What is meant by Boltzmann statistics of ideal gas?

In statistical mechanics and thermodynamics, Maxwell - Boltzmann statistics explains the statistical distribution of the energies of the classic gas molecules over different states in thermal equilibrium. Maxwell Boltzmann statistics helps us to derive the Maxwell - Boltzmann distribution of ideal gas molecules, which is the result of the kinetic theory of gases.

Maxwell- Boltzmann distribution

In statistical mechanics, Maxwell - Boltzmann distribution is a probability distribution. It was first derived by Maxwell in 1870 and in 1880, Boltzmann completed the important investigations into the physical origins of the distribution. By maximizing the entropy of the system, the distribution is derived on the ground. First, it was used to describe the molecule speed in the ideal gas, where they move freely inside a container without interacting with each other. Also, assume that the system of molecules is in thermodynamical equilibrium. The energy of such molecule obeys Maxwell-Boltzmann statistics and the speed distribution is calculated by equating the molecule energies with the kinetic energy.

The lists of derivations are,

• Maximum entropy probability distribution, with some limitations in the conservation of energy.
• The canonical ensemble.

Maxwell-Boltzmann distribution is applied to all three-dimensional molecules velocities, but it depends only on the speed of the molecule. The kinetic theory of gases is applied to all the classical ideal gas molecules, which is a representation of real gases. In real gases, there are different types of effects that make their speed distribution different from the Maxwell-Boltzmann distribution. But, rarefied gases at normal temperature have characters similar to the ideal gas and for those gases, Maxwell -Boltzmann approximation suits well.

Maxwell - Boltzmann distribution function

Let us consider a system which contains large number of molecules, the fraction of the molecule with an infinitesimal element of 3-dimensional velocity space be, $d^{3}v$, which is placed on a vector whose magnitude v is $f\left(v\right) d^{3}v$.

$f\left(v\right) d^{3}v={\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{e}^{-\frac{m{v}^2}{2\pi kT}}{d}^{3}v$       (1)

Here,

m is the mass of the molecule.

k is the  Boltzmann’s constant.

T is thermodynamic temperature.

f(v) is the probability distribution function.

The velocity of space is expressed in cartesian co-ordeinate is,$d^{3}v=d{v}_x,d{v}_y,d{v}_z$.

In spherical coordinate system, the velocities is expressed as,$d^{3}v=v^{2}dv d\Omega$, where $d\Omega$ is the solid angle. 

The Maxwell distribution function for the molecule in only one direction, if it is in the x-direction, it is given by,

$f\left(v_x\right) d^3{v}_x={\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{e}^{-\frac{m{v_x}^2}{2\pi kT}}d{v}_x$

The above products is derived by integrating the three dimensional form given over $v_{y}and v_z.$

Considering the symmetry of f(v), it can be integrated over a solid angle, and their probability distribution function of speeds is given as,

$f\left(v\right) dv={\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4\pi v^2{e}^{-\frac{m{v}^2}{2\pi kT}}dv$       (2)

The probability density function is defined as, the probability per unit speed. It is given by,

$f_p=\sqrt{\frac{2}{\pi }}\frac{x^2{e}^{-{\displaystyle \raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$2a^2$}\right.}}}{a^3}$

This density function is similar to the Maxwell- Boltzmann distribution function but with a distribution parameter a given by :

$a=\sqrt{\frac{kT}{m}}$

The ordinary differential equation which is satisfied by this distribution is given as,

$$\begin{gathered} kTvf\text{'}\left(v\right)+f\left(v\right)(m{v}^2-2kT)=0 \\ f\left(1\right)=\sqrt{\frac{2}{\pi }}{e}^{-\frac{m}{2kT}}{\left(\frac{m}{kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$

Properties of Maxwell – Boltzmann distribution

The properties which are obtained from this distribution are,

• Most probable speed.
• Mean speed.
• Root mean square speed.

This property suits well for ideal gases, monoatomic gases such as helium, and also molecular gases such as oxygen. This is because in spite of their large heat capacity, they have unchanged degrees of freedom and translational kinetic energy.

Most probable speed

The most probable speed is the speed commonly possessed by all the molecules in the system and it corresponds to have the mode f(v). The probable speed is found by differentiating f(v) and equating it to zero, then finding the value v.

$$\begin{gathered} v_p=\frac{df\left(v\right)}{dv}=-8\pi {\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}v{e}^{-\frac{m{v}^2}{2kT}}\left(\frac{m{v}^2}{2kT}-1\right)=0 \\ \frac{m{v}_p^2}{2kT}=1 \\ {v}_p=\sqrt{\frac{2kT}{m}}=\sqrt{\frac{2RT}{M}} \end{gathered}$$

Where, R is the universal gas constant and M is the molar mass of the molecule. The molar mass is calculated as the product of molecule mass and the Avogadro number.$(m=M{N}_A)$

Mean speed

Mean speed (v) is the expected value of the speed distribution. From the equation(2), and substitute b=$\frac{1}{2a^2}=\frac{m}{2kT}$.

Therefore,

$$\begin{gathered} <v>={\int }_0^{\infty }vf\left(v\right)dv \\ <v>=4\pi {\left(\frac{b}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} {\int }_0^{\infty }{v}^3 e^{-b{v}^2}dv \\ <v>=4\pi {\left(\frac{b}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \frac{1}{2b^2} \\ <v>=\sqrt{\frac{4}{\pi b}} \end{gathered}$$

Now, substitute the value of b,

$$\begin{gathered} <v>=\sqrt{\frac{8kT}{m}}=\sqrt{\frac{8RT}{M}} \\ <v>=\frac{2}{\sqrt{\pi }}{v}_p \end{gathered}$$

Root mean square speed

It is the root mean square of the mean squared speed ($v^2$). It corresponds to the speed of the molecule with the kinetic energy.

$$\begin{gathered} <v>=\sqrt{<v^2>}={\left({\int }_0^{\infty }{v}^{2}f\left(v\right)dv\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ <v_{rms}>={\left(4\pi {\left(\frac{b}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} {\int }_0^{\infty }{v}^4 e^{-b{v}^2}dv \right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \\ <v_{rms}>=4\pi {\left(\frac{b}{\pi }\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \frac{3}{8}\sqrt{\frac{\pi }{b^5}} \\ <v_{rms}>=\sqrt{\frac{3}{\pi b}}=\sqrt{\frac{3kT}{m}}=\sqrt{\frac{3RT}{M}} \\ <v_{rms}>=\sqrt{\frac{3}{2}}{v}_p \end{gathered}$$

Maxwell-Boltzmann statistics

In thermodynamics, Maxwell - Boltzmann statistics explains the average number of molecules present in the given single-molecule microstate. Based on certain assumptions, it is concluded that the logarithm of the fraction of particles in a microstate is proportional to the ratio of the energy of that state to the temperature of the system.

The assumptions that are used while making this equation are, the molecules in the system do not interact and they are classical. That means each molecule state in the system is independent of the other molecule states. And also, the molecules are assumed to be in thermal equilibrium.

$-\log \left(\frac{N_i}{N}\right)\alpha \frac{E_i}{T}$

The above equation is rewritten by using a normalizing factor.

$\frac{N_i}{N}=\frac{exp\left(\raisebox{1ex}{$E_i$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.\right)}{{\displaystyle \sum _J^{}}exp\left(\raisebox{1ex}{$E_j$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.\right)}$       (3)

Where,

$N_i$ is the expected number of molecules in the single-particle microstate.

$E_i$ is the energy of the microstate.

N is the total number of molecules in the system.

k is the Boltzmann constant.

T is absolute temperature.

In equation (3), the denominator is the normalizing factor, so that the left-hand side term is added to unity. In other words, it is known as partition function (Z).

Distribution for momentum vector

In order to find the momentum vector distribution, consider the potential energy of the system as zero. Hence, all the energy is in the form of kinetic energy.

The relationship between the momentum and kinetic energy in non-relativistic molecules is,

$E=\frac{p^2}{2m}$

Where $p^2$ is the square of the momentum vector$p = [p_x, p_y, p_z].$

Hence, the equation (3) is given by,

$\frac{N_i}{N}=\frac{1}{Z}exp\left[\frac{p_{i,x}^2+p_{i,y}^2+p_{i,z}^2}{2mkT}\right]$

The factor $\frac{N_i}{N}$ is directly proportional to the probability density function $f_p$ for finding a molecule momentum components. Therefore,

$f_p(p_x,p_y,p_z) \alpha exp\left[\frac{p_x^2+p_y^2+p_z^2}{2mkT}\right]$

On integrating,

$\iint {\int }_0^{\infty } exp\left[-\frac{p_x^2+p_y^2+p_z^2}{2mkT}\right]d{p}_x,d{p}_y,d{p}_z={\left(\sqrt{\pi }\sqrt{2mkT}\right)}^3$

Hence, the normalized distribution function is,

$f_p(p_x,p_y,p_z) ={\left(\sqrt{\pi }\sqrt{2mkT}\right)}^{3}exp\left[\frac{p_x^2+p_y^2+p_z^2}{2mkT}\right]$

Distribution for energy

The energy distribution is given by,

$f_E\left(E\right)dE=f_p\left(p\right)d^{3}p$

Where $d^{3}p$ is infinitesimal phase volume of momenta, related to the energy interval dE. By using the spherical symmetry of the energy momentum dispersion relation, it can be written as,

$E=\frac{\left|p^2\right|}{2m}$

And the equation is rewritten as,

$$\begin{gathered} d^{3}p=4\pi {\left|p\right|}^{2}d\left|p\right| \\ {d}^{3}p=4\pi m\sqrt{2mE}dE \end{gathered}$$

Therefore,

$$\begin{gathered} f_E\left(E\right)dE=\frac{1}{{\left(2\pi mkT\right)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}}{e}^{-\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.}4\pi m\sqrt{2mE}dE \\ {f}_E\left(E\right)dE=2\sqrt{\frac{E}{\pi }}{\left(\frac{1}{kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{e}^{-\raisebox{1ex}{$E$}\!\left/ \!\raisebox{-1ex}{$kT$}\right.} \end{gathered}$$

By considering the gas to be a type of quantum gas, the Maxwell – distribution is also obtained.

Distribution for velocity vector

In the distribution function of the velocity vector, the velocity probability density is directly proportional to the momentum probability density function. It is given by,

$f_v{d}^{3}v=f_p{\left(\frac{dp}{dv}\right)}^3{d}^{3}v$

Using the momentum formula ,p=mv, we get,

$f_v\left(d{v}_x,d{v}_y,d{v}_z\right)={\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}exp\left[\frac{m_x^2+m_y^2+m_z^2}{2kT}\right]$

Context and Applications

This is one of the important topics in statistical mechanics and thermodynamics, especially for

• Bachelors and Masters in Physics
• Bachelors and Masters in Chemistry

Practice Problems

Question 1: Maxwell - Boltzmann law is applicable for ____.

a. Distinguishable particles

b. Indistinguishable particles

c. Particles with half-integral spin

d. None of these

Answer: The correct option is a.

Explanation: This distribution is applicable only for distinguishable particles, such as classical atoms and molecules.

Question 2: For nitrogen molecules, what is the most probable speed for one molecule at 300K?

a. 316m/s

b. 569m/s

c. 422 m/s

d. 648m/s

Answer: The correct option is c.

Explanation:

The molar mass of nitrogen molecule M=$28\times {10}^{-3}kg/mol$

Temperature T=300K

Most probable speed 

$$\begin{gathered} v_p=\sqrt{\frac{2RT}{M}}=\sqrt{\frac{2\times 8.31\times 300}{28\times {10}^{-3}}} \\ {v}_p=421.9m/s \\ {v}_p=422m/s \end{gathered}$$

Question 3: What is the ratio of RMS speed to the most probable speed?

$$\begin{gathered} a.\sqrt{2}:\sqrt{3} \\ b.\sqrt{3}:\sqrt{2} \\ c.1:1 \\ d.3:2 \end{gathered}$$

Answer: The correct option is b.

Explanation: The rms speed is$v_{rms}=\sqrt{\frac{3RT}{M}}$

The most probable speed is $v_p=\sqrt{\frac{2RT}{M}}$

The ratio of RMS speed to the most probable speed is,

$$\begin{gathered} \frac{v_{rms}}{v_p}=\frac{\sqrt{{\displaystyle \frac{3RT}{M}}}}{\sqrt{\frac{2RT}{M}}}=\frac{\sqrt{3}}{\sqrt{2}} \\ {v}_{rms}:v_p=\sqrt{3}:\sqrt{2} \end{gathered}$$

Question 4: What is the RMS speed of methane at the temperature of 300K?

a. 450m/s

b.678m/s

c.574m/s

d.236 m/s

Answer: The correct option is b.

Explanation:

Temperature T=300K

Boltzmann constant k=$1.38\times {10}^{-23}J/K$

mass of methane m=$\frac{M}{N_A}=\frac{16\times {10}^{-3}}{6.023\times {10}^{26}}=2.65\times {10}^{-26}kg$

Therefore,

$$\begin{gathered} v_{rms}=\sqrt{\frac{3kT}{m}} \\ {v}_{rms}=\sqrt{\frac{3\times 1.38\times 30}{2.65}}\times {10}^2 \\ {v}_{rms}=6.78\times {10}^2 m/s \\ {v}_{rms}=678m/s \end{gathered}$$

Question 5: Maxwell - Boltzmann statistics cannot be applied to_____.

a. Photons

b. Atoms

c. Molecule

d. None of these

Answer: The correct option is a.

Explanation: This distribution is applicable only for distinguishable particles. Atoms and molecules are distinguishable particles whereas photons are indistinguishable particles.