1
Gases
The Kinetic Theory of Gases
Basics
1. Sample of gas huge number of atoms (molecules).
2. Atoms are very small compared the space in which they travel.
3. Atoms have translational energy only (ignore rotational, vibrational or other
internal structure).
Molecular View of Pressure
Atoms have translational energy coming motion in three dimensions.
 tr 
1
1
m  v 2x  v 2y  v z2   mv 2
2
2
Assume gas is in a box with lengths, lx, ly, lz.
Assume gas is in thermodynamic equilibrium
- No net energy transfer between walls of box and gas atoms.
Consider the z-component of an atom’s velocity as it hits the wall of the box with
lengths lx and ly.
1. Collision reverses vz.
2. vx and vy remain unchanged.
3. Overall speed and energy are unchanged. (elastic collisions)
Force is the time derivative of momentum.
Fz 
dp z d  mv z  dm
dv
dv


v z  m z  m z  ma z
dt
dt
dt
dt
dt
Find the change in the momentum of the collision with the wall
pz  pf  pi  m  vz   mvz  2mvz
For a single atom i, the time average of the force is:
p z,i   Fz,i dt  Fz,i
 t 2  t1 
Assuming a constant speed, the time change can be related to the distance travelled
by the atom.
t 2  t1 
2lz
vz,i
2
Consider the force on the wall using Newton’s third law.
W
z,i
F
  Fz,i
 2m vz,i

 
 t 2  t1
t 2  t1

 2m vz,i
m vz,i


 2lz vz,i
lz

pz,i
2
Find the total force on the wall by summing over all of the atoms.
W
z
F
 F
W
z,i
N F

W
z,i
2
Nm vz,i
lz
i
Using the average force, calculate the pressure on the wall perpendicular to the
z-direction.
FzW
Nm v z,i
F
p 

A
Az
lx ly
2
lz

Nm v z,i
lx l ylz
2

Nm v z,i
2
V
Consider that choosing z-direction to calculate the pressure is arbitrary.
1. Pressure is the same on any wall.
2. Average molecular velocity is the same in all directions.
v x,i  v y,i  v z,i
vi
2
 v x,i
2
 v y,i
v z,i
2
 p
2

 vz,i
vi
2
 3 vz,i
2
2
3
Nm vi
2
3V
Pressure can be written in terms of the molecular translational energy.
1. For a single molecule,
 tr 
1
m vi
2
2
2. For “N” molecules,
E tr  N tr 
1
Nm vi
2
2
 p
2E tr
3V
 pV 
2
E tr
3
3
Each type of molecule has its own translational energy and therefore its own pressure.
pa 
2E tr,a
pb 
3V
2E tr,b
3V
pc 
2E tr,c
3V
E tr,total  E tr,a  E tr,b  E tr,c
pa  p b  pc 
2E tr,a
3V

2E tr,b
3V

2E tr,c
3V

2E tr
p
3V
This is Dalton’s Law of Partial Pressure.
Molecular View of Temperature
Temperature is a measure of how energy is distributed in a material sample.
1. Higher temperature means more modes of motion are available for a molecule
to move.
- Types of motion include translational, rotational, vibrational and electronic.
2. Lowest temperature implies one mode of motion. (3rd Law of Thermodynamics)
Thus temperature is a function of energy (microscopic or macroscopic).
1. For a gas without internal structure,
T  f  tr  or T  g  E tr 
2. Note the temperature function is different for microscopic and macroscopic
energies.
We can define temperature macroscopically in terms of the ideal gas law.
2
3
E tr  nRT  E tr  nRT
3
2
E
3
 E  E tr  RT
n
2
pV  nRT 
EN 
E tr
3 nRT 3 RT
  tr 

N
2 N
2 NA
NA  Avogadro 's number

4
The corresponding microscopic definition follows from a definition of Boltzmann’s
constant, k.
k
R
8.314 J mol  K

 1.3811023 J K
N A 6.022 1023 mol
1. R is used for macroscopic quantities (moles)
2. k is used for microscopic quantities (a molecule)
 tr 
3 RT 3
 kT
2 NA 2
Velocity Distributions
RMS (root-mean-square) Velocity
1
 N v2  2
v   i 
 i N
1
3
1
 tr  kT  m v
2
2
2

1
 3kT  2  3RT  2
v 
 

 m   M 
Maxwell (Maxwell-Boltzmann) Distribution Function
Introductory assumptions
A gas sample has a broad distribution of speeds.
Let the number of molecules with speed between v and v + dv = dNv
The distribution function relates the fraction of total molecules within a range of
velocities between v and v + dv. (This function is what we want to find!).
G  v  dv 
dN v
N
v2
- Probability of finding molecules with velocity between v1 and v2   G  v  dv
v1

- Probability of finding molecules with any speed is
- G  v  is a probability density.
 G  v  dv  1

5
To find the distribution function for any speed, G  v  , first find the distribution
function for velocities in z-direction only, g  vz  .
dN vz
N
 g  v z  dv z
One-dimensional velocity distribution function
Because our sample is in isotropic space (no preferred direction exists), the
distribution functions for the velocities in the other directions are similar.
Let us examine the product of probability densities in all three directions.
g  vx  g  v y  g  vz     v 
To find g, take derivative w.r.t. vx.
g  v x  g  v y  g  v z  
v
2
 vx
2
 vy
2
 vz
2
  v  v
v v x
 2v dv  2v x dv x  2v y dv y  2v z dv z
v v x

v x
v

  v  v x
v v
g  v x  g  v y  g  v z 
  v 
g  v x 
  v  v x   v 
vx




vx g  vx  vx   v  v v   v 
vx g  vx  g  v y  g  vz  vx g  vx  g  v y  g  vz  v
g  v x  g  v y  g  v z  
Similar relationships can be found for the “y” and “z” directions.
g  v y 
v yg  v y 

g  v z 
  v 

vzg  vz  v   v 
  v 
v   v
g  v y 
g  v x 
g  v z 


b
vx g  vx  vyg  vy  vzg  vz 
Since “b” does not depend on vx, vy or vz; it must be a constant.
g  v x 
bv
dv

x
x

 g  v x  dvx
g  v x 
 bv x
g  vx 

1 2
bv x
1 2
bv x  ln g  v x   C  g  v x   Ce 2
2
1
1
1
  v   g  vx  g  v y  g  vz   Ce 2 Ce 2 Ce 2
bv2x
bv2y
bvz2
Evaluate “C” by letting the total probability equal one.
1
 C3e 2

b v2x  v2y  v2z

1
 C3e 2
b v2
6

 g  v  dv  
x

x


Ce
1 2
bv x
2

1

bv 2x
dv x  C e
1 2
bv x
2

A table of definite integrals would include:
C e 2



0
e ax dx 
2
dv x  1
1 
2 a
 1 2 
b
b
dv x  C  2 
when a 
  1  C 
2
b 
2
2
b 12 bv2x
g  vx  
e
2
3
kT
2
1
1
1
 tr  m v 2  m v 2x  v 2y  v z2  m 3v 2x
2
2
2
3
3
kT
m v 2x  kT  v 2x 
2
2
m
2
Alternatively, calculate v x using the probability density.
To evaluate “b”, recall that  tr 
v

  v g  v x  dv x  
2
x

From a table of integrals,


0
2
x

2n  ax 2
x e
To apply the integral, n = 1 and a 
v 2x 

2
b 2 bv2x
v x e dv x
2

b
v 2x e


2
dx 
 2n !
22n 1 n!
1
2
a
n
1
2
b
,
2
bv2x
2
1


b  2! 2
  1
dv x  2
3 

2
b
 23 1! b  2 
  

 2  
Equate the two calculations of v 2x .
v 2x 
1 kT

b
m
 b
m
kT
Thus the one-dimensional velocity distribution function for an ideal gas is:
2

tr ,x
x
m  mv
m  kT
g  vx  
e 2kT 
e
2kT
2kT
Three-dimensional velocity distribution function
7
The one-dimensional distribution function yields the probability of finding an
atom with a specific component of the velocity, vz, within a differential range,
vz and vz + dvz
To find the three-dimensional distribution function yielding a velocity, v, within a
range of v and v + dv, we need to recognize that an atom with velocity, v, can travel
in any direction.
The velocity of a particle moving in an arbitrary direction can be represented in a
three-dimensional “velocity” space.
vy
}dv
v
vx
vz
An atom with velocity, v, could be anywhere on the sphere with radius, v.
The total probability density of an atom with a velocity between v and v + dv lies
between a sphere of radius, v, and a sphere of radius v + dv in velocity space..
The volume of this shell is the difference between the volume of the two spheres.
4
4
3
Vshell  Vsphere  v  dv   Vsphere  v     v  dv   v3
3
3
4
4
4
4
4
2
3
 v3    3v 2 dv    3v  dv     dv   v3
3
3
3
3
3
4
4
4
2
3
   3v 2 dv    3v  dv     dv 
3
3
3
4
   3v 2 dv   4v 2 dv
3




8
The three-dimensional distribution function is the product of the one-dimensional
distribution functions and the volume of the shell in velocity space.
G  v  dv    v  4v 2dv  g  v x  g  v y  g  v z  4v 2dv 
2
2
dN v
N
2
x
x
x
m  mv
m  mv
m  mv
g  v x  g  v y  g  v z  4v dv 
e 2kT
e 2kT
e 2kT 4v 2dv
2kT
2kT
2kT
2
3
3
2
2
mv
Mv
 m  2  2kT
 M  2  2RT
2
G  v  dv  
4v dv  
4v 2dv
 e
 e
 2kT 
 2RT 
The above equation is known as the Maxwell (Maxwell-Boltzmann) distribution
function.
Graphs of Maxwell-Boltzmann distribution function
.55
0.5
298 K
0.4
398 K
f ( u  2  298)
0.3
f ( u  2  398)
2000 K
f ( u  2  2000)
0.2
0.1
0
0
0
0
2
4
6
8
10
u
12
12
velocity
- The average velocity of the gas molecules increases as the temperature
increases.
- At low velocities, the shape of the curve resembles a quadratic function;
whereas at moderate to high velocities, the shape of the curve resembles a
Gaussian function.
9
4.5
SF6, 298 K
4
3
f ( u  2  298)
CO2, 298 K
f ( u  44  298)
f ( u  146  298)
2
SF6, 2000 K
f ( u  146  2000)
1
0
0
H2, 298 K
0
1
2
0
3
4
u
4.5
velocity
- The average velocity of the gas molecules decreases as the molar mass
increases.
Aside: The speed of sound in a gas medium is proportional to the average
velocity of the gas molecules. Thus, breathing He makes your voice
higher and breathing SF6 makes your voice lower.
Other Gas Velocity Averages
Average speed

3

3
2

2
mv
mv
 m  2  2kT
 m  2  2kT 3
2
v   v G  v  dv   v 
4v dv  4 
v dv
 e
 e
2kT 
 2kT  0
0
0 
From a table of integrals:
3



0
2
x 2n 1e  ax dx 
2
n!
2a n 1
3
1
1
mv
1!
 m  2  2kT 3
 m 2 
  8kT  2  8RT  2
v  4 
e
v dv  4 

 
 
 

11 
 2kT  0
 2kT    m    m   M 
2

  2kT  
Most probable speed
10
The most probable speed, vmp, is the speed with the highest probability; that is, the
speed where the velocity distribution is a maximum.
We can find vmp by taking the derivative of the Maxwell-Boltzmann distribution
function, setting it equal to zero and solving for v.
3
dG  v  d  m  2  mv
2kT
 
4v2dv  0
 e
dv
dv  2kT 
3
2
2
2
d  mv
d  mv
 m 2
2kT 2
4

e
v
dv

0

e 2kT v 2dv  0


dv
dv
 2kT 
2
2
2
2
2
mv
mv


d  mv
2mv  mv
mv3  mv
e 2kT v2dv  
e 2kT v2  e 2kT  2v   0  
e 2kT  2ve 2kT  0
dv
2kT
kT
1
1
mv2
2kT
 2kT  2  2RT  2
 
 2  0  v2 
 vmp  
 

kT
m
 m   M 
RMS speed (from above)
1
vrms 
v
2
1
 3kT  2  3RT  2

 

 m   M 
Comparison of all three averages
vmp  v  vrms
vmp : v : vrms

2 : 8  : 3  1.414 :1.596 :1.732  1:1.128 :1.225
Barometric Distribution Law
11
Find relationship between height and barometric pressure
- assume temperature is constant (very shaky assumption)
- assume acceleration due to gravity is constant
- mixture of gases remains constant with height
z + dz
z
Force (weight) at height z in the atmosphere is the atmospheric
pressure multiplied by an unspecified area.
Fz  pA
Force at height z + dz decreases due to a decrease in pressure.
Fzdz   p  dp  A
The force at height z can be written in terms of the force at z + dz by realizing that at
height z, more atmosphere (from dz) has more weight.
Fz  Fz dz  dF
 Fz  Fz dz  dF   p  dp  A  dF  pA
 pA  Adp  dF  pA  dF  Adp
Consider the weight of the air to be equal to its mass times gravity, F  mg
Then consider that the mass of the air is related to its density and volume, m  V
The volume of air is related to the unspecified area used above V  Ah
Thus the weight can be written as F  Ahg
The change in the force is related to the change is height: dF  gA dz
Thus from above dF   Adp
gA dz  Adp   dp  g dz
The density of the gas can be reconsidered in terms of its molar mass.

m nM pM


V
V
RT
After substitution, we can integrate from sea level to height “z”.
12
pM
g dz  dp 
RT
p
 p
Mg
dp
Mgz

dz




ln
p

ln
p

ln
 
0
0 RT
p
RT
 p0 
p0
z
Thus the barometric distribution law can be written as exponentially decreasing
function as we travel further away from the surface of the earth.
p  p0e

Mgz
RT
If T = 250 K and M = 29 g/mol, then the elevation where the pressure is decreased in
half is

 p 
 p  Mgz
Mgz
 ln    ln  0  
RT
 p  RT
 p0 
RT  p 0 
8.314 J mol  K  250K
 p 
ln   
ln  0   5064 m
3
2
1
Mg  p  29 g mol 1kg 10 g  9.81m s
 p0 
2 
According to the barometric distribution law, the atmospheric pressure in Omaha
(elevation  1000 ft = 305 m)
 z
p  p0e

Mgz
RT
 1atm e

29g mol1kg 103 g9.81m s 2 305m
8.314J molK298K
 1atm e 0.0350
 0.965atm  p  734 mmHg  28.9inHg
Remarkable agreement with actual measurements!
Boyle’s Law
13
Pressure and volume of a gas are inversely proportional to each other, all other factors
being constant (such as temperature and amount).
1
k
p
p
 pV  k
 p1V1  p2V2
V
V
Consider cylinder with piston
Pa
- gas molecules inside cylinder hit piston and
cylinder walls creating pressure
P
- initially pressure inside cylinder is
atmospheric pressure, pa
a
Now apply additional external pressure, pex.
- I.e., push down on piston.
P
P
ext
- pressure inside is greater
- gas molecules hit walls more often
a
- note volume is smaller
P
P
a
ext
Now pull out piston, to create lower pressure inside.
Pext
Pa
Pa Pext
- total pressue (pressure inside) is less than
atmospheric pressure
- volume of gas is larger
Note that the equation pV = k is a hyperbola. Thus for a specific temperature, a plot
of pressure versus volume yields a parabola. Such curves are known as isotherms.
14
isotherms
p
V
The isotherms characterize a substance’s isothermal expansion or
isothermal compression.
A substance’s isothermal compressibility is defined as

1  V 


V  p T,n
Notes:
1. Dividing by the volume makes the isothermal compressibility an intensive
quantity. Intensive quantities are portable and thus can be tabulated for reference.
2. The negative sign in the definition makes the value of the isothermal
compressibility a positive value.
3. The partial derivative in the definition is the slope in the above p-V plot.
Charles’ (Gay-Lussac) Law
15
Volume and temperature are directly proportional to each other, all other factors
being constant.
V
V1 V2
V kT 
k
VT


T
T1 T2
Note that V = kT is a linear equation with zero for a y-intercept. Thus plots of ideal
gas on a volume/temperature graph will yield straight lines that extrapolate to T = 0.
Thus we see one reason why using an absolute temperature scale in thermodynamics
is necessary.
isobars
V
T
An isobar is a thermodynamic path where the pressure is being held constant.
Such isobars were the basis of the ideal gas thermometer mentioned above.
Avogadro’s Law
Volume and amount are directly proportional to each other, all other factors being
constant.
Vn
V k n 
V
k
n

V1 V2

n1 n 2
Avogadro’s law was a brilliant leap of insight. By measuring the behavior of gases,
one could count (!!) the number of atoms with a gas sample.
Ideal Gases
16
Microscopic assumptions
1. gas molecules are point particles (they have no volume)
2. gas molecules do not reaction
3. gas molecules do not have any interaction (no intermolecular forces)
Macroscopic definition
 U 

 0
 V T
- i.e., There is no change in the internal energy of a gas with respect to a change
in the volume at constant temperature. (More later)
Equation of State
The equation of state can be found from combining Boyle’s law, Charles’ law and
Avogadro’s law in the familiar ideal gas equation: pV = nRT or
p
V
RT
V
V
is the molar volume. ( V  Vm in the Laidler text.)
n
Miscellaneous
Pressure and temperature for an ideal gas is directly proportional to each other. Thus
a plot of pressure/temperature plot yield a straight line. These lines are known as
isochores.
Real Gases
Real gases are gases where the nonideality of the gases is considered. Almost always
the actual volume of the gas molecules is accounted for. This volume is known as the
covolume. Also the intermolecular forces are nonzero; thus they will affect the
behavior of the gas, especially at high pressure and low temperature. Dozens of
equations of state for real gases exist. They all have limits of applicability; thus,
before using a real gas equation of state, one should take stock of the pressure and
temperature ranges being used and checking if the equation of state is applicable.
Real Gas Equations of State (Closed forms)
17
Van der Waal’s
p
RT
a
 2
Vb V
b is the covolume
- V – b is the reduced volume
a is a measure of attractions between molecules
- Note that the negative sign in the term illustrates that attractive forces will
reduce the pressure.
- Note the dependence on the square of the molar density, where the molar
density is
mol 
n 1

V V
- The attractive interaction depends on the square of the molar density since the
attractive forces are dependent on the interaction between two molecules.
The equations try to account for subtle interactions, usually empirically, that is, by
curve fitting. Below are more examples.
Beattie-Bridgeman
c 

RT 1 
3 
 VT  V  B  A
p
V2
V2


Redlich-Kwong
p
RT
a

Vb
TV V  b

Bertholet
p
RT
a

V  b TV 2
Dieterici

a
RTe RTV
p
Vb
Virial Expansions

18
Sometimes the experimental equation of state is difficult to fit into a closed form. But
having an equation that relates the different thermodynamic variables to each other is
still desirable. The data may be fitted to a polynomial expression to whatever
accuracy is desired. These polynomial equations of state are also known as virial
expansions. Below are two common ways to express a virial expansion where the
first equation is the most common. Note the coefficients of either virial expansion are
functions of the temperature.
p
RT  B  T  C  T  D  T 



1 
V 
V
V2
V3



and
p
RT
1  B  T  p  C  T  p 2  D  T  p3 

V

The coefficients in the virial expansions above are given specific names
B(T) – second virial coefficient
C(T) – third virial coefficient
D(T) – fourth virial coefficient
Boyle temperature
Let us look closer at the first virial expansion above.
p
RT  B  T  C  T  D  T 



1 
V 
V
V2
V3



If we assume that the series converges, i.e., the coefficients get smaller from B to C to
D etc…, then, for practical applications, often only the first two terms are used.
p
RT  B  T  
1 

V 
V 
Since polynomial coefficient is a function of temperature, we can find the
temperature where the coefficient equals zero. This temperature where the second
virial coefficient equals zero is known as the Boyle temperature. At the Boyle
temperature, the gas behaves approximately as an ideal gas.
Continuing with Real Gases
19
Isotherms for real gases
isotherms
p
Tc
critical
point
(Vc, pc)
gas
gas and
liquid
liquid
supercritical
fluid
V
Since real gas molecules interact, condensation will occur when the pressure
increases along an isotherm as long as the temperature is below a certain value.
The green region is where condensation occurs. Note the pressure remains constant
until all the gas has condensed into liquid.
Note the isotherm does not enter the condensation region, but it is tangent to the
region.
Critical Points
At a given temperature and pressure the distinction between gas and liquid becomes
blurred. The fluid still fills its container like a gas; however, it is unable to be
condensed into a liquid. Such a fluid is called a supercritical fluid. The specific
pressure, volume and temperature where the boundary between gas and liquid
disappears is called the critical point, with the pressure being the critical pressure,
Pc, the volume being the critical volume Vc and the temperature being the critical
temperature, Tc.
The supercritical fluid is neither a gas nor a liquid. On the microscopic level, we can
think of the fluid as maintaining an ill-defined equilibrium between the gas phase and
the liquid phase. Portions of the fluid are aggregating to form liquid regions and
portions are de-coalescing into the gas phase. The “liquid” aggregates have no welldefined size or shape. Thus light scattering in unpredictable ways and the substance
appears milky. (Whereas separately the gas and the liquid may be colorless.) This
milky condition is called critical opalescence. It is a macroscopic indication that we
have a supercritical condition along with the necessary condition that the fluid cannot
be condensed.
Any real gas equation of state needs to satisfy the following mathematical conditions:
20
 p 

  0  isotherm is flat at the critical point
 V T
 2p 
 2   0  isotherm at the critical point is an inflection point
 V T
(Unfortunately, my picture above does not show this.)
Thus we have a couple of key relationships between the pressure, volume and
temperature at the critical point.
Compressibility
By definition, the compressibility of gas is: Z 
PV
.
RT
Note that for an ideal gas, the compressibility is always equal to unity (one).
Deviations away from unity relay information about the nature of the non-ideality of
the real gas.
2.0
Z
repulsive
forces
dominate
H2
1.0
200
400
600
800
P(atm)
CH4
C2H4
attractive
forces
dominate
0.0
Where the compressibility is less than one, attractive forces dominate.
Where the compressibility is greater than one, repulsive forces dominate.
The Law of Corresponding States and Reduced Variables
21
Reduced variables – dimensionless quantities where a thermodynamic value for a
real gas is divided by its corresponding critical value.
reduced pressure – p r 
p
pc
reduced temperature – Tr 
reduced volume – Vr 
V
Vc
T
Tc
The Law of Corresponding States – All gaseous substances appear to behave
identically when described by their reduced variables.
In other words, all real gases with the same reduced pressure and reduced
temperature, irregardless of the specific substance, will have the same reduced
volume.
At a deeper level, what the law of corresponding states is saying is that all of the
nonideality of a gas is characterized by its critical point.
- This is remarkable!
A consequence of the Law of Corresponding States is that two parameter real gas
equations of state can be rewritten is terms of their reduced variables rather that their
“a” and “b” parameters.
Let us examine how to rewrite the van der Waals equation in terms of only reduced
variables.
RT
a
p
 2
Vb V
We will use the relationships between pressure and volume at the critical point to find
expressions for the “a” and “b” parameters.
 p 

 0
 V T
 2p 
 2  0
 V T
RTc
a     RT    a
 p      RT
 2 


 
 

 2 
 V T  V T  V  b V   V T V  b  V T V
Vc  b




 2p     
RT
2a     
RT





 2 
 


2
3
V   V T  V  b
 V T  V T
 Vb


2RTc
6a

 4 0
3
Vc
Vc  b




 p 
From 
  0 , solve for a.
 V T



2

     2a
2
  V T V3


2a
0
3
Vc
22
RTc
V  b
2

c
2a
3
Vc
 a
RTc Vc

3
2 Vc  b

2
 2p 
Substitute this into the  2   0 equation and solve for b.
 V T
2RTc
V  b
3
c
2RTc
V  b
c
a
3


3
3RTc
6a
6  RTc Vc 
 4  4

2
2
Vc
Vc  2 Vc  b  Vc Vc  b


3RTc
V
2
3



 2Vc  3Vc  3b  b  c
2
Vc  b Vc
3
V V b
RTc Vc

c

c
3

V 
2  Vc  c 
3 

2





RTc Vc
3
1
2
2Vc 1  
 3
2

RTc Vc 9RTc Vc

4
8
2
9
Now find the critical values in terms of “a” and “b”.
Vc
 Vc  3b
3
9RTc Vc
8a
8a
8a
a
 Tc 


8
9RVc 9R  3b  27Rb
b
Use the van der Waal’s equation itself to find pc.
 8a 
R

RTc
a
a
8a
a
4a
3a
a
27Rb 

pc 
 2 


 2


2
2
2
2
Vc  b Vc
3b  b
 3b  54b 9b 27b 27b 27b 2
Now make the following substitutions: p = prpc, V = VrVc and T = TrTc into the van
der Waals equation.
23
p
RT
a
 2
Vb V
 pc pr 
RTc Tr
a
 2 2
Vc Vr  b Vc Vr
 8a 
R
 Tr
a
8aTr
a
27Rb 
 a 

 
p 


 2 2
2
2  r
2
2
 3b  Vr  b  3b  Vr 27b 3Vr  1 9b Vr
 27b 

1 
  a   8Tr
p

   r


  2
2
  9b   3 3Vr  1 Vr 
 3
8Tr
3
 pr 
 2
3Vr  1 Vr
 a  p
  2  r
 9b  3




8Tr
1

 2

 3 3Vr  1 Vr



Thus now the van der Waals equation is written only in terms of reduced variables
and has the same form for all real gases. The “a” and “b” parameters are included in
the critical variables that are included in the reduced variables.