Chapter 5
The Gaseous
State
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Contents and Concepts
Gas Laws
We will investigate the quantitative relationships that describe the behavior of gases.
1. Gas Pressure and Its Measurement
2. Empirical Gas Laws
3. The Ideal Gas Law
4. Stoichiometry Problems Involving Gas Volumes
5. Gas Mixtures; Law of Partial Pressures
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Kinetic-Molecular Theory
This section will develop a model of gases as
molecules in constant random motion.
6. Kinetic Theory of Gases
7. Molecular Speeds; Diffusion and Effusion
8. Real Gases
4
5
• Gases assume the volume and shape of their containers.
• Gases are the most compressible state of matter.
• Gases will mix evenly and completely when confined to
the same container.
• Gases have much lower densities than liquids and solids.
Physical Characteristics of Gases
NO2 gas
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Gases differ from liquids and solids:
They are compressible.
Pressure, volume, temperature, and amount
are related.
Some applications
• How does a pressure cooker work?
• How is gas pressure applied in spray cans?
• How does a hot air balloon work?
• Why do we not want our tires to be full during hot summer days?
• Why do balloons deflate when left outside on a cold weather?
8
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,325 Pa
Pressure = Force Area
(force = mass x acceleration)
9
Manometers Used to Measure Gas Pressures
closed-tube open-tube
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Boyle’s Law The volume of a sample of gas at constant
temperature varies inversely with the applied
pressure.
The mathematical relationship:
In equation form:
PV
1
ffii
constant
VPVP
PV
11
P a 1/V
P x V = constant
P1 x V1 = P2 x V2
Boyle’s Law
Constant temperature
Constant amount of gas
12
A sample of chlorine gas occupies a volume of 946 mL at a
pressure of 726 mmHg. What is the pressure of the gas (in
mmHg) if the volume is reduced at constant temperature to 154
mL?
P1 x V1 = P2 x V2
P1 = 726 mmHg
V1 = 946 mL
P2 = ?
V2 = 154 mL
P2 = P1 x V1
V2
726 mmHg x 946 mL 154 mL
= = 4460 mmHg
P x V = constant
• 1. The pressure on a 2.50 L of anesthetic gas changes from 105 kPa to 40.5 kPa. What will be the new volume if the temperature remains constant?
• 2. A gas with a volume of 4.00 L at a pressure of 205 kPa is allowed to expand to a volume of 12.0 L. What is the pressure in the container of the temperature remains constant?
14
Chemistry in Action:
Scuba Diving and the Gas Laws
P V
Depth (ft) Pressure
(atm)
0 1
33 2
66 3
15 As T increases V increases
Variation in Gas Volume with Temperature at Constant Pressure
16
Variation of Gas Volume with Temperature
at Constant Pressure
V a T
V = constant x T
V1/T1 = V2 /T2 T (K) = t (0C) + 273.15
Charles’ &
Gay-Lussac’s
Law
Temperature must be
in Kelvin
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A balloon was immersed in liquid nitrogen (black container) and is shown immediately after being removed. It shrank because air inside contracts in volume.
As the air inside warms, the balloon expands to its orginial size.
Charles’ Law
• Volume vs. Temperature at constant pressure
• Volume is directly proportional to temperature
• Vα T @ constant P
• V=kT; V/T= k
• V1/T1 = V2/T2
• 1. If a sample of gas occupies 6.80 L at 3250 C, what will be its volume at 250C if the pressure does not change?
Note: convert Celsius to Kelvin scale by adding 273 to the given 0C.
• 2. Exactly 5.00 L of air at -50.00C is warmed to 100.00C. What is the new volume if the pressure remains constant?
20
A sample of carbon monoxide gas occupies 3.20 L at 125 0C.
At what temperature will the gas occupy a volume of 1.54 L if
the pressure remains constant?
V1 = 3.20 L
T1 = 398.15 K
V2 = 1.54 L
T2 = ?
T2 = V2 x T1
V1
1.54 L x 398.15 K 3.20 L
= = 192 K
V1 /T1 = V2 /T2
T1 = 125 (0C) + 273.15 (K) = 398.15 K
Gay-Lussac’s Law
• Pressure and Temperature at constant Volume
• P is proportional to T @ constant volume
• PαT @ constant V
• P=kT; P/T=k
• P1/T1 = P2/T2
• A gas has a pressure of 6.58 kPa at 539K. What will be the pressure at 211 K if the volume does not change?
• The pressure in an automobile tire is 198 kPa at 270C. At the end of a trip on a hot sunny day, the pressure has risen to 225 kPa. What is the temperature of the air in the tire?
• Helium gas in a 2.00-L cylinder is unter 1.12 atm pressure. At 36.5 0C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas?
23
Avogadro’s Law
V a number of moles (n)
V = constant x n
V1 / n1 = V2 / n2
Constant temperature
Constant pressure
Avogadro’s Principle
• Volume is proportional to number of moles
• V α n where n is the number of moles
• More moles occupy greater volume
25
Ammonia burns in oxygen to form nitric oxide (NO) and water
vapor. How many volumes of NO are obtained from one volume
of ammonia at the same temperature and pressure?
4NH3 + 5O2 4NO + 6H2O
1 mole NH3 1 mole NO
At constant T and P
1 volume NH3 1 volume NO
Combined Gas Law
• P α T
V
P=k T
V
PV = k
T
• At 0.000C and 1.00 atm pressure a sample of gas occupies 30.0 mL. if the temperature is increased 30.00C and the gas sample is transferred to 20.0 mL container, what is the gas pressure?
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Standard Temperature and Pressure (STP)
The reference condition for gases, chosen by
convention to be exactly 0°C and 1 atm pressure.
The molar volume, Vm, of a gas at STP is 22.4
L/mol.
The volume of the
yellow box is 22.4 L. To
its left is a basketball.
STP
• At STP,
• T=00C or 273 K
• P= 1atm
• At Standard temperature and pressure, molar volume is 22.4 L
• Meaning, 1 mole of any gas occupies 22.4 liters at STP
Ideal Gas Law
• Ideal gas behaves as if there is no IMF present among the molecules of gas.
• PV = R
nT
R = 1atm (22.4L)
1 mole ( 273)
R= .0821 atm-L/mol-K
30
Ideal Gas Equation
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Boyle’s law: P a (at constant n and T) 1 V
V a nT
P
V = constant x = R nT
P
nT
P R is the gas constant
PV = nRT
31
What is the volume (in liters) occupied by 49.8 g of HCl at STP?
PV = nRT
V = nRT
P
T = 0 0C = 273.15 K
P = 1 atm
n = 49.8 g x 1 mol HCl
36.45 g HCl = 1.37 mol
V = 1 atm
1.37 mol x 0.0821 x 273.15 K L•atm
mol•K
V = 30.7 L
Sample problems for Ideal gas law
• If the pressure exerted by the gas at 250C in a volume of 0.044 L is 3.81 atm, how many moles of gas are present?
• Determine the celsius temperature of 2.49 moles of gas contained in a 1.00-L vessek at a pressure of 143 kPa.
33
Density (d) Calculations
d = m V
= PM RT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT
P M = d is the density of the gas in g/L
34
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and 27.0 0C. What is the molar mass of the gas?
dRT
P M = d = m
V
4.65 g
2.10 L = = 2.21
g
L
M = 2.21
g
L
1 atm
x 0.0821 x 300.15 K L•atm
mol•K
M = 54.5 g/mol
Density of gas from ideal gas equation
• D=M/V
• V=M/D
• PV=nRT
• P(Mass/D)= nRT
• Since n= mass/MW
• Then,
• P(Mass/D)=(Mass/MW)RT
• Therefore: P(MW)=DRT
• D=P(MW)
RT
• What is the density of a gas at STP that has a molar mass of 44.0 g/mol?
D=1atm(44.0g/mol)
.0821 (273K)
D= 1.96 g/L
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Stoichiometry and Gas Volumes Use the ideal gas law to find moles from a given
volume, pressure, and temperature, and vice
versa.
37
Gas Stoichiometry
What is the volume of CO2 produced at 37 0C and 1.00 atm
when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
g C6H12O6 mol C6H12O6 mol CO2 V CO2
5.60 g C6H12O6 1 mol C6H12O6
180 g C6H12O6
x 6 mol CO2
1 mol C6H12O6
x = 0.187 mol CO2
V = nRT
P
0.187 mol x 0.0821 x 310.15 K L•atm
mol•K
1.00 atm = = 4.76 L
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Gas Mixtures Dalton found that in a mixture of unreactive gases,
each gas acts as if it were the only gas in the
mixture as far as pressure is concerned.
39
Dalton’s Law of Partial Pressures
V and T are constant
P1 P2 Ptotal = P1 + P2
40
Consider a case in which two gases, A and B, are in a
container of volume V.
PA = nART
V
PB = nBRT
V
nA is the number of moles of A
nB is the number of moles of B
PT = PA + PB XA = nA
nA + nB XB =
nB
nA + nB
PA = XA PT PB = XB PT
Pi = Xi PT mole fraction (Xi ) = ni
nT
41
A sample of natural gas contains 8.24 moles of CH4, 0.421
moles of C2H6, and 0.116 moles of C3H8. If the total pressure
of the gases is 1.37 atm, what is the partial pressure of
propane (C3H8)?
Pi = Xi PT
Xpropane = 0.116
8.24 + 0.421 + 0.116
PT = 1.37 atm
= 0.0132
Ppropane = 0.0132 x 1.37 atm = 0.0181 atm
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A 100.0-mL sample of air exhaled from the
lungs is analyzed and found to contain
0.0830 g N2, 0.0194 g O2, 0.00640 g CO2,
and 0.00441 g water vapor at 35°C. What
is the partial pressure of each component
and the total pressure of the sample?
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mL10
L1mL100.0
K308Kmol
atmL0.08206
Ng28.01
Nmol1Ng0.0830
3
2
22
N2P
mL10
L1mL100.0
K308Kmol
atmL0.08206
Og32.00
Omol1Og0.0194
3
2
22
O2P
mL10
L1mL100.0
K308Kmol
atmL0.08206
COg44.01
COmol1COg0.00640
3
2
22
CO2P
mL10
L1mL100.0
K308Kmol
atmL0.08206
OHg18.01
OHmol1OHg0.00441
3
2
22
OH2P
atm0.749
atm0.153
atm0.0368
atm0.0619
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atm0.7492N P
atm0.1532O P
atm0.03682CO P
atm0.0619OH2P
OHCOON 2222PPPPP
P = 1.00 atm
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The partial pressure of air in the alveoli
(the air sacs in the lungs) is as follows:
nitrogen, 570.0 mmHg; oxygen, 103.0
mmHg; carbon dioxide, 40.0 mmHg; and
water vapor, 47.0 mmHg. What is the mole
fraction of each component of the alveolar
air?
mmHg40.02CO P
mmHg570.02N P
mmHg47.0OH2P
mmHg103.0 2O
P
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OHCOON 2222PPPPP
570.0 mmHg
103.0 mmHg
40.0 mmHg
47.0 mmHg
P = 760.0 mmHg
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Mole fraction of N2
Mole fraction of H2O Mole fraction of CO2
Mole fraction of O2
mmHg760.0
mmHg47.0
mmHg760.0
mmHg40.0
mmHg760.0
mmHg103.0
mmHg760.0
mmHg570.0
Mole fraction N2 = 0.7500
Mole fraction O2 = 0.1355
Mole fraction CO2 = 0.0526
Mole fraction O2 = 0.0618
48
2KClO3 (s) 2KCl (s) + 3O2 (g)
PT = PO + PH O 2 2
Collecting a Gas over Water
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Collecting Gas Over Water Gases are often collected over water. The result is
a mixture of the gas and water vapor.
The total pressure is equal to the sum of the gas
pressure and the vapor pressure of water.
The partial pressure of water depends only on
temperature and is known (Table 5.6).
The pressure of the gas can then be found using
Dalton’s law of partial pressures.
50
Vapor of Water and Temperature
51
?
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You prepare nitrogen gas by heating
ammonium nitrite:
NH4NO2(s) N2(g) + 2H2O(l)
If you collected the nitrogen over water
at 23°C and 727 mmHg, how many
liters of gas would you obtain from 5.68
g NH4NO2?
Molar mass NH4NO2
= 64.06 g/mol
P = 727 mmHg
Pvapor = 21.1 mmHg
Pgas = 706 mmHg
T = 23°C = 296 K
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P = 727 mmHg
Pvapor = 21.1 mmHg
Pgas = 706 mmHg
T = 23°C = 296 K P
nRTV
Molar mass NH4NO2
= 64.06 g/mol
4 2 24 2
4 2 4 2
1mol NH NO 1mol N5.68 g NH NO
64.04 g NH NO 1mol NH NO
= 0.8887 mol N2 gas
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P = 727 mmHg
Pvapor = 21.1 mmHg
Pgas = 706 mmHg
T = 23°C = 296 K
n = 0.0887 mol
P
nRTV
mmHg760
atm1mmHg706
K)(296Kmol
atmL0.08206mol0.0887
V
= 2.32 L of N2
(3 significant figures)
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2K
2
1mvE
Kinetic-Molecular Theory (Kinetic Theory)
A theory, developed by physicists, that is based on
the assumption that a gas consists of molecules in
constant random motion.
Kinetic energy is related to the mass and velocity:
m = mass
v = velocity
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Postulates of the Kinetic Theory
1. Gases are composed of molecules whose sizes are negligible.
2. Molecules move randomly in straight lines in all directions and at various speeds.
3. The forces of attraction or repulsion between two molecules (intermolecular forces) in a gas are very weak or negligible, except when the molecules collide.
4. When molecules collide with each other, the collisions are elastic.
5. The average kinetic energy of a molecule is proportional to the absolute temperature.
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An elastic collision is one in which no kinetic
energy is lost. The collision on the left causes the
ball on the right to swing the same height as the
ball on the left had initially, with essentially no loss
of kinetic energy.
58
Kinetic theory of gases and …
• Compressibility of Gases
• Boyle’s Law
P a collision rate with wall
Collision rate a number density
Number density a 1/V
P a 1/V
• Charles’ Law
P a collision rate with wall
Collision rate a average kinetic energy of gas molecules
Average kinetic energy a T
P a T
59
Kinetic theory of gases and …
• Avogadro’s Law
P a collision rate with wall
Collision rate a number density
Number density a n
P a n
• Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the presence of another gas
Ptotal = SPi
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Molecular Speeds According to kinetic theory, molecular speeds vary
over a wide range of values. The distribution
depends on temperature, so it increases as the
temperature increases.
Root-Mean Square (rms) Molecular Speed, u
A type of average molecular speed, equal to the
speed of a molecule that has the average
molecular kinetic energy
mM
RTu
3
61
The distribution of speeds
for nitrogen gas molecules
at three different temperatures
The distribution of speeds
of three different gases
at the same temperature
urms = 3RT M
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When using the equation
R = 8.3145 J/(mol · K).
T must be in Kelvins
Mm must be in kg/mol
?
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What is the rms speed of carbon
dioxide molecules in a container at
23°C?
mM
RTu
3rms
T = 23°C = 296 K
CO2 molar mass =
0.04401 kg/mol
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mol
kg0.04401
K296Kmol
s
mkg
8.314532
2
rmsu
25
rms 2
m1.68 10
su
2rms
m4.10 10
su
2
2
s
mkgJ
Recall
65
Gas diffusion is the gradual mixing of molecules of one gas
with molecules of another by virtue of their kinetic properties.
NH3
17 g/mol
HCl 36 g/mol
NH4Cl
r1
r2
M2 M1 =
molecular path
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Diffusion
The process whereby
a gas spreads out
through another gas
to occupy the space
uniformly.
Below NH3 diffuses
through air. The
indicator paper tracks
its progress.
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Effusion
The process by which a gas flows through a small
hole in a container. A pinprick in a balloon is one
example of effusion.
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Graham’s Law of Effusion At constant temperature and pressure, the rate of
effusion of gas molecules through a particular hole
is inversely proportional to the square root of the
molecular mass of the gas.
mM
1moleculesofeffusionofrate
?
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Both hydrogen and helium have been used as the buoyant gas in blimps. If a small leak were to occur, which gas would effuse more rapidly and by what factor?
Hydrogen will diffuse more quickly by a factor of 1.4.
2.016
4.002
4.002
1
2.016
1
HeRate
HRate 2
70
Deviations from Ideal Behavior
1 mole of ideal gas
PV = nRT
n = PV RT
= 1.0
Repulsive Forces
Attractive Forces
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Real Gases At high pressure the relationship between pressure
and volume does not follow Boyle’s law. This is
illustrated on the graph below.
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At high pressure, some of the assumptions of the
kinetic theory no longer hold true:
1. At high pressure, the volume of the gas
molecule (Postulate 1) is not negligible.
2. At high pressure, the intermolecular forces
(Postulate 3) are not negligible.
73
Effect of intermolecular forces on the pressure exerted by a gas.
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Van der Waals Equation An equation that is similar to the ideal gas law, but
which includes two constants, a and b, to account
for deviations from ideal behavior.
The term V becomes (V – nb).
The term P becomes (P + n2a/V2).
Values for a and b are found in Table 5.7
nRTnbVV
anP
2
2
75
Van der Waals equation
nonideal gas
P + (V – nb) = nRT an2 V2 ( )
}
corrected
pressure
}
corrected
volume
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Use the van der Waals equation to calculate
the pressure exerted by 2.00 mol CO2 that
has a volume of 10.0 L at 25°C. Compare
this with value with the pressure obtained
from the ideal gas law.
n = 2.00 mol
V = 10.0 L
T = 25°C = 298 K
For CO2:
a = 3.658 L2 atm/mol2
b = 0.04286 L/mol
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n = 2.00 mol
V = 10.0 L
T = 25°C = 298 K
L10.0
K)(298Kmol
atmL0.08206mol2.00
P
= 4.89 atm
(3 significant figures)
V
nRTP
Ideal gas law:
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n = 2.00 mol
V = 10.0 L
T = 25°C = 298 K
For CO2:
a = 3.658 L2 atm/mol2
b = 0.04286 L/mol
Pactual = 4.79 atm
(3 significant figures)
2
2
V
an
nbV
nRTP
2
2
22
L10.0
mol
atmL3.658mol2.00
mol
L0.04286mol2.00L10.0
K298Kmol
atmL0.08206mol2.00
P
atm0.146atm4.933 P