Example 5.1
A Canadian weather report gives the atmospheric pressure as 100.2 kPa. What is the
pressure expressed in the unit torr?
Strategy
To convert a pressure from one unit to another, we need a relationship between the
two units that we can use as a conversion factor. Generally we can find such a
relationship in Table 5.2.
Solution
To convert from kPa to Torr, we need this relationship from Table 5.2:
1 atm = 760 Torr = 101.325 kPa. That is,
With this conversion factor, we find that
Example 5.1 continued
Assessment
Note that the units kPa cancel and that the pressure expressed in the unit torr is a
larger number (by a factor of about 7.5) than the pressure in kilopascals, as
expected from the form of the conversion factor.
Exercise 5.1A
Carry out the following conversions of pressure units.
(a) 0.947 atm to mmHg
(c) 29.95 in. Hg to Torr
(b) 98.2 kPa to Torr
(d) 768 Torr to atm
Exercise 5.1B
What is the pressure, in kilopascals and in atmospheres, if a force of 1.00 x 102 N is
exerted on an area of 5.00 cm2?
Example 5.2
Calculate the height of a column of water (d = 1.00 g/cm3) that exerts the same pressure
as a column of mercury (d = 13.6 g/cm3) 760 mm high.
Strategy
With the above equation, we can establish the pressures exerted by the columns of
mercury and water:
When we equate these two pressures, we will find that there is only one unknown,
for which we can solve.
Solution
We start by equating the two
liquid pressures.
Next we cancel the factor g and
substitute the known numerical
values into the resulting expression.
Then we solve the expression for
Example 5.2 continued
Assessment
Because water has a much lower density than mercury (about 1/14), it should take a
much taller water column to produce the same pressure as a mercury column, and
that is what we found. To measure normal atmospheric pressure with a water-filled
barometer, we would have to use one that is more than 10 m tall—as tall as a threestory building!
Exercise 5.2A
Calculate the height of a column of carbon tetrachloride, CCl 4 (d = 1.59 g/cm3), that
exerts the same pressure as a column of mercury (d = 13.6 g/cm3) 760 mm high.
Exercise 5.2B
A diver reaches a depth of 30.0 m. What is the pressure in atmospheres that is exerted by
this depth of water? What is the total pressure the diver experiences at this depth?
Explain.
Example 5.3 A Conceptual Example
Without doing calculations, arrange the drawings in Figure 5.5 so that the pressures
denoted in red are in increasing order.
Analysis and Conclusions
The pressure in (a) is expressed as the depth of the liquid mercury, that is,
745 mmHg. The pressure of helium in the open-end manometer (b) is slightly
greater than Pbar, which is 762 Torr = 762 mmHg and, therefore, greater than the
pressure in (a). Although there is much less mercury in (c) than in (a), the pressure
of the mercury column in (c) is greater than in (a) because of its greater height: 75.0
cm = 750 mm. Finally, in the closed-end manometer in (d), the difference in the two
mercury levels, 735 mm, is the actual gas pressure.
(d)
PNe
735 mmHg
<
<
<
(a)
PHg(l)
745 mmHg
<
<
<
(c)
Patm
750 mmHg
<
<
<
(b)
PHe
above 762 mmHg
Example 5.3 continued
Exercise 5.3A
Place a barometric pressure of (e) 101 kPa into the order of increasing pressures
established in Example 5.3. If possible, also place a barometric pressure of (f) 103 kPa in
the order. If not possible, explain why.
Exercise 5.3B
Use principles from this section to explain (a) how a beverage is transferred from glass to
mouth when one uses a soda straw and (b) why an old-fashioned hand-operated pump
cannot raise water from a well if the water level is more than about 30 feet below the
pump.
Example 5.4
A helium-filled party balloon has a volume of 4.50 L at sea level, where the atmospheric
pressure is 748 Torr. Assuming that the temperature remains constant, what will be the
volume of the balloon when it is taken to a mountain resort at an altitude of 2500 m,
where the atmospheric pressure is 557 Torr?
Strategy
We can solve the mathematical expression of Boyle’s law for Vfinal and calculate its
value from the information given.
Solution
Using Equation (5.2), we derive the equation for Vfinal:
Then we can substitute known values for the three variables on the right side of the
equation:
Assessment
Because the final pressure is less than the initial pressure, we expect the final
volume to be greater than the initial volume, and it is.
Example 5.4 continued
Exercise 5.4A
A sample of helium occupies 535 mL at 988 Torr and 25 °C. If the sample is transferred
to a 1.05-L flask at the same temperature, what will be the gas pressure in the flask?
Exercise 5.4B
A sample of air occupies 73.3 mL at 98.7 kPa and 0 °C. What volume will the air occupy
at 4.02 atm and the same temperature?
Example 5.5 An Estimation Example
A gas is enclosed in a cylinder fitted with a piston. The volume of the gas is 2.00 L at 398
Torr. The piston is moved to increase the gas pressure to 5.15 atm. Which of the
following is a reasonable value for the volume of the gas at the greater pressure?
0.20 L
0.40 L
1.00 L
16.0 L
Analysis and Conclusions
The initial pressure (398 Torr) is about 0.5 atm. An increase in pressure to 5.15 atm
is about a tenfold increase. As a result, the volume should drop to about one-tenth
of its initial value of 2.00 L. A final volume of 0.20 L is the most reasonable
estimate. (The calculated value is 0.203 L.)
Exercise 5.5A
A gas is enclosed in a 10.2-L tank at 1208 Torr. Which of the following is a reasonable
value for the pressure when the gas is transferred to a 30.0-L tank?
0.40 atm
25 lb/in2
400 mmHg
3600 Torr
Exercise 5.5B
Which of the following is a reasonable estimate of the pressure-volume product of gas
in the 30.0-L tank of Exercise 5.5A? (a) 3.6 x 104 Torr L, (b) 4 x 103 mmHg L,
(c) 1.6 x 104 kg/m2, (d) 1.6 kPa m3
Example 5.6
A balloon indoors, where the temperature is 27 °C, has a volume of 2.00 L. What will
its volume be outdoors, where the temperature is –23 °C? (Assume no change in the
gas pressure.)
Strategy
We can use Charles’s law to determine how the change in temperature affects the
volume, but we must consider the temperature change on the absolute (Kelvin)
temperature scale.
Solution
First, we convert both temperatures to the Kelvin scale, using Equation (5.3):
Now we apply Charles’s law in the form of Equation (5.5) and solve for Vfinal:
Assessment
Because the final temperature is lower than the initial temperature, we expect the
final volume to be smaller than the initial volume, and it is.
Example 5.6 continued
Exercise 5.6A
A sample of hydrogen gas occupies 692 L at 602 °C. If the pressure is held constant, what
volume will the gas occupy after cooling to 23 °C?
Exercise 5.6B
The balloon described in Example 5.6 must be at a particular temperature in order to have
a volume of 2.25 L. Assuming that the pressure remains constant, find this temperature in
kelvins and in degrees Celsius.
Example 5.7 An Estimation Example
A sample of nitrogen gas occupies a volume of 2.50 L at –120 °C and 1.00 atm pressure.
To which of the following approximate temperatures should the gas be heated in order to
double its volume while maintaining a constant pressure?
–240 °C
–60 °C
–12 °C
30 °C
Analysis and Conclusions
The initial temperature is about 150 K (–120 + 273.15 ≈ 150 K). The final
temperature must be twice the initial temperature—about 300 K—if the volume is
to double. The Celsius temperature corresponding to 300 K is about 30 °C (30 +
273.15 ≈ 300 K). Notice that the four temperatures from which to choose are all
multiples or fractions of –12 °C, but this has nothing to do with our answer because
the relationship between volume and temperature must be based on the Kelvin scale
and not the Celsius scale.
Exercise 5.7A
If the gas in Example 5.7—initially 2.50 L at –120 °C and 1.00 atm—is brought to a
temperature of 180 °C while a constant pressure is maintained, which of the following is
the approximate final volume?
(a) 3.75 L
(b) 5.0 L
(c) 7.5 L
(d) 10.0 L
Exercise 5.7B
A sample of gas is initially 1.50 L at 50 °C and 0.60 atm pressure. If this gas is brought to
a final temperature of 150 °C while the volume is held constant, which of the following is
a close estimate of the final pressure?
(a) 1.8 atm
(b) 0.8 atm
(c) 0.5 atm
(d) 0.2 atm
Example 5.8
Calculate the volume occupied by 4.11 kg of methane gas, at CH4(g), at STP.
Strategy
We must first convert the mass of gas to an amount in moles and then use the molar
volume as a conversion factor to get the volume of the gas at STP.
Solution
We can do all this in a single setup, where the conversion factor based on molar
volume is shown in red.
Exercise 5.8A
What is the mass of propane, C3H8, in a 50.0-L container of the gas at STP?
Exercise 5.8B
Solid carbon dioxide, called dry ice, is useful in maintaining frozen foods because it
vaporizes to CO2(g) rather than melting to a liquid. How many liters of CO2(g), measured
at STP, will be produced by the vaporization of a block of dry ice
(d = 1.56 g/cm3) that measures 12.0 in. x 12.0 in. x 2.00 in.?
Example 5.9
The flasks pictured in Figure 5.11 contain O2(g), the one on the left at STP and the one on
the right at 100 °C. What is the pressure at 100 °C?
Strategy
We start with Equation (5.8),
where the subscript 1 represents the initial condition (STP) and the subscript 2
represents the final condition. Then we solve for the final pressure, P2.
Solution
Assessment
Because the final temperature is higher than the initial temperature, we expect the
final pressure to be greater than the initial pressure; and it is. Note that, although we
need to know that the amount and volume of O2(g) remain constant, we do not need
to know the value of either. Also, we do not need to know that the gas is O 2; it
could be some other gas.
Example 5.9 continued
Exercise 5.9A
Aerosol containers often carry the warning that they should not be heated. Suppose an
aerosol container was filled with a gas at 2.5 atm and 22 °C. If there is the possibility that
the container may rupture if the pressure exceeds 8.0 atm, at what temperature is rupture
likely to occur?
Exercise 5.9B
In the manner we used to derive Amontons’s law relating gas pressure and temperature,
derive an expression relating the pressure and amount of gas when both the temperature
and volume remain constant. Rationalize your result in terms of the kinetic-molecular
theory and a sketch in the manner of Figure 5.11.
Example 5.10
What is the pressure exerted by 0.508 mol O2 in a 15.0-L container at 303 K?
Strategy
Knowing n, V, and T of a gas, we can use the ideal gas law to solve for its pressure.
Solution
First, we solve the ideal gas law, Equation (5.9), for pressure,
P, by dividing both sides of the equation by the volume,
V, and then canceling V on the left side of the equation.
Because data are given in the units mol, L, and K, we can use R = 0.0821 L
atm mol–1 K–1 and substitute the data directly into the equation.
Our result will be a pressure in atmospheres.
Example 5.10 continued
Assessment
An important check is for the proper cancellation of units, leading to an answer in
atm. Also, compare the O2(g) with one mole of gas at STP. The 0.508 mol O2(g), if
confined to 22.4 L at 273 K, would exert a pressure of about 0.5 atm. Because the
gas is confined to 15.0 L, its pressure should exceed 0.5 atm, but still be less than 1
atm. The pressure increase on raising the temperature to 303 K is small (increasing
by 303/273). The answer (0.842 atm) is reasonable.
Exercise 5.10A
How many moles of nitrogen gas, N2, are there in a sample that occupies 35.0 L at a
pressure of 3.15 atm and a temperature of 852 K?
Exercise 5.10B
If volume and temperature are held constant, how many moles of N2(g) should be added
to the gas described in Exercise 5.10A to increase the pressure to 5.00 atm?
Example 5.11
What is the volume occupied by 16.0 g ethane gas (C2H6) at 720 Torr and 18 °C?
Strategy
We will need to solve the ideal gas equation for volume, V, and use an appropriate
unit for each variable in the equation.
Solution
We can use the units that are consistent with the value R = 0.0821 L atm mol–1 K–1,
as shown here.
Finally, we can substitute the converted data for the variables n, T, and P into the
ideal gas equation, which we have solved for V.
Example 5.11 continued
Assessment
For 0.5 mol C2H6 at STP, the volume would be 11.2 L. The actual amount of C2H6
is slightly more than 0.5 mol; the temperature is somewhat higher than 273 K; and
the pressure (720 Torr) is somewhat less than 1 atm. All of these factors would
make the answer somewhat larger than 11.2 L, and we find that it is (13.4 L).
Exercise 5.11A
What is the temperature, in degrees Celsius, at which 15.0 g O2 exerts a pressure of 785
Torr in a volume of 5.00 L?
Exercise 5.11B
How many grams of N2(g), at 25.0 °C and 734 Torr, will occupy the same volume as 25.0
g O2(g) at 30.0 °C and 755 Torr?
Example 5.12
If 0.550 g of a gas occupies 0.200 L at 0.968 atm and 289 K, what is the molecular mass
of the gas?
Strategy
Here, the mass (m) of gas is given, and in an approach similar to that used in
Example 5.11, we will solve the ideal gas equation for the number of moles (n). The
molar mass is M = m/n, and the molecular mass (in atomic mass units) is
numerically equal to the molar mass (in g/mol).
Solution
First, let’s calculate the amount of gas, in moles, from the ideal gas equation.
Now we can use the known mass and the calculated number of moles to determine
the molar mass.
The molar mass of the gas is 67.4 g/mol, and the molecular mass is therefore
67.4 u.
Example 5.12 continued
Assessment
Our answers have the correct units for molar mass and molecular mass. Moreover,
the values seem to be in the correct range. The molecular mass of a gaseous
substance at about STP cannot be less than 1 u (that of H atoms), nor is it likely to
be very large (say, greater than 100–200 u). At about STP, substances that have high
molecular masses are most likely to be liquids or solids (as discussed in Chapter
11).
Exercise 5.12A
Calculate the molar mass of a gas if 0.440 g occupies 179 mL at 741 mmHg and 86 °C.
Exercise 5.12B
If 8.0 g CH4(g) occupies the same volume at 0 °C as 8.0 g O2(g) at STP, what is the
pressure of the CH4(g)?
Example 5.13
Calculate the molecular mass of a liquid that, when vaporized at 100 °C and 755 Torr,
yields 185 mL of vapor that has a mass of 0.523 g.
Strategy
We can use Equation (5.10) to calculate molar mass from our knowledge of m, T, P,
and V. Then we can establish molecular mass from molar mass.
Solution
We must convert from mL to L and from °C to K, but we can leave the pressure
in Torr and use the corresponding value of R from Table 5.3, that is, 62.364 L
Torr mol–1 K–1.
Now we can substitute these data into Equation (5.10).
The molar mass is 87.1 g/mol, and the molecular mass is 87.1 u.
Example 5.13 continued
Exercise 5.13A
Calculate the molecular mass of a liquid that, when vaporized at 98 °C and 715 mmHg,
yields 121 mL of vapor having a mass of 0.471 g.
Exercise 5.13B
Diacetyl, a substance that contributes to the characteristic flavor and aroma of butter,
consists of 55.80% C, 7.03% H, and 37.17% O by mass. In the gaseous state at 100 °C
and 747 mmHg, a 0.3060-g sample of diacetyl occupies a volume of 111 mL. Establish
the molecular formula of diacetyl.
Example 5.14
Calculate the density of methane gas, CH4, in grams per liter at 25 °C and 0.978 atm.
Strategy
We could do this as a two-step problem in which we (1) determine the number of
moles of CH4 in 1.00 L of the gas at the stated temperature and pressure and then
(2) determine the mass of this gas. To do so, however, is really the same as solving
Equation (5.11). So let’s do it the (simpler) latter way.
Solution
Exercise 5.14A
Calculate the density of ethane gas (C2H6) in grams per liter at 15 °C and 748 Torr.
Exercise 5.14B
What is the molar mass of a gaseous alkane hydrocarbon having a density of 2.42 g/L at
20.0 °C and 762 Torr? What is the molecular formula of the gas?
Example 5.15
Under what pressure must O2(g) be maintained at 25 °C to have a density of 1.50 g/L?
Strategy
We can use Equation (5.11), d = MP/RT, to solve for any one of the five quantities
(here, P) if the other four are known.
Solution
First, we solve Equation (5.11) for the unknown pressure:
Then, we substitute the known quantities into the right side:
Assessment
To ensure that we have used a correct variation of the ideal gas equation, we can
use unit cancellation as a check. Here, all units cancel except for the desired atm.
Exercise 5.15A
To what temperature must propane gas (C3H8) at 785 Torr be heated to have a density of
1.51 g/L?
Exercise 5.15B
At what temperature will the density of a sample of O2(g) at 725 Torr be the same as the
density of NH3(g) at 22.5 °C and 1.45 atm?
Example 5.16
How many liters of O2(g) are consumed for every 10.0 L of CO2(g) produced in the
combustion of liquid pentane, C5H12, if all volumes are measured at STP?
Strategy
The fact that pentane and water are liquids does not affect our calculation because
the substances of interest, CO2 and O2, are both gases. The actual temperatures and
pressures before, during, and after the combustion reaction are not relevant, as long
as each gas volume is measured at the same temperature and pressure. The
condition of STP is relevant only in establishing the particular temperature and
pressure at which the two gas volumes are compared.
Solution
We begin by writing a balanced equation for the combustion reaction:
The fundamental equivalence in the reaction is
Based on the equal volume-equal numbers hypothesis, we can substitute a volume
unit for the unit mol:
This equivalence provides the stoichiometric factor we need. Thus,
Example 5.16 continued
Exercise 5.16A
What volume of oxygen gas is required to burn 0.556 L of propane gas, C 3H8, if both gas
volumes are measured at the same temperature and the same pressure?
Exercise 5.16B
What volume of O2(g) is consumed in the combustion of 125 g of gaseous dimethyl ether,
(CH3)2O, if both gas volumes are measured at the temperature and pressure at which the
density of dimethyl ether is 1.81 g/L?
Example 5.17
In the chemical reaction used in automotive air-bag safety systems, N2(g) is produced by
the decomposition of sodium azide, NaN3, at a somewhat elevated temperature:
What volume of N2(g), measured at 25 °C and 0.980 atm, is produced by the
decomposition of 62.5 g NaN3?
Strategy
For this reaction, we must relate the number of moles of N2(g) to the number of
moles of NaN3(s), using their molar ratio. Then, using the ideal gas equation, we
can determine the volume of nitrogen from the number of moles of nitrogen.
Solution
First, we can determine the amount of N2(g) produced when 62.5 g NaN3
decomposes. Here we use the molar mass (1 mol NaN3/65.01 g NaN3) and a
stoichiometric factor from the balanced equation (3 mol N2/2 mol NaN3) as
conversion factors:
Then, we can use the ideal gas equation to determine the volume of 1.44 mol N 2(g)
under the stated conditions:
The solution is also outlined in the stoichiometry diagram of Figure 5.13.
Example 5.17 continued
Assessment
We can establish an approximate answer rather easily. Because the quantity of
NaN3 is about 1 mol, the number of moles of N2 should be about equal to the
stoichiometric factor, that is, 3/2 = 1.5. Because the gas conditions are not far
removed from STP, the 1.5 mol of gas should occupy a volume of about
1.5 x 22.4 L = 33.6 L. This is close to our calculated value of 35.9 L.
Exercise 5.17A
Quicklime (CaO), used in the construction industry, is manufactured by heating limestone
(CaCO3) to decompose it:
How many liters of CO2(g) at 825 °C and 754 Torr are produced in the decomposition of
45.8 kg CaCO3(s)?
Exercise 5.17B
How many liters of cyclopentane, C5H10 (d = 0.7445 g/mL), must be burned in excess
O2(g) to produce 1.00 x 106 L of CO2(g) measured at 25.0 °C and 736 Torr?
Example 5.18
A 1.00-L sample of dry air at 25 °C contains 0.0319 mol N2, 0.00856 mol O2,
0.000381 mol Ar, and 0.00002 mol CO2. Calculate the partial pressure of N2(g) in the
mixture.
Strategy
We are given a rather complete description of the composition of air, but since each
component gas expands to fill the 1.00-L volume, the basic question becomes,
“What is the pressure exerted by 0.0319 mol N2 in a 1.00-L container at 25 °C?”
Solution
We apply Equation 5.13 for N2(g) in the dry air mixture,
Exercise 5.18A
Calculate the partial pressure of each of the other components of the air sample in
Example 5.18. What is the total pressure exerted by all the gases in the sample?
Exercise 5.18B
What is the total pressure exerted by a mixture of 4.05 g N2, 3.15 g H2, and 6.05 g He
when the mixture is confined to a 6.10-L container at 25 °C?
Example 5.19
The main components of dry air, by volume, are N2, 78.08%; O2, 20.95%; Ar, 0.93%; and
CO2, 0.04%. What is the partial pressure of each gas in a sample of air at 1.000 atm?
Strategy
Volume percent is the same as mole percent, and from the mole percents we can
write the mole fractions. Thus, 78.08 volume percent N2 is the same as 78.08 mole
percent N2, which, in turn, is the same as a 0.7808 mole fraction of N2. We can
apply this reasoning to the other gases in the mixture, too: The mole fraction for O 2
is 0.2095; for Ar, 0.0093; and for CO2, 0.0004.
Solution
Each partial pressure can be obtained using Equation (5.16):
Exercise 5.19A
A sample of expired (exhaled) air is composed, by volume, of the following main
gaseous components: N2, 74.1%; O2, 15.0%; H2O, 6.0%; Ar, 0.9%; and CO2, 4.0%. What
is the partial pressure of each gas in the expired air at 37 °C and 1.000 atm?
Exercise 5.19B
A 75.0-L sample of natural gas at 23.5 °C contains methane (CH4) at a partial pressure of
505 Torr, ethane (C2H6) at 201 Torr, propane (C3H8) at 43 Torr, and 10.5 g of butane
(C4H10). Calculate the mole fraction of each component.
Example 5.20 A Conceptual Example
Describe what must be done to change the gaseous mixture of hydrogen and helium
shown in Figure 5.15a to the conditions illustrated in Figure 5.15b.
Analysis and Conclusions
First, we assess the situation in (a): Because the chosen volume is 22.4 L, the
temperature is 273 K (0 °C), and the total amount of gas is 1.00 mol, the total
pressure must be 1.00 atm. This condition is the familiar molar volume of an ideal
gas at STP (22.4 L).
Next, we appraise the situation in (b): The volume and temperature remain as
they were in (a), but the pressure increases to 3.00 atm. This tripling of the pressure
while the volume and temperature are fixed requires a tripling of the amount of gas.
Therefore, there must be 3.00 mol gas in (b) instead of the 1.00 mol in (a). Because
is 2.00 atm, the partial pressure of the
the partial pressure
other gas or gases must be 1.00 atm. The corresponding mole fractions are 2/3 for
H2 and 1/3 for the other gas or gases. The 3.00-mol mixture of gases in (b) must
consist of 2.00 mol H2 and 1.00 mol of another gas or gases.
In order to get from (a), where we have 0.50 mol H2 and 0.50 mol He, to (b),
where we have 2.00 mol H2 and 1.00 mol of some other gas or gases, we must add
1.50 mol H2 and 0.50 mol of the other gas or gases to the flask. Note that this 0.50
mol could be 0.50 mol He, but it could also be any single gas or combination of
gases other than hydrogen.
Example 5.20 continued
Exercise 5.20A
Why is it not possible to achieve the change between (a) and (b) in Figure 5.15 by adding
hydrogen gas alone? Why can it not be achieved by adding helium gas alone? Why is it
necessary that some hydrogen be added but not necessary that any helium be added?
Exercise 5.20B
Which of the following actions would you take to change the pressure of 2.2 L of N 2 at
STP to 2.0 atm at 400 °C while maintaining a constant volume: (a) add 0.10 mol N 2;
(b) add 0.020 mol N2; (c) release 0.10 mol N2; (d) release 0.46 L of N2, measured
at STP?
Example 5.21
Hydrogen produced in the following reaction is collected over water at 23 °C when the
barometric pressure is 742 Torr:
What volume of the “wet” gas will be collected in the reaction of 1.50 g Al(s) with excess
HCl(aq)?
Strategy
The “wet” gas is the mixture of hydrogen gas and water vapor. However, because
the two gases are found in the same collection vessel, they occupy the same
volume, which means we need to calculate the volume of only one of them.
Although we can easily get the pressure of the water vapor from Table 5.4, that
information alone is not enough to calculate the volume of the water vapor using
the ideal gas law because, in order to work the problem that way, we also need to
know the number of moles of water vapor, and we do not. We can, however, gather
the data necessary to determine the volume of hydrogen, by using Dalton’s law of
partial pressures and the ideal gas law.
Solution
To begin, we know the temperature, 23 °C, and we can calculate the partial pressure
of hydrogen using Dalton’s law of partial pressures.
Example 5.21 continued
Solution continued
We can determine the number of moles of hydrogen from the stoichiometry of the
reaction.
With this information, we can solve the ideal gas equation for V and then calculate
the volume of hydrogen, which is equal to the volume of the “wet” gas.
Exercise 5.21A
Hydrogen gas is collected over water at 18 °C. The total pressure inside the collection jar
is set at the barometric pressure of 738 Torr. If the volume of the gas is 246 mL, what
mass of hydrogen is collected? What is the mass of the wet gas? (Hint: What is the mass
of the water vapor?)
Exercise 5.21B
A sample of KClO3 was heated to decompose it to potassium chloride and oxygen.
The O2 was collected over water at 21 °C and a barometric pressure of 746 mmHg.
A 155-mL volume of the gaseous mixture was obtained. What mass of KClO3 was
decomposed?
Example 5.22
Without doing detailed calculations, determine which of the following is a likely value for
urms of O2 molecules at 0 °C, if urms of H2 at 0 °C is 1838 m/s.
(a) 115 m/s
(b) 460 m/s
(c) 1838 m/s
(d) 7352 m/s
(e) 29,400 m/s
Analysis and Conclusions
We could calculate urms for O2 at 0 °C from Equation (5.23), but let’s see how we
can come up with a reasonable answer with a minimum of calculation.
We must not be misled by response (c)—that the H2 and O2 molecules travel at
the same speed. Because they are at the same temperature, they do have equal
average translational kinetic energies 1/2(mu). However, because O2 molecules
have a greater mass (m) than H2 molecules, the O2 molecules must have a lesser
velocity (u). Thus O2 molecules, on average, should move more slowly. This fact
eliminates responses (d) and (e), as well as (c).
The molar mass ratio of O2 to H2 is 32/2 = 16. Response (a), 115 m/s, is onesixteenth of 1838 m/s, but this answer is incorrect because the molecular speeds are
inversely related not to the molar mass ratio but to the square root of this ratio:
Thus the urms speed of molecules is one-quarter that for
hydrogen, or about 460 m/s. The correct response is (b).
Exercise 5.22A
Without doing detailed calculations, determine which of the following gases has the
greatest urms:
(a) CH4(g) at 0 °C,
(b) O2(g) at 250 °C,
(c) SO2(g) at 750 °C,
(d) NH3(g) at 35 °C.
Example 5.22 continued
Exercise 5.22B
At which temperature(s) listed here will urms of O2 be greater than urms of H2 at 0 °C, that
is, greater than 1838 m/s?
(a) 1000 K
(b) 2000 K
(c) 3000 K
(d) 4000 K
(e) 5000 K
Example 5.23
If compared under the same conditions, how much faster than helium does hydrogen
effuse through a tiny hole?
Strategy
To answer this question, we can set up a ratio of the two effusion rates in
accordance with Graham’s law (Equation 5.24).
Solution
If we place the rate of effusion of hydrogen,
, in the numerator on the left, its
molar mass,
, must go into the denominator under the square-root sign on the
right. The situation is reversed for helium:
The ratio of the two rates is 1.41, which means
1.41 times faster than helium.
= 1.41 x
Hydrogen effuses
Exercise 5.23A
Which effuses faster, N2 or Ar, when the two gases are compared under the same
conditions? How much faster?
Exercise 5.23B
Which effuses faster, nitrogen monoxide or dimethyl ether, (CH3)2O, when the two gases
are compared under the same conditions? How much faster, expressed as a percentage?
Example 5.24
One percent of a measured amount of Ar(g) escapes through a tiny hole in 77.3 s. One
percent of the same amount of an unknown gas escapes under the same conditions in 97.6
s. Calculate the molar mass of the unknown gas.
Strategy
Here we are given information regarding effusion times. From Equation (5.25), we
can relate effusion times to effusion rates and, in turn, establish a relationship
between effusion time and molar mass. Thus, the molar mass of the unknown gas
can be calculated from the ratio of the effusion times.
Solution
First, let’s get the appropriate form for an expression relating effusion times and
molar masses. To do this, we can combine Equation (5.25), relating effusion time to
effusion rate, and Equation (5.24), relating effusion rate to molar mass.
This result tells us that effusion time is directly proportional to the square root of
molar mass.
Now the only variable that is not known is Munk.
Example 5.24 continued
Solution continued
We solve for Munk by squaring both sides of the final equation.
Assessment
We can see that this is a reasonable answer. Because the unknown gas effuses more
slowly than does Ar, the unknown molar mass must be greater than that
of Ar.
Exercise 5.24A
Two percent of a sample of N2(g) effuses from a tiny opening in 57 s. Two percent of the
same amount of an unknown gas escapes under the same conditions in 83 s. Calculate the
molar mass of the unknown gas.
Exercise 5.24B
Five percent of a sample of O2(g) effuses from an orifice in 123 s. How long should it
take five percent of the same amount of butane, C4H10, to effuse under the same
conditions?
Cumulative Example
Two cylinders of gas are used in welding. One cylinder is 1.2 m high and 18 cm in
diameter, containing oxygen gas at 2550 psi and 19 °C. The other is 0.76 m high and 28
cm in diameter, containing acetylene gas (C2H2) at 320 psi and 19 °C. Assuming
complete combustion, which tank will be emptied, leaving unreacted gas in the other?
Strategy
Our basic task is to determine the limiting reactant in the combustion, that is, C 2H2
or O2. For this, we must compare the numbers of moles of reactants in the two
cylinders to the requirements dictated by the stoichiometric relationship between
them (that is, the stoichiometric factor). We obtain the stoichiometric factor from
the balanced equation for the combustion reaction. To determine the number of
moles of O2 and of C2H2 available for reaction, we must use the ideal gas law; but
to obtain the necessary gas volumes we need first to calculate the volumes of the
cylinders from the cylinder dimensions. When we have both the stoichiometric
factor and the numbers of moles of the two gases, we can determine the limiting
reactant. The cylinder containing the limiting reactant will be emptied in the
reaction.
Solution
We begin with the unbalanced equation for the combustion, which forms carbon
dioxide and water.
Stoichiometric coefficients are added to balance the equation.
Cumulative Example continued
Solution continued
Beginning with the oxygen cylinder, the volume in cubic centimeters is calculated
from the data given.
We now solve for the number of moles of oxygen that occupies this volume. Table
5.2 gives the conversion factor we need between psi (pounds per square inch) and
atmospheres.
We repeat the calculations of volume and number of moles for acetylene.
Because the question merely asks which gas is completely consumed, we do not
need to determine the amount of product. We just need to find which reactant is
limiting.
Cumulative Example continued
Solution continued
A simple calculation of the number of moles of C2H2 needed to react with all the
available O2 will suffice.
We have available only 43 mol of C2H2, meaning all the C2H2 will be used up.
Assessment
We can check several parts of our work. The volume of the oxygen cylinder
(30.5 L) is somewhat more than the molar volume at STP (22.4 L), and the pressure
is much higher than 150 atm, telling us there should be much more than 150 mol of
oxygen gas in the cylinder. Likewise, the acetylene cylinder holds about two molar
volumes at roughly 20 atm, or approximately 40 mol of gas. The amount of C 2H2
that could be consumed by the available O2 is more than 60 mol (2/5 x 150), but
with only about 40 mol C2H2 available, the acetylene tank will be emptied.