CHEM344
Part I
201402
Molecular Kinetic
Theory of Gases
 Introduction
 The Assumption of the Kinetic Theory of Gases
1. The gases are composed of small particles called molecules, for any one
gas all molecules are of the same size and mass but differ from gas to
another.
2. The molecules are in continuous random motion, they travel in a straight
line and during their motion colloid each other and with the wall of
container.
3. The pressure of the gas is rises from the sum of the forces of collision of
the molecules with the walls of the container.
4. Since the pressure of the gas does not change with time at any given
pressure and temperature. The molecular collision must involve no energy
loss due to the fraction, that is to say molecular collision are perfectly
elastic.
5. The absolute temperature is a quantity proportional to the average kinetic
energy of all molecules in a gas.
6. At law pressure the average distance between molecules are largely,
compared with molecular diameter and hence the attractive forces
between molecules which depends on the distance of molecules separation
may be neglected. (For Ideal Gas)
7. Finally, since the molecules are small compared with the distance between
them, their volume may be neglected compared with the volume of the
container. (For Ideal Gas)
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CHEM344
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 The Theoretical explanation of the kinetic theory of
gases (Kinetic Equation)
PV  31 mn u 2
Where
P
V
m
pressure
volume
mass
n`
n
u
No. of molecules = n * N
No. of moles & N Avogadro’s No.
velocity (square of the root mean
square velocity)
 Detection of the velocity of the gas
Kinetic Equation
PV  31 mn u 2
General Gas Equation
So
PV  nRT
mn u 2  nRT
n   nN
 31 mnNu 2  nRT
 mN  M
 31 Mu 2  RT
1
3
u
3RT
M
 Relation between the velocity and the density 
u
3RT
M
PV  nRT
m
PV  RT
M
PM  RT

PM 
Page 2 of 23
m
RT
V
CHEM344
RT 
u
u 
PM

Part I
201402
3PM
M
3P

Ex.
Calculate the velocity of hydrogen molecule at 0 C
u
3RT
M
3 * 8.314 * 10 7 * 273
u
2
u  184359 .7cmS 1
T = 273 K
M = 2*1 g/mol
R =8.314 * 107 erg mol-1 K-1
 Kinetic Energy of Translation
The only type of energy we have notice to gas molecules is that due to the
molecular motion along three coordinate axis (x, y and z) is called the kinetic
energy of translation.
The amount of energy is obtained as follows
From kinetic theory and gas laws
PV  31 mn u 2
PV  nRT
1
 2  nRT
3 mn u
2
1
 nRT
3 mnNu
2
1
 nRT  1
3 nMu
The kinetic equation of translation is given by
E k  21 nMu 2  2
From 1 & 2
Page 3 of 23
CHEM344
E k  23 nRT
For 1 mol of gas
Part I
201402
E k  23 RT
 Notes
1. The translational energy of an ideal gas is completely independent of
the nature of gas or the pressure of the gas and depends only on the
absolute temperature
e.g. at 300 K all ideal gases will contain 853.65 Cal per mole
E k  23 RT
E k  23 * 1.897 * 300
E k  853.65 Cal
2. The average kinetic energy per molecule can be calculated as follows
E k  23 RT
RT
E k  23
N
E k  23 kT
Where
k= 1.3805*10-16 erg molecule-1 K-1
Per mol
Per molecule
(Boltzmann Constant)
3. According to the kinetic theory of gases, an increase in temperature
of a body is equivalent to an increase in the average translational Ek
for the molecules i.e. T = F(Ek)
So
E k  23 RT
E
T  23 k R
4. The kinetic theory interprets of absolute zero temperature is that
the complete stop of all molecular motion (the zero point of K. E.)
This picture has been some what changed by quantum theory which
require small residue of energy even at absolute zero temperature
5. The average translational K. E. may be resolved into three component
in the three degrees of freedom corresponding to the velocities
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CHEM344
Part I
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parallel to the three coordinate axis (Principle of Equipartition of
Energy)
E k  21 nMu 2
E k  21 nMu x2  21 nMu y2  21 nMu z2
 Degrees of Freedom
 Definitions
 The degrees of freedom are the No. of coordinates that are necessary
to specify the position of a particle in space
 For spherical model, it has three degrees of freedom, that will be the
case of mono atomic molecule e.g. He atom
 For a molecule that consists of N`atom e.g (O2 N`=2) or (CO2 N`=3),
three coordinates are necessary to specify the position of each atom,
there for this molecules will possess a total of 3N` degrees of freedom
z
z
x
3 D.F.
y
(x2,y2,z2)
z
(x1,y1,z1)
(x,y,z)
y
(x2,y2,z2)
(x1,y1,z1) (x3,y3,z3)
x
x
y
6 D.F.
9 D.F.
 Since each atom in the molecule is not free to move independently from
the other (the molecule translate as a hole), so one can describe the
translational of a hole molecule in term of 3 coordinate of the center of
mass
 The center of mass is the point in molecule where all the mass are
considered to be centered
z
y

m1
x
3 T.D.F.
Page 5 of 23
Center
of mass
m2
CHEM344
Part I
201402
 The remaining 3N`-3 degrees of freedom must corresponds to other
modes of motion with diatomic and polyatomic molecules
 This motion are called internal modes of motion and consists of
rotational and vibrational motion
 For diatomic molecule if considering that the 2 atoms are connected by
a rigid rod (this is called the dumbbell model), can be rotated about its
center of mass in two perpendicular direction and therefore it has 2
rotational degrees of freedom
z
y
z
x
x
y
z-direction
z
x
y
y-direction
x-direction
2 R.D.F.
 As the same a linear polyatomic molecule CO2 will possess 2 rotational
degrees of freedom
 For non linear polyatomic molecule SO2, can rotate about 3 perpendicular axis, So it has 3 rotational degrees of freedom
z
S
O
C
O
CO2
O
SO2
O
y
x
x-direction
3 R.D.F.
 The remaining (3N`-5) for linear and (3N`-6) for non linear molecules
must be due to another form of motion called the vibrational motion
 This vibrational motion results in stretching and compressing of the
bond
Equilibrium
Stretching
Comperssing
Vibrational Motions
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 From this motion the molecule will possess two vibrational energies
called (Vibrational K. E. & Vibrational P. E.)
 Mathematical form for K. E.
1. Translational Kinetic Energy
mu x2
(E k )Trans 
2
m is mass of molecule
2. Rotational Kinetic Energy
Iw x2
(E k )Rot 
2
I is moment of inertia
3. Vibrational Kinetic Energy
(E k )Vib 
ux is velocity in x direction
wx is angular velocity
u 2
2
 is reduced mass

m1m2
m1  m2
u is vibrational velocity of atoms
4. Vibrational Potential Energy
(E p )Vib 
kx 2
2
k is force constant of bond
x is displacement of atoms from
equilibrium position
 Note the similarity in the mathematical forms of this energy terms
 The Principle of equipartition of Energy
 When a gas takes up heat energy, this energy distributes itself equally
in each of the independent ways by which the molecules can absorb
energy
 i.e. heat distributed into translational kinetic energy, rotational kinetic
energy, vibrational kinetic energy and vibrational potential energy
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CHEM344
Part I
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 Summary of calculated degrees of freedom
Example
Total
Degrees of
Freedom
Translational
Degrees of
Freedom
Rotational
Degrees of
Freedom
Vibrational
Degrees of
Freedom
O2
Polyatomic
Linear
CO2
Polyatomic
Nonlinear
SO2
3N`
(3)
3N`
(6)
3N`
(9)
3N`
(9)
3
3
3
3
0
2
2
3
0
1
3N`-5
3N`-6
Monoatomic
Diatomic
He
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CHEM344
Part I
201402
 Determination of Heat Capacity
 The heat capacity per unit volume is the heat required to raise the
temperature of 1 mol of substance 1 C.
 There are two types of the heat capacity depending on the substance is
heated at constant volume or at constant pressure.
 When hated at constant volume, all the supplied heat tends to increase
the internal energy of substance and the heat capacity is called Cv.
 When hated at constant pressure, the supplied heat not only increases
the internal energy of substance, but also to make possible expansion of
the substance against the extension of pressure, so the heat capacity
is called Cp.
 Note that Cp is larger than Cv
C p  Cv  R
 How the principle of equipartition of energy predicts the
observed values of heat capacity of all types of gases
1. For monoatomic gas
 There are only 3 degrees of freedom, they are all translational
 Each degrees of freedom contributes ½ RT for one mole of the
internal energy
 So the total energy will be
E  23 RT
 The change in internal energy for 1 mol of gas when the
temperature changes by 1 C is equal to
E  E 2  E1  23 R (T2 T1 )
E  23 R
 That is the heat capacity at constant volume Cv. i.e.
Cv  23 R  23 * 1.987  2.18 cal / mol deg
C p  Cv  R  23 R  R  52 R
 52 * 1.987  4.97 cal / mol deg
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CHEM344
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201402
 The experimental value of the Cp is 4.97 cal/mol deg, the same
as calculated above
2. For diatomic gas
 In case of the diatomic molecule in addition of 3 translational
degrees of freedom there are 2 rotational degrees of freedom
and one vibrational degrees of freedom which has both kinetic
and potential energy
 So
KET
KE R
KEV
PEV
E 
 23 RT
 22 RT
 21 RT
 21 RT
7
2
RT
 And from Cv definition
Cv  72 R  72 * 1.987  6.95 cal/mol deg
C p  Cv  R  29 R  9.94 cal/mol deg
C p 92 9
 
   1.286
Cv 7 2 7
 Note that the theoretical value of Cv just calculated above are
higher than the experimental values at 25 C, this is because
the vibrational modes don’t contributes to the total energy at
law temperature.
3. Linear triatomic gas molecule
 We have 3N DF i.e. 9 DF, contributes to 3 TDF, 2 RDF, (3N-5)
i.e. 4 VDF, So each contributes to the total energy
ET  23 RT
E R  22 RT
EV  4RT
E  132 RT
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CHEM344
Part I
201402
 And the heat capacity will be
Cv  132 R  12.92 cal/mol deg
C p  152 R  14.90 cal/mol deg
 
Cp
Cv

15
 1.15
13
4. Nonlinear triatomic gas molecule
 We have 3N DF i.e. 9 DF, contributes to 3 TDF, 3 RDF, (3N-6)
i.e. 3 VDF, So each contributes to the total energy
ET  23 RT
E R  23 RT
EV  3RT
E  6RT
 And the heat capacity will be
Cv  6R  11.92 cal/mol deg
C p  7R  13.91 cal/mol deg
 
Cp
Cv

7
 1.17
6
 Calculate the heat capacity at constant volume and pressure for the
acetylene gas which is linear and ammonia gas which is non linear
 For acetylene, we have 3N DF i.e. 12 DF, contributes to 3 TDF, 2 RDF,
(3N-5) i.e. 7 VDF, So each contributes to the total energy
ET  23 RT
E R  22 RT
EV  7RT
E  192 RT
 And the heat capacity will be
Cv  192 R  18.88 cal/mol deg
C p  212 R  20.86 cal/mol deg
 
Cp
Cv

21
 1.11
19
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CHEM344
Part I
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 For ammonia, we have 3N DF i.e. 12 DF, contributes to 3 TDF, 3 RDF,
(3N-6) i.e. 6 VDF, So each contributes to the total energy
ET  23 RT
E R  23 RT
EV  6RT
E  9RT
 And the heat capacity will be
Cv  9R  17.88 cal/mol deg
C p  10R  19.87 cal/mol deg
 
Cp
Cv

10
 1.11
9
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CHEM344
Part I
201402
 Maxwell-Boltzman Distribution of
Molecular Velocities
 Maxwell Distribution of molecular velocity
 The values of velocity of any individual molecule is never known
 One can only obtain a distribution of molecular velocities for the whole
collection of molecules at definite temperature from which the average
velocities can be obtained
 This distribution of molecular velocity can be represented graphically by
plotting the fraction of molecules having a given velocity (nu) against the
velocity (u) at a definite temperature
 This graph is known as Maxwell distribution molecular velocity curve
 The shape of the curve has been predicted theoretically and proven
experimentally
nu
 The total area under the curve gives
maximum
the total number of molecules, i.e. the
No. of molecules having all velocities
 The area under the curve between u1
and u2 gives the total No. of molecules
large No.
having velocities between these values,
small No.
also the same case for u3 and u4
 It is noticed from the curve that the
u1 u2
u3 u4
up u urms
u
greatest portion of the molecules having
velocities called the most probable velocity (up)
 It is the velocity possess by greatest No. of molecules which is given by
the top or maximum of the curve
 On both sides of up the curve drops off sharply, indicating that very low
and very high velocities possessed only by small fraction of molecules
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CHEM344
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 Comparison between the distribution of molecular velocities at two
different temperatures
 The total No. of molecules is the same at both temperature and the total
area under the two curves are equal
nu
T2>T1
 Increasing temperature, increases
the K.E. of the molecules, and thereT1
T2
fore a greater fraction of molecules
would be expected to have higher
velocities at higher temperature,
this is evident from the curve
 The curve at high temperature (T2)
has its maximum shifted to the higher
up
up
u
velocity as compared to that for T1
 Boltzmann distribution law
 The molecules in a gas are in continuous random motion, during which they
enter into repeated collisions with each other. This collisions causes
redistribution of energy
 The mathematical expression that relate the No. (ni) of the molecules
having an energy (Ei) relative to the total No. of molecules (nT) at a given
temperature is called the Boltzmann distribution law
ni = nT e
-Ei
RT
for 1 mol of gas
Example
Calculate the No. of molecules in a mole of an ideal gas (He) having at least 4
times the thermal energy at 25 C.
ET  32 RT
Ei  4* 32 RT = 6RT
For 1 mol nT = NA
ni = NAe
-6RT
RT
 ni 
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CHEM344
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 Root mean square velocity, Average velocity and most probable
velocity
 The Most Probable Velocity
 It is the velocity possess by greatest No. of molecules in a gas
 The Average Velocity (ū)
 In general if ni molecules have a velocity ui so
u=
1
nT
nT
i ni ui
 And the average of the square of the velocity is
u =
2
1
nT
nT
i ni ui2
 The Root Mean Square Velocity (urms)
 From above equation
urms = u 2
 It can be determined from kinetic gas equation and ideal gas equation as
PV  31 mn u 2
PV  nRT
1
 2  nRT
3 mn u
 31 mnNu 2  nRT
 mN  M
 31 Mu 2  RT
urms =
3RT
M
urms = 1.7
RT
M
n   nN
 The equations of ū and up can be determined by advanced methods gives
u = 1.6
RT
M
&
u p = 1.4
Page 15 of 23
RT
M
CHEM344
Part I
201402
urms : u : u p
1.7 : 1.6 : 1.4
Example
1. Suppose that a gas contains 5 molecules with velocity of 2 m/sec, 10
molecules with velocity 3 m/sec and 4 molecules with velocity 6 m/sec,
determine the urms, ū, up.
u=
1
nT
nT
i ni ui
nT = 5 + 10 + 4 = 19
u=
1
19
5 * 2 + 10 * 3 + 4 * 6  
u =
2
u2 =
1
19
1
nT
nT
64
= 3.4 m/sec
19
i ni ui2
5 * 4 + 10 * 9 + 4 * 36  
254
= 13.4
19
urms = u 2  13.4 = 3.7 m/sec
From definition of the up so it will be 3 m/sec (the largest No. of molecules
(10) has velocity 3 m/sec)
up = 3 m/sec
Note that
urms : u
: up
1.7 : 1.6 : 1.4
3.7 : 3.4 : 3
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CHEM344
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2. Calculate urms, ū, up for O2 gas at 25 C, at what temperature would H2 gas
has the same values of the velocities
RT
urms = 1.7
M
8.314 * 10 7 erg mol -1K -1 * 298 K
urms = 1.7
 481.95 m sec -1
-1
32 g mol
urms
: u
erg = g cm2 sec -2
: up
1.7
: 1.6
: 1.4
481.95 : 453.6 : 396.9
For H2
8.314 * 10 7 erg mol -1K -1 *T K
48195 cm sec = 1.7
2 g mol -1
-1
T = 18.6 K
 Frequency of collision
 For Bimolecular gaseous reactions the frequency of collision (Z) can be
determined from the kinetic theory of gases
 For similar molecules e.g.
2HI  H2 + I2
Z = 0.921 2  2 n 2
3RT
M
 For different molecules e.g.
H2 + I2  2HI
2




2 8  RT  M1 + M2 
2 
Z = n1n2  1
n

M1 M2
 2

Where
n

M
is No. of molecules per cubic centimeter at temperature T
is molecular diameter
is molecular weight
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CHEM344
Part I
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Example
Find the molecular collision no. for the reaction (2HIH2+I2) at 556 K, if the
collision diameter for HI is 3.5*10-8 cm and the molecular weight is 127.6
NA (6.023 * 10 23 )
No. of molecules n per cm =
 2.69 * 10 16
1000 * 22.4
3
Z = 0.921 2  2 n 2
Z = 0.921 2 *3.14 * 3.5 * 10
3RT
M
 2.69 * 10 
-8 2
16 2
3* 8.314* 10 7 * 556
127.6
 Mean Free Path
 The path occupy by a molecule between two successive collisions with
other molecules
 The No. of collision between molecules is
2  2 nu
 If is the average velocity so the average distance between two collision
(mean free path L) is
L
u
1

2  2 nu
2  2 n
 Viscosity of gases
 The kinetic theory of gases describe the viscosity of gases to a transfer
of momentum from one moving plane to another
  31 uL 
 The viscosity of gases are in units of micropoise (10-6 poise)
 Calculation of the mean free path and collision diameter from the
viscosity coefficient
 Mean free path
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  31 uL 
L
3
u
RT
M
1.88 
L
 RT M
u = 1.6
 Collision Diameter
1
2  2 n
3
L
u
L
u=
8RT
M
2 =
2  RT
3 n  M
1
2
3
From 1, 2 and 3
Example
The viscosity of H2 gas at zero C is 8.4*10-5 poise. Determine the mean free
path, collision diameter of the molecules at this temperature and 1 atm
pressure, also determine the No. of collisions in 1 sec in 1 cm3 and 1 C.
L
1.88 
 RT M
T = 273 K
 = 8.41*10-5 poise
R = 8.314*107 erg/mol K
M = 2 g/mol
NA = 6.023*1023 molec/mol
2 gmol -1
M
-5
-3
=

=
8.9
*
10
g
cm
22400 22400 cm3 mol -1
1.88 * 8.41 * 10 -5
L 
 cm
-5 8.314* 10 7 * 273
8.9 * 10
2
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CHEM344
Part I
n=
NA
 2.69 * 10 16 molec/cm3
22400
 2 =
2  RT

3 n  M
Z = 0.921 2  2 n 2
201402
cm 2
3R * 274

M
molecule
 The Behavior of Real Gases
 The properties of real gases deviate from the ideal gas due to
 The attraction forces between the molecule
 The volume of the gas molecules
 These factors were not taken into account in ideal gas law
 The assumption of kinetic theory of gases that must be modified are
 The molecules exerts no forces on each other
 The volume of molecules are neglected compared to the total volume
of the container
 These two assumption are incorrect which can be seen in the following
H2
figure
O2
PV
 If a gas obey Boils law, it is defined as an
CO2
ideal gas
40
 For a given mass of an ideal gas at constant
30
Ideal
temperature, the pressure volume product 20
(PV) should be constant for all pressure
10
 This means that if one plot the pressure
200 400 600
P (atm)
volume (PV) # P the line would be horizontal line
 The result of different gases (Real Gases) over a wide range of
pressure and temperature indicate that all gases showed deviation from
this ideal behavior and the deviation increase at higher pressure and
lower temperature
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CHEM344
Part I
 The Compressibility Factor
201402
N2 (0 C)
 The following plot indicates deviation
PV
CO2 (40 C)
RT
from ideal gas behavior
H2 (0 C)
 In this graph the Y axis is the ratio of
1.4
(PV/RT) where V is the volume occupied
1.0
Ideal
by 1 mole of gas at constant temperature 0.6
0.2
and pressure
 This ratio is called the compressibility
200 400 600
P (atm)
factor
 Since the compressibility factor equal 1 for 1 mole of an ideal gas in the
case of real gases the compressibility factor will vary and hence its
deviation from unity is an index of deviation from ideal behavior
 The assumption of kinetic theory of gases that all molecules are like
(that they all hard sphere) does not appear to be a case
 The molecules differ in their ability to attract each other, this
difference must be like to a difference in nature of the molecule and
the bond formed
 For both N2 and CO2 the curve first decreases and then increases again
as the pressure increase
 This decrease in compressibility factor below unity is more pronounced
in CO2 than N2
 This more pronounced minimum for CO2 means that for a given increase
in pressure, there is much greater volume decrease in CO2 than N2
 This greater decrease in volume for a given applied pressure result in
PV becoming less than RT and the ratio PV/RT will be less than 1
 The increase in compressibility factor with increasing pressure
indicates that the volume decrease is much less at this higher pressure
and that makes the product PV grater than RT
 Effect of Temperature on N2 gas
 As the temperature increases the minimum in the curve becomes less
pronounced, until the temperature is reached at which the
compressibility factor remains above unity for all P
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Part I
201402
 As the temperature increase the minimum become smaller and at the
same time the position of the minimum moves to low pressure
PV
RT
-70 C
20 C
400 C
1.4
1.0
Ideal
0.6
0.2
200 400 600
P (atm)
 Van der Waals Equation of State
1. Correction arise due to the volume of molecule
 The volume for a real gas Vr is greater than the ideal volume Vi by
an amount equal to the effective volume of molecules
Vi =Vr - nb
 This volume is called the excluded volume and given by b
 The excluded volume is not equal to the actual volume of molecule,
but is four times the molecular volume for spherical volume
Vpair = 34  2r  = 8  34  r 3 
3
excluded volume
Ve = 21 * 8  34  r 3  = 4VM
r r
PV
i i = Pi V - nb  = nRT
2r
Surface
2. The correction that will arise from the intermolecular forces
 The measured pressure Pr is less than the ideal pressure
predicted by simple kinetic theory of gases
A
B
Net inward Force on
molecule B at surface
No Net Force on molecule
A in the interior of gas
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CHEM344
Part I
201402
 It is found that the inward force is proportional the density of
the gas
Total inward force   2
Total inward force = a  2
 The density of the gas is proportional to the number of moles of
molecules per unit volume at any T and P
=
n
V
 So the total inward force which is correction factor of pressure is
2
n 
Total inward force = a  
V 
 And thus
an 2
Pi = Pr + 2
V
 Finally the van der Waals equation becomes

an 2 
 Pr + 2  Vr - nb  = nRT
V 

 For 1 mole of gas
a 

P
+
V - b  = RT
 r
2  r
V


 Note that the constants a and b are called the van der Waals
constants, they are dependent on (P, V, T) and so can be
determined experimentally from P, V, and T data
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