5: The Structure of the Atom
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432 Chapter 13 Gases
Pext
Pext
Increase in
temperature
a
A gas confined in a
cylinder with a movable
piston. The gas pressure
Pgas is just balanced by
the external pressure Pext.
That is, Pgas ϭ Pext.
Figure 13.13
b
The temperature of the
gas is increased at
constant pressure Pext.
The increased particle
motions at the higher
temperature push back
the piston, increasing the
volume of the gas.
▲
The Relationship Between Volume and Temperature
Now picture the gas in a container with a movable piston. As shown in Figure 13.13a, the gas pressure Pgas is just balanced by an external pressure Pext.
What happens when we heat the gas to a higher temperature? As the temperature increases, the particles move faster, causing the gas pressure to increase. As soon as the gas pressure Pgas becomes greater than Pext (the pressure
holding the piston), the piston moves up until Pgas ϭ Pext. Therefore, the KM
model predicts that the volume of the gas will increase as we raise its temperature at a constant pressure (Figure 13.13b). This agrees with experimental observations (as summarized by Charles’s law).
EXAMPLE 13.14
Using the Kinetic Molecular Theory
to Explain Gas Law Observations
Use the KM theory to predict what will happen to the pressure of a gas when
its volume is decreased (n and T constant). Does this prediction agree with
the experimental observations?
SOLUTION
When we decrease the gas’s volume (make the container smaller), the particles hit the walls more often because they do not have to travel so far between the walls. This would suggest an increase in pressure. This prediction
on the basis of the model is in agreement with experimental observations of
gas behavior (as summarized by Boyle’s law). ■
In this section we have seen that the predictions of the kinetic molecular theory generally fit the behavior observed for gases. This makes it a useful and successful model.
13.10
OBJECTIVES:
Gas Stoichiometry
To understand the molar volume of an ideal gas. • To learn the definition
of STP. • To use these concepts and the ideal gas equation.
We have seen repeatedly in this chapter just how useful the ideal gas equation is. For example, if we know the pressure, volume, and temperature for
a given sample of gas, we can calculate the number of moles present:
13.10 Gas Stoichiometry
433
n ϭ PV/RT. This fact makes it possible to do stoichiometric calculations for
reactions involving gases. We will illustrate this process in Example 13.15.
EXAMPLE 13.15
Gas Stoichiometry: Calculating Volume
Calculate the volume of oxygen gas produced at 1.00 atm and 25 °C by the
complete decomposition of 10.5 g of potassium chlorate. The balanced equation for the reaction is
2KClO3(s) S 2KCl(s) ϩ 3O2(g)
SOLUTION
Where Are We Going?
We want to determine the volume of oxygen gas collected by the decomposition of KClO3.
What Do We Know?
• We know the temperature and pressure of oxygen gas.
• We know the mass of KClO3.
• The balanced equation: 2KClO3(s) S 2KCl(s) ϭ 3O2(g).
• Ideal gas law: PV ϭ nRT.
What Information Do We Need?
• R ϭ 0.08206 L atm/mol K.
• We need the number of moles of oxygen gas.
• Molar mass of KClO3.
How Do We Get There?
This is a stoichiometry problem very much like the type we considered in
Chapter 9. The only difference is that in this case, we want to calculate the
volume of a gaseous product rather than the number of grams. To do so, we
can use the relationship between moles and volume given by the ideal gas
law.
We’ll summarize the steps required to do this problem in the following
schematic:
Grams
of
KClO3
MATH SKILL BUILDER
10.5
122.6 ϭ 0.085644
0.085644
0.0856
Round
off
0.0856 ϭ 8.56 ϫ 10Ϫ2
1
Moles
of
KClO3
2
Moles
of
O2
3
Volume
of
O2
Step 1 To find the moles of KClO3 in 10.5 g, we use the molar mass of
KClO3 (122.6 g).
10.5 g KClO3 ϫ
1 mol KClO3
ϭ 8.56 ϫ 10Ϫ2 mol KClO3
122.6 g KClO3
Step 2 To find the moles of O2 produced, we use the mole ratio of O2 to
KClO3 derived from the balanced equation.
8.56 ϫ 10Ϫ2 mol KClO3 ϫ
3 mol O2
ϭ 1.28 ϫ 10Ϫ1 mol O2
2 mol KClO3
434 Chapter 13 Gases
Step 3 To find the volume of oxygen produced, we use the ideal gas law
PV ϭ nRT, where
P ϭ 1.00 atm
Vϭ?
n ϭ 1.28 ϫ 10Ϫ1 mol, the moles of O2 we calculated
R ϭ 0.08206 L atm/K mol
T ϭ 25 °C ϭ 25 ϩ 273 ϭ 298 K
Solving the ideal gas law for V gives
Vϭ
nRT
ϭ
P
(1.28 ϫ 10Ϫ1 mol) a0.08206
L atm
b (298 K)
K mol
1.00 atm
ϭ 3.13 L
Thus 3.13 L of O2 will be produced.
Self-Check EXERCISE 13.11 Calculate the volume of hydrogen produced at 1.50 atm and 19 °C by the reaction of 26.5 g of zinc with excess hydrochloric acid according to the balanced equation
Zn(s) ϩ 2HCl(aq) S ZnCl2(aq) ϩ H2(g)
See Problems 13.85 through 13.92. ■
In dealing with the stoichiometry of reactions involving gases, it is useful to define the volume occupied by 1 mole of a gas under certain specified
conditions. For 1 mole of an ideal gas at 0 °C (273 K) and 1 atm, the volume
of the gas given by the ideal gas law is
Vϭ
STP: 0 °C and 1 atm
EXAMPLE 13.16
(1.00 mol)(0.08206 L atmրK mol)(273 K)
nRT
ϭ
ϭ 22.4 L
P
1.00 atm
This volume of 22.4 L is called the molar volume of an ideal gas.
The conditions 0 °C and 1 atm are called standard temperature
and pressure (abbreviated STP). Properties of gases are often given under
these conditions. Remember, the molar volume of an ideal gas is 22.4 L at
STP. That is, 22.4 L contains 1 mole of an ideal gas at STP.
Gas Stoichiometry: Calculations Involving Gases at STP
A sample of nitrogen gas has a volume of 1.75 L at STP. How many moles of
N2 are present?
SOLUTION
Where Are We Going?
We want to determine the number of moles of nitrogen gas.
What Do We Know?
• The nitrogen gas has a volume of 1.75 L at STP.
What Information Do We Need?
• STP ϭ 1.00 atm, 0 °C.
• At STP, 1 mole of an ideal gas occupies a volume of 22.4 L.
How Do We Get There?
We could solve this problem by using the ideal gas equation, but we can take
a shortcut by using the molar volume of an ideal gas at STP. Because 1 mole
13.10 Gas Stoichiometry
435
of an ideal gas at STP has a volume of 22.4 L, a 1.75-L sample of N2 at STP
contains considerably less than 1 mol. We can find how many moles by using the equivalence statement
1.000 mol ϭ 22.4 L (STP)
which leads to the conversion factor we need:
1.75 L N2 ϫ
1.000 mol N2
ϭ 7.81 ϫ 10Ϫ2 mol N2
22.4 L N2
Self-Check EXERCISE 13.12 Ammonia is commonly used as a fertilizer to provide a source of nitrogen
for plants. A sample of NH3(g) occupies a volume of 5.00 L at 25 °C and
15.0 atm. What volume will this sample occupy at STP?
See Problems 13.95 through 13.98. ■
Standard conditions (STP) and molar volume are also useful in carrying
out stoichiometric calculations on reactions involving gases, as shown in
Example 13.17.
EXAMPLE 13.17
Gas Stoichiometry: Reactions Involving Gases at STP
Quicklime, CaO, is produced by heating calcium carbonate, CaCO3. Calculate the volume of CO2 produced at STP from the decomposition of 152 g of
CaCO3 according to the reaction
CaCO3(s) S CaO(s) ϩ CO2(g)
SOLUTION
Where Are We Going?
We want to determine the volume of carbon dioxide produced from 152 g of
CaCO3.
What Do We Know?
• We know the temperature and pressure of carbon dioxide gas (STP).
• We know the mass of CaCO3.
• The balanced equation: CaCO3(s) S CaO(s) ϭ CO2(g).
What Information Do We Need?
• STP ϭ 1.00 atm, 0 °C.
• At STP, 1 mole of an ideal gas occupies a volume of 22.4 L.
• We need the number of moles of carbon dioxide gas.
• Molar mass of CaCO3.
How Do We Get There?
The strategy for solving this problem is summarized by the following
schematic:
Grams
of
CaCO3
1
Moles
of
CaCO3
2
Moles
of
O2
3
Volume
of
O2