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Inside an enclosure containing one gram molecule of a gas there are of the order of 10 molecules in interaction. In this context "the behaviour of a single molecule" is a meaningless concept. Yet we know, for example, that the energy of the molecules will not be the same for each of them; interaction means variation of energy, and some will have higher or lower energy than the average. What we can attempt to get at is . This is the problem of the .

Looked at from the point of view of the distribution of molecules over the range of possible energies, nothing seems to prevent the widest possible spread. There is no reason why there should be any gaps, i.e. a range of energies in which are no molecules.

On the other hand, if any particular molecule has momentarily reached a very low or very high (relatively speaking) state of energy, then there is about a hundred per cent probability that the next collision will bring ittowards the average state. The probability of producing very high or very low energy molecules is very small, and hence at any one time only a very small proportion of the molecules will be in the extreme energy ranges.

From these qualitative representations, we can form some anticipation of the general shape of the energy distribution (see fig. 8.1).

The next step is to find a which will enable us to produce a definite shape for the curve. A definite shape is vulnerable to the facts of experimentation i.e. it can be exposed as wrong -which the qualitative picture is not.

It is useless to set about applying Newton's laws to 10 interacting particles. It is out of the question. instead, we ignore the mechanics of interaction and look only at the of molecules in each "". The whole energy continuum is conceptually divided up into a series of very small "volumes" of energy; these are the energy cells.

To illustrate the notion of energy cells, we can take the case of particles with kinetic energy. The energies of these particles can be represented in a or a . At any moment, each particle will have a definite velocity and this can be represented as a . We can imagine a representation of the velocities of all particles in a three dimensional velocity space (see fig. 8.2).

All velocity vectors originate at the same point. If their terminations are represented by points and we ignore the lines, we end up with a "cloud" of dots which directly represents the velocity distribution of the particles. An energy cell will then be represented by some element of volume of the velocity If there are two molecules in one of the energy cells then there are the following possible arrangements: space. The number of points appearing in the element of volume is the number of particles in the corresponding energy cell. In the case of particles with kinetic energy, the volume elements chosen are those corresponding to the range to + . These are spherical shells about the origin since particles with the same magnitude of velocity have the same energy irrespective of direction (see fig. 8.3). Thus an energy cell in this case has the "volume" 4

Our task is to find the distribution of particles amongst the different energy cells. We have to derive a general expression for , the number of particles in the energy cell, which has a "volume" , and corresponds to the energy

i .

(i)Probability: To illustrate the meaning of probability before explaining it further, we take a simple case of three energy cells corresponding to energies

and three particles A, B and C.

If there is one molecule to each energy cell, there are the following possible arrangements:

And if there are two molecules in one of the energy cells then there are the following possible arrangements:

And if there are three molecules in one of the energy cells:

Now, if we imagine that the molecules change from energy cell to energy cell without rhyme or reason, then the first situation (with one molecule to each energy cell) is going to arise far more often than the other two. We say that the first situation . The greater the number of ways of arranging the molecules for a particular energy distribution, the more probable it is.

There is a general formula for calculating the number of arrangements, W:

N is the total number (of molecules, in this case), is the number in cell 1, is the number in cell 2, etc.; and is the "volume" of the energy cell

, etc. The intrinsic probability of one molecule being in a cell of volume g, is g i.e. it depends on the "volume" of the energy cell. You should notice that we always when dealing with combinations of probabilities and numbers of arrangements, so that in this case the probability of two molecules being in the energy cell with "volume" g is g x g or g, and so on.

In our simple example, g = g = g = 1. We put it equal to one because we are interested only in comparing probabilities. It is on the basis of this standard formula from statistical theory that calculations of energy distribution can be made.

(ii) : The number of particles remains constant (N) and the total energy remains constant (U). Thus we have two constraints on the situation, which can be expressed; the number of particles is constant:

n _i = N [8.2]

the sum of the energies of all particles equals the internal energy, which is constant:

n _i

i = U [8.3]

(iii)

We can apply these definitions and techniques to the simple case of space distribution. Only one of the two constraints is relevant:

= N. We picture a unit volume divided into cells of size , , etc. In these there are respectively, , , etc. molecules. The statistical formula [8.1] is the starting point. It gives us:

log W = log N! +

( n ₁ 1og g ₁ - log n₁! )

When is very large we can use the approximation log ! = (log - 1), and then:

log W = constant +

n i log( gi / ni )

If the space is continuous - so that it does not matter how small we make our energy cells - the maximum value of log W can be found by using the methods of calculus:

log( g / n ) = constant, or ( n / g ) = another constant.

The number of molecules in each cell is directly proportional to the size of the cell. In other words, we have a .

(iv)

We have the additional constraint that

n _i

_i = n ₁

₁ + n ₂

₂ + ... = U, a constant [8.3]

This additional constraint changes the result for the maximum probability, and the outcome is:

og ( g ₁ / n ₁ ) =

where

and

are both constants, and

is a function of the internal energy (which is constant). Thus:

( / ) = NA

where A is a constant, and N is the total number of particles.

There is a fall-off in the number of molecules with higher energies that follows an exponential law - rather like the fall-off of density in the atmosphere with height. So we can say that a given energy state

has a weighting of e - (

i ).

In Study Unit 5, we concerned ourselves with only the average kinetic energy of molecules. Now we can see something about the spread of energies. For an ideal gas, N is the total number of molecules and U is the sum of their kinetic energies. The energy

is then expressed as /; and , the size of an energy cell, is 4

(see section above). We can then write down (using equation [8.4]) an expression for the

Fig. 8.4 shows the exponential shape of the relation between the particle density ( / ), and the corresponding energy state

. Fig. 8.5 shows the velocity distribution for the molecules of an ideal gas.

In arriving at the expression for the velocity distribution, we have assumed that we can treat the available energy states as continuous with each other. We also assumed that there was no restriction on the number of molecules which could be in any one energy state at a time. However, in Study Unit 7 we learned that energy is ultimately quantized-that is, there will be energy levels available.

In fact, the number of available energy states is vastly greater than the number of molecules. Thus, in spite of quantization, we can ignore the discrete energy structure, and we can also ignore any consequences resulting from molecules being in the same energy state at the same time.

When particles do exist in the same energy state, then they come into interaction with each other in a special way. The interaction depends on the kind of particle involved-atom, electron, photon, etc. In the INVESTIGATION, we shall come across problems involving cases where this interaction becomes important and produces statistics which are .

There are enough equations for us to be able to work out what

is in terms of U, the internal energy of an ideal gas. The calculation is not given here, since only the result is of importance to us:

= ( (³/₂)(^N/_U) )

Now, from the kinetic theory of an ideal gas which was discussed in Study Unit 5, we know that:

U = ³/₂ NkT [8.6]

where k, Boltzmann's constant, is the ratio of the gas constant (R) to the number of molecules in one gram molecule of a gas (N). The we can see that:

= 1 / kT [8.7]

This result turns out to be universal. Any two systems in thermal equilibrium with each other will have the same value for

In the kinetic theory of an ideal gas, every molecule has degrees of freedom, corresponding to the three independent directions of motion in space. In the random interactions of the molecules of the gas, no one of these degrees of freedom can be favoured above any of the others, nor can any group of molecules be favoured above any other group. So the available energy is partitioned equally-over time-between the degrees of freedom and between all the molecules.

Consider the expression for the total energy of one gram molecule of an ideal gas (see Study Unit 5):

U = ³/₂ RT

This can be put in the form:

U = 3N( ¹/₂ )( ^R/_N )T

where N is the total number of molecules in one gram molecule-the same for every gas. Then, putting / = :

It would thus appear very convenient to take it that for any degree of freedom of any kind whatsoever for a particle, there should correspond an average energy /T.

We now know that thermal equilibrium corresponds to, (a) maximum entropy S of the system, and (b) maximum number of arrangements W corresponding to the energy distribution of its particles. Can these two be linked together in a quantitative way? They can - by means of a simple formula:

S = k log W [8.8]

The derivation of this formula involves techniques with which you might not be familiar, but it has such importance that we should know about it at this point.

The energy distribution of a few particles is subject to random fluctuations. The greater the number of particles, the more strongly one particular energy distribution manifests itself. The equilibrium state of maximum entropy then corresponds to the absence of any tendency to change. As the temperature increases, the number of arrangements corresponding to maximum entropy increases by leaps and bounds.