Download The kinetic theory of the gases

THE KINETIC THEORY OF GASES
Mono-atomic
Fig.1
1
2
3
Average kinetic
energy of a
single particle
Fig.2
INTERNAL ENERGY U and EQUATION OF STATE
For a mono-atomic gas, we will assume that the total energy U is equal to the
number of atoms multiplied by the average kinetic energy of each.
(We are disregarding any possibility of excitation or motion inside the atoms
itself. The kinetic energy of the center of mass motion is all the energy there is
in the atom.)
U = N <½ m v2>
(2)
The kinetic theory of gases thus gives as an equation of state for the gas
PV = (2/3) U
To gain some generality, we express the expression above as
PV = (γ - 1) U
(γ = 5/3 for the case of a mono-atomic gas)
For non mono-atomic gases γ is different than 5/3.
(3)
Once the equation of state of a system is known, different
thermodynamic processes can be studies in detail. Here let’s
consider an adiabatic process
ADIABATIC PROCESS Q = 0
As an application of the expression above for the equation of
state, let’s consider an adiabatic process, one in which the gas
container is thermally well isolated, i.e. no heat-transfer occurs.
No heat-transfer is added or removed to the gas, Q = 0
As the piston is pressed, work is done on the gas and the internal
energy increases. We would like to predict the changes in the gas
volume as a consequence of changes in pressure.
Q=0
Insulation
Pressure
P
F External
force
Fig. 3 Gas isolated from heat transfer.
 Using the first law of thermodynamics, dU = Q - W, we
obtain

dU = - W
When the volume changes by dV, the work done by the gas W
is PdV. Therefore,
dU = - (P dV )
(4)
 On the other hand, according to the equation of state
PV= (-1)U, which gives,
dU = d [PV / (-1) ]
= [ 1/ (-1) ] [P dV + V dP ]
(5)
 From (4) and (5),
- (P dV ) = [1/ (-1)] [P dV + V dP ]
- (-1) (P dV ) = P dV + V dP
-  (P dV ) = VdP
Dividing by PV
-  (dV) /V = (dP) / P
-  lnV + const = ln P
const =  ln V + ln P
= ln (P V 
This implies,
P V = Const (adiabatic process) (6)
( = 5/3 for the case of a mono-atomic gas.)
Q=0
Insulation
P
Pressure
F
adiabatic
process
P
V
Fig. 4 Gas undergoing an adiabatic process.
V
What happens at THERMAL EQUILIBRIUM
conditions
Case: Two different gases in the same container
Consider a gas mixture of two types of molecules, 1 and 2,
If molecules of type-1 were standing still, they would be speed by
their collision with the faster moving molecules of type-2. If the
type-1 molecules were to pick up too much speed, then, again
through collisions, they would pass the energy back to the
molecules of type-1. The question is to find the rule that
determines the relative speed of the two types of molecules.
m1
Insulation
m2
Fig. 5 A mixture of molecules of different masses.
As demonstrated in the appendix to these notes, the rule is,
2
< ½ m1v1 > = < ½
m2 v22 >
(7)
(the heavier atoms move slower than the lighter ones.)
Case: Two different gases in different containers
Figure 6 shows two different gases, initially occupying equal
volumes. A movable piston separate the gases. What is the relative
speed of the molecules once a thermal equilibrium is reached?
Insulation
1
2
Movable
piston
Fig. 6
 Initially, we could argue that the final equilibrium state is
reached when the pressure at both side is the same.
According to expression (1), P = (2/3) (N/V) <½ mv2>; thus a
condition of MECHANICAL EQUILIBRIUM, would imply,
(2/3) (N1/V1) <½ m1v12> = (2/3) (N2/V2) <½ m2v22>
It turns out, as argued in the Feynman lectures Vol-1, chapter
39), this condition is not sufficient. Something else must
happen.
 The condition that must be satisfied is,
<½ m1v12> = <½ m2v22>
In the two cases mentioned above, the mean kinetic
energy of the molecules turns out to be a property of
the thermal equilibrium condition, and not of the
particular gas.!!!
That is, the notion that we had about the gas reaching a common
“temperature” after letting them to interact for awhile mean that
the gas reaches a situation in which the average kinetic energy per
molecule is the same.
Definition of temperature
We have found that the average kinetic energy is a property of
the “temperature” and not of the particular gases used. This
suggests using the average kinetic energy itself as the temperature.
Indeed, the simplest approach could be to call the average energy
itself “the temperature”. The higher the average kinetic energy the
higher the temperature. This would constitute our absolute
temperature scale.
[This possibility is not at all disconnected form the discussion of
defining an absolute temperature scale within the context of
discussing the concept of entropy, as will be seen in the next
chapter.]
Unfortunately, the scale of temperature has been chosen
differently. A constant conversion factor k is introduced between
the average kinetic energy of a molecule < ½ mv2 > and the , thus
defined, absolute temperature scale T.
< ½ mv2 > ≡ 3/2 k T
Definition of
absolute temperature T
(8)
The unit of the absolute temperature T is denominated the Kelvin
(K)
k = 1.38 x 10-23 J/K
Boltzmann's constant
[ T(absolute) = T(in degrees Celcius) + 273.15 = TC + 273.15
TF = (9/5) TC + 32
]
The Ideal Gas Law
Now we can put the new definition of temperature into Eq (1).
The expression PV = (2/3) U = (2/3) N < ½ m v2 > becomes,
PV = N k T
(9)
This result indicates, equal volumes of different gases, at the same
pressure and temperature, have the same number of molecules,
because of Newton’s laws.
In terms of moles n = N/No (where No≡ 6.02 x 1023), expression (9)
become, PV = n No k T . Making No k = 8.317 Jmol-1K-1=R, one
obtains,
PV = n R T
(10)
Question: Is expression (9) still valid when considering diatomic molecules?
First, let’s put expression (9) again in terms of the average kinetic
energy per molecule,
PV = (2/3) U = (2/3) N < ½ m v2 > (mono-atomic gas)
Notice that, during its derivation we were invoking the collision of
mono-atomic molecules of mass m that and translation motion of
speed v against a wall.
If the molecules were, for example, diatomic, what would its
internal motion (i.e. vibrational oscillations between the two
atoms) affect its translational motion? Apparently none, since all
what matter is the net impact against the walls of the container
(what the molecule does in between collisions does not affect the
pressure the molecules exert on the walls. That pressure results
only from the collisions of the molecules against the walls.)
Since the translational motion of the molecules is characterized by
the velocity of its center of mass vCM, if we repeat the derivation of
the kinetic theory of the gases, we will obtain,
PV = (2/3) N < ½ M vCM2 > (di-atomic gas)
where M is the mass of the di-atomic gas.)
(11)
An extra step (argued in the Feynman Lectures on Physics, Vol-1,
Chapter 39) lead to the following result.
< ½ M vCM2 > = (3/2) kT
(12)
Hence, from (12) and (11),
PV = N k T (gas composed of
di-atomic molecules)
(13)
Internal energy of a gas composed of di-atomic
molecules
m1
m2
We have learned that the
Condition of thermal
equilibrium implies
<½ m1v12> = <½ m2v22>
In this case, the atoms are
joined
What is the condition of
thermal equilibrium in
this case?
Answer:
<½ m1v12> = <½ m2v22>
That is, whether or not are forces involved, the average kinetic
energy of an atom is the same (it depends only on the temperature.)
Summary (diatomic molecules)
 One consideration
<½ m1v12> = <½ m2v22> = (3/2) kT
Total kinetic energy =
<½ m1v12> + <½ m2v22> = 3 kT
 Alternative consideration
rCM
KCM = < ½ M vCM2 > = (3/2) kT
Total kinetic energy =
KCM + Kinternal-motion
Since the total kinetic energy = 3kT then the above two results
imply,
Kinternal-motion = (3/2) kT

Internal motion
Kinternal-motion = 2 ( ½ k T) + 1 ( ½ k T) = (3/2) kT