## Energy Distribution

### Task number: 3944

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

\[\rho(v)=4\pi\left({\frac{m}{2\pi kT}}\right)^{\frac{3}{2}}v^{2}e^{-{\frac{mv^2}{2kT}}},\]*ρ*(*v*) for velocity given bywhere

*m*is mass,*T*thermodynamic temperature and*k*Boltzmann constant.The probability of finding a particle in an elementary velocity interval (

*v*;*v*+ d*v*) is*ρ(v)*d*v*. The probability of finding a particle in elementary energy interval (*E*;*E*+ d*E*) is given by*ρ(E)*d*E*.We know the relationship between energy

\[E=\frac{1}{2}mv^{2}.\]*E*and velocity*v*From here we can first determine velocity using energy and by differention we obtain a relationship between d

*v*and d*E*. We substitute both to the above mentioned distribution function.#### Hint 3

The most probable energy

*E*_{max}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 Kgas temperature *E*_{max}= ?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

\[p(E_{1}\,<\,E\,<\,E_{2})\,=\,\int_{E_{1}}^{E_{2}}{\rho(E)}\,dE.\]*ρ*(*E*) with such attribute, that the probability of finding a particle in the energy interval from*E*_{1}to*E*_{2}is given byDuring this we use the known distribution function for velocity

\[\rho(v)=4\pi\left({\frac{m}{2\pi kT}}\right)^{\frac{3}{2}}v^{2}e^{-{\frac{mv^2}{2kT}}},\]*ρ*(*v*) given bywhere

*m*is mass,*T*thermodynamic temperature and*k*Boltzmann constant.The probability of finding a particle in an elementary velocity interval (

*v*;*v*+ d*v*) is*ρ(v)*d*v*. This corresponds to the probability of finding the particle in the elementary energy interval (*E*;*E*+ d*E*) given by*ρ(E)*d*E*.Between energy

\[E=\frac{1}{2}mv^{2}.\]*E*and velocity*v*there is a known relationshipWe 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

\[\rho(E)\mathrm{d}E\,=\,4\pi\sqrt{\frac{m}{2}}\left(\frac{1}{\pi kT} \right)^{\frac{3}{2}}Ee^{\frac{-E}{kT}}\frac{\mathrm{d}E}{\sqrt{E}\sqrt{2m}}\]*v*and its elementary change d*v*we obtainand after adjustment

\[\rho(E)=2\pi\left(\frac{1}{\pi kT}\right)^{\frac{3}{2}}\sqrt{E}e^{\frac{-E}{kT}}.\]We have managed to find a distribution function for energy. The most probable value of energy

\[\frac{\mathrm{d}\rho(E)}{\mathrm{d}E}\,=\,2\pi \left(\frac{1}{\pi kT}\right)^{\frac{3}{2}}\left(\frac{1}{2\sqrt{E}}e^{\frac{-E}{kT}}+\sqrt{E}\left(\frac{-1}{kT}\right)e^{\frac{-E}{kT}}\right).\]*E*_{max}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):We adjust the derivative

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

\[\left(\frac{1}{2\sqrt{E_{max}}}-\frac{\sqrt{E_{max}}}{kT}\right)=0.\]After adjusting this equation

\[\frac{1}{2\sqrt{E_{max}}}=\frac{\sqrt{E_{max}}}{kT}\]and determining the most probable energy value

\[E_{max}=\frac{kT}{2}.\]*E*_{max}we finally obtainNotice that the wrong solution mentioned in

*Hint 1*led to twice as big a value of the most probable velocity.#### Numerical Solution

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

The most probable energy value is approximately 2·10

^{− 21}J.