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Thermodynamics
Lecture 6
Ideal Gas Behavior
Non-ideal behavior
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Macroscopic 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
motion (3/2RT = 1/2M<u2>) M is molar
mass
Greater average velocity
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
molar energy
by the number of moles, n, to obtain:
3 nRT = 1 nM <u 2>
2
2
energy
Mass and molar mass
If m is the mass and M is the molar mass
of a particle then we can also write:
nM = Nm (N is the number of particles)
In other words nNA = N where NA is
Avagadro’s number.
3 nRT = 1 Nm <u 2>
2
2
Kinetic Model of Gases
Assumptions:
1. A gas consists of molecules that move randomly.
2. The size of the molecules is negligible.
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:
s12 + s22 + s32 + ... + sN2
<u > = c =
N
2
2
s1 + s2 + s3 + ... + sN
c=
N
The root-mean-square speed
The ideal gas equation of state is consistent with an
interpretation of temperature as proportional to the kinetic
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
= 481.8 m / s
The Maxwell Distribution
Not all molecules have the same speed. Maxwell assumed
that the distribution of speeds was Gaussian.
F(s) = 4 M
2RT
3/2
2
Ms
s exp –
RT
2
As temperature increases the r.m.s. speed increases and
the width of the distribution increases. Moreover, the
functions is a normalized distribution. This just means
that the integral over the distribution function is equal to 1.

F(s)ds = 1
0
The Maxwell Distribution
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
Energy has units of Joules
1 J = 1 Nm
Work and energy have the same units.
Work is defined as the result of a force
acting through a distance.
We can also define chemical energy in
terms of the energy per mole.
1 kJ/mol
1 kcal/mol = 4.184 kJ/mol
Thermal Energy
Thermal energy can be defined as RT.
Its magnitude depends on temperature.
R = 8.31 J/mol-K or 1.98 cal/mol-K
At 298 K, RT = 2476 J/mol (2.476 kJ/mol)
Thermal energy can also be expressed on a
per molecule basis. The molecular
equivalent of R is the Boltzmann constant, kB.
R = NAk B
NA = 6.022 x 1023 molecules/mol
Extensive and Intensive Variables
Extensive variables are proportional to the
size of the system.
Extensive variables: volume, mass, energy
Intensive variables do not depend on the
size of the system.
Intensive variables: pressure, temperature,
density
Equation of state relates P, V and T
The ideal gas equation of state is
PV = nRT
An equation of state relates macroscopic
properties which result from the average
behavior of a large number of particles.
P
Macroscopic
Microscopic
Microsopic 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 Dp = 2mux.
Transit time
c
ux
b
area = bc
a
The time between collisions with one wall is
Dt = 2a/ux.
This is also the round trip time.
Transit time
c
Round trip
distance is 2a
area = bc
ux
b
a
The time between collision is Dt = 2a/ux.
velocity = distance/time.
time = distance/velocity.
The pressure on the wall
force = rate of change of momentum
Dp 2mu x mu x2
F=
=
= a
Dt 2a/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
F
x
x
P=
=
=
V
bc abc
Average properties
Pressure does not result from a single
particle striking the wall but from many
particles. Thus, the velocity is the average
velocity times the number of particles.
Nm 
u 2x
P=
V
PV = Nm 
2
ux
Average properties
There are three dimensions so the velocity along
the x-direction is 1/3 the total.
1
2
=
u
 3 
u 2x
Nm 
u2
PV =
3
From the kinetic theory of gases
1 Nm
3
2
u = nRT
2
2
Putting the results together
When we combine of microscopic view of
pressure with the kinetic theory of gases
result we find the ideal gas law.
PV = nRT
This approach assumes that the molecules
have no size (take up no space) and that
they have no interactions.
Pressure of a dense fluid
For a dense fluid (or a liquid) such as water
we can think of the pressure arising from
the weight of the column of fluid above the
point where the measurement is made.
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
=
=
=
= rgh
A
Ah
V
where r is the density r = m/V.
The dependence of atomspheric
pressure on altitude
We can think of the atmosphere is a fluid,
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 density of an ideal gas is:
r = m/V = nM/V = MP/RT
The dependence of pressure on elevation
is:
MPg
dP = – rg dh = –
dh
RT
We need to collect variables of integration on the
same side of the equation.
dP = – Mg dh
P
RT
The barometric pressure formula
Then we integrate (assuming P0=1 at h=0):
P
P0 = 1
dP = – Mg
P
RT
h
dh
0
Mgh
P
ln
=–
P0
RT
Mgh
P = P0exp –
RT
Mgh
or P = exp –
RT
atm
Imperfect Gases
(Non-ideal or Real Gases)
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The Compression 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
CompressionFactor =
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 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. 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
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. From the preceding
considerations we see the B < 0 for ammonia, ethene, methane
and B > 0 for hydrogen.
The Virial Equation of State
We write Z in complete form as:
PVm
= 1 + B + C2 + ...
RT
Vm Vm
An then solve for the pressure:
P = RT 1 + B + C2 + ...
Vm
Vm Vm
This expression is known as the virial equation of 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).
Secondly, they have attractive forces between molecules.
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<s
r>s
u(r)
A hard sphere potential is the easiest potential to parameterize.
The hard sphere diameter corresponds to the interatomic
spacing in a closest packed geometry such as that shown
for the noble gas argon.
The diameter can be estimated
Ar Ar Ar Ar
s 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
Ar Ar Ar Ar
of its simplicity.
s
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.
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.
The volume is nb
When the gas is
completely compressed.
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 simulations. LJ
12
6
s
s
V (R) = 4
–
R
R
The potential has a long-range attractive tail –1/r6, and
negative well depth , and a steeply rising repulsive wall
at R = s. Typically the parameter s is related to the
hard sphere diameter of the molecule. For a monoatomic
condensed phase s 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. Thus,  ~ kT or 1.38x10-23 J/K(120 K) = 1.65x10-21 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 < s.
This is the repulsive term in the potential that arises from
close contacts between molecules. The minimum is found
for Rmin = 21/6 s. The well depth is  in units of energy.

Rmin
The van der Waal’s Equation of State
The microscopic terms  and s in the L-J
potential can be related to the a and b
parameters in the van der Waal’s equation of
state below.
2
n
nRT
P=
– a
V – nb V 2
The significance of b is the same as for the
hard sphere potential. The parameter a is
related to the attractive force between
molecules. It tends to reduce the pressure
compared to an ideal gas.
The van der Waal’s Equation of State
in terms of molar volume
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. 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 = 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
of the argon phase diagram
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
the derivative of the vdW equation of
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 =
The critical point is a flat inflection point. This means
that both the first and second derivatives on the curve
vanish at that point. These derivatives are:
The solution of these equations (and the van der Waal;s
equation itself leads to the following values for the critical
constants:
The Law of Corresponding States
The critical values, Pc, Vc and Tc, can be used to calculate a
critical compression factor.
The critical compression factor is predicted to be the same
for all gases. This fact leads to the principle of
corresponding states. This principle states that all gases
will have identical compression curves if they are
normalized to their critical constants. We define the
reduced, pressure, volume and temperature as: