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Particles in a box

Particles in a Box

Uri Lachish, Rehovoth

urila@zahav.net.il

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' (1)

and the law of conservation of energy states that:

mvm2 +MvM2 = mvm'2 + MvM'2 (2)

From these two laws it is found that:

vm - vM = vM' - vm' (3)

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 (4)

(1 + m/M) vm' – (1 – m/M) vm = (m/M) 2vM (5)

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 (6)

vm' = -(vm – 2vM) (7)

As should he expected, the velocity of the macroscopic body is not

affected by the collision, while the magnitude of the velocity of the

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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 (8)

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:

(1/6)(N/V)AvΔt (9)

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 (10)

So, it is found that the pressure of the gas on the wall is:

p = F/A = (1/3) (N/V) Av2 (11)

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

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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 (12)

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) (13)

By integrating this equation, it is found that:

v/v0 = (V0/V)1/3 (14)

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or by the use of eqn. (11):

p/p0 = (V0/V)5/3 (15)

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.

Quantum expansion

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) – ei(kx-wt)) (16)

where k = /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) (17)

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 (18)

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

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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 successive states.

References

1. Sears. F. W., and Salinger, G. L., "Thermodynamics. Kinetic

Theory and Statistical Thermodynamics., "3rd Ed., Addison-

Wesley, Reading, Massachusetts, 1975, p. 262.

2. 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)

3. Tolman. R. C., "The Principles of Statistical Mechanics", Oxford

University Press, New York. 1938, p. 386.

4. Shiff, L. I., "Quantum Mechanic, "3rd Ed., McGraw-Hill, New

York, 1968, p. 289.

5. Born, M., "Atomic Physics." 7th Ed., Blackie. London. 1965, pp.

12-24.

On the net: January 2021

Particles in a box DOI: 10.13140/RG.2.2.18559.12960

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