Chemistry 331
Lecture 6
Ideal Gas Behavior
NC State University
Macroscopic
p variables P, T
Pressure is a force per unit area (P= F/A)
The force arises from the
change in momentum as
particles hit an object and
change direction.
Temperature derives from molecular
i molar
l
motion (3/2RT = 1/2M<u2>) M is
mass
Greater average
g velocity
y
results in a higher
temperature.
u is the velocity
Mass and molar mass
We can multiply the equation:
3 RT = 1 M <u 2>
2
2
by the number of moles, n, to obtain:
3 nRT = 1 nM <u 2>
2
2
If m is the mass and M is the molar of a
particle then we can also write:
p
nM = Nm (N is the number of particles)
Mass and molar mass
In other words nNA = N where NA is
Avagadro’s
Avagadro
s number
number.
3 nRT = 1 Nm <u 2>
2
2
Kinetic Model of Gases
Assumptions:
1. A gas consists of molecules that move randomly.
2 Th
2.
The size
i off th
the molecules
l
l iis negligible.
li ibl
3. There are no interactions between the gas molecules.
Because there are such large numbers of gas molecules
in any system we will interested in average quantities.
We have written average with an angle bracket.
For example, the average speed is:
s 12 + s 22 + s 32 + ... + s N2
<u > = c =
N
2
2
c=
s 1 + s 2 + s 3 + ... + s N
N
We use s for speed
and
d c ffor mean speed.
d
Velocity and Speed
When we considered the derivation of pressure using a
kinetic model we used the fact that the gas exchanges
momentum with the wall of the container. Therefore, the
vector (directional) quantity velocity was appropriate.
However, in the energy expression the velocity enters as
the square and so the sign of the velocity does not matter.
In essence it is the average speed that is relevant for the
energy Another way to say this is the energy is a scalar
energy.
scalar.
E = 1 m<u 2> = 1 mv 2 = 1 mc 2
2
2
2
p = mu = mv
All of these notations
mean the same thing.
The root-mean-square
q
speed
p
The ideal gas equation of state is consistent with an
p
of temperature
p
as p
proportional
p
to the kinetic
interpretation
energy of a gas.
1 M u 2 = RT
3
If we solve for <u2> we have the mean-square speed.
 u 2  = 3RT
M
If we take the square root of both sides we have the r.m.s.
speed.
u
2 1/2
=
3RT
M
Example
The r.m.s. speed of oxygen at 25 oC (298 K) is 482 m/s.
Note: M is converted to kg/mol!
u 
2 1/2
3 8.31 J/mol–K 298 K
=
0.032 kg/mol
g
= 481.8 m / s
The Maxwell Distribution
Not all molecules have the same speed. Maxwell assumed
that the distribution of speeds was Gaussian
Gaussian.
F(s) = 4 M
2RT
3/2
2
Ms
s exp –
RT
2
As temperature
p
increases the r.m.s. speed
p
increases and
the width of the distribution increases. Moreover, the
fucntions is a normalized distribution. This just means
that the integral over the distribution function is equal to 1
1.

F(s)ds = 1
0
See the MAPLE
worksheets for examples.
Units of Pressure
Force has units of Newtons
F = ma (kg m/s2)
Pressure has units of Newtons/meter2
P= F/A = (kg m/s2/m2 = kg/s2/m)
These units are also called Pascals (Pa).
1 bar = 105 Pa = 105 N/m2.
1 atm = 1.01325 x 105 Pa
Units of Energy
gy
Energy has units of Joules
1 J = 1 Nm
N
Work and energy have the same units.
Work is defined as the result of a force
g through
g a distance.
acting
We can also define chemical energy in
terms of the energy per mole
mole.
1 kJ/mol
1 kcal/mol
k l/ l = 4.184
4 184 kJ/mol
kJ/ l
Thermal Energy
gy
Thermal energy can be defined as RT.
I magnitude
Its
i d d
depends
d on temperature.
R = 8.31 J/mol-K or 1.98 cal/mol-K
At 298 K
K, RT = 2476 J/mol (2.476
(2 476 kJ/mol)
Thi iis th
This
the molar
l energy available
il bl tto a
system at 298 K.
Thermal Energy
gy
Thermal energy can also be expressed on a
per molecule basis
basis. The molecular
equivalent of R is the Boltzmann constant, k.
R = NAk
NA = 6.022 x 1023 molecules/mol
At 298 K, RT = 2476 J/mol (2.476 kJ/mol)
RT = 4.11 x 10-21 J
Extensive and Intensive Variables
Extensive variables are proportional to the
size of the system.
Intensive variables do not depend on the
size of the system.
system
Extensive
E
t
i variables:
i bl
volume,
l
mass, energy
Intensive variables: pressure, temperature,
density
Equation of state relates P, V and T
The ideal gas equation of state is
PV = nRT
RT
An equation of state relates macroscopic
properties which result from the average
behavior of a large number of particles.
P
Macroscopic
Microscopic
Microsopic
p view of momentum
c
ux
b
area = bc
a
A particle with velocity ux strikes a wall.
The momentum of the particle changes from mux
to –mux. The momentum change is p = 2mux.
Transit time
c
ux
b
area = bc
a
The time between collision is t = 2a/ux.
Transit time
c
Round trip
distance is 2a
area = bc
ux
b
a
The time between collision is t = 2a/ux.
velocityy = distance/time.
time = distance/velocity.
The p
pressure on the wall
force = rate of change of momentum
 p 2mu x mu x2
F=
=
= a
t 2a/u
a/u x
The pressure is the force per unit area.
The area is A = bc and the
volume of the box is V = abc
2
2
mu
mu
x
x
F
P=
=
=
V
bc abc
Average
g p
properties
p
Pressure does not result from a single
particle
ti l striking
t iki th
the wallll b
butt ffrom many
particles. Thus, the velocity is the average
velocity
l it titimes th
the number
b off particles.
ti l
Nm 
u 2x
P=
V
PV = Nm 
2
ux
Average
g p
properties
p
There are three dimensions so the velocity along
the xx-direction
direction is 1/3 the total
total.
1
2
=
u
 3 
u 2x
Nm 
u2
PV =
3
From the kinetic theory of gases
1 Nm
u 2 = 3 nRT
2
2
Putting
g the results together
g
When we combine of microscopic view of
pressure with
ith th
the ki
kinetic
ti th
theory off gases
result we find the ideal gas law.
PV = nRT
This approach applies to a monatomic gas
like neon or argon. What about internal
motions of molecules?
RT is a natural energy
gy scale
We can rewrite the ideal gas law in terms
off the
th molar
l volume
l
V = V/n
The ideal gas law has the form
PV = RT
The molar volume at standard T and P
V = RT =
P
8.31 J/mol–K 298 K
1 013  10 5 N/m 2
1.013
= 0.0244 m 3 = 24.4 L
Microscopic
p variables
Monatomic gases: translation
Pressure and temperature can be described
solely in terms of the ballistic motion of the gas.
Diatomic gases: translation,
translation vibration,
vibration rotation
Center of mass
Quantized energy
gy levels
The constant h, known as Planck’s constant
gives the scale for quantized energy levels
levels.
h = 6.626 x 10-34 J
Translation – particle in a box
Vibration – harmonic oscillator
Rotation – rigid rotator
The energy levels for each of these is
obtained by solution of the Schrödinger
equation.
The energy
gy level spacing
p
g
The constant h, known as Planck’s constant
gives the scale for quantized energy levels
levels.
h = 6.626 x 10-34 J
Motion
# Formula
Vibration v (v + 1/2)h
Rotation J [h2/82I]J(J+1)
Translation n [h2/8ma2]n2
kJ/mol
1 – 20
10-3 – 1
10-11
We will see how to obtain these in the second half of the course.
Levels are thermally
y ppopulated
p
E
RT
Vib ti
Vibration
R t ti
Rotation
T
Translation
l ti
Key points regarding the
microscopic view
Translational energy levels are so densely
spaced that these can be treated using
classical methods.
We can treat particles as ideal even though
they have vibrations and rotations. The
dynamics of the gas are not affected.
We will see that the heat capacity of the
gas is affected by the “internal”
internal degrees of
freedom.
Key points regarding the
microscopic view
The kinetic energy of a large number of
individual particles is proportional to the
temperature of the system. As the system
heats up we can picture the molecules
moving more rapidly.
Pressure results from the net momentum
transfer between the particles and wall of
the container.
Pressure of a dense fluid
For a dense fluid (or a liquid) such as water
we can think
thi k off the
th pressure arising
i i ffrom
the weight of the column of fluid above the
point
i t where
h
th
the measurementt is
i made.
d
The force is due to the mass of water m
(kg) accelerated by gravity (g = 9.8 m/s2).
P=
F
A
mg
mgh
mgh
=
=
=
= gh
A
Ah
V
where  is the density  = m/V.
The dependence of atomspheric
pressure on altitude
We can think
W
thi k off the
th atmosphere
t
h
is
i a fluid,
fl id
but it is not dense. Moreover, unlike water
the density of the atmosphere decreases
with altitude. Thus, at high elevations both
the pressure and the density are
decreased. To obtain the dependence of
pressure on height h above the earth’s
surface we use the ideal gas law to define
the density of an ideal gas.
The dependence of atomspheric
pressure on altitude
The d
Th
density
it off an id
ideall gas iis:
 = m/V = nM/V = MP/RT
The dependence of pressure on elevation
is:
MPg
dP = – g dh = –
dh
RT
We need to collect variables of integration
g
on the
same side of the equation.
dP = – Mg dh
P
RT
The barometric p
pressure formula
Then we integrate (assuming P0=1 at h=0):
P
dP = – Mg
RT
P0 = 1 P
Mgh
P
ln
=–
P0
RT
h
dh
0
Mgh
Mgh
P = P0exp –
or P = exp –
atm
RT
RT
Isotherms
We can plot the pressure as a function of the
volume as shown below. Each of the curves
on the plot has a constant temperature.
Partial pressure
For any gas in a mixture of gases the partial
pressure is defined as:
p
Pj = xjP
where xj is the mole fraction of component j
and P is the total pressure.
Th mole
The
l ffraction
ti iis d
defined
fi d as:
nj
xj =
ni
i
The molar volume
We can rewrite the ideal gas law in terms
off the
th molar
l volume
l
V
Vm = n
The ideal gas law has the form
PVm = RT
Th molar
The
l volume
l
att standard
t d d T and
dP
Vm = RT =
P
8.31 J/mol–K 298 K
1.013  10 N/m 2
5
= 0.0244
0 0244 m 3 = 24.4
24 4 L
The Compression
p
Factor
One way to represent the relationship between ideal and real
gases is to plot the deviation from ideality as the gas is
compressed (i.e. as the pressure is increased).
The compression factor is defined as:
Molar volume of gas
Compression Factor =
Molar volume of perfect gas
Written in symbols this becomes:
Z=
Vm
perfect
Vm
=
PVm
RT
Note that perfect gases are also called ideal gases.
Imperfect gases are sometimes called real gases.
The Compression
p
Factor
A plot of the compression factor reveals that many gases
exhibit Z < 1 for low pressure. This indicates that attractive
forces dominate under these conditions.
As the pressure increases Z crosses 1 and eventually becomes
positive for all gases
gases. This indicates that the finite molecular
volume leads to repulsions between closely packed gas
molecules. These repulsions are not including the ideal gas
model.
Attractive
Region
Repulsive
Region
The Virial Expansion
p
One way to represent the deviation of a gas from ideal (or
perfect) behavior is to expand the compression factor in
powers of the inverse molar volume. Such an expansion is
known as a virial expansion.
Z = 1 + B + C2 + ...
Vm Vm
The coefficients B, C etc. are known as virial coefficients.
For example, B is the second virial coefficient.
Virial coefficients depend on temperature
temperature. From the preceding
considerations we see the B < 0 for ammonia, ethene, methane
and B > 0 for hydrogen.
The Virial Equation
q
of State
We write Z in complete form as:
PVm
= 1 + B + C2 + ...
Vm Vm
RT
An then solve for the pressure:
P = RT 1 + B + C2 + ...
Vm
Vm Vm
This expression is known as the virial equation of state
state.
Note that if B, C etc. are all equal to zero this is just the ideal
gas law. However, if these are not zero then this equation
contains corrections to ideal behavior.
Relating the microscopic to
the macroscopic
Real gases differ from ideal gases in two ways.
First they have a real size (extent). The excluded volume
results in a repulsion between particles and larger pressure
than the corresponding ideal gas (positive contribution to
compressibility).
S
Secondly,
dl th
they h
have attractive
tt ti fforces between
b t
molecules.
l
l
These are dispersive forces that arise from a potential energy
due to induced-dipole induced-dipole interactions.
We can relate the potential energy of a particle to the terms
in the virial expansion or other equation of state. While we
will not do this using math in this course we will consider the
graphical form of the potential energy functions.
Hard Sphere Potential
u(r) = 
u(r) =0
r<
r>
u(r)
A hard sphere potential is the easiest potential to parameterize.
The hard sphere diameter corresponds to the interatomic
spacing
i iin a closest
l
t packed
k d geometry
t such
h as th
thatt shown
h
for the noble gas argon.
The diameter can be estimated
Ar Ar Ar Ar
 Ar Ar Ar Ar
from the density of argon in
the solid state. The hard sphere
Ar Ar Ar Ar
potential is widely used because
A Ar
Ar
A Ar
A Ar
A
of its simplicity.

r
The Hard Sphere Equation of State
As a first correction to the ideal gas law
we can consider the fact that a gas has
finite extent. Thus, as we begin to decrease
the volume available to the gas the pressure
increases more than we would expect due
to the repulsions between the spheres of
finite molar volume, b, of the spheres.
nRT
P=
V – nb
Gas molecule
of volume B
The Hard Sphere Model
Low density: These are ideal gas conditions
The Hard Sphere Model
Increasing density: the volume is V
b is the molar volume of the spheres
spheres.
The Hard Sphere Model
Increasing density
The Hard Sphere Model
Increasing density
The Hard Sphere Model
High density: At sufficiently high density
the gas becomes a high density fluid or
a liquid.
The Hard Sphere Model
Limiting density: at this density the hard
spheres have condensed into an ordered
lattice. They are a solid. The “gas” cannot
be compressed further
further.
If we think about the density in each of these cases we can
see that it increases to a maximum value.
Lennard-Jones Potential Function
The Lennard-Jones potential is a most commonly used
potential function for non-bonding interactions in atomistic
computer
t simulations.
i l ti
12
6
LJ


V (R) = 4
–
R
R
The potential has a long
long-range
range attractive tail –1/r
1/r6, and
negative well depth , and a steeply rising repulsive wall
at R = . Typically the parameter  is related to the
hard sphere diameter of the molecule. For a monoatomic
condensed phase  is determined either from the solid
state or from an estimate of the packing in dense liquids.
The well depth e is related to the heat of vaporization of
a monatomic fluid. For example, liquid argon boils at ~120K
at 1 atm.
atm Thus
Thus,  ~ kT or 1
1.38x10
38x10-23 J/K(120 K) = 1
1.65x10
65x10-21 J.
J
This also corresponds to 1.03 kJ/mol.
Graphical Representation L-J Potential
The L-J potential function has a steep rise when r < .
This is the repulsive term in the potential that arises from
close
l
contacts
t t between
b t
molecules.
l
l
Th
The minimum
i i
iis ffound
d
for Rmin = 21/6 . The well depth is  in units of energy.

Rmin
The van der Waal’s Equation of State
The microscopic terms  and  in the L-J
potential can be related to the a and b
parameters in the van der Waal’s equation of
state below
below.
2
n
nRT
P=
– a
2
V – nb
V
The significance of b is the same as for the
hard sphere potential
potential. The parameter a is
related to the attractive force between
molecules It tends to reduce the pressure
molecules.
compared to an ideal gas.
The van der Waal’s Equation of State
i tterms off molar
in
l volume
l
Recall that Vm = V/n so that the vdW equation
of state becomes:
P = RT – a2
Vm – b Vm
We can plot this function for a variety of
different temperatures. As we saw for the
ideal gas these are isotherms.
isotherms At sufficiently
high temperature the isotherms of the vdW
equation of state resemble those of the ideal
gas.
The argon phase diagram
For argon
Tc = 150.8 K
Pc = 48.7 bar = 4934.5 Pa
Vc = 74.9 cm3/mol
Critical Point
Significance of the critical point
Note that the vdW isotherms look very different
from those of the ideal gas below the critical
point. Below the critical point there are two
possible phases, liquid and gas. The liquid
phase is found at small molar volumes.
The gas phase is observed at larger molar
volumes. The shape of the isotherms is not
physically reasonable in the transition region
between the phases. Note that the implication
is that there is a sudden change in volume
for the phase transition from liquid to gas.
View of the liquid region
off the
th argon phase
h
diagram
di
Liquid
Phase Equilibrium Region
Critical Parameters
The critical parameters can be derived in terms
of the vdW a and b parameters as well as the
gas constant R.
The derivation can use calculus since
q
of
the derivative of the vdW equation
state is zero at the critical point.
Given that this is also an inflection point
the second derivative is also zero.
a
2
27b
8a
Tc =
27Rb
Vc = 3b
Pc =