## Energy Distribution

Determine the most probable kinetic energy of a translational motion of ideal gas particles at temperature of 290.15 K.

• #### Hint 1 – Beware of Wrong Solution

This task is deceptive because it takes an easy path to a seemingly logical solution that is unfortunately wrong. The problem is that we know the relation for the most probable velocity

$v_{max}=\sqrt{\frac{2kT}{m}}$

and we could anticipate that the maximum value of energy distribution will be the same as the maximum value of velocity distribution and therefore the following will apply:

$E_{max}=\frac{1}{2}mv^{2}_{max}=\frac{1}{2}m \frac{2kT}{m}=kT.$

However, this reasoning is wrong! It is wrong because the energy distribution is in different form than the velocity distribution and its maxima do not match.

• #### Hint 2

First find the distribution function ρ(E) for energy.

Use a distribution function ρ(v) for velocity given by

$\rho(v)=4\pi\left({\frac{m}{2\pi kT}}\right)^{\frac{3}{2}}v^{2}e^{-{\frac{mv^2}{2kT}}},$

where m is mass, T thermodynamic temperature and k Boltzmann constant.

The probability of finding a particle in an elementary velocity interval (v;v + dv) is ρ(v)dv. The probability of finding a particle in elementary energy interval (E;E + dE) is given by ρ(E)dE.

We know the relationship between energy E and velocity v

$E=\frac{1}{2}mv^{2}.$

From here we can first determine velocity using energy and by differention we obtain a relationship between dv and dE. We substitute both to the above mentioned distribution function.

• #### Hint 3

The most probable energy Emax is defined as the maximum value of determined distribution function ρ(E).

How do you find a maximum value of a function?

• #### Given Values

 T = 290.15 K gas temperature Emax = ? most probable kinetic energy value

Table values:

 k = 1.38 ·10−23 J K−1 Boltzmann constant
• #### Analysis

First we find the distribution function for energy. While doing so we use the distribution function for velocity and a known relationship between energy and velocity.

Then we will determine the most probable energy value as the maximum value of distribution function for energy. This means that we will make a derivative of this function and make it equal to zero.

• #### Solution

First we need to find the distribution function for energy, i.e. the function ρ(E) with such attribute, that the probability of finding a particle in the energy interval from E1 to E2 is given by

$p(E_{1}\,<\,E\,<\,E_{2})\,=\,\int_{E_{1}}^{E_{2}}{\rho(E)}\,dE.$

During this we use the known distribution function for velocity ρ(v) given by

$\rho(v)=4\pi\left({\frac{m}{2\pi kT}}\right)^{\frac{3}{2}}v^{2}e^{-{\frac{mv^2}{2kT}}},$

where m is mass, T thermodynamic temperature and k Boltzmann constant.

The probability of finding a particle in an elementary velocity interval (v;v + dv) is ρ(v)dv. This corresponds to the probability of finding the particle in the elementary energy interval (E;E + dE) given by ρ(E)dE.

Between energy E and velocity v there is a known relationship

$E=\frac{1}{2}mv^{2}.$

We differentiate it

$\mathrm{d}E=mv\mathrm{d}v.$

and determine the elementary change of velocity

$\mathrm{d}v=\frac{\mathrm{d}E}{mv},$

and after substituting

$v=\sqrt{\frac{2E}{m}}$

we obtain

$\mathrm{d}v=\frac{\mathrm{d}E}{m\sqrt{\frac{2E}{m}}}=\frac{\mathrm{d}E}{\sqrt{E}\sqrt{2m}}.$

Now we can write

$\rho(v)\mathrm{d}v\,=\,4\pi\left(\frac{m}{2\pi kt}\right)^{\frac{3}{2}}v^{2}e^{-\frac{mv^{2}}{2kT}}\mathrm{d}v.$

After substituting the relations for velocity v and its elementary change dv we obtain

$\rho(E)\mathrm{d}E\,=\,4\pi\sqrt{\frac{m}{2}}\left(\frac{1}{\pi kT} \right)^{\frac{3}{2}}Ee^{\frac{-E}{kT}}\frac{\mathrm{d}E}{\sqrt{E}\sqrt{2m}}$

$\rho(E)=2\pi\left(\frac{1}{\pi kT}\right)^{\frac{3}{2}}\sqrt{E}e^{\frac{-E}{kT}}.$

We have managed to find a distribution function for energy. The most probable value of energy Emax can be determined as a maximum value of this function. This is done by making a derivative with respect to energy E (we use the rule for making a derivative of product and a composite function):

$\frac{\mathrm{d}\rho(E)}{\mathrm{d}E}\,=\,2\pi \left(\frac{1}{\pi kT}\right)^{\frac{3}{2}}\left(\frac{1}{2\sqrt{E}}e^{\frac{-E}{kT}}+\sqrt{E}\left(\frac{-1}{kT}\right)e^{\frac{-E}{kT}}\right).$

$\frac{\mathrm{d}\rho(E)}{\mathrm{d}E}\,=\,2\pi\left(\frac{1}{\pi kT}\right)^{\frac{3}{2}}e^{\frac{-E}{kT}}\left(\frac{1}{2\sqrt{E}}-\frac{\sqrt{E}}{kt}\right)$

and make it equal to zero

$\frac{\mathrm{d}\rho(E)}{\mathrm{d}E}=0\;\;\;\Rightarrow \;\;\;2\pi\left(\frac{1}{\pi kT}\right)^{\frac{3}{2}}e^{\frac{-E}{kT}}\left(\frac{1}{2\sqrt{E}}-\frac{\sqrt{E}}{kt}\right)=0.$

From here we obtain a condition

$\left(\frac{1}{2\sqrt{E_{max}}}-\frac{\sqrt{E_{max}}}{kT}\right)=0.$

$\frac{1}{2\sqrt{E_{max}}}=\frac{\sqrt{E_{max}}}{kT}$

and determining the most probable energy value Emax we finally obtain

$E_{max}=\frac{kT}{2}.$

Notice that the wrong solution mentioned in Hint 1 led to twice as big a value of the most probable velocity.

• #### Numerical Solution

$E_{max}=\frac{kT}{2}=\frac{1.34\cdot{10^{-23}}\cdot{290.15}}{2}\,\mathrm{J}\dot{=}2\cdot{10^{-21}}\mathrm{J}$   