Particle in a box
Elastic Collisions Between Gas Particles and a Wall
Consider the elastic collision between two bodies of masses m and M moving
along a straight line 1, 2. The velocities of these bodies before the collision
are vm and vM respectively, and after the collision they
are vm' and vM'. The law of
conservation of linear momentum states that:
mvm +MvM = mvm' + MvM'
and the law of conservation of energy states that:
mvm2 +MvM2 =
mvm'2 + MvM'2
From these two laws it is found that:
vm - vM = vM' - vm'
Namely, the magnitude of the relative velocity is conserved during the collision,
while the direction is reversed. It is also found from eqns. (1) and (2) that:
(1 + m/M) vM' – (1 – m/M)
vM = (m/M) 2vm
(1 + m/M) vm' – (1 – m/M)
vm = (m/M) 2vM
Now, assume that m is a microscopic body while M is macroscopic.
Thus, in the limit m/M = 0, it is found that:
vM' = vM
vm' = -(vm – 2vM)
As should he expected, the velocity of the macroscopic body is
not affected by the collision, while the magnitude of the velocity
of the microscopic body is changed by the amount 2vM and its direction
is reversed. If the velocity of the macroscopic body is much slower
than that of the microscopic one, vM << vm, then the amount of kinetic
energy (ΔEk) which is transferred upon a collision will be:
ΔEk = 2mvmvM
This amount is proportional to the velocity of the macroscopic body and
is independent of its mass. Energy is transferred from the microscopic
body to the macroscopic one when they are moving in the same direction,
and it is transferred in the reverse direction when they are moving
towards each other.
These results will now be applied to analyzing the properties of a gas
confined in a box. For simplicity, it is assumed that the gas is
monoatomic (thus it has no internal degrees of freedom) and that its
particles do not interact with each other.
The gas pressure
The calculation of the pressure of the gas on the walls is well
known and will be repeated briefly.
The atoms which may collide with a wall surface of area A in a time
interval Δt, are contained in the volume specified by base area A and
height vΔt, where v is their average velocity. Since they are free to move
in all directions, only 1/6 of them will actually strike the wall.
Therefore, the average number of collisions with the wall during the
interval Δt is:
N is the number of atoms in the box and V is its volume. For each
collision a momentum 2mv is given to the wall and, therefore, the
overall momentum transferred to the wall during Δt is:
FΔt = (1/3) (N/V) Av2Δt
So, it is found that the pressure of the gas on the wall is:
p = F/A = (1/3) (N/V) v2
The state equation of an ideal gas pV = NkT, may he derived from eqn. (11)
by using the thermodynamic relationship εk = (1/2)kT, where
εk is the
average kinetic energy per degree of freedom of the gas particle.
An exact calculation which takes the real distribution of the velocities
into account, yields the same results.
If the wall A is moving with velocity vM, then every atom that
collides with its surface, will transfer to it the kinetic energy 2mvvM
(eqn. (8)). The overall energy which is transferred to the wall during Δt is
(by eqn. (9) and eqn. (11)):
(1/3) (N/V) mv2AvMΔt = pdV
where dV = AvMΔt is the change in the volume of the box. (A similar
calculation in three dimensions is found in ref. (1).)
It is seen by this calculation that the process of elastic collisions,
between the atoms of the gas and the moving wall, is the mechanism by which
the thermal kinetic energy of the atoms is transformed into macroscopic
mechanical work. The cumulative effect of many collisions causes the
pushing of the wall, but during each single collision the transfer of
energy is determined only by the wall instantaneous movement. These
conclusions are results of the laws of conservation of momentum and energy.
Adiabatic Expansion of a Gas
When a gas is heated (or cooled) through a wall, atoms which strike its
surface will leave it with a higher (or lower) average velocity. In this
case the velocity of the atom after the collision does not depend wholly on
its velocity before the collision, in contrast to the process of elastic
collisions, where the relationship between these velocities is unique
(eqn. (7)). Therefore, elastic collisions cannot transfer heat between
the gas and the walls, and if a process consists of only such
collisions, it will be adiabatic.
The equations that describe adiabatic expansion of a gas are derived as follows:
It is seen from eqn. (9) that the probability that a single atom will
collide with the wall during the time interval Δt is (l/6)(l/V)AvΔt.
For each such a collision the atom loses (or gains) the velocity 2vM (eqn. (1)),
so the change of velocity, dv, during Δt is:
dv = -(v/3)(vMAΔt/V) = -(v/3)(dV/V)
By integrating this equation, it is found that:
v/v0 = (V0/V)1/3
or by the use of eqn. (11):
p/p0 = (V0/V)5/3
This is the equation of adiabatic expansion of a monoatomic gas.
By use of the ideal gas equation, pV = NkT, it can be written also in the
form: T/T0 = (V0/V)2/3, or,
p/p0 = (T/T0)5/2
The calculation can be extended to molecular gases by taking into account transfer of energy
between the translational and internal degrees of freedom of the molecules.
The quantum mechanical calculation of adiabatic expansion of a quantum gas
is direct 3, and is based on the "adiabatic theorem" 4, which states that
under certain conditions a slow and continuous change in the parameters of
a system causes a similar continuous change in the wave functions and energy
levels, and it does not induce transitions between different states.
In order to demonstrate the similarity between the classical and quantum
mechanical cases, the adiabatic theorem will be verified for a one dimensional
infinite square well potential 5.
The ground state of a particle in an infinite square potential,
which extends from x = 0 to x = L, can be written as combination of two
travelling waves in opposite directions:
ψ(x, t) = (i/(2L)1/2)(e-i(kx + wt)
where k = π/L and w = hp2/2mL2. Now,
if the boundary at x = L is moving
slowly with velocity vM, then when the waves are reflected from it, their
wave number (and frequency) will be shifted according to the Doppler effect:
k' = k (1 – 2vM/c)
where c = w/k is the velocity of the waves.
During the time interval Δt each wave is reflected cΔt/2L
times by the moving boundary, so the overall shift of the wave number will be:
Δk = -(kvMΔt/L) = -πΔL/L2
If the square well expands from L1 to L2, then by eqn. (18)
the wave number will be shifted from π/L1 to π/L2.
Therefore, the ground state function at L1, will he transformed into the ground
state function at L2. (By similar calculation,
this result is valid also for the higher states.) In this case the condition for
adiabatic process is that in a single reflection the Doppler shift of the
wavenumber (or frequency) will be small compared to the difference between
- Sears. F. W., and Salinger, G. L., "Thermodynamics. Kinetic Theory and Statistical
Thermodynamics", 3rd Ed., Addison-Wesley, Reading, Massachusetts, 1975, p. 262.
- Lachish. U., "Derivation of Some Basic Properties of Ideal Gases and Solutions from Processes of Elastic Collisions", J. Chem. Ed., vol. 55 (6), p. 369-371 (1978)
- Tolman. R. C., "The Principles of Statistical Mechanics", Oxford University Press, New York. 1938, p. 386.
- Shiff, L. I., "Quantum Mechanic", 3rd Ed., McGraw-Hill, New York, 1968, p. 289.
- Born, M., "Atomic Physics", 7th Ed., Blackie. London. 1965, pp. 12-24.
On the net: January 2021
Particles in a box
The Sun and the Moon a Riddle in the Sky
Marble the Uniform Earth Image