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Unformatted text preview: Chapter Five GASES
• Properties of Gases
The Gas Laws
The Ideal Gas Law
The Kinetic Molecular Theory of Gases
Nonideal Gases PROPERTIES OF GASES
3. List the properties that are characteristic of gases.
Interconvert pressure readings among the common units used to express pressure.
Calculate the pressure of a gas tapped in an open-tube manometer. Characteristics of Gases. All matter exists as either a solid, a liquid, or a gas, depending on the
temperature and the pressure. Of the three states, the simplest to describe is the gaseous state. Gases have certain
5. Expansion. Gases expand indefinitely to fill the space available to them.
Indefinite shape. Gases fill all parts of a container evenly and so have no definite shape of their own.
Compressibility. Gases are the most compressible of the states of matter.
Mixing. Two or more gases will mix evenly and completely when confined to the same container.
Low density. Gases have much lower densities than liquids and solids. Typically, densities of gases are
about 1/1000 those of liquids and solids. Gas Pressure. One of the most obvious properties of a gas, to someone who has inflated bicycle or
automobile tires, is its pressure. P ressure is force per unit area. Any exertion of force applied to an area can be
expressed as pressure.
Pressure = force
area Some bicycle tires must be inflated to 60 psi. That's 60 pounds per square inch of tire surface. Recall that
weight is the force with which an object is pulled vertically downward by gravity. Thus, a pound is a measure of
force (whether due to gravity or not), and pounds per square inch is a unit of pressure.
A large number of units are used to measure pressure. The s tandard atmosphere is the average pressure
exerted by Earth's atmosphere at sea level. In common units this is 14.7 psi. This pressure is enough to support
a column of mercury 760 mm high at 0°C. The pressure that supports one mm of Hg is called a torr. In the SI,
pressure is measured in a unit called the pascal (Pa), which is defined as the pressure from a force of one newton
exerted on one square meter of area. 85
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1 Pa = 1 N/m2
It takes 1.013 × 10 5 Pa to equal 1 atm. Since pascals are so small, the kilopascal is sometimes more convenient
1 kPa = 10 3 Pa
Table 5.1 Pressure Units in Terms of the Standard Atmosphere.
1 atmosphere =
= 760.0 mm Hg = 760.0 torr
1.013 × 10 5 Pa = 101.3 kPa
33.9 ft H2 O
14.7 pounds per square inch (psi) Pressure Measurement with an Open-Tube Manometer. A manometer is a device used to measure
the pressure of gases, rather than that of the atmosphere. In an open-tube manometer (Example 5.2) the pressure
of the trapped gas is the sum of the pressure of a mercury column of height h, plus the pressure of the
atmosphere. Example 5.1 illustrates pressure measurement with a manometer.
EXAMPLE 5.1 Pressure Measurement
What is the pressure of the gas trapped in the J-tube shown below if the atmospheric pressure is 0.803 atm?
the air Gas 42 mm Hg •Method of Solution
The pressure of the gas is sufficient to support a column of mercury 42 mm high and hold back the pressure of
the atmosphere as well. Therefore the pressure of the gas is:
P gas = P h + P atm
The pressure from the mercury column is just the difference in heights of the two mercury surfaces, 42 mm Hg.
The atmospheric pressure is 610 mm Hg.
P gas = 42 mm Hg + 610 mm Hg
P gas = 652 mm Hg
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EXAMPLE 5.2 Converting Pressure Units
The atmospheric pressure on Mars is about 0.22 inches of Hg as compared to about 30 inches of Hg on Earth.
Express this pressure in mm Hg, atmospheres, and pascals.
•Method of Solution
Treat this like any other factor label problem. First state the problem:
? mm Hg = 0.22 in Hg
Recall that there are 2.54 cm/in, therefore there are 25.4 mm/in.
The unit factor needed is
Converting inches to millimeters gives:
? mm Hg = 0.22 in Hg × 2 5.4 mm
1 in = 5.6 mm Hg
The number of atmospheres is
1 a tm
760 mm Hg ? atm = 5.6 mm Hg ×
= 7 .4 × 10 –3 atm
The pressure in pascals is ? pascals = 7.4 × 10 –3 atm × 1 .013 × 1 0 5 P a
1 atm = 750 Pa
1. 2. Back Convert a pressure of 645 mm Hg into its value in
Do the following unit conversions.
a. 125 mm Hg to torr
b. 725 mm Hg to kilopascals
c. 1500 psi to atmospheres Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 8 8 / Gases THE GAS LAWS
3. State Boyle's, Charles's, and Avogadro's laws and know what variables are held constant for each law. Solve
problems involving these laws.
Solve problems involving the combined gas law.
Solve problems using Dalton's law of partial pressures. The Pressure-Volume Relationship. Experiments carried out in the 17th and 18th centuries led to
three gas laws the provide a basis for our understanding of the behavior of gases. It is helpful when discussing
the volume changes of a gas to visualize that we are considering a gas trapped in a container with a movable
barrier so that its volume is not fixed. See Figures 5.5 and 5.6 of the text, and the open-tube manometer above.
Boyle's law states that the volume of a given amount of gas at a constant temperature is inversely
proportional to the pressure applied to the gas. In other words, as the pressure appied to a gas increases its
volume decreases. Mathematically we can write this in two ways:
P or V = k1
P where k is a constant.
Rearranging yields a mathematical statement of Boyle's law. The product of P × V is a constant for a fixed
amount of gas at a constant temperature.
PV = k 1 (Boyle's Law)
For changes in pressure from P 1 t o P 2 , we can apply Boyle's law. Since the product PV is a constant, its
value at the initial pressure P1 and inital volume V 1 will equal PV at the final pressure P2 and final volume
P 1 V1 = P 2 V2 (n and T are constant) The Temperature-Volume Relationship. C harles's law states that at constant pressure the volume
of a gas is directly proportional to its absolute temperature. Mathematically we can write this in two ways:
V α T or V = k2 T
where k is a constant. This equation shows that V is proportional to the absolute temperature T.
Rearranging we get another form of Charles's law:
Since V/T is equal to a constant, then V1 /T1 at an initial temperature T 1 and volume V1 will equal the quotient
V2 /T2 at a final temperature T2 and volume V2 .
(n and P are constant)
T2 The Volume-Amount Relationship. A vogadro's law says that at the same temperature and pressure,
equal volumes of different gases contain the same number of molecules (or moles). This is another way of
saying that the volume of a gas is proportional to the number of moles present. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Gases / 8 9
V = nk3 (P and T are constant)
A particularly useful quantity is the volume occupied by one mole of ideal gas called the m olar volume .
Since the volume of a specific amount of gas varies with the temperature and the pressure, a set of standard
conditions have been defined. S tandard temperature and pressure (abbreviated STP) are 0°C and 1 atm
pressure. One mole of a gas at STP will occupy 22.4 L.
In addition to these gas laws, a useful relationship for changes in pressure, volume, and temperature of a
fixed number of moles of gas is called t he combined gas law.
P 1 V1
P 2 V2
T2 Dalton's Law of Partial Pressures. Each component gas of a mixture of gases uniformly fills the
containing vessel. Each component exerts the same pressure as it would if it occupied that volume alone. Thus,
the total pressure of a mixture is the sum of the individual pressures, called the p artial pressures , of each
P t = PA + P B + P C + …
Also, the total number of moles nt is the sum of the number of moles of each component:
nt = nA + n B + n C + …
The ratio nA/nt , the ratio of the number of moles of component A to the total number of moles of all
components, is called the mole fraction of component A. The symbol XA is used for the mole fraction of
component A. According to Dalton's law, the partial pressure of a gas is proportional to its mole fraction in the
PA = nA
P = XAP t
nt t Whenever gases are collected by displacement of water, the total gas pressure is the sum of the partial
pressure of the collected gas and the partial pressure of water vapor.
P t = Pgas + P H2 O
Consequently, the partial pressure of the collected gas is Pgas = P t – P H2 O, where the partial pressure of water
depends on the temperature and corresponds to the vapor pressure of water.
EXAMPLE 5.3 Gas Law Calculation
When 2.0 L of chlorine gas (Cl2 ) at STP is warmed at constant pressure to 100°C, what is the new volume?
•Method of Solution
The volume of a fixed amount of a gas at constant pressure is proportional to the absolute temperature according
to Charles's law, where T 1 = 273 K and T2 = (100°C + 273°C) K = 373 K.
T2 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9 0 / Gases
V2 = V1 × T2
, where V 1 , the initial volume, is 2.0 L.
T1 Substituting, we get:
2.0 L × 3 73 K
= 2.7 L
273 K •Comment
Note that we expect V2 > V1 because of the temperature increase. This provides us with a simple check of our
answer. If we calculated that V2 < V1 , we would know our answer was wrong.
EXAMPLE 5.4 Combined Gas Law Calculation
Given 10.0 L of neon gas at 5°C and 630 mm Hg, calculate the new volume at 400°C and 2.5 atm.
•Method of Solution
First note that both the pressure and the temperature of the gas are changed, but that the number of moles is
constant. Write the ideal gas equation with all the constant terms on one side.
= nR = a constant
We see that PV/T is a constant. Therefore
P 1 V1 P 2 V2
where the subscript 2 refers to the final state, and the subscript 1 refers to the initial state.
Rearranging to solve for V2 gives
V2 = V1 ×
1 a tm
where P1 = 630 mm Hg ×
= 0.829 atm
760 mm Hg
P 2 = 2.5 atm
T1 = (273°C + 5°C) K = 278 K
T2 = (273°C + 400°C) K = 673 K
V2 = 10.0 L × 0 .829 atm
6 73 K
278 K = 8.0 L
Note that both P1 and P2 must be expressed in the same units and both temperatures in kelvins.
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EXAMPLE 5.5 Partial Pressures
When oxygen gas is collected over water at 30°C and the total pressure is 645 mm Hg:
a. What is the partial pressure of oxygen? Given the vapor pressure of water at 30°C is 31.8 mm Hg.
b. What are the mole fractions of oxygen and water vapor?
•Method of Solution for (a)
Mixtures of gases obey Dalton's law of partial pressures which says that the total pressure is the sum of the
partial pressures of oxygen and water vapor.
P t = P O2 + P H2 O
•Calculation for (a)
P O2 = P t – P H2 O = 645 mm Hg – 31.8 mm Hg
P O2 = 613 mm Hg
•Method of Solution for (b)
Recall that the partial pressures of O2 and H2 O are related to their mole fractions,
P O2 = XO2 P t P H2 O = XH2 OP t •Calculation for (b)
Therefore, on rearranging, we get:
P H2 O
X H2 O =
XO2 = 613/645 = 0.950, and XH2 O = 31.8/645 = 0.049
Note also that the sum of the mole fractions is 1.0, within the number of significant figures, given:
XO2 + XH2 O = 0.950 + 0.049 = 0.999
3. 4. Two liters of oxygen gas at –15°C are heated and the volume expands. At what temperature will the volume
reach 2.31 L? 5. A 20.0 mL sample of a gas is enclosed in a gas-tight syringe at 50.0°C. What will the volume of the gas
be at the same pressure after the syringe has been immersed in ice water? 6. Back A sample of gas has a volume of 200 cm3 at 25°C and 700 mm Hg. If the pressure is reduced to 280 mm
Hg, what volume would the gas occupy at the same temperature? If 2.00 L of oxygen at –15°C are allowed to warm to 25°C at constant pressure, what is the new volume of
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7. The gas pressure in an aerosol can is 1.5 atm at 25°C. What pressure would develop in the can if it were
heated to 450°C? 8. A sample of gas occupies 155 mL at 21.5°C, and at 305 mm Hg. What is the pressure of the gas sample
when it is placed in a 5.00 × 10 2 mL flask at a temperature of –10°C? 9. Hydrogen and helium are mixed in a 20.0 L flask at room temperature (20°C). The partial pressure of
hydrogen is 250 mm Hg and that of helium is 75 mm Hg. How many grams of H2 and He are present? THE IDEAL GAS LAW
3. Apply the ideal gas equation to calculate one of its variables.
Determine the density of an ideal gas of known molecular mass.
Determine the molecular mass of an ideal gas of given density. The Ideal Gas Equation. The properties of a gas that we are most often concerned with are pressure
(P), volume (V), absolute temperature (T), and amount (number of moles, n). The experimentally observed
relationship between these properties is called the i deal gas law :
PV = nRT
and R is the ideal gas constant, R = 0.0821 L. atm/K. mol. The dots appearing in the units are to remind us that
both L and atm are in the numerator, and K and mol are in the denominator. The units of R may also be written,
R = 0.0821 L atm K–1 mol –1 .
The ideal gas law has great importance in the study of gases. It does not contain information that is
characteristic of any particular gas. Rather, it is a generalization applicable to most gases, at pressures up to
about 10 atm, and at temperatures above 0°C. An ideal gas is one whose behavior agrees with that predicted by
the ideal gas law. A calculation using the ideal gas law is shown in Example 5.6. Gas Density. The density of a gaseous compound is d = m/V, where m is the mass of the gas in grams.
Recall the number of moles n of gas is n = m/M , where M is the molar mass. Substituting for n in the ideal
gas equation gives
PV = m
M Rearranging gives
and so the density of a gas that follows the ideal gas law is:
RT Note that the density is directly proportional to its molar mass, as we noted earlier. Note also that the density
increases as gas pressure increases, and density decreases as temperature increases. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Gases / 9 3
The Molar Mass of a Gas. Since the density of a gas is directly proportional to its molar mass, then a
measurement of the density of a gas allows the calculation of its molar mass. By rearranging the above equation,
P The molar mass of an unknown gaseous substance can be calculated, which will help in its identification.
EXAMPLE 5.6 Ideal Gas Law
What volume will be occupied by 0.833 mole of fluorine (F2 ) at 645 mm Hg and 15.0°C?
•Method of Solution
The ideal gas law relates the volume of a gas to its temperature, pressure and number of moles.
P First convert the pressure into atmospheres:
1 a tm
= 0.849 atm
760 mm Hg
(0.833 mol)(0.0821 L. atm/K. mol)(288 K)
= 23.2 L
0.849 atm P = 645 mm Hg × Comment
Even though it takes longer to write out the units for each term in the equation, doing so is very helpful because
when the units cancel and leave you with the desired units (in this case liters) you are more likely to get the
EXAMPLE 5.7 Molar Mass of a Gas
A gaseous compound has a density of 1.69 g/L at 25°C and 714 torr. What is its molar mass?
•Method of Solution
The density of an ideal gas is directly proportional to its molecular mass. We use the equation
RT • Calculation
Convert pressure into units of atmospheres.
1 a tm
P = 714 torr ×
= 0.939 atm
Rearranging and substituting, we get:
dRT (1.69 g/L)(0.0821 L. atm/K. mol)(298 K)
= 44.0 g/mol
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10. What is the pressure in atmospheres of 1.20 × 10 4 moles of methane, CH4 , when stored at 22°C in a
3.00 × 10 3 L tank?
11. 1.75 g sample of CO2 is contained in a 7.50 × 10 2 mL flask at 35°C. What is the pressure of the gas?
12. How many moles of CO2 gas are required to fill a 5.00 L balloon to a pressure of 1.05 atm at 5.0°C?
13. How many grams of helium are required to fill a 10.0 L balloon to a pressure of 1250 torr at 25°C?
14. How many O2 molecules occupy a 1.00 L flask at 75°C and 777 mm Hg?
15. What is the density of H 2 (g) at 35°C and 650 torr?
16. Which one of the following gases will have the greatest density when they are all compared at the same
temperature and pressure?
O2 CO2 NO3 CF4 17. When 2.96 g of mercuric chloride is vaporized in a 1.00 liter bulb at 680 K, the pressure is 450 mm. What
is the molar mass and molecular formula of mercuric chloride?
18. Determine the molar mass of Freon-11 gas if a sample weighing 0.597 g occupies 1.00 × 10 2 mL at 95°C,
and 1000 mm Hg. GAS STOICHIOMETRY
1. Calculate the volume of any gaseous reactant consumed or of any gaseous product generated in a chemical
reaction, given the mass of any species involved in the reaction. Mole Relationships. Earlier we saw that stoichiometry involves relationships between masses and moles
of the reactants and the products of a chemical reaction. The ideal gas equation furnishes a way to convert
between moles and the volume of a gas. The reaction of a known number of moles of a given reactant will yield
a specific volume of a gaseous product. When the balanced equation is known, the volume of the product can be
calculated by the following path:
mass of a
given reactant → moles of
reactant moles of
→ gaseous product → volume of the product
gas at P and T _______________________________________________________________________________
EXAMPLE 5.8 Volume of a Gaseous Product
Calculate the volume of ammonia gas, measured at 645 torr and 21°C, that is produced by the complete reaction
of 25.0 g quicklime, CaO, with excess ammonium chloride, NH4 Cl, solution. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Gases / 9 5
•Method of Solution
First write the balanced chemical equation.
CaO(s) + 2NH4 Cl(aq) → 2NH3 (g) + CaCl2 (aq) + H2 O(l)
Here we must determine the number of moles of NH3 formed in the reaction and convert this into the volume of
an ideal gas. Three steps are required: (1) convert g CaO to moles CaO, (2) convert moles CaO to moles NH3
produced, and (3) use the ideal gas equation to calculate the volume of NH3 .
grams CaO → moles CaO → moles NH 3 → volume NH 3
We already know how to find conversion factors 1 and 2; therefore let's first find the number of moles of NH3
Stating the problem:
? mol NH 3 = 25.0 g CaO
? mol NH 3 = 25.0 g CaO × 2 m ol NH 3
1 m ol CaO
= 0.891 mol NH 3
56.1 g CaO
1 mol CaO The volume of 0.891 mol NH 3 can be calculated from the ideal gas equation:
P Convert pressure into units of atm before substituting into the ideal gas equation.
P = 645 torr × 1 a tm
= 0.849 atm
760 torr 0.891 m ol (0.0821 L . atm/K. mol) (294 K)
= 25.3 L
19. The discovery of oxygen resulted from the decomposition of mercury(II) oxide.
2HgO → 2Hg + O2 (g)
a. What volume of oxygen will be produced by the decomposition of 25.2 grams of the oxide, if the gas is
measured at STP?
b. How many grams of mercury(II) oxide must be decomposed to yield 10.8 L of O2 gas at 1 atm and 298 K.
20. Sodium azide decomposes according to the equation.
2NaN3 (s) → 2Na(s) + 3N 2 (g)
What volume of N2 at 1.1 atm and 50.0°C will be produced by the decomposition of 5.0 g NaN3 ?
21. Consider the reaction of 20.0 g calcium oxide with carbon dioxide.
CaO(s) + CO 2 (g) → CaCO 3 (s)
If you have 5.5 L of CO2 at 7.50 atm and 22°C, will you have enough carbon dioxide to react with all the
CaO? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 9 6 / Gases KINETIC MOLECULAR THEORY OF GASES
2. Describe the assumptions on which the kinetic molecular theory is based.
Calculate the root-mean-square speeds of given molecular gases. Postulates of the Kinetic Molecular Theory. Up to this point in the chapter we have emphasized the
laws of gas behavior. Theories are developed to explain observed behavior and properties, and are used to make
predictions of behavior under new circumstances. The theory that has been very successful in accounting for the
properties of gases is called the k inetic molecular theory of gases .
The main assumptions that explain the behavior of an ideal gas areas follows:
4. Gases consist of molecular particles. These molecules are much smaller than the distances between them.
The volume occupied by a molecule is considered negligible.
Gas molecules are in constant motion. They experience frequent collisions with each other and with the
Although energy may be transferred from one molecule to another in a collision, the total energy of all
molecules remains constant.
The average kinetic energy of a molecule in a gas is
KE = 2 m u 2 __
where m is the mass of the molecule and u is its speed. The quantity u2 is the mean square speed. The
average kinetic energy is proportional to the absolute temperature:
KE ∝ T
KE = CT
where C is a proportionality constant.
5. Gas molecules behave as independent particles; all attractive and repulsive intermolecular forces are
negligible. The manner in which the kinetic theory explains many of the properties of gases and the gas laws is
discussed in Section 5.7 of the text. According to the kinetic theory, gas molecules are in constant, random
motion. The pressure exerted by a gas on the container walls is assumed to arise from impacts on the walls by
gas molecules. During impacts, force is exerted on the wall area. The gas pressure depends on the volume, the
temperature, and the amount of gas. The kinetic molecular theory explains these observations in the following
1. Boyle's law V = k1 /P: As the volume of the container is increased, there are fewer impacts of gas
molecules with the wall per second; thus, the gas pressure is lowered.
2. Charles's law V = k2 T: As the temperature of a gas increases, the average kinetic energy and speed of the
gas molecules increase. Thus, the collision frequency and the force of impact with the wall increase. If one
of the walls is a movable piston, then the volume of the container will expand until balanced by the
3. Gay-Lussac's law P = kT: When a gas is heated in a container with a fixed volume, the gas molecules
will impact more forcefully with the wall, thus increasing the pressure.
4. Avogadro's law V = k3 n: As more gas molecules are added to the container, the number of impacts per
second with the wall increases, and so the pressure increases correspondingly. If the volume of the container
is not fixed, then the volume will increase. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Gases / 9 7
Speeds of Gas Molecules. Experiments show that the molecules of a gas do not all move at the same
speed. Instead, the speeds are distributed over a range. The distribution, called the M axwell speed
distribution (Figure 5.15 in the textbook), always has the same general shape. The highest point on the curve
marks the most probable speed. Some molecules move with speeds much less than the most probable speed,
while others move with speeds much greater than this speed. According to Maxwell, the r oot-mean-square
speed, u rms , of gas molecules with molar mass M at absolute temperature T is given by the equation:
where R = 8.314 J/K . mol, and the molar mass M must be expressed in kg/mole in order for the units to come
out in m/s. This equation reveals that gas molecules, on the average, travel at very high speeds. For instance, at
25°C the root mean square speed of He atoms is 1.36 × 10 3 m/s, and for N 2 molecules it is 5.15 × 10 2 m /s.
These values correspond to 3040 miles per hour and 1150 mph, respectively. The speed of sound in a gas is
about the same as the root mean square speed of the gas molecules. The speed of sound through He gas is much
greater than through air.
The kinetic molecular theory predicts that the average kinetic energy of the molecules of a gas is determined
only by the temperature. The average kinetic energy is:
KE = m u 2 = CT
and has the same value for all gases at the same temperature. This means that gas molecules with higher molar
masses will have slower average speeds, while lighter molecules will have higher average speeds.
EXAMPLE 5.9 Root-Mean-Square Speed
Calculate the root-mean-square speed of gaseous argon atoms at STP.
•Method of Solution
The root-mean-square speed is given by the equation
urms = 3RT
M where R = 8.314 J/K·mol. T is the absolute temperature 273 K; and M is the molar mass of Ar in kilograms,
0.0399 kg/mol. Note that standard pressure is irrelevant here because the average kinetic energy depends only on
the temperature. Before substituting into the equation, we recall that 1 J = 1 kg m2 /s 2 .
urms = 3 ( 8.314 J/K . mol)(273 K)
1 k g m 2 /s 2
–3 k g/mol)
(39.9 × 1 0 urms = (17.1 × 10 4 m 2 /s 2 )1/2 = 413 m/s
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22. Calculate the root-mean-square speed of ozone molecules (O3 ) in the stratosphere where the temperature is –
23. If NO2 molecules have a root-mean-square speed of 290 m/s. What is the temperature corresponding to this
24. If the root-mean-square speed of an O2 molecule is 4.2 × 10 2 m/s at 25°C, what is the average speed of a
Cl 2 molecule at the same temperature? NONIDEAL GASES
2. List two molecular properties responsible for deviations from ideal gas behavior. And identify the
conditions under which gases will be most likely to behave in a nonideal manner.
Calculate the pressure of a real gas as predicted by the van der Waals equation. Nonideal Gas Behavior. According to the kinetic theory, molecules of a gas are quite independent of
one another and do not exert attractive or repulsive forces on one another. Such molecules would obey the ideal
gas law. For an ideal gas, a plot of PV/nRT versus P at constant temperature should be a horizontal line, with
PV/nRT = 1 at all values of P . As Figure 5.19 of the text shows, this is true for real gases only at low
pressures (less than ~10 atm).
Another way to observe nonideality of gases is to lower the temperature. Cooling a gas decreases the
average kinetic energy of its molecules. All gases when cooled enough will condense to the liquid. This
suggests that intermolecular forces of attraction exist between molecules of real gases. Condensation occurs
when the average kinetic energy is not great enough for molecules to break away from intermolecular attractive
forces. The same forces account for nonideal behavior in gases.
Van der Waals Equation. Van der Waals modified the ideal gas law to account for two molecular
properties that affect gas behavior: intermolecular forces and finite molecular volume. According to van der
Waals, the pressure of a real gas will be lower than that of an ideal gas because attraction to neighboring
molecules tends to lessen the impact a real molecule makes with the wall. We can express this in mathematical
P real = Pideal – an2
V2 where an2 /V is the appropriate correction. Both n and V are defined as before. The proportionality constant a is
different for each gas. The magnitude of a is a measure of the strength of intermolecular attractive forces, and
increases with increasing molecular mass and increasing complexity of molecular structure (Table 5.4 in the
The other correction concerns the volume of the gas. According to the assumption of the kinetic theory,
molecules are volumeless points. Since every real molecule does occupy a small, though finite, volume, a
certain volume within a container of volume V is excluded, or not available, to gas molecules. The effective
volume is (V – nb), where b is a constant that represents the volume of a molecule. The term nb is a correction
for the finite volume of a mole of molecules. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Gases / 9 9
The van der Waals equation is written in the form of the ideal gas equation.
P ideal Veffective = nRT P real + an2 (V – nb) = nRT V2 corrected
EXAMPLE 5.10 Van der Waals Equation
The molar volume of isopentane, C5 H12 , is 1.0 L at 503 K and 30.0 atm:
a. Does isopentane behave as an ideal gas?
b. Given that a = 17.0 L2 atm/mol2 and b = 0.136 L/mol, calculate the pressure of isopentane as predicted by
the van der Waals equation.
•Method of Solution
a. According to the ideal gas equation the pressure of 1 mole of a gas at 503 K that occupies 1.0 L is:
(1.0 mol)(0.082 L·atm/K·mol)(503 K)
1.0 L P = 41.3 atm
This calculated result differs considerably from the observed pressure of 30 atm. In fact, the percent error
which is the difference between the two values divided by the actual pressure is:
% error = 41.3 atm – 30.0 atm
30.0 atm % error = 37.7%
We conclude that under these conditions C5 H12 behaves in a nonideal manner.
b. In this case, write the van der Waals equation P real + an2 (V – nb) = nRT V2 and substitute into it, but first calculate the correction terms.
( 17.0 L 2 ·atm/mol2 )(1.0 mol)2
= 17.0 atm
1.0 L 2
nb = 1.0 mol (0.136 L/mol) = 0.136 L
nRT = (1.0 mol)(0.0821 L·atm/K·mol)(503 K) = 41.3 L atm
Now substitute using the van der Waals equation.
(P + 17.0 atm)(1.0 L – 0.136 L) = 41.3 L atm
(P + 17.0 atm)(0.864) = 41.3 atm
P + 17.0 atm = 47.8 atm
P = 30.8 atm
Thus, the pressure calculated by the van der Waals equation is much closer to the actual value of 30.0 atm.
The percent error is only 2.6 percent.
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25. 0.50 mol CCl4 gas is introduced into a 10.0 L flask. What fraction of the total volume of the flask is
occupied by CCl4 molecules? See table 5.4 in the text for the value of the van der Waals constant b.
26. Which gas should have the higher strength of attractive forces between its molecules: ammonia or carbon
________________________________________________________________________________ CONCEPTUAL QUESTIONS
1. Why is the density of a gas much lower than that of a solid or liqiud? 2. Refer to the figure in Example 5.1 of the gas trapped in the end of the J-tube. If water was in the tube
instead of mercury what height h of water would produce the same pressure as 42 mm Hg? 3. For each of the following properties state whether it is characteristic of all gases, some gases, or no gases.
b. follows Boyles law at 100 atm and 20°C..
c. follows Boyles law at 1 atm and –200°C.
d. more compressible than water. 4. Which has more molecules: a. 1.0 L of O2 gas at 20°C and 2.0 atm or b. 1.0 L of SF4 gas at 20°C and
2.0 atm? Which one has more mass? 5. How does kinetic molecular theory explain Charles's law? 6. Compare the average kinetic energy of Xe molecules to that of He molecules at the same temperature. 7. Compare the average speed of Xe molecules to that of He molecules at the same temperature. PRACTICE TEST
1. A barometer reads 695 mm Hg. Calculate the pressure in units of:
a. atm b. torr c. Pa d. psi
2. The pressure of H2 gas in a 0.50 L cylinder is 1775 psi at 70°F. What volume would the gas occupy at 1
atm and the same temperature?
3. If 30.0 L of oxygen is cooled at constant pressure from 200°C to 0°C, what would the new volume of
oxygen? Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Gases / 1 01
4. What is the pressure of the gas trapped in the apparatus shown below when the atmospheric pressure is
the air Gas 68 mm Hg 5. A balloon has a volume of 1500 L of He at 0.925 atm and 23°C. At an altitude of 20 km the temperature is
–50°C and the pressure is 0.25 bar. What is the volume of this balloon at 20 km?
6. Given that the Martian atmosphere is mostly CO2 , at a pressure of 5.5 mm Hg at a temperature of –
31.4°C. What is the density of the atmosphere?
7. How many grams of chlorine (Cl 2 ) occupy a 0.716 L cylinder when the pressure is 10.9 atm at 30°C?
8. Fill in the blank spaces in the table.
h. V n 7.25 atm
451 torr 150 mL 2.50 × 10–3 mol
152 kPa 120 mL
2.00 × 102 mol
10.0 mL0.0625 mol
25°C T 325 K
301 K 9. Calculate the volume occupied by 15.2 g of CO2 at 0.74 atm and 24°C.
10. What is the density of uranium hexafluoride gas, UF6 , at STP?
11. When 1.48 g of mercuric chloride is vaporized in a 1.00 L bulb at 680 K, the pressure is 225 mm Hg.
What is the molar mass and molecular formula of mercuric chloride vapor?
12. Determine the molecular mass of chloroform gas if a sample weighing 0.495 g is collected as a vapor (gas)
in a flask of volume 127 cm3 at 98°C. The pressure of the chloroform vapor at this temperature in the flask
was determined to be 754 mm Hg. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 02 / Gases
13. A 150 mL sample of O2 gas is collected over water at 20°C and 758 torr. What volume will the same
sample of oxygen occupy at STP when it is dry?
14. A sample of nitrogen gas is bubbled through liquid water at 25°C and then collected in a volume of 750
cm3 . The total pressure of the gas, which is saturated with water vapor, is found to be 740 mm at 25°C.
The vapor pressure of water at this temperature is 24 mm. How many moles of nitrogen are in the sample?
15. The volume of carbon monoxide gas (CO) collected over water at 25°C was 680 cm3 with a total pressure
of 752 mm Hg. The vapor pressure of water at 25°C is 23.8 mm Hg. Determine the partial pressure and
mole fraction of CO in the container.
16. The partial pressures of N2 , O 2 , and Ar in dry air are 570, 153, and 6 torr, respectively. What are the mole
fractions of these three gases?
17. A mixture of 40.0 g of O2 and 40.0 g of He has a total pressure of 0.900 atm. What are the partial
pressures of O2 and He in the mixture?
18. a. What volume of CO2 at 1 atm and 225°C would be produced by the reaction of 12.0 g NaHCO3 ?
2NaHCO3 + H2 SO4 → Na2 SO4 + 2CO2 + 2H2 O
b. On cooling to 20°C, what volume would the CO2 occupy?
19. How many liters of ammonia at 10 atm and 500°C can be produced by the reaction of 6.0 g of hydrogen
with excess N2 ?
3H2 (g) + N 2 (g) → 4NO(g) + 6H 2 O(l)
20. In the oxidation of ammonia
4NH3 (g) + 5O 2 (g) → 4NO(g) + 6H 2 O(l)
how many liters of O2 , measured at 18°C and 1.10 atm, must be used to produce 50 liters of NO at the
21. How many liters of oxygen, measured at STP, are required for the complete combustion of 72.0 grams of
hexane (C6 H14 ), a component of gasoline?
22. If the root-mean-square speed of an N2 molecule is 475 m/s at 25°C, what is the root-mean-square speed of
a He molecule at 25°C?
23. In each of the following pairs, which gas would you expect to deviate more than the value PV/nRT = 1
expected for an ideal gas?
a. N 2 or SF6
b. He or O 2
c. CO 2 or SO 2
24. Calculate the pressure of 200 moles of NH3 in a 10.0 L container at 500°C using (a) the ideal gas law, and
(b) the Van der Waals equation. General Problems
25. A 0.356 g sample of XH2 (s) reacts with water according to the following equation:
XH2 (s) + 2H 2 O(l) → X(OH)2 (s) + 2H 2 (s)
The hydrogen evolved is collected over water at 23°C and occupies a volume of 431 mL at 746 mm Hg
total pressure. Find the number of moles of H2 produced and the atomic mass of X.
Vapor pressure of H2 O = 21 mm Hg. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website Gases / 1 03
26. A H2 gas thermometer has a volume of 100.0 cm3 when immersed in an ice-water bath at 0°C. When
immersed in boiling liquid Cl 2 , the volume of the H2 at the same pressure was 87.2 cm 3 . Determine the
temperature of the boiling point of Cl2 in K and °C.
27. a. The buoyant force on a balloon is proportional to the mass of air it displaces. What mass of air is
displaced by a weather balloon at an altitude where the pressure is 212 mm Hg and the temperature is –
35°C? Initially the balloon contained 10 moles of He.
b. The lift of the balloon is the difference between its mass and that of the displaced air. What is the lift of
the above balloon?
28. A certain noble gas compound contains 68.8 percent Kr and 31.2 percent F. Its density at STP is 5.44 g/L.
What is the molecular formula of the compound? ANSWERS
26. Back a. 125 torr b. 86.0 kPa
a. 0.849 atm b. 96.5 kPa c. 102 psi
0.55 g H2 and 0.33 g He
2.15 × 10 22 molecules
279 g/mol, HgCl2
a. 1.30 L b. 191 g
155 K (–118°C)
2.8 × 10 2 m/s
carbon tetrachloride (according to van der Waals constant a) Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 1 04 / Gases
6. 7. Because of the large distances between molecules in the gaseous state as compared to the liquid state.
Since Hg is 13.6 times more dense than water, it would take a column of water 13.6 times higher than Hg
to exert the same pressure. Using the definition of 1 atm from Table 5.1 we see that 760 mm Hg = 33.9 ft
of H2 O. Therefore 42 mm Hg = 1.87 ft H2 O. That's 22.4 inches H2 O.
a. some gases b. no gases c. no gases
d. all gases
Equal volumes of two gases at the same temperature and pressure contain the same number of molecules.
There are more grams of SF4 because SF4 molecules have a higher molecular mass than that of O 2
Hint. See Application to the Gas Laws in Section 5.7 of the text.
The average kinetic energy of molecules is directly proportional to the absolute temperature. Therefore,
since both gases are at the same temperature, the average kinetic energy of Xe molecules and He molecules
will be the same.
He molecules, being lighter than Xe molecules, have a higher average speed than Xe molecules. Practice Test
38. a. 0.914 atm b. 695 torr c. 9.26 × 10 4 Pa d. 13.4 psi
790 mm Hg
a. 336 K
b. 434 K
c. 6.39 mol
d. 7.36 × 10 –3 mol
e. 197 L
f. 3960 L
g. 1.38 atm
h. 153 atm
279 g/mol, HgCl2
0.0289 mol N2
CO: 728 mm Hg; X = 0.968
0.782 N 2 , 0.210 O2 , 0.008 Ar
P O2 = 0.1 atm PHe = 0.8 atm
a. 5.84 L b. 3.44 L
C 6 H14 + 9 1/2 O2 → 6CO 2 + 7H2 O; 178 L O2
a. SF6 b . O 2 c. SO2
a. 1270 atm b. 3250 atm
0.0169 mol H2 , 40.1 g/mol
238 K and –35°C
a. V = 700 L, thus 290 g air. b. 290 g air – 40 g He = 250 g lift.
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