CHAPTER 5
The Gaseous State
CHAPTER TERMS AND DEFINITIONS
Numbers in parentheses after definitions give the text sections in which the terms are explained. Starred
terms are italicized in the text. Where a term does not fall directly under a text section heading,
additional information is given for you to locate it.
pressure
force exerted per unit area of surface (5.1)
acceleration*
change of speed per unit time (5.1)
pascal (Pa)
SI unit of pressure; 1 Pa = 1 kg/(m ∙ s2) (5.1)
barometer
device for measuring the pressure of the atmosphere (5.1)
manometer
device that measures the pressure of a gas or liquid in a sealed vessel (5.1)
millimeters of mercury (mmHg or torr) traditional unit of pressure equal to that exerted by a 1mm column of mercury at 0.00C in a barometer or manometer (5.1)
atmosphere (atm)
exact (5.1)
bar
traditional unit of pressure equal to exactly 760 mmHg; 1 atm = 101.325 kPa,
a unit of pressure equal to 1  105 Pa, slight less than 1 atm (5.1)
compressibility*
ability to be squeezed into a smaller volume by the application of pressure (5.2)
Boyle’s law
the volume of a sample of gas at a given temperature varies inversely with the applied
pressure (5.2)
linearly* term describing how one variable changes with the change in another variable if a plot of
the two variables gives a straight line (5.2)
extrapolate*
Kelvin scale*
273.15 (5.2)
to extend a line beyond the plotted data points (5.2)
absolute temperature scale on which the units (kelvins, K) are given by K = C +
Charles’s law
the volume occupied by any sample of gas at a constant pressure is directly
proportional to the absolute temperature (5.2)
law of combining volumes*
the volumes of reactant gases at a given pressure and temperature are
in ratios of small whole numbers (5.2)
Avogadro’s law
equal volumes of any two gases at the same temperature and pressure contain the
same number of molecules (5.2)
molar gas volume (Vm)
pressure (5.2)
volume occupied by one mole of any gas at a given temperature and
standard temperature and pressure (STP)
be 0C and 1 atm pressure (5.2)
molar gas constant (R)
reference conditions for gases chosen by convention to
constant of proportionality relating the molar volume of a gas to T/P (5.3)
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Chapter 5: The Gaseous State
105
ideal gas law
mathematical expression combining all the gas laws and relating the volume (V),
pressure (P), Kelvin temperature (T), and moles (n) of a gas to the molar gas constant R; PV = nRT (5.3)
Amontons’s law* the pressure of a given amount of gas at a fixed volume is proportional to the
absolute temperature (Example 5.5)
partial pressure
pressure exerted by a particular gas in a gas mixture (5.5)
Dalton’s law of partial pressures
the sum of the partial pressures of all the different gases in a
mixture is equal to the total pressure of the mixture (5.5)
mole fraction
fraction of moles of a component gas in the total moles of a gas mixture (5.5)
vapor pressure* partial pressure of the molecules of a substance in the gaseous state in the presence
of the liquid (or solid) substance (5.5)
kinetic-molecular theory of gases (kinetic theory)
random motion (5.6, introductory section)
postulates*
idea that a gas consists of molecules in constant
basic statements from which all conclusions or predictions of a theory are deduced (5.6)
ideal gas* gas that follows the ideal gas law; its molecules have essentially no volume of their own
and no attraction for each other (5.6)
intermolecular forces*
forces of attraction or repulsion between molecules (5.6)
root-mean-square (rms) molecular speed (u) type of average molecular speed, or the speed of a
molecule that has the average molecular kinetic energy; can be shown to equal
u=
3RT
Mm
where R is the molar gas constant, T is the kelvin temperature, and Mm is the molar mass for the gas
(5.7)
diffusion
(5.7)
process whereby a gas spreads out through another gas to occupy the space uniformly
effusion escape of a gas through a small hole into a vacuum at the same velocity it had in the
container (5.7)
Graham’s law of effusion the rate of effusion of gas molecules from a particular hole is inversely
proportional to the square root of the molecular mass of the gas at constant T and P (5.7)
enrichment*
process used to increase the percentage of one isotope in a sample (5.7)
van der Waals equation equation similar to the ideal gas law, but it includes two constants, a and b,
to account for deviations from ideal behavior (5.8)
CHAPTER DIAGNOSTIC TEST
1.
The measured pressure of a gas is 745 mmHg. What is this pressure in atmospheres?
2.
Calculate the final pressure of a gas when 15.0 L of the gas at 743 mmHg is transferred to a 39.2L container at the same temperature.
3.
At 1.50 atm and 23C, a gas occupies 13.5 L. What is its volume in liters at 1.04 atm and 5C?
4.
If 2.35 L CO2 at some pressure and temperature contains 0.0648 mol CO2, how many moles of
helium atoms are there in 2.25 L of helium at the same pressure and temperature?
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106
Chapter 5: The Gaseous State
5.
Starting with the ideal gas law, derive the relationship between the pressure and temperature of a
gas.
6.
What is the temperature of an ideal gas at 531 mmHg if the density is 4.60 g/L and its molecular
mass is 63.8 g/mol?
a.
155C
b.
243 K
c.
6.88 K
d.
118 K
e.
None of the above
7.
What volume of CO2 can be prepared from the reaction of 152 g CaCO3 with 1.15 mol HCl at
15.0C and 765 mmHg?
8.
A 5.000-L sample of gas at 125C has the following composition: 0.765 g N2, 0.843 g O2, and
0.684 g H2O.
9.
a.
What is the mole fraction of each gas in the mixture?
b.
What is the partial pressure of each component gas in the mixture?
The total pressure of a mixture of gases at 301 K was 973 mmHg. Analysis of this mixture
showed the composition to be 3.33 mol H2, 1.69 mol NO, and 0.488 mol O2. Calculate the partial
pressure of each gas.
10. A sample of N2 was collected over water at 26.1C and a pressure of 1.07 atm. The observed
volume of gas was 0.783 L. Calculate the mass of the dry N2. (Vapor pressure of water at 26.1C
is 25.4 mmHg.)
11. Which of the following statements about the kinetic theory of gases is (are) correct?
a.
Molecules are viewed as point masses, and their masses can be neglected.
b.
Weak van der Waals forces are responsible for the collisions between gas molecules.
c.
An increase in temperature causes an increase in the number of gas molecules in a sample.
d.
The pressure of a gas decreases as the volume of individual gas molecules increases.
e.
The average kinetic energy of gas molecules is proportional to the absolute temperature of
the gas.
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Chapter 5: The Gaseous State
107
12. Which of the following statements about the following figure is (are) correct?
a.
At a higher temperature, there are more gas molecules.
b.
A greater fraction of higher-speed molecules is observed at a lower temperature.
c.
A higher temperature is associated with molecules having a higher speed.
d.
A temperature increase causes an increase in the more probable speeds and in the average
speeds of molecules.
e.
None of the above are correct.
13. The rate of effusion of Kr compared with that of N2 ( rKr / rN ) is
2
a.
1.73.
b.
0.578.
c.
2.99.
d.
0.334.
e.
none of the above.
14. Calculate the rms molecular speed of CO2 molecules at room temperature, 25C.
ANSWERS TO CHAPTER DIAGNOSTIC TEST
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
9.80  101 atm (5.1, PS Sk. 1)
2.
284 mmHg (5.2, PS Sk. 2)
3.
17.6 L (5.2, PS Sk. 2)
4.
0.0620 mol (5.2)
5.
P  T or P = CT, constant n and V, where C is a constant. (5.3, PS Sk. 3)
6.
d (5.3, PS Sk. 5)
7.
13.5 L (5.4, PS Sk. 6)
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108
Chapter 5: The Gaseous State
8.
9.
a.
0.298, 0.287, 0.415
b.
0.179 atm N2, 0.172 atm O2, 0.248 atm H2O (5.5, PS Sk. 7)
PH2 = 588 mmHg, PNO = 299 mmHg, PO2 = 86.2 mmHg (5.5, PS Sk. 7)
10. 0.925 g (5.5, PS Sk. 8)
11. e (5.6)
12. d (5.7)
13. b (5.7, PS Sk. 10)
14. 411 m/s (5.7, PS Sk. 9)
SUMMARY OF CHAPTER TOPICS
To be able to solve the problems dealing with gases, you will have to know the gas laws presented in
the chapter. Table 5.1 is presented to help you learn them.
Table 5.1 Gas Laws
Name of Law
Equation
Conditions
Boyle’s law
PV = constant, or Pf Vf = PiVi
Constant T, n
Charles’s law
V
V
V
= constant, or f = i
Tf
Ti
T
Constant P, n
(Amontons’s law)
P
P
P
= constant, or f = i
T
Tf
Ti
Constant V, n
Combined gas law
Pf Vf
PV
= i i
Tf
Ti
For fixed amount of gas
Avogadro’s law
Vm = specific constant
Ideal gas law
PV = nRT
Dalton’s law of partial
pressures
P = PA + PB + PC + …
Graham’s law
van der Waals equation
Rate of effusion of gas
Rate  1/ M m
(P +
n2 a
) (V  nb) = nRT
V2
Depending on T, P;
independent of gas
Same container,
constant T and P
Moderate pressures
5.1 Gas Pressure and Its Measurement
In this section you are introduced to two instruments, the barometer and the manometer. Note how each
instrument is constructed and exactly what each measures.
Learning Objectives

Define pressure and its units.

Convert units of pressure. (Example 5.1)
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Chapter 5: The Gaseous State
109
Problem-Solving Skill
1.
Converting between units of pressure. Given the pressure of a gas in one unit of pressure (Pa,
mmHg, or atm), be able to convert to another unit of pressure (Example5.1).
Exercise 5.1
A gas in a container had a measured pressure of 57 kPa. Calculate the pressure in units of atm and
mmHg.
Wanted: pressure in atm and mmHg
Given:
pressure in units of kPa (57 kPa)
Known:
760 mmHg = 1.01325  105 Pa = 1 atm; 1 kPa = 103 Pa
Solution:
57 kPa 
1 atm
103 Pa

= 0.56 atm
1.01325  105 Pa
1 kPa
57 kPa 
760 mmHg
103 Pa

= 430 mmHg
1.01325  105 Pa
1 kPa
5.2 Empirical Gas Laws
Learning Objectives

Express Boyle’s law in words and as an equation.

Use Boyle’s law. (Example 5.2)

Express Charles’s law in words and as an equation.

Use Charles’s law. (Example 5.3)

Express the combined gas law as an equation.

Use the combined gas law. (Example 5.4)

State Avogadro’s law.

Define standard temperature and pressure (STP).
Problem-Solving Skill
2.
Using the empirical gas laws. Given an initial volume occupied by a gas, calculate the final
volume when the pressure changes at fixed temperature (Example 5.2), when the temperature
changes at fixed pressure (Example 5.3), and when both pressure and temperature change
(Example 5.4).
It is important to note that volume V is the volume occupied by the gas, not the volume of the gas
molecules themselves. This will be discussed at greater length in Section 5.6, on kinetic theory.
You can nicely illustrate the inverse relationship between pressure and volume if you blow up a
balloon, tie it, and then rather abruptly sit on it! The weight of your body decreases the volume, but the
resulting increase in the gas pressure inside bursts the balloon.
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110
Chapter 5: The Gaseous State
With eight gas laws to memorize, it is easy to get them mixed up at exam time. The following
information may help you to get the right answers when using Boyle’s law. In performing the
calculations, you will always multiply one quantity, say, P, by a ratio of the two values of the other
quantity, in this case a ratio of Vi and Vf. Observe whether the volume increases or decreases. If Vf is
greater than Vi, then, according to Boyle’s law, the pressure must decrease. Therefore, you must put the
smaller volume on top when you write the ratio. (The ratio will be less than 1.) See the note just after
the solution to Exercise 5.2.
Exercise 5.2
A volume of carbon dioxide gas, CO2, equal to 20.0 L was collected at 23C and 1.00 atm pressure.
What would be the volume of carbon dioxide if it were collected at 23C and 0.830 atm?
Given: Vi = 20.0 L
Pi = 1.00 atm
Vf = ?
Pf = 0.830 atm
(T and n remain constant.)
Known: Boyle’s law states that ViPi = VfPf.
Solution: Vf = Vi 
Pi
1.00 atm 
= 20.0 L  
 = 24 L
Pf
 0.830 atm 
Pressure decreases, so volume increases. Thus the ratio by which the initial volume is multiplied
must be greater than 1. (The larger value of P goes on top.)
Note that beginning in the section on Charles’s law, we use lowercase t for reporting temperatures in
degrees Celsius and capital T for reporting temperatures in kelvins. Be sure to use kelvins whenever
you work problems involving gases.
Using both Charles’s law and the combined gas law (combining Boyle’s and Charles’s laws), it is again
useful to think about forming the ratios of Pi and Pf , Vi and Vf , or Ti and Tf so that the answers will be
larger or smaller, as the laws state. The notes included with the following two exercise solutions stress
this point.
You will need to memorize the values of STP (0C = 273.15 K and 1 atm P). Also memorize the molar
gas volume (of any gas) at STP, 22.41 L/mol.
Exercise 5.3
If we expect a chemical reaction to produce 4.38 dm3 of oxygen, O2, at 19C and 101 kPa, what will be
the volume at 25C and 101 kPa?
Given: Vi = 4.38 dm3
Ti = 19C + 273 = 292 K
Vf = ?
Tf = 25C + 273 = 298 K
(P and n remain constant.)
Known: Charles’s law states that
Solution: Vf = Vi 
Vi
V
= f .
Ti
Tf
Tf
 298 K 
3
= 4.38 dm3  
 = 4.47 dm
Ti
 292 K 
Note that T increases, so V increases. The ratio of temperatures must be greater than 1. (The larger
value of T goes on top.)
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Chapter 5: The Gaseous State
111
Exercise 5.4
A balloon contains 5.41 dm3 of helium, He, at 24C and 101.5 kPa. Suppose that the gas in the balloon
is heated to 35C. If the helium pressure is now 102.8 kPa, what is the volume of the gas?
Given: Vi = 5.41 dm3
Pi = 101.5 kPa
Ti = 24˚C (297 K)
Vf = ?
Pf = 102.8 kPa
Tf = 35˚C (308 K)
(n remains constant.)
Known: The combined gas laws give
Vi Pi
VP
= f f .
Ti
Tf
Solution:
 Pi 
 Tf 
308 K 
 101.5 kPa 
3
 
    = 5.41 dm  


 102.8 kPa 
 297 K 
 Pf 
 Ti 
Vf = Vi  
= 5.54 dm3
Note that the pressure increases, so the pressure ratio must be less than 1 to give a decrease in the
volume. Also note that the temperature increases, so the temperature ratio must be greater than 1
to give an increase in volume.
5.3 The Ideal Gas Law
Learning Objectives

State what makes a gas an ideal gas.

Learn the ideal gas law equation.

Derive the empirical gas laws from the ideal gas law. (Example 5.5)

Use the ideal gas law. (Example 5.6)

Calculate gas density. (Example 5.7)

Determine the molecular mass of a vapor. (Example 5.8)

Use an equation to calculate gas density.
Problem-Solving Skills
3.
Deriving empirical gas laws from the ideal gas law. Starting from the ideal gas law, derive the
relationship between any two variables (Example 5.5).
4.
Using the ideal gas law. Given any three of the variables P, V, T, and n for a gas, calculate the
fourth from the ideal gas law (Example 5.6).
5.
Relating gas density and molecular mass. Given the molecular mass, calculate the density of a
gas for a particular temperature and pressure (Example 5.7), or given the gas density, calculate the
molecular mass (Example 5.8).
In solving problems using the ideal gas law, you must have your values in the units in which the gas
constant R is reported. The usual value of R is 0.0821 L ∙ atm /(K ∙ mol). Thus V must be in liters, P in
atmospheres, T in kelvins (as the capital T indicates), and the amount of gas in moles.
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112
Chapter 5: The Gaseous State
Exercise 5.5
Show that the moles of gas are proportional to the pressure for constant volume and temperature.
Known:
The ideal gas law includes variables P, V, n, and T.
Solution: Using PV = nRT, solve for n:
V 
n = P 

RT


Since R, T, and V are constant, we can write
n = P ∙ C (where C is a constant) or n  P
Exercise 5.6
What is the pressure in a 50.0-L gas cylinder that contains 3.03 kg of oxygen, O2, at 23C?
Wanted: P
Given:
V = 50.0 L; n =
3.03  103 g
= 94.69 mol O2
32.0 g/mol
T = 23C + 273 = 296 K
Known: These variables are related by the ideal gas law, which we solve for P; R = 0.0821 L ∙
atm /(K ∙ mol).
Solution:
P=
(94.69 mol)(0.0821 L • atm)(296 K)
nRT
=
= 46.0 atm
V
(50.0 L)(K • mol)
Exercise 5.7
Calculate the density of helium, He, in grams per liter at 21C and 752 mmHg. The density of air under
these conditions is 1.188 g/L. What is the difference in mass between 1 liter of air and 1 liter of helium?
(This mass difference is equivalent to the buoyant, or lifting, force of helium per liter.)
Wanted: density of He (g/L); mass difference of 1 L air and 1 L He
Given:
t = 21C, P = 752 mmHg, air density = 1.188 g/L
Known: Since density is mass per unit volume, calculating moles from mass of He in 1 L, using
PV = nRT and the atomic mass of He, gives the density. Then subtract that from the value of the
density of air.
Solution: Solve for moles He in 1 L:
n=?
P = 752 mmHg 
1 atm
= 0.9895 atm
760 mmHg
T = (21C + 273) K = 294 K
V = 1 L (exact)
n=
PV
(0.9895 atm)(1 L)(K • mol)
=
= 0.04100 mol He
(294 K)(0.0821 L • atm)
RT
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Chapter 5: The Gaseous State
113
Then solve for grams He, which is also the value of the density of He:
0.04100 mol He 
4.00 g He
= 0.1640 g He; density of He = 0.1640 g/L
mol He
Subtract densities to find the difference in mass:
1.188 g (per L air) – 0.1640 g (per L He) = 1.024 g
Exercise 5.8
A sample of a gaseous substance at 25C and 0.862 atm has a density of 2.26 g/L. What is the
molecular mass of the substance?
Wanted:
molecular mass
Given:
T = 25C + 273 = 298 K; P = 0.862 atm; density = 2.26 g/L
Known: Molecular mass (in amus) is the same number as molar mass (g/mol). We can use V =
1 L (exact). We can use the ideal gas law to solve for moles and then form a ratio of 2.26 g over
the calculated number of moles to get g/mol.
Solution: Find moles of vapor:
n=
PV
(0.862 atm)(1 L)(K • mol)
=
= 0.03523 mol
(298 K)(0.0821 L • atm)
RT
Find Mm:
Mm =
grams vapor
2.26 g
=
= 64.1 g/mol
moles vapor
0.03523 mol
Thus the molecular mass is 64.1 amu.
A Chemist Looks at: Nitric Oxide Gas and Biological Signaling
Questions for Study
1.
What is unusual about NO being a biologically important molecule?
2.
How does NO play a positive biological role in the body?
3.
Discuss the role of NO in blood pressure regulation.
4.
How does nitroglycerine work in relieving angina attacks?
Answers to Questions for Study
1.
Whereas most biologically important molecules are large molecular species, NO is a small
diatomic molecule.
2.
NO molecules released by white blood cells surround bacterial or tumor cells and destroy them by
interfering with cell processes.
3.
The cells lining the interior walls of blood vessels release NO, which diffuses into the blood and
muscle tissue of the vessel wall. In the tissue, NO relaxes and dilates the tissue. This causes a
lowering of blood pressure.
4.
The decomposition of nitroglycerine in the body produces NO, which relaxes blood vessels. Thus
blood can flow more freely.
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114
Chapter 5: The Gaseous State
5.4 Stoichiometry Problems Involving Gas Volumes
Learning Objective

Solving stoichioimetry problems involving gas volumes. (Example 5.9)
Problem-Solving Skill
6.
Solving stoichiometry problems involving gas volumes. Given the volume (or mass) of one
substance in a reaction, calculate the mass (or volume) of another produced or used up (Example
5.9).
Now that you are familiar with the properties and behavior of gases, you can use the stoichiometry you
learned in Chapter 4 to calculate volumes of gaseous reactants and products in chemical equations.
Stoichiometry problems involving gases have one step that causes problems for most students. This step
is the conversion between moles of gas and the volume occupied by 1 mole of gas. For any gas at STP,
the molar volume is always 22.41 L/mol. In most cases, however, the gas is not at STP; you then must
use the ideal gas law either to find the moles of gas when volume is given or to find the volume once
the moles of gas are known.
Exercise 5.9
How many liters of chlorine gas, Cl2, can be obtained at 40C and 787 mmHg from 9.41 g of hydrogen
chloride, HCl, according to the following equation?
2KMnO4(s) + 16HCl(aq)  8H2O(l) + 2KCl(aq) + 2MnCl2(aq) + 5Cl2(g)
Wanted:
VCl2
Given:
P = 787 mmHg 
1 atm
= 1.036 atm
760 mmHg
T = 40C + 273 = 313 K (Assume 40C is exact.)
Known: We can get moles of Cl2 from the reaction stoichiometry and then use the ideal gas law
to get the volume.
Solution: Find moles of Cl2:
n = 9.41 g HCl 
5 mol Cl 2
1 mol HCl

= 0.08057 mol Cl2
16 mol HCl
36.5 g HCl
Find VCl :
2
V=
nRT
(0.08057 mol)(0.0821 L • atm)(313 K)
=
= 2.00 L
(1.036 atm)(K • mol)
P
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Chapter 5: The Gaseous State
115
5.5 Gas Mixtures; Law of Partial Pressures
Learning Objectives

Learn the equation for Dalton’s law of partial pressures.

Define the mole fraction of a gas.

Calculate the partial pressure and mole fractions of a gas in a mixture. (Example 5.10)

Describe how gases are collected over water and how to determine the vapor pressure of
water.

Calculate the amount of gas collected over water. (Example 5.11)
Problem-Solving Skills
7.
Calculating partial pressures and mole fractions of a gas in a mixture. Given the masses of
gases in a mixture, calculate the partial pressures and mole fractions (Example5.10).
8.
Calculating the amount of gas collected over water. Given the volume, total pressure, and
temperature of gas collected over water, calculate the mass of the dry gas (Example 5.11).
As Dalton observed, each gas in a mixture of gases that don’t react with each other exerts a pressure
independent of any or all other gases present. So the pressure of a gas does not change if another gas is
added or removed, by chemical reaction, for instance. Neither does the volume change because volume
refers to the space the gas occupies—the volume of the container. The total pressure does change,
though, because it is the sum of all the partial pressures.
Exercise 5.10
A 10.0-L flask contains 1.031 g O2 and 0.572 g CO2 at 18C. What are the partial pressures of oxygen
and carbon dioxide? What is the total pressure? What is the mole fraction of oxygen in the mixture?
Known:
We can find the moles of each gas using the molar mass and then find the pressure of
each using the ideal gas law; mole fraction O2 =
nO2
ntotal
=
PO2
Ptotal
Solution: Moles of O2:
1.031 g O2 
PO2 =
nO2 RT
V
=
1 mol O 2
= 0.03222 mol O2
32.0 g O 2
0.03222 mol  0.0821 L • atm  (18  273) K
(10.0 L)(K • mol)
= 0.07698 atm O2
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.
116
Chapter 5: The Gaseous State
Moles of CO2:
0.572 g CO2 
nCO2 RT
PCO2 =
=
V
1 mol CO2
= 0.01300 mol CO2
44.0 g CO2
0.01300 mol  0.0821 L • atm  (18  273) K
(10.0 L)(K • mol)
= 0.03105 atm CO2
Total pressure:
P = PO + PCO = 0.07698 atm + 0.03105 atm = 0.1080 atm
2
2
Mole fraction of O2:
nO2
ntotal
=
0.03222 mol O2
= 0.713
0.03222 mol O2  0.01300 mol CO2
In working problems with gases collected over water, remember that it is the pressure, not the
volume, that must be adjusted for the presence of the water vapor.
Exercise 5.11
Oxygen can be prepared by heating potassium chlorate, KClO3, with manganese dioxide as a catalyst.
The reaction is

2KClO3(s)  2KCl(s) + 3O2(g)
How many moles of O2 would be obtained from 1.300 g KClO3? If this amount of O2 were collected
over water at 23C and at a total pressure of 745 mmHg, what volume would it occupy?
Wanted: moles of O2, V of O2 collected
Given:
balanced equation; 1.300 g KClO3; t = 23C; P = 745 mmHg
Known:
We can find moles from the equation and molar masses;
P = PH O + PO , so PO = P  PH O
2
2
2
2
we can find the volume using the ideal gas law.
Solution: Find moles of O2:
1.300 g KClO3 
1 mol KClO3
3 mol O 2

= 0.015905 = 0.01591 mol O2
122.6 g KClO3
2 mol KClO3
Find PO in atm (the PH O at 23C from text Table 5.6 is 21.1 mmHg):
2
2
PO2 = 745 mmHg  21.1 mmHg = 723.9 mmHg
723.9 mmHg 
1 atm
= 0.9525 atm
760 mmHg
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Chapter 5: The Gaseous State
117
Find VO :
2
VO2
=
0.01591 mol O2  0.0821 L • atm  (23  273) K
nRT
=
(0.9524 atm)(K • mol)
P
= 0.406 L
5.6 Kinetic Theory of an Ideal Gas
You might ask why we develop a model for an ideal gas when no such gas exists. In fact, most gases
encountered under normal laboratory conditions, at 1 atm pressure and at temperatures of 0 to 100C,
are very well described by this model. Most gases act ideally under these conditions. In Section 5.8 we
will see how to describe a gas that does not behave ideally.
Learning Objectives

List the five postulates of the kinetic theory.

Provide a qualitative description of the gas laws based on the kinetic theory.
5.7 Molecular Speeds; Diffusion and Effusion
Learning Objectives

Describe how the root-mean-square (rms) molecular speed of gas molecules varies with
temperature.

Describe the molecular-speed distribution of gas molecules at different temperatures.

Calculate the rms speed of gas molecules. (Example 5.12)

Define effusion and diffusion.

Describe how individual gas molecules move undergoing diffusion.

Calculate the ratio of effusion rates of gases. (Example 5.13)
Problem-Solving Skills
9.
Calculating the rms speed of gas molecules. Given the molecular mass and temperature of a gas,
calculate the rms molecular speed (Example 5.12).
10. Calculating the ratio of effusion rates of gases. Given the molecular masses of two gases,
calculate the ratio of rates of effusion (Example 5.13), or given the relative effusion rates of a
known and an unknown gas, obtain the molecular mass of the unknown gas (as in Exercise 5.15).
Exercise 5.12
What is the rms speed (in m/s) of a carbon tetrachloride molecule at 22C?
Known: u =
3RT
Mm
(We must use the value for R with SI
units and molar mass in kilograms.)
Solution:
1/ 2
 3  8.31 kg • m2  (22  273) K 
u=  2

3
 (s • K • mol)(154.0  10 kg/mol) 
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118
Chapter 5: The Gaseous State
Exercise 5.13
At what temperature do hydrogen molecules, H2, have the same rms speed as nitrogen molecules, N2, at
455°C? At what temperature do hydrogen molecules have the same average kinetic energy?
Solution: To find T for N2, set speeds equal:
uH2 = uN2
3RT
=
M mH 2
3R (455  273) K
M mN2
To solve for T, square both sides:
3RT
3R (728 K)
=
3
2.02  10
28.0  103
T=
728 K  2.02
= 52.5 K
28.0
Because kinetic energy is proportional to T, all molecules would have the same average kinetic
energy at the same temperature, whether at 728 K or 52.5 K.
Exercise 5.14
If it takes 3.52 s for 10.0 mL of helium to effuse through a hole in a container at a particular
temperature and pressure, how long would it take for 10.0 mL of oxygen, O2, to effuse from the same
hole at the same temperature and pressure? (Note that the rate of effusion can be given in terms of
volume of gas effused per second.)
Solution:
Rate He
=
Rate O 2
M mO 2
M mHe
=
32.0 g/mol
=
4.00 g/mol
8.00 = 2.83
Setting the first term equal to 2.83 and solving for rate O2 and then substituting the given values
for the rate of He gives
Rate O2 =
rate He
10.0 mL/3.52 s
2.84 mL/s
=
=
= 1.00 mL/s
2.83
2.83
2.83
Time O2 = 10.0 mL 
s
= 10.0 s
1.00 mL
Exercise 5.15
If it takes 4.67 times as long for a particular gas to effuse as it takes hydrogen under the same
conditions, what is the molecular mass of the gas? (Note that the rate of effusion is inversely
proportional to the time it takes for a gas to effuse.)
Known:
The time it takes a gas to effuse is inversely proportional to the rate; Graham’s law.
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Chapter 5: The Gaseous State
119
Solution:
a statement of Graham’s law
rate H 2
Time unknown gas
=
=
rate unknown gas
Time H 2
Mmunknown gas
MmH2
= 4.67
Squaring the last two terms gives
M munknown gas
2.02 g/mol
= 21.81
M munknown gas = 44.1 amu
5.8 Real Gases
Learning Objectives

Explain how and why a real gas is different from an ideal gas.

Use the van der Waals equation. (Example 5.14)
Problem-Solving Skill
11. Using the van der Waals equation. Given n, T, V, and the van der Waals constants a and b for a
gas, calculate the pressure from the van der Waals equation (Example 5.14).
Remember that V in the ideal gas law is the volume of the container. If we do not subtract the correction
factor from the V term, the V we use will be too large because the molecular volume actually takes up
some of the space a gas occupies. This molecular volume is negligible when the pressure is low, but it
becomes significant at high pressures.
It is easier to see why we add the correction for pressure rather than subtract it if we rearrange the
equation to solve for P:
P=
nRT
n2 a
 2
(V  nb)
V
This shows that pressure is one term minus another (minus the correction factor). The correction factor
thus gives an equation that describes reality: the actual pressure is less than ideal. This correction is
unnecessary when the temperature is high. When the temperature is low, however, the molecules do not
move as quickly and the molecular attraction becomes significant.
Exercise 5.16
Use the van der Waals equation to calculate the pressure of 1.000 mol of ethane, C2H6, that has a
volume of 22.41 L at 0.0C. Compare the result with the value predicted by the ideal gas law.
Wanted: van der Waals pressure; compare with ideal pressure
Given:
1.000 mol C2H6, V = 22.41 L, t = 0.0C
Known: van der Waals equation; R = 0.08206 L ∙ atm/ (Kmol) ; Table 5.7 gives a = 5.570 L2 ∙
atm/mol2; b = 0.06499 L/mol; T = (0.0C + 273.2) = 273.2 K; Pideal of 1.000 mol of gas at STP =
1.000 atm.
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120
Chapter 5: The Gaseous State
Solution: Rearrange the van der Waals equation to solve for pressure and substitute in values.
P=
=
nRT
n2 a
 2
(V  nb)
V
(1.000 mol)[0.08206 L • atm/(K • mol)](273.2 K)
22.41 L  (1.000 mol  0.06499 L/mol)

(1.0002 mol2 )(5.570 L2 • atm/mol 2 )
22.412 L2
= 1.0033 atm – 0.010910 atm = 0.992 atm
This value is 0.008 atm lower than the ideal pressure of 1.000 atm.
ADDITIONAL PROBLEMS
1.
Calculate the density of CO2 at exactly 100°C and standard pressure.
2.
Use the van der Waals equation,

n2 a 
P 
 (V  nb) = nRT
V2 

to calculate the pressure of 0.750 mol CO2(g) inside a 1.85-L steel cylinder at 273.2 K. The values
for a and b are 3.66 L2 ∙ atm/mol2 and 0.0429 L/mol, respectively.
3.
A sample of nitrogen gas occupies 46.8 L at 28°C and 745 mmHg. What volume would it occupy
at 28°C and 855 mmHg?
4.
What would be the volume of a 255-L sample of gas if it were heated, at constant pressure, from
28°C to 42°C?
5.
At 90.0°C, XeF4 is a gas. When 0.394 g of this noble-gas compound is introduced into a 0.91-L
gas bulb at 90.0°C, the pressure is 47.3 mmHg. What pressure would this same amount of gas
exhibit when it is transferred to a 2.13-L bulb at 120.0°C?
6.
The first step in the industrial production of NH3 using natural gas as the source of hydrogen can
be represented by the following equation:
CH4(g) + H2O(g)  CO(g) + 3H2(g)
Assume that natural gas is pure methane, CH4. Calculate the kilograms of hydrogen gas produced
from 1.200  103 m3 of natural gas at 745 mmHg and 25°C.
7.
CoF3 is often used to fluorinate organic compounds. During the fluorination of (CH3)3N, one of
the gaseous products was isolated and purified. At 90.1°C and 105.7mmHg, 1.46 g of the
compound occupied a volume of 1.83 L. What is the molecular mass of this fluorination product?
8.
Calculate
9.
a.
The mole fraction of O2 in a container that contains CO2 at 285 mmHg, O2 at 305mmHg,
and N2 at 245 mmHg.
b.
The root mean square (rms) molecular speed of N2 at 25°C.
c.
The ratio of the rates of effusion of H2 and Rn through a fine pinhole.
How many grams of oxygen gas are contained in a 250.0-mL bottle of the gas collected over
water at 25°C and 755 mmHg?
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Chapter 5: The Gaseous State
121
10. The height of the mercury column in a barometer is independent of the diameter or bore of the
barometer tube. Explain why this is so if pressure is defined as force F per unit area A and area is
defined as ¼πd2.
ANSWERS TO ADDITIONAL PROBLEMS
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
Since n = m/Mm,
m
 RT
Mm
PV =
Rearrange to solve for m/V (density):
PM m
(1 atm)  (44.0 g/mol)
m
=
=
= 1.44 g/L (5.3, PS Sk. 5)
V
[0.0821 L • atm/(K • mol)]  373 K
RT
2.
Rearrange the van der Waals equation to give
nRT
n2 a
 2
V  nb
V
P=
Substituting the respective values for n, R, T, V, a, and b gives
P=
(0.750 mol)[0.08206 L • atm/(K • mol)](273.2 K)
[1.85 L  0.750 mol (0.0429 L/mol)]
–
(0.750 mol) 2 (3.66 L2 • atm/mol 2 )
(1.85 L) 2
= 8.65 atm (5.8, PS Sk. 11)
3.
46.8 L 
745 mmHg
= 40.8 L (5.2, PS Sk. 2)
855 mmHg
4.
255 L 
315 K
= 267 L (5.2, PS Sk. 2)
301 K
5.
47.3 mmHg 
6.
nCH4 =
393 K
0.91 L

= 22 mmHg (5.2)
363 K
2.13 L
PV
0.9803 atm K • mol  1.200  106 L
=
= 4.808  104 mol CH4
RT
0.0821 L • atm  298 K
Calculate the kilograms of H2:
4.808  104 mol CH4 
3 mol H 2
2.02 g H 2
kg H

 3 2 = 2.91  102 kg H2
1 mol CH 4
mol H 2
10 g H 2
(5.4, PS Sk. 6)
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122
7.
Chapter 5: The Gaseous State
n=
PV
(0.1391 atm)(1.83 L)
=
= 8.541  10–3 mol
[0.0821 L • atm/(K • mol)](363 K)
RT
Find Mm:
Mm =
grams
1.46 g
=
= 1.71  102 g/mol (5.3)
mole
8.541  103 mol
8.
a.
305 mmHg
= 0.365 (5.5, PS Sk. 7)
(285  305  245) mmHg
b.
u=
c.
9.
rH 2
rRn
3  8.31 kg • m2  298 K
= 515 m/s (5.7, PS Sk. 9)
s 2 • K • mol  0.0280 kg/mol
=
222 g/mol
= 10.5 (5.7, PS Sk. 10)
2.02 g/mol
PO2 = (755 – 23.8) mmHg = 731.2 mmHg 
1 atm
= 0.9621 atm
760 mmHg
Solve PV = (m/Mm)RT for m and substitute in values.
m=
PVM m
0.9621 atm  0.2500 L  K • mol  32.0 g/mol
=
298 K  (0.821 L • atm)
RT
= 0.315 g O2 (5.5, PS Sk. 8)
10. By definition, the pressure P is the force F of the liquid column per unit area A:
P = F/A
The downward force of the liquid may be expressed as the product of the mass m of the liquid and
the acceleration due to gravity g. The pressure expression can then be written
P=
mg
F
=
A
A
Because A = ¼πd2, it looks as though the pressure does depend on the diameter or bore of the
tube. However, the mass of a liquid can be expressed as the product of the density d and liquid
volume V, where volume is the product of column height h and cross-sectional area A. This gives
P=
mg
dVg
dhAg
=
=
= dhg
A
A
A
Because g is a constant, 9.81 ms2, the indicated liquid pressure depends only on the density of the
liquid and on the liquid column height, neither of which depends on the diameter or bore of the
tube. (5.1)
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Chapter 5: The Gaseous State
123
CHAPTER POST-TEST
1.
Indicate whether each of the following statements is true or false. If the statement is false, change
it so that it is true.
a.
The accompanying graph depicts the change in temperature with volume at constant
pressure for an ideal gas. True/False:
________________________
________________________
________________________
________________________
________________________.
b.
At constant pressure, the density of a gas will decrease as the absolute temperature increases.
True/False: ________________________________________________________________
__________________________________________________________________________
c.
At standard temperature and pressure, 1 mole of H2 gas will occupy the same volume as 1
mole of O2 gas, but the equal volumes will have different masses. True/False:
__________________________________________________________________________
__________________________________________________________________________
d.
One assumption of kinetic-molecular theory is that gas molecules have virtually no volume
and are attracted to each other by van der Waals forces. True/False:
__________________________________________________________________________
__________________________________________________________________________
e.
If CO2 and CO molecules are at the same temperature, the average kinetic energy of each
sample of gas molecules would be inversely proportional to the ratio of their molecular
masses. True/False:__________________________________________________________
__________________________________________________________________________
f.
The rate of effusion of NH3 is about one-half that of HCl. True/False:
__________________________________________________________________________
2.
Calculate the value of the ideal gas constant R in units of (mL ∙ mmHg)/(K ∙ mol).
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124
3.
Chapter 5: The Gaseous State
One mole of phosgene, COCl2, at 115°C and 645 mmHg will occupy a volume of
a.
37.5 L.
b.
18.6 L.
c.
27.0 L.
d.
22.4 L.
e.
none of the above.
4.
A 5.46-g sample of CO has a density of 1.209 g/L at 9°C and 1 atm pressure. Calculate the density
of 4.73 g of CO at 28.0°C and 739 mmHg.
5.
The combustion of butane, C4H10, gives CO2 and H2O:
13O2 + 2C4H10 
 8CO2 + 10H2O
If 93.5 L O2 and 10.3 L C4H10 react to give 33.0 L CO2 at 769 mmHg and 554°C, the percentage
conversion of C4H10 to CO2 was
a.
100%.
b.
57.5%.
c.
80.1%.
d.
78.0%.
e.
There are insufficient data to calculate.
6.
Titanium reacts with chlorine to give a clear, colorless liquid. The elemental analysis of this
compound shows a composition of 25.25% Ti and 74.75% Cl. The molecular mass was
determined in the gas phase from the following data: mass of sample = 1.63 g, temperature =
171°C, pressure = 736 mmHg, volume = 0.323 L. What is the molecular formula of this titanium–
chlorine compound?
7.
Fill in the correct words below. The combustion of propane is represented by the following
equation:
C3H8(g) + 5O2(g)

 3CO2(g) + 4H2O(g)
If the reaction is carried out in a closed vessel at a constant temperature and volume:
8.
a.
The pressure ____________________ as the reaction proceeds to completion.
(decreases/increases)
b.
The total number of molecules after the reaction is completed is ___________________ the
number of molecules before the reaction is initiated.
(greater than/less than)
A closed container contains an equal number of N2 and O2 molecules at 30°C and standard
pressure. Assuming ideal-gas behavior, which of the following statements is (are) correct?
a.
As the temperature increases, the average number of N2 and O2 molecules increases.
b.
The partial pressure of N2 is the same as that of O2.
c.
The N2 molecules make a greater contribution to the pressure because of their smaller mass.
d.
If the pressure is decreased, the frequency of impacts of the N2 and O2 molecules with the
walls of the container increases.
e.
None of the above are correct statements.
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Chapter 5: The Gaseous State
9.
125
Which of the following statements about Figures A and B is (are) incorrect?
a.
In each figure, a given gas is at a constant temperature with only P and V varying.
b.
The PV product for 1 mole of an ideal gas at constant temperature is a constant of value RT.
c.
The deviations from the dashed horizontal line in Figures A and B for each curve indicate
the relative deviation from ideal-gas behavior for the respective gas.
d.
The minimum in the CO2 plot results when the volume increases at a faster rate than the
increase in pressure.
e.
Real-gas behavior approaches that of an ideal gas at low P.
Figure A
Figure B
10. Which of the following gases would have the largest value for b in the van der Waals equation for
a nonideal gas?
a.
O2
b.
CO2
c.
C4H10
d.
Ar
e.
HF
ANSWERS TO CHAPTER POST-TEST
If you missed an answer, study the text section and problem-solving skill (PS Sk.) given in parentheses
after the answer.
1.
a.
False. The graph depicts the change in pressure with volume at constant temperature. (5.2,
PS Sk. 2)
b.
True. (5.3)
c.
True. (5.2)
d.
False. One assumption of kinetic theory is that gas molecules have virtually no volume and
are not attracted to each other. (5.6)
e.
False. If CO2 and CO molecules are at the same temperature, the average kinetic energy of
each sample of gas molecules would be the same. (5.6)
f.
False. The rate of effusion of NH3 is about one and one-half times that of HCl. (5.7, PS Sk.
10)
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126
Chapter 5: The Gaseous State
2.
6.24  104 (mL ∙ mmHg)/(K ∙ mol) (5.3)
3.
a (5.3, PS Sk. 4)
4.
1.10 g/L (5.3, PS Sk. 5)
5.
c (5.4, PS Sk. 6)
6.
TiCl4 (5.3, PS Sk. 5)
7.
a.
increases
b.
greater than (5.5)
8.
b (5.6)
9.
d (5.8)
10. c (5.8)
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