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U N I T III
KINETIC MOLECULAR
DESCRIPTION OF THE
STATES OF MATTER
CHAPTER 9
The Gaseous State
CHAPTER 10
Solids, Liquids, and Phase Transitions
CHAPTER 11
Solutions
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Gas
Liquid
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Solid
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9
THE GASEOUS STATE
CHAPTER
9.1 The Chemistry of Gases
9.2 Pressure and Temperature of Gases
9.3 The Ideal Gas Law
9.4 Mixtures of Gases
9.5 The Kinetic Theory of Gases
9.6 Real Gases: Intermolecular Forces
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Li + H2O → LiOH + H2
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9.1 THE CHEMISTRY OF GASES
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CaCO3(s) + 2HCl(g) →
CaCl2(s) + CO2(g) + H2O(l)
K2SO3(s) + 2HCl(g) →
2KCl(s) + SO2(g) + H2O(l)
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9.2 PRESSURE AND TEMPERATURE OF
GASES
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 Pressure
Evangelista Torricelli
Torricelli’s barometer
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(ITA, 1608-1647)
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Force exerted by the mercury column at its base
F = mg
Pressure : P 
F mg mg m


 gh   gh
A
A V /h V
 = 13.5951 g cm–3 → density of Hg(l) at 0 oC
g = 9.80665 m s–2 → gravitational acceleration
h = 76 cm → height of mercury column
1 atmosphere pressure (1 atm) is
1.01325 x 105 kg/ms2 or Pa; the SI unit
= 101.325 kPa
1.01325 bar (1 bar = 105 Pa
760 mm Hg (at 273 K)
760 torr (at any temperature)
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 Pressure and Boyle’s Law
~ Experiments on the compression and expansion of air.
“The spring of the Air and Its Effects” (1661).
 Boyle’s J-tube experiment
Trapped air at the closed end of the J-tube:
h (in mm)
P  1 atm 
760 mm atm 1
Add Hg and measure the volume of air (V):
P
or
C
1
C 
V
V 
Robert Boyle
PV  C
(UK, 1627-1691)
C: a constant at constant T and fixed amount of gas
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Fig. 9.3 Boyle’s J-tube.
(a)Same Hg height on two sides
Pconfined = Pair
(b) Hg added
 Difference in Hg heights, h
 Vconfined compressed
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(a) P vs. V  hyperbola
(b) P vs. 1/V  straight line passing through the origin (slope: C)
(c) PV vs. P  straight line independent of P
(parallel to P-axis, intercept C on PV-axis)
The value of C at 0oC and for 1 mol of gas,
~ good for all gases at very low pressure
C = PV = 22.414 L atm
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 Temperature and Charles’s law
V  constant  T (at constant n and P)
V

t  273.15o C   1
 V0 
t
T


V  V0  1 

V

0

o
273.15
 273.15 C 
T (Kelvin)  273.15  t (Celsius)
Jacques Charles
(France,1746-1823)
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Fig. 9.5 Volume of a gas confined at constant P increases as T increases.
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Fig. 9.6 The volume of a sample of a gas is a function
of temperature at constant pressure.
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◈ Absolute Temperature Scale, K
Kelvin, Lord William Thomson
(UK, 1824-1907)
0 K : ‘absolute zero’ temperature
273.15 K : triple point of water
0 K = - 273.15 oC  0 oC = 273.15 K
Kelvin scale: absolute, thermodynamic temperature
scale - ‘absolute zero’ is the temperature at which all
thermal motion ceases in the classical description of
thermodynamics.
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9.3 THE IDEAL GAS LAW
Boyle’s law:
V  1/P (at constant T and n)
Charles’ law:
V  T (at constant P and n)
Avogadro’s hypothesis: V  n (at constant T and P)
PV = nRT
√ Equation of state
√ Limiting law for real gases
as P  0
√ Universal gas constant, R
R = 8.314 J·K–1·mol–1
= 8.206 x 10–2 L·atm·K–1
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V
nT
P
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EXAMPLE 9.5
Concentrated nitric acid acts on copper
to give nitrogen dioxide and dissolved copper ions according
to the balanced chemical equation
Cu(s) + 4H+(aq) + 2NO3-(aq) → 2NO2(g) + Cu2+(aq) + 2H2O(l)
Suppose that 6.80 g copper is consumed in this reaction, and
that the NO2 is collected at a pressure of 0.970 atm and a
temperature of 45oC. Calculate the volume of NO2 produced.
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9.4 MIXTURES OF GASES
▶ Partial pressure (Pi) of the ith gas in a mixture of gases →
pressure that the ith gas would exert if it occupied the container
alone
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◆ Dalton’s Law of Partial Pressures
The total pressure of a mixture of gases is the
sum of the partial pressures of its component.
P  PA  PB     Pi
i
▶ Mole fraction of the component A is xA
nA
xA 
,
xA  xB    1
nA  nB  
PA 
nRT
RT
nA P
nA RT
, P
  nA  nB  

 PA 
 xA P
V
V
V
nA  nB  
 PA = xAP
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EXAMPLE 9.6
When NO2 is cooled to room temperature, some of it reacts
to form a dimer, N2O4, through the reaction
2NO2(g) → N2O4(g)
Suppose 15.2 g of NO2 is placed in a 10.0 L flask at high temperature and the
Flask is cooled to 25oC. The total pressure is measured to be 0.500 atm.
What partial pressures and mole fractions of NO2 and N2O4 are present?
(a)
(Dalton’s Law)
(b)
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9.5 THE KINETIC THEORY OF GASES
1. A gas consists of a collection of molecules in
continuous random motion.
2. Gas molecules are infinitesimally small (mass) points.
3. The molecules move in straight lines until they collide.
4. The molecules do not influence one another except
during collisions.
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- Collision with walls: consider molecules
traveling only in one dimension, x, with a
velocity of vx.
The change in momentum (final – initial)
of one molecule: -2mvx = 2mvx momentum
change for the wall
All the molecules within a distance vxDt of the wall
and traveling toward it will strike the wall during the
interval Dt.
If the wall has area A, all the particles in a volume
AvxDt will reach the wall if they are traveling toward
it.
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The number of molecules in the volume AvxDt is
that fraction of the total volume V, multiplied by
the total number of molecules:
The average number of collisions with the wall
during the interval Dt is half the number in the
volume AvxDt:
The total momentum change = number of collisions
× individual wall momentum change
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Force = rate of change of momentum =
(total momentum change)/Dt
 mean-square speed
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- Kinetic energy of NA molecules,
- average kinetic energy per molecule,
kB = R/NA
- root-mean-square speed
M = molar mass = NAm
Root mean square speeds of some
gases at 25 oC
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Maxwell-Boltzmann distribution of speed
DD
NN
N
N
3/2
= f (v)Dv with
Fraction of
molecules with
speeds between
v and v + Dv
 M  2  M v2 /2 RT
f (v)  4 
 ve
 2 RT 
3/2
or
 m 
2  mv2 /2 k BT
f (v)  4 
 ve
 2 k BT 
Boltzmann constant:
kB  R / NA  1.38066 1023 J  K1
James Clerk Maxwell
(Scotland, 1831-1879)
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Ludwig Eduard Boltzmann
(Austria, 1844-1906)
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Fig. 9.13 A device for measuring the distribution of molecular speeds.
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Fig. 9.14 Maxwell-Boltzmann distribution of molecular
speeds in N2 at three different temperatures.
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Different speeds associated with a collection
of gas molecules or atoms
1. Most probable speed (vmp)
df(v)
2kBT or 2RT
= 0
vmp =
dv
M
m
v = vmp
_
2. Average speed (v)
oo
_
8kBT or 8RT
v = vf(v)dv =
M
m
0
_
3. Mean square speed (v2)
_
oo
3kBT or 3RT
2
v = v2f(v)dv =
m
M
0
4. Root mean square speed (vrms)
vrms =
_
v2
=
3kBT or
m
3RT
M
_
vmp < v < vrms= 1.000:1.128:1.225
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vmp
v
vrms
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9.6 REAL GASES: INTERMOLECULAR
FORCES
▶ Compression (or Compressibility)
factor, Z
z is a measure of deviation from ideality
z =
Vm
ideal
Vm
=
Vm
RT/P
=
PVm
RT
=
PV
nRT
Vm = molar volume = V/n
- For an ideal gas, z = 1
- For real gases, z deviates from
1 as P increases:
z < 1 when attractive forces dominate
z > 1 for repulsive forces dominate
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z = PV/nRT
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◈ The Van der Waals Equation of State
▶ Corrections to the ideal equation of state
- Attraction at long distance:
Reduction in collision frequency  n / V
Reduction in intensity of collision  n / V
Pideal
n2
 Pa 2
V
- Repulsion at short distance:
No overlap of molecules → excluded volume effect
Reduction in free volume  n
Videal  V  bn
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▶ Van der Waals equation:

n 
 P  a 2  V  nb   nRT
V 

2
a: atm L2 mol-2
b: L mol-1
R: L atm mol-1 K-1
Rearranging this equation to solve for P gives the form of the equation
that is most often used in calculations:
nRT _ an2
P =
_
V2
V nb
See Slide 37
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Influence of van der Waals parameters a and b on
the compressibility factor z
Repulsive forces (through b) increase z above 1.
Attractive forces (through a) reduce z.
The Boyle temperature TB
1
nb
+...
Substituting the approximation
~ 1 +
_
V
1
nb/V
(at low n/V) into the equation above gives,
z ~1 +
a n
b_
+....
RT V
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The temperature at which the coefficient in brackets is 0 (so z = 1)
is the Boyle temperature (TB)
a
TB =
Rb
For T > TB repulsive forces dominate and z > 1
For T < TB attractive forces dominate and z < 1
-Note that the Constant b is the volume excluded by 1 mol of molecules
and should be close to Vm, the volume per mole in the liquid state.
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Example of the use of the van der Waals
equation for the calculation of pressure
A 10.0-L tank containing 25 mol of O2 is stored in a diving supply
shop at 25 oC. Use the data in table 4.5 and the van der Waals
equation to estimate the pressure in the tank.
Solution
Rewriting the van der Waals equation so as to solve for P,
2
nRT _
n2
_
n
(V nb) = nRT becomes P =
P + a
a 2
_
2
V
V
(V nb)
P=
(25 mol) x (0.08206 L atm K-1 mol-1) x (298 K)
_
10.0 L
(25 mol) x (3.19 x 10-2 L mol -1)
_ (1.364 L2 atm mol-2) x
(25 mol)2
(10.0 L)2
= 58 atm
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 Intermolecular Forces
 Lennard-Jones Potential:
  12   6 
VLJ ( R)  4      
 R  
 R 
(He)
(He)
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(Ar)
(Ar)
where  is the depth and  is the
distance at which V(R) passes
through zero.
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10 Problem Sets
For Chapter 9,
6, 18, 30, 40, 42, 50, 64, 72, 76, 92
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