Maxwell–Boltzmann Distribution

Last Updated : 04 Feb, 2024

From the kinetic theory of gases, we have learnt that all the particles in air travel at different speeds and the speed of each particle are due to the collisions between the particles present in the air. Thus, we cannot tell the speed of each particle in the gas or air. Instead, we can tell the number of particles or in other words, we can say that the distribution of particles with a particular speed in gas at a certain temperature can be known. James Maxwell and Ludwig Boltzmann showed the distribution of the particles having different speeds in an ideal gas. Let us look further into Maxwell Boltzmann’s distribution.

Maxwell Boltzmann Distribution

The Maxwell Boltzmann distribution can be studied with the help of a graph given below in this article.

The graph shows the number of molecules possessing a certain speed on the Y-axis and their respective speeds on the X-axis. We can see that the maximum speed is only possessed by a very small number of molecules whereas most of the molecules possess a speed called most probable speed.

Let us see the Maxwell-Boltzmann distribution curve for an ideal gas at two different temperatures T₁ and T₂ where T₁>T₂.

Maxwell Boltzmann Distribution graph for two temperature T1 and T2

It can be seen from the above graph that as the temperature rises, the curve begins to flatten and broaden. Also, the particles that possess the most probable speed reduce but the area under the graph increases which shows that more particles gain energy and are set in motion.

Derivation of the Maxwell Boltzmann Formula

Consider an isolated system containing a gas which contains a number of particles. Let us assume that there are n number of particles and the volume and energy of the gas are fixed which is denoted by V and U respectively. Suppose the system has n energy levels which are e₀, e₁, e₂…….., e_r. Now we need to calculate the number of particles having a certain energy level.

We know that different particles have different energy. Suppose the number of particles with energy e0 is n0, with energy e1 is n1 and so on. Thus the number of particles at different energy levels are n₁, n₂, ………., n_r.

Let W be the number of ways to attain a certain energy level. This can be calculated using,

$W = \frac{n!}{n_0!n_1!\dots n_r!}$

Rewriting the above equation using, we get

$ln(n!) \cong n \space ln(n)-n+\frac{1}{2}ln(2\pi n)$

When n becomes very large, this equation can be written as,

$ln(n!) \cong n \space ln(n)-n$

This equation can be written as follows in terms of W

$ln W \cong n\space ln(n)-\Sigma n_i\space ln(n_i) - (n-\Sigma n_i)$

We know that $\Sigma n_i = n$

Thus, $ln W \cong n\space ln(n)-\Sigma n_i\space ln(n_i)$

Differentiating the above equation with respect to $dn_i$

$d(lnW) = -\Sigma (1+ln(n_i))dn_i = -\Sigma ln(n_i)dn_i$

Using Lagrange’s multipliers,

$d(lnW)= -\Sigma ln(n_i)dn_i-\alpha\Sigma dn_i -\beta\Sigma dn_i$

The above equation will attain maximum value when d(lnW)=0

$0=-\Sigma ln(n_i)dn_i-\alpha\Sigma dn_i -\beta\Sigma dn_i$

$0=-\Sigma (ln(n_i)+\alpha +\beta e_i)\ dn_i$

$ln(n_i)=-\alpha-\beta e_i$

Taking antilogarithm on both sides,

$n_i = e^{-\alpha}e^{-\beta e_i}$

Using the 0th law of thermodynamics we can calculate the value of $\beta$ as

$\beta = -1/k_bT$

where
k_b is the Boltzmann constant.

This equation is used to plot the graph shown above and is not used for solving numerical problems.

Different Types of Speed from Maxwell Boltzmann Distribution

The Maxwell-Boltzmann distribution gives information about various types of speeds possessed by the molecules of gas. In general, a molecule’s speed can be studied under three heading

Most-Probable Speed
Average Speed
Root Mean Square Speed

Let us see the different types of speeds that can be calculated from the Maxwell-Boltzmann distribution.

Most-Probable Speed

This is the speed that is possessed by the majority of the molecules or the maximum number of molecules in a gas. It is represented as v_{mp} and can be calculated as,

$v_{mp}=\sqrt{\frac{2RT}{M}}$

where,
R is the universal gas constant,
T is the absolute temperature of the gas in Kelvin
M is the molar mass of the gas

Average Speed

It is defined as the average speed of all the molecules in a gas. It can be found by dividing the sum of the speeds of all the molecules by the total number of molecules in the gas. It is represented as $v_{avg}$ . It is calculated using,

$v_{avg}=\sqrt{\frac{8RT}{\pi M}}$

where,
R is the universal gas constant,
T is the absolute temperature of the gas in Kelvin
M is the molar mass of the gas

Root Mean Square Speed

It is the square root of the average of the squares of the speed of all the molecules in the gas. It is represented as $v_{rms}$ . It can be calculated using,

$v_{rms}=\sqrt{\frac{3RT}{M}}$

where,
R is the universal gas constant,
T is the absolute temperature of the gas in Kelvin
M is the molar mass of the gas

Read More

Kinetic Energy and Molecular Speeds
Molecular Speed Formula
Molecular Orbital Theory

Solved Examples

Example 1: Calculate the average speed of CO₂ gas at 300K.

Solution:

T = 300 K

M = 44 g

$v_{avg}=\sqrt{\frac{8RT}{\pi M}}$

$= \sqrt{\frac{8×8.314×300}{3.14×44}}$

= 12.01 m/s

Thus, the average speed of the gas is 12.01 m/s.

Example 2: Calculate the V_rms speed of the NO gas at 600K.

Solution:

T = 600 K

M = 30 g

$v_{rms}=\sqrt{\frac{3RT}{M}}$

$= \sqrt{\frac{3×8.314×600}{30}}$

= 22.33 m/s

Thus, the V_rms speed of the gas is 22.33 m/s.

Example 3: Calculate the most probable speed of the gas with a molar mass of 20g at 100K.

Solution:

T = 100 K

M = 20 g

$v_{mp}=\sqrt{\frac{2RT}{M}}$

$= \sqrt{\frac{2×8.314×100}{20}}$

= 9.12 m/s

Thus, the most probable speed of the gas is 9.12 m/s.

Example 4: Which of the following gases have a higher V_rms speed, O₂ at 200K or N₂ at 400K?

Solution:

For O₂

T = 200 K

M = 32 g

$v_{rms}=\sqrt{\frac{3RT}{M}}$

$= \sqrt{\frac{3×8.314×200}{32}}$

= 12.48 m/s

Thus, the most probable speed of the O2 is 12.48 m/s.

For N₂

T = 400 K

M = 28 g

$v_{rms}=\sqrt{\frac{3RT}{M}}$

$= \sqrt{\frac{3×8.314×400}{28}}$

= 18.87 m/s

Thus, the most probable speed of the N2 is 18.87 m/s.

V_rms speed of N₂ is higher than O₂ at the given conditions.

Example 5: Calculate the temperature of the SO₂ gas if its average speed is 5m/s.

Solution:

M = 64g

$v_{avg}=50ms^{-1}$

$v_{avg}=\sqrt{\frac{8RT}{\pi M}}$

$5 =\sqrt{\frac{8×8.314×T}{3.14×64}}$

$25 =\frac{8×8.314×T}{3.14×64}$

T = 75.53K

Thus, the temperature of the SO₂ gas is 75.53K.

FAQs on Maxwell Boltzmann Distribution

Question 1: What is the use of Maxwell-Boltzmann?

Answer:

The Maxwell-Boltzmann distribution shows the number of particles that possess a particular speed in the gas at a given temperature.

Question 2: What are the different types of speed that can be seen from the distribution?

Answer:

The different types of speed that can be seen from the distribution are room mean square speed, average speed and most probable speed.

Question 3: State the equation for Maxwell Boltzmann distribution.

The equation for Maxwell Boltzmann distribution is,

$n_i = e^{-\alpha}e^{-\beta e_i}$

where
n_i is the number of particles at a certain energy level
e_i is the energy level.
$\alpha$ and $\beta$ are constants.

Question 4: What is the value of the Boltzmann constant?

Answer:

The value of Boltzmann constant is 1.308 ×10^-23 JK^-1

Question 5: What is the effect of increasing the temperature on the Maxwell-Boltzmann distribution?

Answer:

On increasing the temperature, the number of particles having the most probable speed decreases and the curve moves a little downwards and more towards the X-axis which means that the maximum speed of the particles increases.

Suggest improvement

Normal Distribution

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Maxwell–Boltzmann Distribution