User:Inconspicuum/Physics (A Level)/Boltzmann Factor

Particles in a gas lose and gain energy at random due to collisions with each other. On average, over a large number of particles, the proportion of particles which have at least a certain amount of energy &epsilon; is constant. This is known as the Boltzmann factor. It is a value between 0 and 1. The Boltzmann factor is given by the formula:

$$\frac{n}{n_0} = e^{\frac{-\epsilon}{kT}}$$,

where n is the number of particles with kinetic energy above an energy level &epsilon;, n0 is the total number of particles in the gas, T is the temperature of the gas (in kelvin) and k is the Boltzmann constant (1.38 x 10-23 JK-1).

This energy could be any sort of energy that a particle can have - it could be gravitational potential energy, or kinetic energy, for example.

Derivation
In the atmosphere, particles are pulled downwards by gravity. They gain and lose gravitational potential energy (mgh) due to collisions with each other. First, let's consider a small chunk of the atmosphere. It has horizontal cross-sectional area A, height dh, molecular density (the number of molecules per. unit volume) n and all the molecules have mass m. Let the number of particles in the chunk be N.

$$n = \frac{N}{V} = \frac{N}{A\; dh}$$

Therefore:

$$V = A\;dh$$ (which makes sense, if you think about it)

By definition:

$$N = nV = nA\; dh$$

The total mass &Sigma; m is the mass of one molecule (m) multiplied by the number of molecules (N):

$$\Sigma m = mN = mnA\; dh$$

Then work out the weight of the chunk:

$$W = g\Sigma m = nmgA\; dh$$

The downwards pressure P is force per. unit area, so:

$$P = \frac{W}{A} = \frac{nmgA\; dh}{A} = nmg\; dh$$

We know that, as we go up in the atmosphere, the pressure decreases. So, across our little chunk there is a difference in pressure dP given by:

$$dP = -nmg\; dh$$ (1) In other words, the pressure is decreasing (-) and it is the result of the weight of this little chunk of atmosphere.

We also know that:

$$PV = NkT$$

So:

$$P = \frac{NkT}{V}$$

But:

$$n = \frac{N}{V}$$

So, by substitution:

$$P = nkT$$

So, for our little chunk:

$$dP = kT\; dn$$ (2)

If we equate (1) and (2):

$$dP = -nmg\; dh = kT\; dn$$

Rearrange to get:

$$\frac{dn}{dh} = \frac{-nmg}{kT}$$

$$\frac{dh}{dn} = \frac{-kT}{nmg}$$

Integrate between the limits n0 and n:

$$h = \frac{-kT}{mg}\int_{n_0}^n\frac{1}{n}\;dn = \frac{-kT}{mg}\left [ \ln{n} \right ]_{n_0}^n = \frac{-kT}{mg}\left ( \ln{n} - \ln{n_0}\right ) = \frac{-kT}{mg}\ln{\frac{n}{n_0}}$$

$$ln{\frac{n}{n_0}} = \frac{-mgh}{kT}$$

$$\frac{n}{n_0} = e^{\frac{-mgh}{kT}}$$

Since we are dealing with gravitational potential energy, &epsilon; = mgh, so:

$$\frac{n}{n_0} = e^{\frac{-\epsilon }{kT}}$$

A Graph of this Function
This topic comes up in Q10 494 June 2010. The Values used for various things in that question are

$$k=1.4 \times 10^{-23} JK^{-1},~g = 9.8,~m=4.9 \times 10^{-26} Kg,~T=290K$$