
relationship is shown in Equation 10.8, where PA and nA are the partial pressure and
amount (in moles), respectively, of gas A in a mixture of gases, and ntotal is the total
amount (in moles) of gas in the mixture:
PA
n
5 A
(10.8)
Ptotal
ntotal
The ratio nA/ntotal is the mole fraction of gas A in a mixture of gases, and it is given the
symbol χA. Rearranging Equation 10.8 to solve for the partial pressure of gas A,
PA 5 χAPtotal
Unit 10 Gases
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(10.9)
Table 10.4.1 Vapor Pressure of Water
Temperature
(ºC)
Vapor pressure
of H2O (mm Hg)
19
16.48
20
17.54
21
18.65
22
19.83
23
21.07
24
22.38
25
23.76
26
25.21
27
26.74
28
28.35
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Note that mole fraction is a unitless quantity. Also, the sum of the mole fractions for
all gases in a mixture is equal to 1. For a mixture containing gases A, B, and C, for
example,
χA 1 χB 1 χC 5 1
Example Problem 10.4.3 Use mole fraction in calculations involving gases.
A gas mixture contains the noble gases Ne, Ar, and Kr. The total pressure of the mixture is
2.46 atm, and the partial pressure of Ar is 1.44 atm. If a total of 18.0 mol of gas is present,
what amount of Ar is present?
Solution:
You are asked to calculate the amount of gas present in a mixture.
You are given the total pressure of the mixture, the total amount of gas in the mixture, and
the partial pressure of one of the gases in the mixture.
The amount (in moles) of Ar is equal to the total moles of gas in the mixture times its mole
fraction. The first step, then, is to determine the mole fraction of Ar in the mixture. Solving
Equation 10.9 for mole fraction of Ar,
PAr 5 χArPtotal
Ar 5
PAr
1.44 atm
5
5 0.585
Ptotal
2.46 atm
Use the mole fraction of Ar and the total amount of gases in the mixture to calculate moles of
Ar in the mixture.
n
nAr
0.585 5 Ar 5
ntotal
18.0 mol
nAr 5 10.5 mol
Video Solution
Tutored Practice
Problem 10.4.3
10.4c Gas Laws and Stoichiometry
We investigated stoichiometric relationships for systems involving pure substances and solutions in Stoichiometry (Unit 3) and Chemical Reactions and Solution Stoichiometry (Unit 4),
where the amount of a reactant or product was determined from mass data or from volume
and concentration data. We now have the tools needed to include the gas properties of pressure, temperature, and volume in the stoichiometric relationships derived in those units.
Interactive Figure 10.4.2 gives a schematic representation of the relationship between
gas properties and the amount of a reactant or product.
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Interactive Figure 10.4.2
Use gas properties in stoichiometry calculations.
gA
n=
Gas
properties
⎛ mol A⎞
3 ⎝ gA ⎠
Grams of
B
Multiply by
1/molar mass of A.
PV
RT
mol A
Moles of
A
vol (L) A 3 ⎛ mol A ⎞
⎝ L ⎠
⎛ mol B⎞
3 ⎝ mol A⎠
Multiply by the
stoichiometric ratio.
Multiply by
concentration of A.
Volume (L)
of solution A
Multiply by
mol B
molar mass of B.
⎛ gB ⎞
3 ⎝ mol B⎠
n=
PV
RT
Gas
properties
Moles of
B
Divide by
mol B
concentration of B.
⎛ L ⎞
3 ⎝ mol B⎠
Volume (L)
of solution B
© 2013 Cengage Learning
Grams of
A
Schematic flow chart of stoichiometric relationships
Example Problem 10.4.4 Use gas laws in stoichiometry calculations.
A sample of O2 with a pressure of 1.42 atm and a volume of 250. mL is allowed to react with
excess SO2 at 129 ºC.
2 SO2(g) 1 O2(g) S 2 SO3(g)
Calculate the pressure of the SO3 produced in the reaction if it is transferred to a 1.00-L flask
and cooled to 35.0 ºC.
Solution:
You are asked to calculate the pressure of a gas produced in a chemical reaction.
You are given the balanced equation for the reaction; the pressure, volume, and temperature
of a reactant; and the volume and temperature of the gas produced in the reaction.
Step 1. Calculate the amount of reactant (O2) available using the ideal gas law.
c
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b
Example Problem 10.4.4 (continued)
nO2 5
11.42 atm2 10.250 L2
PV
5
5 0.0108 mol
10.082057 L # atm /K # mol2 1129 1 273.15 K2
RT
Step 2. Use the amount of limiting reactant (O2) and the balanced equation to calculate the
amount of SO3 produced.
2 mol SO3
0.0108 mol O2 3
5 0.0216 mol SO3
1 mol O2
Step 3. Use the ideal gas law and the new volume and temperature conditions to calculate
the pressure of SO3.
PSO3 5
10.0216 mol SO32 10.082057 L # atm /K # mol2 135.0 1 273.15 K2
nRT
5
5 0.546 atm
V
1.00 L
Video Solution
Tutored Practice
Problem 10.4.4
Section 10.4 Mastery
10.5 Kinetic Molecular Theory
10.5a Kinetic Molecular Theory and the Gas Laws
According to the kinetic molecular theory,
●
●
●
●
gases consist of molecules whose separation is much larger than the molecules
themselves;
the molecules of a gas are in continuous, random, and rapid motion;
the average kinetic energy of gas molecules is determined by the gas temperature, and
all gas molecules at the same temperature, regardless of mass, have the same average
kinetic energy; and
gas molecules collide with one another and with the walls of their container, but they
do so without loss of energy in “perfectly elastic” collisions.
Note that when we talk about the behavior of gas molecules, we include monoatomic gaseous atoms in our definition of molecules.
Relating the Kinetic Molecular Theory to the Gas Laws
For any theory to be recognized as useful, it must be consistent with experimental observations. The kinetic molecular theory therefore must be consistent with and help explain
the well-known gas laws. At the molecular level, the concept of pressure is considered in
terms of collisions between gas molecules and the inside walls of a container. Each collision between a moving gas molecule and the static wall involves the imparting of a force
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pushing on the inside of the wall. The more collisions there are and the more energetic
the collisions on average, the greater the force and the higher the pressure. Changing the
number of molecule-wall collisions by, for example, lowering the temperature of a sample
of gas (as shown in Interactive Figure 10.5.1) results in a change in the other properties
of the gas.
Interactive Figure 10.5.1
Charles D. Winters
Relate kinetic molecular theory to the gas laws.
When air-filled balloons are placed in liquid nitrogen (–196 ºC), the gas volume
decreases dramatically
P and n
If the number of gas molecules inside a container is doubled, the number of molecule–wall
collisions will exactly double (assuming volume and temperature are constant). This means
that twice as much force will push against the wall and the pressure is twice as great.
P and T
As the temperature of a gas in a container increases, the molecules move more rapidly.
This does two things: it leads to more frequent collisions between the molecules and the
walls of the container, and it also results in more energetic collisions between the molecules and container walls. Therefore, as temperature increases, there are more frequent
and energetic collisions on the inside of the walls and a greater pressure (assuming constant amount of gas and volume).
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P and V
As the volume of a container is increased, the gas molecules take longer to move across the
inside of the container before hitting the wall on the other side. This means that the frequency of collisions decreases, resulting in a lower pressure (assuming constant amount of
gas and temperature).
10.5b Molecular Speed, Mass, and Temperature
Gas molecules move through space at very high speeds. As you saw in Thermochemistry
(Unit 5), the kinetic energy (KE) of a moving object is directly related to its mass (m) and
speed (v).
1
mv2
2
For a collection of gas molecules, the mean (average) kinetic energy of one mole of gas
particles is directly related to the average of the square of the gas velocity (NA 5 Avogadro’s
number).
1
(10.10)
KE 5 NA mv2
2
As described in the third postulate of the kinetic molecular theory, it can also be shown that
kinetic energy of one mole of gas particles is directly related to the absolute temperature of the
gas [Equation 10.11, where the ideal gas constant has units of J/K·mol (R 5 8.314 J/K·mol)].
3
(10.11)
KE 5 RT
2
KE 5
Combining Equations 10.10 and 10.11, we see that the average velocity of molecules in
a sample of gas is inversely related to the mass of gas molecules (m) and directly related
to the temperature of the gas (T).
1
3
(10.12)
NA mv2 5 RT
2
2
Taking the square root of the average of the square of the gas velocity gives the root
mean square (rms) speed of a gas. The rms speed is not the same as the average speed of
a sample of gas particles, but the two values are similar. Rearranging Equation 10.12 to solve
for rms speed gives another important relationship between molecular speed and mass:
3RT
v2 5
NAm
vrms 5 "v 2 5
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3RT
(10.13)
Å M
In Equation 10.13, R 5 8.3145 J/K·mol and M 5 molar mass of the gas. Notice that the
thermodynamic units of R require molar mass to be expressed in units of kilograms per
mole when using this equation (1 J 5 1 kg·m2/s2).
vrms 5
Example Problem 10.5.1 Calculate root mean square (rms) speed.
Calculate the rms speed of NH3 molecules at 21.5 ºC.
Solution:
You are asked to calculate the rms speed for a gas at a given temperature.
You are given the identity and temperature of the gas.
Use Equation 10.13, with molar mass in units of kilograms per mole.
vrms
3RT
3 18.3145 J /K # mol2 121.5 1 273.15 K2
5
5
5 657 m /s
Å M
Å
0.01703 kg /mol
Video Solution
Tutored Practice
Problem 10.5.1
Boltzmann Distribution Plots
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Interactive Figure 10.5.2
Explore Boltzmann distribution plots.
Most probable speed
for O2 at 25 °C
Number of molecules
Average speed
for O2 at 25 °C
0
200
© 2013 Cengage Learning
Not all molecules in a sample of gas move at the same speed. Just like people or cars, a
collection of gas molecules shows a range of speeds. The distribution of speeds is called a
Boltzmann distribution. Interactive Figure 10.5.2 shows a Boltzmann distribution plot
for O2 at 25 ºC. Each point in a Boltzmann distribution plot gives the number of gas molecules moving at a particular speed. This Boltzmann distribution for O2 starts out with low
numbers of molecules at low speeds, increases to a maximum at around 400 m/s, and then
decreases smoothly to very low numbers at about 1000 m/s. This means that in a sample of
O2 gas at 25 ºC, very few O2 molecules are moving slower than 100 m/s, many molecules are
moving at speeds between 300 and 600 m/s, and very few molecules are moving faster than
900 m/s. The peak around 450 m/s indicates the most probable speed at which the molecules are moving at this temperature. It does not mean that most of the molecules are moving at that speed.
Boltzmann distribution plots for a number of different gases, each at the same temperature, are shown in Figure 10.5.3. The heights of the curves in this figure differ because the
area under each curve represents the total number of molecules in the sample. In the case
of O2, the gas molecules have a relatively narrow range of speeds and therefore more molecules moving at any particular speed. Helium has a wide range of speeds, and therefore
the curve is stretched out, with few molecules moving any particular speed.
400
600
800 1000 1200
Molecular speed (m/s)
Boltzmann distribution plot for O2 at 25 ºC
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Number of molecules
O2 (32 g/mol)
All at the same
temperature
N2 (28 g/mol)
H2O (18 g/mol)
0
500
1000
1500
Molecular speed (m/s)
2000
© 2013 Cengage Learning
He (4 g/mol)
Figure 10.5.3 Boltzmann distribution plots for four different gases at 25 ºC
Notice the relationship between the Boltzmann distribution plots and the molar mass
of the gases in Figure 10.5.3. The peak in the O2 curve is farthest to the left, meaning it has
the slowest-moving molecules. The peak in the H2O curve is farther to the right, which
means that H2O gas molecules move, on average, faster than O2 molecules at the same
temperature. The helium curve is shifted well to the right and has very fast-moving molecules.
Recall that average molecular speed depends on both the average kinetic energy and on the
mass of the moving particles. According to the kinetic molecular theory, all gases at the
same temperature have the same kinetic energy. Therefore, if a gas has a smaller mass, it
must have a larger average velocity.
Molecular speed changes with temperature, as shown in Interactive Figure 10.5.4. The
curve for O2 at the higher temperature is shifted to the right, indicating that O2 molecules
move faster, on average, at the higher temperature. As the temperature increases, the average kinetic energy increases. Because the mass is constant, as average kinetic energy increases, average molecular speed must increase.
In summary, Boltzmann distributions give us the following information:
1. Gases move with a range of speeds at a given temperature.
2. Gases move faster, on average, at higher temperatures.
3. Heavier gases move more slowly, on average, than lighter gases at the same
temperature.
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Interactive Figure 10.5.4
Investigate how temperature affects Boltzmann plots.
At 25 °C more molecules are
moving at about 400 m/s than
at any other speed.
O2 at 25 °C
O2 at 1000 °C
0
200
400
600
Many more molecules
are moving at 1600 m/s
when the sample is at
1000 °C than when it
is at 25 °C.
800 1000 1200 1400
Molecular speed (m/s)
1600
1800
© 2013 Cengage Learning
Number of molecules
Very few
molecules
have very
low
speeds
Boltzmann distribution plots for O2 at 25 ºC and 1000 ºC
10.5c Gas Diffusion and Effusion
As you saw in a previous example problem, an ammonia molecule moves with an rms speed
of about 650 m/s at room temperature. This is the equivalent of almost 1500 miles per hour!
If you open a bottle of ammonia, however, it can take minutes for the smell to travel across a
room. Why do odors, which are the result of volatile molecules, take so long to travel? The
answer lies in gas diffusion. As a gas molecule moves in an air-filled room, it is constantly
colliding with other molecules that block its path. It therefore takes much more time for a
gas sample to get from one place to another than it would if there were nothing in its way.
This gas process is called diffusion, the mixing of gases, and is illustrated in Interactive
Figure 10.5.5.
A process related to diffusion is effusion, the movement of gas molecules through a
small opening into a vacuum (Interactive Figure 10.5.6). Graham’s law of effusion states
that the rate of effusion of a gas is inversely related to the square root of its molar mass.
That is, lighter gases move faster and effuse more rapidly compared with heavier gases,
which move more slowly and effuse more slowly.
rate of effusion ~
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1
"M
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Interactive Figure 10.5.5
© 2013 Cengage Learning
Explore gas diffusion.
Before mixing
After mixing
Gas diffusion
Interactive Figure 10.5.6
Explore gas effusion.
© 2013 Cengage Learning
Vacuum
Before effusion
During effusion
Gas effusion
A useful form of Graham’s law is used to determine the molar mass of an unknown gas.
The effusion rate of the unknown gas is compared with that of a gas with a known molar
mass. The ratio of the effusion rates for the two gases is inversely related to the square root
of the ratio of the molar masses.
M2
rate1
5
rate2 Å M1
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Example Problem 10.5.2 Use Graham’s law of effusion to calculate molar mass.
A sample of ethane, C2H6, effuses through a small hole at a rate of 3.6 3 1026 mol/h. An unknown gas, under the same conditions, effuses at a rate of 1.3 3 1026 mol/hr. Calculate the
molar mass of the unknown gas.
Solution:
You are asked to calculate the molar mass of an unknown gas.
You are given the effusion rate and identity of a gas and the effusion rate of the unknown gas
measured under the same conditions.
Use Equation 10.14 and the molar mass of ethane to calculate the molar mass of the unknown gas.
rate1
M2
5
rate2 Å M1
3.6 3 1026 mol/h
M2
5
1.3 3 1026 mol/h Å 30.07 g /mol
M2 5 230 g/mol
Is your answer reasonable? The unknown gas effuses at a slower rate than ethane, so its
molar mass should be greater than that of ethane.
Video Solution
Tutored Practice
Problem 10.5.2
10.5d Nonideal Gases
According to the kinetic molecular theory, an ideal gas is assumed to experience only perfectly elastic collisions and take up no actual volume. So, although gas molecules have
volume, it is assumed that each individual molecule occupies the entire volume of its container and that the other molecules do not take up any of the container volume. At room
temperature and pressures at or below 1 atm, most gases behave ideally. However, at high
pressures or low temperatures, gases deviate from ideal behavior.
For an ideal gas, PV/nRT 5 1 at any pressure. Therefore, one way to show deviation from
ideal behavior is to plot the ratio PV/nRT as a function of pressure (Interactive Figure 10.5.7).
All three of the gases shown in Interactive Figure 10.5.7 deviate from ideal behavior at high
pressures.
Two types of deviations occur from ideal behavior.
1. Deviations due to gas volume
At high pressures, the concentration of the gas in a container is very high and as a result, molecules, on average, are closer together than they are at lower pressures. At these
high pressures, the gas molecules begin to occupy a significant amount of the container
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Interactive Figure 10.5.7
Explore deviations from ideal gas behavior.
2.5
CH4
2.0
H2
1.0
0.5
0
Ideal gas
0
200
400
600
Pressure (atm)
800
© 2013 Cengage Learning
PV
nRT
N2
1.5
A plot of PV/nRT versus pressure
volume; thus, the predicted volume occupied by the gas is less than the actual (container) volume. Because the volume occupied by the gas is less than the container volume,
the correction is subtracted from the container volume.
The amount of volume occupied by the gas molecules depends on the amount of gas
present (n), and a constant (b) that represents how large the gas molecules are. Therefore,
under nonideal conditions,
V 5 Vcontainer 2 nb
2. Deviations due to molecular interactions
Under ideal conditions, gas molecules have perfectly elastic conditions. When the temperature decreases, however, gas molecules can interact for a short time after they collide.
When this happens, there are fewer effective particles in the container (some molecules
form small clusters, decreasing the number of particle-wall collisions) and these molecular
interactions decrease the force with which molecules collide with the container walls. Thus,
the predicted pressure of the gas is greater than the actual (measured) pressure.
Because the predicted pressure is greater than measured pressure, the pressure correction
is added to the measured pressure.
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The strength of these molecular interactions depends greatly on the number of collisions
and therefore depends on the amount of gas present (n) and the volume (V) of the container.
Gases differ in these interactions, and this is reflected in a constant, a, which is specific to a
given gas at a particular temperature. Under nonideal conditions,
P 5 Pmeasured 1
n2a
2
Vcontainer
These two deviations from ideal behavior are combined into a separate, more sophisticated gas law, the van der Waals equation. Table 10.5.1 shows some constants for common gases.
n2a
(10.15)
aPmeasured 1 2
b 1V 2 nb2 5 nRT
Vmeasured
Notice that the values of b increase roughly with increasing size of the gas molecules. The
values of a are related to the tendency of molecules of a given species to interact at the
molecular level.
Example Problem 10.5.3 Calculate pressure using the ideal gas law
and the van der Waals equation.
Table 10.5.1 Van der Waals Constants
a (L2·atm/mol2)
b (L/mol)
H2
0.244
0.0266
He
0.034
0.0237
N2
1.39
0.0391
NH3
4.17
0.0371
CO2
3.59
0.0427
CH4
2.25
0.0428
Gas
A 1.78-mol sample of ammonia gas is maintained in a 2.50-L container at 302 K. Calculate the
pressure of the gas using both the ideal gas law and the van der Waals equation (van der Waals
constants are listed in Table 10.5.1).
Solution:
You are asked to calculate the pressure of a gas sample assuming ideal and nonideal behavior.
You are given the identity, amount, volume, and temperature of the gas.
First use the ideal gas law to calculate the pressure in the flask.
P5
11.78 mol2 10.082057 L # atm /K # mol2 1302 K2
nRT
5
5 17.6 atm
V
2.50 L
Compare this pressure with that calculated using the van der Waals equation.
aPmeasured 1
c Pmeasured 1
n2a
b
2
Vmeasured
1V 2 nb2 5 nRT
11.78 mol2 2 14.17 L2 # atm /mol22
d 3 2.50 L 2 11.78 mol2 10.0371 L /mol2 4
12.50 L2 2
5 11.78 mol2 10.082057 L # atm /K # mol2 1302 K2
Video Solution
Tutored Practice
Problem 10.5.3
Pmeasured 5 16.0 atm
The actual pressure in the container is about 10% less than that calculated using the ideal gas law.
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Unit Recap
Key Concepts
10.1 Properties of Gases
●
Gas pressure is a measure of the force exerted on a given area of surface (10.1b).
●
Gas pressure is measured using a barometer or a manometer (10.1b).
●
The SI unit of pressure is the pascal (Pa), but common pressure units are atmosphere,
millimeters of mercury, and bar (10.1b).
10.2 Historical Gas Laws
●
●
●
●
Boyle’s law states that the pressure and volume of a gas sample are inversely related
when the amount of gas and temperature are held constant (10.2a).
Charles’s law states that the temperature and volume of a gas sample are directly related when the pressure and the amount of gas are held constant (10.2b).
Absolute zero (0 K, 2273.15 ºC) is the temperature at which gas volume is equal to
zero (10.2b).
Avogadro’s law states that the volume and amount (in moles) of a gas are directly related when pressure and temperature are held constant (10.2c).
10.3 The Combined and Ideal Gas Laws
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The combined gas law is used to calculate the new pressure, volume, amount, or temperature of a gas when one or more of these properties is changed (10.3a).
The ideal gas law incorporates the three historical gas laws and a constant, which is given
the symbol R and called the universal gas constant or the ideal gas constant (10.3b).
An ideal gas is a gas whose variables (P, V, n, and T) vary according to the ideal gas
law (10.3b).
Standard temperature and pressure (STP) conditions are defined as 273.15 K (0 ºC)
and 1 atm (10.3b).
At STP, one mole of an ideal gas has a standard molar volume of 22.4 L (10.3b).
10.4 Partial Pressure and Gas Law Stoichiometry
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Dalton’s law of partial pressures states that the pressure of a gas mixture is equal to the
sum of the pressures due to the individual gases of the sample, called partial pressures
(10.4a).
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10.5 Kinetic Molecular Theory
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According to the kinetic molecular theory, (1) gases consist of molecules whose separation is much larger than the molecules themselves; (2) the molecules of a gas are in
continuous, random, and rapid motion; (3) the average kinetic energy of gas molecules
is determined by the gas temperature, and all gas molecules at the same temperature,
regardless of mass, have the same average kinetic energy; and (4) gas molecules collide
with one another and with the walls of their container, but they do so without loss of
energy in “perfectly elastic” collisions (10.5a).
The average velocity of molecules in a sample of gas is inversely related to the mass of
gas molecules and directly related to the temperature of the gas (10.5b).
The root mean square (rms) speed of a gas is directly related to the temperature of the
gas and inversely related to the molar mass of the gas (10.5b).
A Boltzmann distribution plot shows the distribution of speeds in a sample of a gas at
a given temperature (10.5b).
Gases mix in a process called diffusion (10.5c).
Effusion is the movement of gas molecules through a small opening into a vacuum
(10.5c).
Graham’s law of effusion states that the rate of effusion of a gas is inversely related to
the square root of its molar mass (10.5c).
At high pressures or low temperatures, gases deviate from ideal behavior (10.5d).
The van der Waals equation is used to calculate gas pressure under nonideal conditions
(10.5d).
Key Equations
P1V15 P2V2
V1
V
5 2
T1
T2
Ptotal 5 PA 1 PB 1 PC
(10.7)
(10.2)
PA
n
5 A
Ptotal
ntotal
(10.8)
PA 5 χAPtotal
(10.9)
V1
V
5 2
n1
n2
(10.3)
P1V1
PV
5 2 2
n1T1
n2T2
(10.4)
PV 5 nRT
dRT
M5
P
Unit 10 Gases
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(10.1)
KE 5 NA
(10.5)
(10.6)
KE 5
1
mv2
2
3
RT
2
(10.10)
(10.11)
NA
1
3
mv2 5 RT
2
2
vrms 5
3RT
Å M
M2
rate1
5
rate2 Å M1
aPmeasured 1
(10.12)
(10.13)
(10.14)
n2a
2
Vmeasured
b 1V 2 nb2 5 nRT (10.15)
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Key Terms
10.1 Properties of Gases
pressure
barometer
millimeters of mercury (mm Hg)
torr
pascal (Pa)
standard atmosphere (atm)
bar
10.2 Historical Gas Laws
Boyle’s law
Charles’s law
Avogadro’s law
Unit 10
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10.3 The Combined and Ideal Gas Laws
combined gas law
ideal gas law
ideal gas constant (R)
ideal gas
standard temperature and pressure (STP)
standard molar volume
10.4 Partial Pressure and Gas Law
Stoichiometry
Dalton’s law of partial pressures
vapor pressure
mole fraction
10.5 Kinetic Molecular Theory
kinetic molecular theory
root mean square (rms) speed
Boltzmann distribution
diffusion
effusion
Graham’s law of effusion
van der Waals equation
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