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Thermal Physics- 4CCP1471
2017/2018
Lecture 2
The Kinetic Theory of Gases
Pressure, Temperature, and Internal Energy
Dr Shahriar Sajjadi
Department of Physics
King’s College London
6th Floor, Strand
Building, Room S6.19
1. Definition and Assumptions
The kinetic theory of gases explains
Macroscopic properties of gases
pressure, temperature, volume,
and specific heat
by relating
them to
Microscopic properties of gases
molecular composition and
motion
The three variables associated with a gas, volume-temperature-pressure, are all
consequences of the motion of atoms:
 The volume is the results of the freedom the atoms have to spread through the
container.
 The pressure is a result of the collisions of the atoms with the container’s walls.
 The temperature is related to the kinetic energy or the speed of the atoms.
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Definition and Assumptions
The average distance
separating the gas particles is
large compared to their size
Molecules collide with each
other and walls, and all these
collisions are perfectly elastic
The number of molecules is
so large that statistical
treatment can be applied
Assumptions
There is no interaction among
particles (no attraction and
repulsion) and they are in
constant random walk
The particles have the same
mass
Molecules obey Newton’s law
of motion
2. Pressure – Particle Velocity Relation
System:
n moles of a gas
V volume of the cubical box where molecules are contained (L is the dimension)
T the temperature of the wall and gas molecules.
Objective: We look for the connection between the
pressure p (=F/A) exerted by the gas on the walls
and the speeds of the molecules.
Assumptions: For simplicity, we assume that the
molecules only collide with the walls.
o The linear momentum of a particle is a vector quantity that is defined as (m is
the mass of the particle and v is its velocity):


p  mv
kg.m/s
Because m is always a positive scalar quantity, then p and v have the same
direction.
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Pressure – Particle Velocity Relation
19-3 Pressure,
Temperature,
rms
o Newton
expressed his
second law of and
motion
in Speed
terms of momentum, that is:
“The time rate of change of the momentum of a particle is equal to the net force
acting on the particle”


dp
Fnet 
dt



p x  (mvx )  (mv x )  2mvx
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Pressure – Particle Velocity Relation
19-3 Pressure, Temperature, and rms Speed
o The momentum Δpx delivered to the wall by the molecule.
p x  2mv x
o The time between collisions, Δt, is:
t  2 L / v x
o The average rate at which momentum is delivered to the shaded wall by a
single molecule is
p x 2mv x mv x2


t 2 L / v x
L
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Pressure – Particle Velocity Relation
19-3 Pressure,
Temperature, and rms Speed
nd
o From Newton’s 2 Newton’s law (i.e., the rate at which the momentum is
delivered to the wall is the force acting on that wall), the total force exerted by
all molecules (N):
2
F
mv x21 / L  mvx22 / L  ......  mv xN
/L
p  2x 
2
L
L
where indices 1, 2, 3, 4,……….N indicate different particles in the system.
m
2
p   3 (v x21  vx22  ..........  v xN
)
L 
o We can now introduce (vx2)mean, which is the average value of the square of the x
component of the molecules speeds, into the equation:
where
n
m
N=nNA
L3
 nmN A  v 2x1  v 2x 2  ..........  v 2xN   nmN A  2
p


(v )
3
  L3  x mean
nN A
 L 

( v 2x ) mean
the number of moles
the mass of the particle
the number of molecules
the volume of the box.
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Pressure – Particle Velocity Relation
mNA = M (molar mass)→
o For any molecule,
p
nM (v x2 ) mean
V
y
v2 = v2x + v2y + v2z
vz
o Because of many molecules being present all moving randomly:
(v x2 ) mean  (v y2 ) mean  (v z2 ) mean 
p
v
vy
z
vx
(v x2 ) mean  1 / 3(v 2 ) mean
nM (v 2 ) mean
3V
o The root-mean-square speed of the molecules is vrms. So:
v2rms= (v2 )mean = (v2x )mean + (v2y )mean + (v2z )mean
vrms  (v 2 ) mean
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Pressure – Particle Velocity Relation
19-3 Pressure, Temperature, and rms Speed
o So the pressure relation is:
p
2
nMv rms
3V
 pressure increases with the speed of the molecules
 pressure increases with the number of molecules or moles n
 pressure is inversely proportional to the volume V.
o The relation for pV is (mass of the gas mg = nM):
1
1
2
2
pV  nMv rms
 m g v rms
3
3
 The equation above suggests that the product pV only depends on
particles speed (and quantity of the the gas).
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3. Temperature – Particle Velocity Relation
p
2
nMv rms
3V
o Using the ideal gas equation of state, p=nRT/V:
RT 
2
Mvrms
3
o An expression for T:
vrms 
T
3RT
M
2
Mvrms
3R
 This means the temperature of a gas T is proportional to velocity squared of
the molecules v2rms.
 This also means larger molecules move at lower speed at a given temperature.
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4. Internal Energy and Molar Specific Heats
o The internal energy of a gas, U, includes contributions from the following
motion of molecules
1. Translational,
2. Rotational,
3. Vibrational
o All molecules can store internal energy in forms of translational kinetic
energy.
o Large molecules, diatomic above, can store energy in forms of rotational and
vibration motions too.
o The rotational and vibrational motions of molecules are activated by collisions,
and therefore are coupled to the translational motion of the molecules.
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Internal Energy
 1. Translational Kinetic Energy
o The translational kinetic energy of a molecule with mass m moving
at velocity v at any instant is:
εtrans = 1/2mv2
o Because the speed of a particle can change when it collides with other
molecules, its average translational kinetics energy over time is :
 trans ,mean  (1 / 2mv 2 ) mean
Because
vrms 
3RT
M
M/m =NA
R/NA = kB
2
 1/ 2m(v 2 ) mean  1 / 2mv rms
1
2
 trans ,mean  ( m )
 trans ,mean 
3RT
M
3
k BT
2
 At a given temperature T, all ideal gas molecules - no matter what their mass is 12
have the same average translational kinetic energy, namely 3/2kBT.
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Internal Energy
 2,3. Rotational and Vibrational Energy
o There is no simple treatment for these energies. However, for these modes of
motion there is a shortcut which simplifies the treatment.
o The branch of physics known as “statistical mechanics” has shown that
For a large number of particles obeying the laws of Newtonian mechanics,
the available energy is, on average, shared equally by each
independent degree of freedom (f).
o A generalization of this result, known as the theory of equipartition of energy
states that:
Each degree of freedom contributes ½kBT to the energy of the system,
where possible degrees of freedom are those associated with translation,
rotation, and vibration of molecules.
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Internal Energy
 1,2,3. Internal Energy
o The internal energy of a system with degree of freedom f is:
f
f
f
 mean  k BT
U  ( nN A ) mean  (nN A ) k BT  nRT
2
2
2
o The change in internal energy U of the system from T1 to T2 therefore is:
f
U  nRT
2
for n mole
o The thermodynamic definition for U is (by definition Cv 
U  nCv T
1 dU
):
n dT
 Then one can conclude from the comparison:
Cv 
f
R
2
C p  Cv  R 

For ideal gases
f
f
R  R  (1 ) R
2
2
C p (2  f )

Cv
f
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Degree of Freedom
f for Monoatomic Gases
o Monoatomic gases can only translate.
o For monoatomic gases which have a degree of freedom of f = 3 (moving xyz
direction) :
f=1
We already know:
x axis
 trans ,mean  1 m( v 2 ) mean  3 k B T
2
f=1
2
y axis
(v 2x ) mean  (v 2y ) mean  (v z2 ) mean  1 (v 2 ) mean
3
f=1
z axis
1
1
m(vx2 ) mean  k BT
2
2
1
1
2
m(v y ) mean  k BT
2
2
1
1
m(vz2 ) mean  k BT
2
2
1
3
m ( v 2 ) mean  k BT
2
2
o Therefore we can know verify that each translational degree of freedom contributes
an equal amount of energy, ½kBT, to the gas (number of molecules N =nNA).
Translational (f=3)
1
3
3
U  3 N ( k BT )  nN ( k T )  nRT
2
2
2
A
∆U=nCv∆T
Cv 
B
3
R  12.2 J/mol.K
2
3
 U  nRT
2
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Degree of Freedom
f for Diatomic Gases
o Let’s consider a diatomic gas whose molecules have the shape of a dumbbell.
In this model, the center of the mass of molecules can translate in the x, y,
and z directions.
o In addition the molecule can rotate about three mutually perpendicular axes,
but one of them (y) can be neglected. So U is:
translational
rotational
1
1
5
5
U  3N ( k BT )  2 N ( k BT )  N ( k BT )  nRT
2
2
2
2
Cv 
1 dU
1 d 5
5

( nRT )  R  20.8 J/mol.K
n dT
n dT 2
2
C p  Cv  R 
7
R  29.1 J/mol.K
2

Cp
Cv

(7 / 2) 7
  1.40
(5 / 2) 5
 These results surprisingly agree quite well with most of the data for diatomic
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molecules without vibration contribution being taken into account.
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Degree of Freedom
o If the model of vibration / oscillation can be assumed by the two atoms joined
by an imaginary spring, then vibration adds two more degrees of freedom,
which correspond to:
the kinetics energy and
potential energy.
So
translational
rotational
vibrational
1
1
1
U  3 N ( k BT )  2 N ( k BT )  2 N ( k BT )  N ( 7 k BT )  7 nRT
2
2
2
2
2
Cv 
1 dU 1 d 7
7

( nRT )  R  29.1 J/mol.K
n dT n dT 2
2
 This value is inconsistent with experimental data for molecules such as H2,
O2 and N2 at room temperature. But it becomes more consistent with
increasing temperature.
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Degree of Freedom
Hydrogen
CV/R -T graph (diatomic) hydrogen shows:
o There are temperature thresholds above which these contributions can begin.
o Because rotational and oscillatory motions begin at certain energies, only
translation is possible at very low temperatures.
o As the temperature increases, rotational motion can begin. At still higher
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temperatures, oscillatory motion can begin.
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Degree of Freedom
If f is the number of degrees of freedom, then a general expression for CV is:
f 
Cv    R  4.16 f J/mol K
2
Note: For polyatomic molecules, the kinetic prediction of Cv is far from real values.
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Question 1. Calculation of Vrms
The speed of 22 particles are as follows. What are a) vmean, and b)vrms?
Ni
vi (cm/s)
2
1.0
4
2.0
6
3.0
8
5.0
2
6.0
Solution
a) The average speed is:
v mean 
 n v  2(1.0)  4(2.0)  6(3.0)  8(5.0)  2(6.0) cm s -1  3.6 cm s -1
2 4 68 2
n
i i
i
b) The rms speed is:
v rms 
v rms 
n v / n
2
i
2(1.0)
i
2
i

 4( 2.0) 2  6(3.0) 2  8(5.0) 2  2(6.0) 2
cm s -1  4.0 cm s -1
2 468 2
 The rms speed is more favoured towards large velocities and is greater than v20mean.
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Question 2. Vrms of air (oxygen) molecules
What is the vrms of oxygen at T =300 K?
Solution
vrms 
v rms 
3RT
M
3  (8.3Jmol -1K 1 )(300K )
0.032kgmol 1
 483ms 1
 So molecules such as air molecules move at the speed of
500 m/s = 0.5 km/s = 1800 km/h.
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Question 3. The speed of H2 molecules in outer space
The lowest possible temperature in outer space is 2.7 K. What is the rms speed,
average speed and most probable speed of hydrogen molecules at this temperature?
Solution
vrms 
3RT
M
v rms 
vmean 
vmp 
v mean 
8RT
M
3(8.31 J/mol.K)(2.7 K)
2.02  10-3 kg/mol
v mp 
2 RT
M
 1.8  10 2 m/s
8 RT
8(8.31 J/mol.K)(2.7 K)

 1.7  10 2 m/s
M
 (2.02  10 -3 kg/mol)
2 RT
2(8.31 J/mol.K)(2.7 K)

 1.5  10 2 m/s
M
2.02  10 -3 kg/mol
 So even at outer space (low T) molecules move relatively fast (170 m/s).
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Question 4. Calculation of kinetics energy of gases
a)
Determine the average value of the translational kinetic energy of the
molecules of an ideal gas at 20oC.
b) What is the translational kinetic energy per mole of an ideal gas ?
Solution
a) The translational kinetic energy of a molecule (εtrans) is:
 trans,mean 
3
3
k BT  (1.38 10  23 J/K)( 293 K )  6.07 10 - 21 J
2
2
 So the energy content (internal energy) of individual molecules is
not significant.
b) The total translational kinetic energy Etrans (per mole) is:
E trans per mole  N A trans  (6.02  1023 )(6.07 10 21 J)  3.65  103 J
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End of Lecture
Main references used for this lecture are:
“Atkin’s Physical Chemistry”, P. Atkins and J. de Paula, 10th Ed., Oxford University Press. Chapter 1.
“Principles of Physics”, J. Walker, D. Halliday., and R. Resnick, Ed. 20, Chapter 19.
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