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 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
\[\rho(E)\mathrm{d}E\,=\,4\pi\sqrt{\frac{m}{2}}\left(\frac{1}{\pi kT} \right)^{\frac{3}{2}}Ee^{\frac{-E}{kT}}\frac{\mathrm{d}E}{\sqrt{E}\sqrt{2m}}\]and 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 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):
\[\frac{\mathrm{d}\rho(E)}{\mathrm{d}E}\,=\,2\pi \left(\frac{1}{\pi kT}\right)^{\frac{3}{2}}\left(\frac{1}{2\sqrt{E}}e^{\frac{-E}{kT}}+\sqrt{E}\left(\frac{-1}{kT}\right)e^{\frac{-E}{kT}}\right).\]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 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
\[E_{max}=\frac{kT}{2}=\frac{1.34\cdot{10^{-23}}\cdot{290.15}}{2}\,\mathrm{J}\dot{=}2\cdot{10^{-21}}\mathrm{J}\]Answer
The most probable energy value is approximately 2·10− 21 J.
