Download The Ideal Gas Equation

Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 11
Gases
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
11
Gases
11.1 Properties of Gases
11.2 The Kinetic Molecular Theory of Gases
Molecular Speed
Diffusion and Effusion
11.3 Gas Pressure
Definition and Units of Pressure
Calculation of Pressure
Measurement of Pressure
11.4 The Gas Laws
Boyle’s Law: The Pressure-Volume Relationship
Charles’s and Gay-Lussac’s Law: The Temperature-Volume Relationship
Avogadro’s Law: The Amount-Volume Relationship
The Gas Laws and Kinetic Molecular Theory
The Combined Gas Law: The Pressure-Temperature-Amount-Volume
Relationship
11.5 The Ideal Gas Equation
Applications of the Ideal Gas Equation
11
Gases
11.6 Real Gases
Factors That Cause Deviation from Ideal Behavior
The van der Waals Equation
van der Waals Constants
11.7 Gas Mixtures
Dalton’s Law of Partial Pressures
Mole Fractions
11.6 Reactions with Gaseous Reactants and Products
Calculating the Required Volume of a Gaseous Reactant
Determining the Amount of Reactant Consumed Using Change in Pressure
Using Partial Pressures to Solve Problems
11.1
Properties of Gases
Gases differ from solids and liquids in the following ways:
1) A sample of gas assumes both the shape and volume of the
container.
2) Gases are compressible.
3) The densities of gases are much smaller than those of liquids and
solids and are highly variable depending on temperature and
pressure.
4) Gases form homogeneous mixtures (solutions) with one another
in any proportion.
11.2
The Kinetic Molecular Theory
The kinetic molecular theory explains how the molecular nature of gases
gives rise to their macroscopic properties.
The basic assumptions of the kinetic molecular theory are as follows:
1) A gas is composed of particles that are separated by large distances.
The volume occupied by individual molecules is negligible.
2) Gas molecules are constantly in random motion, moving in straight
paths, colliding with perfectly elastic collisions.
3) Gas molecules do not exert attractive or repulsive forces on one
another.
4) The average kinetic energy of a gas molecules in a sample is
proportional to the absolute temperature:
Ek  T
The Kinetic Molecular Theory
Gases are compressible because molecules in the gas phase are
separated by large distances (assumption 1).
Pressure is the result of the
collisions of gas molecules with
the walls of their container
(assumption 2).
Decreasing volume increases
the frequency of collisions.
Pressure increases as collision
frequency increases.
The Kinetic Molecular Theory
Heating a sample of gas increases its average kinetic energy
(assumption 4).
Gas molecules must move faster.
Faster molecules collide more frequently and at a greater speed.
Pressure increases as collision frequency increases.
The Kinetic Molecular Theory
The total kinetic energy of a mole of gas is equal to:
3
RT
2
The average kinetic energy of one molecule is:
1
mu 2
2
m is the mass
u 2 is the mean square speed
For one mole of gas:
1
 3
N A  mu 2   RT
2
 2
rearrange and
Take the square root
(m x NA = M)
urms 
3RT
M
The Kinetic Molecular Theory
The root-mean-square (rms) speed (urms) is the speed of a molecule
with the average kinetic energy in a gas sample.
urms is directly proportional to temperature
urms 
3RT
M
The Kinetic Molecular Theory
The root-mean-square (rms) speed (urms) is the speed of a molecule
with the average kinetic energy in a gas sample.
urms is inversely proportional to the square root of M.
urms 
3RT
M
The Kinetic Molecular Theory
When two gases are at the same temperature, it is possible to
compare the the urms values of the different gases.
urms (1)
M2

urms (2)
M1
Worked Example 11.1
Determine how much faster a helium atom moves, on average, than a carbon
dioxide molecule at the same temperature.
urms (1)
M2

urms (2)
M1
Strategy Use equation
and the molar masses of He and CO2 to
determine the ratio of their root-mean-square speeds. When solving a problem
such as this, it is generally best to label the lighter of the two molecules as
Think
It Remember
that2.the
relationship
between
molar
molecule
1 andAbout
the heavier
as molecule
This
ensures that
the result
will be
molecular
speed iseasy
reciprocal.
A CO2 molecule has
greatermass
than and
1, which
is relatively
to interpret.
approximately 10 times the mass of an He atom. Therefore, we
should
He atom,
onand
average,
to be
moving
approximately
Solution
Theexpect
molaran
masses
of He
CO2 are
4.003
and 44.02
g/mol,
√10 times (~3.2 times) as fast as a CO2 molecule.
respectively.
44.02 g
1 mol
urms(He)
=
= 3.316
urms(CO2)
4.003 g
1 mol
On average, He atoms move 3.316 times as fast as CO2 molecules at the same
temperature.
The Kinetic Molecular Theory
Diffusion is the mixing of gases as the result of random motion and
frequent collisions.
Effusion is the escape of gas molecules from a container to a region
of vacuum.
The Kinetic Molecular Theory
Graham’s law states that the rate of diffusion or effusion of a gas is
inversely proportional to the square root of its molar mass.
Rate 
1
M
The Kinetic Molecular Theory
Determine the molar mass and identity of a diatomic gas that moves
4.67 times as fast as CO2.
Solution:
Step 1: Use the equation below to determine the molar mass of the
unknown gas:
urms (1)
M2

urms (2)
M1
urms(unknown gas) = 4.67 x urms(CO2)
urms (unknown)
44.02 g/mol
 4.67 
urms (CO2 )
M  unknown 
M = 2.018 g/mol
The gas must be H2.
11.3 Gas Pressure
Pressure is defined as the force applied per unit area:
force
pressure =
area
The SI unit of force is the newton (N), where
1 N = 1 kg•m/s2
The SI unit of pressure is the pascal (Pa), defined as 1 newton per
square meter.
1 Pa = 1 N/m2
Gas Pressure
Gas Pressure
A barometer is an instrument that is used to measure atmospheric
pressure.
Standard atmospheric pressure
(1 atm) was originally defined
as the pressure that would
support a column of mercury
exactly 760 mm high.
P  hdg
h = height in m
d = is the density in kg/m3
g = is the gravitational constant (9.80665 m/s2)
Gas Pressure
A barometer is an instrument that is used to measure atmospheric
pressure.
1 atm* = 101,325 Pa
= 760 mmHg*
= 760 torr*
= 1.01325 bar
= 14.7 psi
* Represents an exact number
Gas Pressure
A manometer is a device used to measure pressures other than
atmospheric pressure.
Worked Example 11.2
Calculate the pressure exerted by a column of mercury 70.0 cm high. Express the
pressure in pascals, in atmospheres, and in bars. The density of mercury is
13.5951 g/cm3.
Strategy Use P = hdg to calculate pressure. Remember that the height must be
expressed in meters and density must be expressed in kg/m3.
Solution
1m
h = 70.0 cm ×
= 0.700 m
100 cm
3
d=
13.5951 g
1 kg
100 cm
×
×
cm3
1000 g
1m
g = 9.80665 m/s2
= 1.35951×104 kg/m3
Worked Example 11.2 (cont.)
Solution
9.80665 m
1.35951×104 kg
pressure = 0.700 m ×
×
= 9.33×104 kg/m∙s2
2
3
s
m
9.33×104 Pa
9.33×104
1 atm
Pa ×
= 0.921 atm
101,325 Pa
0.921 atm ×
1.01325 bar
= 0.933 bar
1 atm
Think About It Make sure your units cancel properly in this type of problem.
Common errors include forgetting to express height in meters and density in
kg/m3. You can avoid these errors by becoming familiar with the value of
atmospheric pressure in the various units. A column of mercury slightly less than
760 mm is equivalent to slightly less than 101,325 Pa, slightly less than 1 atm,
and slightly less than 1 bar.
11.4
The Gas Laws
(a)
Boyle’s law states that the
pressure of a fixed amount
of gas at a constant
temperature is inversely
proportional to the volume
of the gas.
1
V
P
P1V1 = P2V2
at constant temperature
(b)
(c)
P (mmHg)
760
1520
2280
V (mL)
100
50
33
The Gas Laws
Calculate the volume of a sample of gas at 5.75 atm if it occupies
5.14 L at 2.49 atm. (Assume constant temperature.)
Solution:
Step 1: Use the relationship below to solve for V2:
P1V1 = P2V2
P1  V1 2.49 atm  5.14 L
V2 

 2.26 L
P2
5.75 atm
Worked Example 11.3
If a skin diver takes a breath at the surface, filling his lungs with 5.82 L of air,
what volume will the air in his lungs occupy when he dives to a depth where the
pressure of 1.92 atm? (Assume constant temperature and that the pressure at the
surface is exactly 1 atm.)
Strategy Use P1V1 = P2V2 to solve for V2.
Solution P1 = 1.00 atm, V1 = 5.82 L, and P2 = 1.92 atm.
V2 =
P1 × V1
1.00 atm × 5.82 L
=
= 3.03 L
P2
1.92 atm
Think About It At higher pressure, the volume should be smaller. Therefore,
the answer makes sense.
The Gas Laws
Charles’s and Gay-Lussac’s law, (or simply Charles’s Law) states
that the volume of a gas maintained at constant pressure is directly
proportional to the absolute temperature of the gas.
Higher temperature
Lower temperature
The Gas Laws
Charles’s and Gay-Lussac’s law, (or simply Charles’s Law) states
that the volume of a gas maintained at constant pressure is directly
proportional to the absolute temperature of the gas.
V T
V1
V2

T1
T2
at constant pressure
Worked Example 11.4
A sample of argon gas that originally occupied 14.6 L at 25°C was heated to
50.0°C at constant pressure. What is its new volume?
Strategy Use V1/T1 = V2/T2 to solve for V2. Remember that temperatures must
be expressed in kelvin.
Solution T1 = 298.15 K, V1 = 14.6 L, and T2 = 323.15 K.
V2 =
V1 × T2
14.6 L × 323.15 K
=
= 15.8 L
T1
298.15 K
Think About It When temperature increases at constant pressure, the volume of
a gas sample increases.
The Gas Laws
Avogadro’s law states that the volume of a sample of gas is directly
proportional to the number of moles in the sample at constant
temperature and pressure.
V n
V1
V2

n1
n2
at constant temperature and pressure
Worked Example 11.5
If we combine 3.0 L of NO and 1.5 L of O2, and they react according to the
balanced equation 2NO(g) + O2(g) → 2NO2(g), what volume of NO2 will be
produced? (Assume that the reactants and products are all at the same temperature
and pressure.)
Strategy Apply Avogadro’s law to determine the volume of a gaseous product.
Solution Because volume is proportional to the number of moles, the balanced
equation determines in what volume ratio the reactants combine and the ratio of
product volume to reactant volume. The amounts of reactants given are
stoichiometric amounts.
According to the balanced equation, the volume of NO2 formed will be equal to
the volume of NO that reacts. Therefore, 3.0 L of NO2 will form.
Think About It When temperature increases at constant pressure, the volume
of a gas sample increases.
The Gas Laws
A sample of gas originally occupies 29.1 L at 0.0°C. What is its new
volume when it is heated to 15.0°C? (Assume constant pressure.)
Solution:
Step 1:Use the relationship below to solve for V2: (Remember that
temperatures must be expressed in kelvin.
V1
V2

T1
T2
V1  T2 29.1 L  288.15 K
V2 

 30.7 L
T1
273.15 K
The Gas Laws
What volume in liters of water vapor will be produced when 34 L of
H2 and 17 L of O2 react according to the equation below:
2H2(g) + O2(g) → 2H2O(g)
Assume constant pressure and temperature.
Solution:
Step 1:Because volume is proportional to the number of moles, the
balanced equation determines in what volume ratio the reactants
combine and the ratio of product volume to reactant volume. The
amounts of reactants given are stoichiometeric amounts.
34 L of H2O will form.
The Gas Laws
Cooling at constant volume:
pressure decreases
Heating at constant
volume: pressure increases
The Gas Laws
Cooling at constant pressure:
volume decreases
Heating at constant pressure:
volume increases
The Gas Laws
The presence of additional molecules causes an increase in pressure.
The Gas Laws
The combined gas law can be used to solve problems where any or
all of the variables changes.
P1V1
P2V2

n1T1
n2T2
The Gas Laws
The volume of a bubble starting at the bottom of a lake at 4.55°C
increases by a factor of 10 as it rises to the surface where the
temperature is 18.45°C and the air pressure is 0.965 atm. Assume
the density of the lake water is 1.00 g/mL. Determine the depth of
the lake.
Solution:
Step 1:Use the combined gas law to find the pressure at the bottom
of the lake; assume constant moles of gas.
P1V1
P2V2

n1T1
n2T2
P2V2T1  0.965 atm  10 V1  277.7 K 
P1 

 9.19 atm
T2V1
 291.6 K V1 
The Gas Laws
The volume of a bubble starting at the bottom of a lake at 4.55°C
increases by a factor of 10 as it rises to the surface where the
temperature is 18.45°C and the air pressure is 0.965 atm. Assume
the density of the lake water is 1.00 g/mL. Determine the depth of
the lake.
Solution:
Step 2:Use the equation P = hdg to determine the depth of the lake.
The pressure represents the difference in pressure from the
surface to the bottom of the lake.
Pressure exerted by the lake = Pbottom of lake – Pair
P = 9.19 atm – 0.965 atm = 8.225 atm
The Gas Laws
The volume of a bubble starting at the bottom of a lake at 4.55°C
increases by a factor of 10 as it rises to the surface where the
temperature is 18.45°C and the air pressure is 0.965 atm. Assume
the density of the lake water is 1.00 g/mL. Determine the depth of
the lake.
Solution:
Step 2 (Cont): Convert pressure to pascals and density to kg/m3.
8.225 atm ×
101,325 Pa
= 833,398 Pa
1 atm
3
1g
1 kg
1000 kg
 100 cm 
×
×
=


cm3
1000 g
m3
 1m 
P = hdg
833,398 Pa = h(1000 kg/m3)(9.81 m/s2)
h = 85.0 m
Worked Example 11.6
If a child releases a 6.25-L helium balloon in the parking lot of an amusement
park where the temperature is 28.50°C and the air pressure is 757.2 mmHg,
what will the volume of the balloon be when it has risen to an altitude where the
temperature is -34.35°C and the air pressure is 366.4 mmHg?
Strategy In this case, because there is a fixed amount of gas, we use P1V1/T1 =
P2V2/T2. The only value we don’t know is V2. Temperatures must be expressed in
kelvins. We can use any units of pressure, as long as we are consistent.
Solution T1 = 301.65 K, T2 = 238.80 K.
V2 =
P1T2V1 757.2 mmHg × 238.80 K × 6.25 L
=
= 10.2 L
P2T1
366.4 mmHg × 301.65 K
Think About It Note that the solution is essentially multiplying the original
volume by the ratio of P1 and P2, and by the ratio of T2 to T1. The effect of
decreasing external pressure is to increase the balloon volume. The effect of
decreasing temperature is to decrease the volume. In this case, the effect of
decreasing pressure predominates and the balloon volume increases significantly.
11.5
The Ideal Gas Equation
The gas laws can be combined into a general equation that
describes the physical behavior of all gases.
1
V
P
V T
V n
Boyle’s law
Charles’s law
Avogadro’s law
nT
V
P
nT
V R
P
rearrangement
PV = nRT
R is the proportionality constant, called the gas constant.
The Ideal Gas Equation
The ideal gas equation (below) describes the relationship among
the four variables P, V, n, and T.
PV = nRT
An ideal gas is a hypothetical sample of gas whose pressurevolume-temperature behavior is predicted accurately by the ideal
gas equation.
The Ideal Gas Equation
The gas constant (R) is the proportionality constant and its value
and units depend on the units in which P and V are expressed.
PV = nRT
The Ideal Gas Equation
Standard Temperature and Pressure (STP) are a special set of
conditions where:
Pressure is 1 atm
Temperature is 0°C (273.15 K)
The volume occupied by one mole of an ideal gas is then 22.41 L:
PV = nRT
(1 mol)(0.08206 L  atm/K  mol)(273.15 K)
V
= 22.41 L
1 atm
Worked Example 11.7
Calculate the volume of a mole of ideal gas at room temperature (25°C) and 1
atm.
Strategy Convert the temperature in °C to kelvins, and use the ideal gas
equation to solve for the unknown volume.
Solution The data given are n = 1 mol, T = 298.15 K, and P = 1.00 atm.
Because the pressure is expressed in atmospheres, we use R = 0.08206
L∙atm/K∙mol in order to solve for volume in liters.
V=
nRT (1 mol)(0.08206 L∙atm/K∙mol)(298.15 K)
=
= 24.5 L
P
1 atm
Think About It With the pressure held constant, we should expect the volume
to increase with increased temperature. Room temperature is higher than the
standard temperature for gases (0°C), so the molar volume at room temperature
(25°C) should be higher than the molar volume at 0°C–and it is.
The Ideal Gas Equation
Using algebraic manipulation, it is possible to solve for variables
other than those that appear explicitly in the ideal gas equation.
PV = nRT
n P

V RT
n P
M 
M
V RT
PM
d
RT
d is the density (in g/L)
M is the molar mass (in g/mol)
The Ideal Gas Equation
What pressure would be required for helium at 25°C to have the
same density as carbon dioxide at 25°C and 1.00 atm?
Solution:
Step 1:Use the equation below to calculate the density of CO2 at
25°C and 1 atm.
PM
d
RT
d =
1.00 atm  44.01 g/mol 
 0.08206 Latm/mol K  298.15 K 
= 1.7966 g/L
The Ideal Gas Equation
What pressure would be required for helium at 25°C to have the
same density as carbon dioxide at 25°C and 1 atm?
Solution:
Step 2:Use the density of CO2 found in step 1 to calculate the
pressure for He at 25°C.
1.7966 g/L =
P = 11.0 atm
 P  4.003 g/mol 
 0.08206 Latm/mol K  298.15 K 
Worked Example 11.8
Carbon dioxide is effective in fire extinguishers partly because its density is
greater than that of air, so CO2 can smother the flames by depriving them of
oxygen. (Air has a density of approximately 1.2 g/L at room temperature and 1
atm.) Calculate the density of CO2 at room temperature (25°C) and 1.0 atm.
Strategy Use d = PM/RT to solve for density. Because the pressure is expressed
in atm, we should use R = 0.08206 L∙atm/K∙mol. Remember the express
temperature in kelvins.
Solution The molar mass of CO2 is 44.01 g/mol.
d=
PM
(1 atm)(44.01 g/mol)
=
= 24.5 L
RT
0.08206 L∙atm/K∙mol)(298.15 K)
Think About It The calculated density of CO2 is greater than that of air under
the same conditions (as expected). Although it may seem tedious, it is a good idea
to write units for each and every entry in a problem such as this. Unit cancellation
is very useful for detecting errors in your reasoning or your solution setup.
11.6
Real Gases
The van der Waals equation is useful for gases that do not behave
ideally.
Experimentally
measured pressure
Container volume

an2 
 P  2  V  nb   nRT
V 

corrected
corrected
pressure term volume term
Real Gases
The van der Waals equation is useful for gases that do not behave
ideally.

an2 
 P  2  V  nb   nRT
V 

Real Gases
Calculate the pressure exerted by 0.35 mole of oxygen gas in a
volume of 6.50 L at 32°C using (a) the ideal gas equation and(b) the
van der Waals equation.
Solution:
Step 1:Use the ideal gas equation to calculate the pressure of O2.
PV = nRT
P
nRT  0.35 mol  0.08206 Latm/mol K 305.15 K 

 1.3 atm
V
6.50 L
Real Gases
Calculate the pressure exerted by 0.35 mole of oxygen gas in a
volume of 6.50 L at 32°C using (a) the ideal gas equation and(b) the
van der Waals equation.
Solution:
Step 2:Use table 11.6 to find the values of a and b for O2.
Real Gases
Calculate the pressure exerted by 0.35 mole of oxygen gas in a
volume of 6.50 L at 32°C using (a) the ideal gas equation and (b)
the van der Waals equation.
Solution
Step 3:Use the van der Waals equation to calculate P.

an2 
 P  2  V  nb   nRT
V 

nRT
an2 0.35 mol  0.08206 Latm/mol K 305.15 K  1.36atmL2 /mol 2 (0.35 mol)2
P



2
6.50 L   0.35 mol  0.0318 L/mol  
V  nb  V 2
 6.50 L 
P = 1.3 atm
Worked Example 11.10
A sample of 3.50 moles of NH3 gas occupies 5.20 L at 47°C. Calculate the
pressure of the gas (in atm) using (a) the ideal gas equation and (b) the van der
Waals equation.
Strategy (a) Use the ideal gas equation, PV = nRT. (b) Use (P + an2/V2)(V – nb)
= nRT and a and b values for NH3 from Table 11.5
Solution T = 320.15 K, a = 4.17 atm∙L/mol2, and b = 0.0371 L/mol.
(a) P =
(3.50 mol)(0.08206 L∙atm/K∙mol)(320.15 K)
nRT
=
= 17.7 atm
5.20 L
V
(b) Evaluating the correction terms in the van der Waals equation, we get
an2
(4.17 atm∙L/mol2)(3.50 mol)2
=
= 1.89 atm
V2
(5.20 L)2
nb = (3.50 mol)
0.0371 L
= 0.130 L
mol
Worked Example 11.10 (cont.)
Solution Finally, substituting these results into the van der Waals equation, we
get
(P + 1.89 atm)(5.20 L – 0.130 L) = (3.50 mol)(0.08206 L∙atm/K∙mol)(320.15 K)
P = 16.2 atm
Think About It As is often the case, the pressure exerted by the real gas is
lower than predicted by the ideal gas equation.
Worked Example 11.11
Consider a gas sample consisting of molecules with radius r. (a) Determine the
excluded volume defined by two molecules and (b) calculate the excluded volume
per mole (b) for the gas. Compare the excluded volume per mole with the volume
actually occupied by a mole of the molecules.
Strategy (a) Use the formula for volume of a sphere to calculate the volume of
a sphere of radius 2r. This is excluded volume defined by two molecules.
(b) Multiply the excluded volume per molecule by Avogardo’s number to
Think About It The excluded volume per mole is four times the
determine excluded volume per mole.
volume actually occupied by a mole of molecules.
4
Solution The equation for volume of a sphere is 3 πr3; NA = 6.022×1023.
4
4
(a) Excluded volume defined by two molecules = 3 π(2r)3 = 8( 3 πr3).
4
4
8( 3 πr3)
NA molecules
4NA( 3 πr3)
(b) Excluded volume per mole =
×
=
2 molecules
1 mole
1 mole
The volume actually occupied by a mole of molecules with a radius r is
4
4NA( 3 πr3).
11.7
Gas Mixtures
When two or more gases are placed in a container, each gas behaves
as though it occupies the container alone.
1.00 mole of N2 in a 5.00 L container at 0°C exerts a pressure of
4.48 atm.
PN2 
(1mol)(0.08206 L  atm/K  mol)(273.15 K)
 4.48 atm
5.00 L
Addition of 1.00 mole of O2 in the same container exerts an
additional 4.48 atm of pressure.
PO2 
(1mol)(0.08206 L  atm/K  mol)(273.15 K)
 4.48 atm
5.00 L
The total pressure of the mixture is the sum of the partial pressures
(Pi):
Ptotal = PN2 + PO2 = 4.48 atm + 4.48 atm = 8.96 atm
Gas Mixtures
Dalton’s law of partial pressure states that the total pressure
exerted by a gas mixture is the sum of the partial pressures exerted
by each component of the mixture:
Ptotal = Pi
Gas Mixtures
Determine the partial pressures and the total pressure in a 2.50-L
vessel containing the following mixture of gases at 15.8°C: 0.0194
mol He, 0.0411 mol H2, and 0.169 mol Ne.
Solution:
Step 1:Since each gas behaves independently, calculate the partial
pressure of each using the ideal gas equation:
PHe
0.0194 mol He  0.08206 Latm/mol K  288.95 K 


 0.184 atm
2.50 L
PH2 
PNe
 0.0411 mol H2 0.08206 Latm/mol K 288.95 K   0.390 atm
2.50 L
0.169 mol Ne  0.08206 Latm/mol K 288.95 K 


 1.60 atm
2.50 L
Gas Mixtures
Determine the partial pressures and the total pressure in a 2.50-L
vessel containing the following mixture of gases at 15.8°C: 0.0194
mol He, 0.0411 mol H2, and 0.169 mol Ne.
Solution:
Step 2:Use the equation below to calculate total pressure.
Ptotal = Pi
Ptotal = 0.184 atm + 0.390 atm + 1.60 atm = 2.17 atm
Gas Mixtures
Each component of a gas mixture exerts a pressure independent of
the other components. The total pressure is the sum of the partial
pressures.
Worked Example 11.12
A 1.00-L vessel contains 0.215 mole of N2 gas and 0.0118 mole of H2 gas at
25.5°C. Determine the partial pressure of each component and the total pressure
in the vessel.
Strategy Use the ideal gas equation to find the partial pressure of each
component
ofAbout
the mixture,
and
sum
the twoinpartial
pressures
to find
Think
It The
total
pressure
the vessel
can also
be the total
pressure.
determined by summing the number of moles of mixture
components (0.215 + 0.0118 = 0.227 mol) and solving the ideal gas
Solution
T = 298.65
equation
for PtotalK:
(0.215 mol)(0.08206 L∙atm/K∙mol)(298.65 K)
PPN2 = = (0.227 mol)(0.08206 L∙atm/K∙mol)(298.65 K)= 5.27
atm
=
5.56
atm
1.00
L
total
1.00 L
(0.0118 mol)(0.08206 L∙atm/K∙mol)(298.65 K)
PH2 =
= 0.289 atm
1.00 L
Ptotal = PN2 + PH2 = 5.27 atm + 0.289 atm = 5.56 atm
Gas Mixtures
The relative amounts of the components in a gas mixture can be
specified using mole fractions.
ni
i =
ntotal
Χi is the mole fraction.
ni is the moles of a certain component
ntotal is the total number of moles.
There are three things to remember about mole fractions:
1) The mole fraction of a mixture component is always less than 1.
2) The sum of mole fractions for all components of a mixture is
always 1.
3) Mole fractions are dimensionless.
Worked Example 11.13
In 1999, the FDA approved the use of nitric oxide (NO) to treat and prevent lung
disease, which occurs commonly in premature infants. The nitric oxide used in
this therapy is supplied to hospitals in the form of a N2/NO mixture. Calculate the
mole fraction of NO in a 10.00-L gas cylinder at room temperature (25°C) that
contains 6.022 mol N2 and in which the total pressure is 14.75 atm.
Strategy
UseAbout
the ideal
gascheck
equation
calculate
the total
of moles in the
Think
It To
yourtowork,
determine
χNnumber
by
subtracting
2
cylinder.
N2 mole
from the
total and
to determine
moles of NO. Divide
χNOSubtract
from 1.moles
Usingof
each
fraction
the total pressure,
moles NO
by total
get moleoffraction.
calculate
themoles
partialtopressure
each component using χi = Pi/Ptotal
and verify that they sum to the total pressure.
Solution The temperature is 298.15 K.
total moles =
PV
(14.75 atm)(10.00 L)
=
= 6.029 mol
RT (0.08206 L∙atm/K∙mol)(298.15 K)
mol NO = total moles – N2 = 6.029 – 6.022 = 0.007 mol NO
χNO =
nNO
0.007 mol NO
=
= 0.001
ntotal
6.029 mol
11.8
Reactions with Gaseous Reactants and Products
What mass (in grams) of Na2O2 is necessary to consume 1.00 L of
CO2 at STP?
2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g)
Solution:
Step 1:Convert 1.00 L of CO2 at STP to moles using the ideal gas
equation.
PV = nRT
(1 atm)(1.00 L) = n(0.08206 Latm/mol K)(273.15 K)
n = 0.04461 moles CO2
Reactions with Gaseous Reactants and Products
What mass (in grams) of Na2O2 is necessary to consume 1.00 L of
CO2 at STP?
2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g)
Solution:
Step 2:Determine the stoichiometric amount of Na2O2.
 2 mol Na 2O2   77.98 g 
0.04461 moles CO2 × 
× 
 = 3.48 g
 2 mol CO2   mol 
Worked Example 11.14
Sodium peroxide (Na2O2) is used to remove carbon dioxide from (and add
oxygen to) the air supply in spacecrafts. It works by reacting with CO2 in the air
to produce sodium carbonate (Na2CO3) and O2.
2Na2O2(s) + 2CO2(g) → 2Na2CO3(s) + O2(g)
What volume (in liters) of CO2 (at STP) will react with a kilogram of Na2O2?
Strategy Convert the given mass of Na2O2 to moles, use the balanced equation
to determine the stoichiometric amount of CO2, and then use the ideal gas
equation to convert moles of CO2 to liters.
Worked Example 11.14 (cont.)
Solution The molar mass of Na2O2 is 77.98 g/mol (1 kg = 1000 g). (Treat the
specified mass of NaO2 as an exact number.)
1000 g Na2O2 ×
1 mol Na2O2
= 12.82 mol Na2O2
77.98 g Na2O2
12.82 mol Na2O2 ×
VCO2 =
2 mol CO2
= 12.82 mol Na2O2
2 mol Na2O2
(12.82 mol CO2)(0.08206 L∙atm/K∙mol)(273.15 K)
= 287.4 L CO2
1 atm
Think About It The answer seems like an enormous volume of CO2. If you
check the cancellation of units carefully in ideal gas equation problems, however,
with practice you will develop a sense of whether such a calculated volume is
reasonable.
Reactions with Gaseous Reactants and Products
Although there are no empirical gas laws that focus on the
relationship between n and P, it is possible to rearrange the ideal
gas equation to find the relationship.
PV = nRT
rearrangement
V
nP
RT
The change in pressure in a reaction vessel can be used to
determine how many moles of gaseous reactant are consumed:
V
n   P 
RT
Worked Example 11.15
Another air-purification method for enclosed spaces involves the use of
“scrubbers” containing aqueous lithium hydroxide, which react with carbon
dioxide to produce lithium carbonate and water:
2LiOH(aq) + CO2(g) → Li2CO3(s) + H2O(l)
Consider the air supply in a submarine with a total volume of 2.5×105 L. The
pressure of 0.9970 atm, and the temperature 25°C. If the pressure in the
Think
About
It Careful
units isdioxide
essential.
Note
that
submarine
drops
to 0.9891
atm ascancellation
the result ofofcarbon
being
consumed
by
this amount of CO2 corresponds to 162 moles or 3.9 kg of LiOH.
an aqueous lithium hydroxide
scrubber, how many moles of CO2 are consumed.
(It’s a good idea to verify this yourself.
Strategy Use Δn = ΔP×(V/RT) to determine Δn, the number of moles CO2
consumed.
Solution ΔP = 0.9970 atm – 0.9891 atm = 7.9×10-3 atm, V = 2.5×105 L, and
T = 298.15 K.
2.5×105 L
-3
ΔnCO2 = 7.9×10 atm ×
= 81 moles CO2
(0.08206 L∙atm/K∙mol) × (298.15 K)
consumed
Reactions with Gaseous Reactants and Products
The volume of gas produced by a chemical reaction can be measured
using an apparatus like the one shown below.
When gas is collected over water in this manner, the total pressure is
the sum of two partial pressures:
Ptotal = Pcollected gas + PH2O
Reactions with Gaseous Reactants and Products
The vapor pressure of water is known at various temperatures.
Gas Mixtures
Calculate the mass of O2 produced by the decomposition of KClO3
when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm.
Solution:
Step 1:Use Table 11.5 to determine the vapor pressure of water at
30.0°C.
Reactions with Gaseous Reactants and Products
Calculate the mass of O2 produced by the decomposition of KClO3
when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm.
Solution:
Step 2:Convert the vapor pressure of water at 30.0°C to atm and then
use Dalton’s law to calculate the partial pressure of O2.
PH O @ 30.0 C = 31.8 torr ×
2
1 atm
= 0.041842 atm
760 torr
Ptotal = PO2 + PH2O
PO2 = Ptotal – PH2O
PO2 = 1.015 atm – 0.041842 atm = 0.973158 atm
Reactions with Gaseous Reactants and Products
Calculate the mass of O2 produced by the decomposition of KClO3
when 821 mL of O2 is collected over water at 30.0°C and 1.015 atm.
Solution:
Step 3:Convert to moles of O2 using the ideal gas equation and then
find mass.
PV = nRT
n
 0.973158 atm  0.821 L 
PV

 0.032117 moles O 2
RT  0.08206 Latm/mol K  303.15 K 
 32.00 g 
0.032117 moles O 2  
  1.03 g O 2
 mole 
Worked Example 11.16
Calcium metal reacts with water to produce hydrogen gas:
Ca(s) + H2O(l) → Ca(OH)2(aq) + H2(g)
Determine the mass of H2 produced at 25°Cand 0.967 atm when 525 mL of the
gas is collected over water as shown in Figure 11.23.
Strategy Use Dalton’s law of partial pressures to determine the partial pressure
Think
Check unit
cancellation
carefully,
of H2, use
the About
ideal gasIt equation
to determine
moles
of H2, and remember
then use the molar
densities
gases
are careful
relatively
low. The
mass of
mass ofthat
H2 the
to convert
toof
mass.
(Pay
attention
to units.
Atmospheric
approximately
half a liter ofwhereas
hydrogen
or near
room of
temperature
pressure
is given in atmospheres,
theatvapor
pressure
water is tabulated
in torr.)and 1 atm should be a very small number.
Solution PH2 = Ptotal – PH2O = 0.967 atm – 0.0313 atm = 0.936 atm
moles of H2 =
(0.9357 atm)(0.525 L)
= 2.01×10-2 mol
(0.08206 L∙atm/K∙mol)(298.15 K)
moles of H2 = (2.008×10-2 mol)(2.016 g/mol) = 0.0405 g H2
11
Chapter Summary: Key Points
Characteristics of Gases
Gas Pressure: Definition and Units
Calculation of Pressure
Measurement of Pressure
Boyle’s Law: The Pressure-Volume Relationship
Charles’s and Gay-Lussac’s Law: The Temperature-Volume Relationship
Avogadro’s Law: The Amount-Volume Relationship
Deriving the Ideal Gas Equation from the Empirical Gas Laws
Applications of the Ideal Gas Equation
Calculating the Required Volume of a Gaseous Reactant
Determining the Amount of Reactant Consumed Using Change in Pressure
Predicting the Volume of a Gaseous Product
Dalton’s Law of Partial Pressures
Mole Fractions
Using Partial Pressures to Solve Problems
Application of the Gas Laws
Molecular Speed
Diffusion and Effusion
Factors That Cause Deviation from Ideal Behavior
The van der Waals Equation