Table of contents 
Introduction 
Energy distribution of Ideal gas in three dimension 
Derivation of MaxwellBoltzmann Distribution in three dimension 
Energy distribution in different dimension 
MaxwellBoltzmann Distribution Law Distribution of Molecular Velocity in perfect gas 
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In statistical mechanics, the Maxwell–Boltzmann distribution describes particle speeds in gases, where the particles move freely without interacting with one another, except for very brief elastic collision in which they may exchange momentum and kinetic energy, but do not change their respective states of intermolecular excitation, as a function of the temperature of the system, the mass of the particle, and speed of the particle. Particle in this context refers to the gaseous atoms or molecules – no difference is made between the two in its development and result.
Maxwell –Boltzmann system constituent identical particles that are distinguishable in nature which means we can distinguish them by name, color, put any number or any level on particle
For example, if we want to identify two distinguishable particles, we can say that first particle is A and second particle is B . In another way, we can also identify the colour of particles as red for first particle and black for second particle. There is no any restriction on number of particles which can occupy any energy level.
Quantum mechanically, the wave function of particle will not overlap to each other because mean separation of particles is more than the thermal wavelength, which is identified by λ. (where λ = is defined as the thermal wavelength)
Number of ways (W) that n_{i} number of distinguishable particle which can be adjusted in g_{i} number of quantum is
Suppose there are states with energies E_{1},E_{2},E_{3},.....E_{l} and degeneracy of each state g_{1}, g_{2}, g_{3},......g_{l }respectively the i^{th } level can be shown schematically
If there is N numbers of distinguishable particles out of these n_{1}, n_{2}, n_{3},....n_{l} particles is adjusted in energy level E_{1},E_{2},E_{3},.....E_{l }respectively.
Total number of particle is constant n_{1+} n_{2+} n_{3+},....+n_{l} = N,
Total energy of configuration is constant n_{1E1+} n_{2E2}_{+},....
Now, number of ways selecting n_{1} out of N particles then distribute it in energy state E_{1} which degeneracy is g_{1 }then
The total number W of distinct ways of obtaining the distribution of n_{1}, n_{2}, n_{3},....n_{l }particles among the energy states E_{1},E_{2},E_{3},.....E_{l}
W =
=
Example 13: Two distinguishable particles have to be adjusted in a state whose degeneracy is three
(a) How many ways the particles can be adjusted?
(b) Show all arrangement.
(a) N = 2,n = 2, g = 3 and number of microstate is W = .
(b) Total number of arrangement for 2 distinguishable in state whose degeneracy is 3.
Example 14: If two distinguishable particle have to adjusted in two quantum level with ground state energy E and first excited state having energy 2E having degeneracy 1 and 2.
(a) What will number of ways that total energy of distribution is 2E i.e. both the particles in ground state.
(b) What will number of ways that total energy of distribution is 3E i.e., one the particle in ground state and other in first excited state .
(c) What will number of ways that total energy of distribution is 4e i.e. both the particle in first excited state.
(a) N = 2, g_{1} = 2, g_{2} = 1, n_{1} = 2, n_{2} = 0 total energy U =
W = = 4 given distribution is shown in figure.
(b) N = 2, g_{1} = 2, g_{2} = 1, n_{1} = 1, n_{2} = 1 total energy U =
W = = 4 given distribution is shown in figure.
(c) N = 2, g_{1} = 2, g_{2} = 1, n_{1} = 1, n_{2} = 1 total energy U =
W = = 1 the given distribution is shown in figure
Entropy (S) is measurement of randomness of system. It is function of number of microstate (which is number of ways to achieve any energy (E). Statistically it is seen so . And Law of nature reveal that at equilibrium the entropy is maximum so
For distinguishable particle. Taking in for equation W =
Using sterling approximation
=
where α and β are used as LaGrange’s multiplier which is using to make equation dimensionless where dimension of β is inverse of energy can be related to equilibrium
temperature T so can be identified later can be related to fugacity
Equating the coefficient of dn_{i} then
So, Maxwell distribution is given by
Calculation of α in three dimensional case
Total number of particles in ideal gas N =
where energy levels are continuous then N =
for three dimensional case g (E) =
N =
N =
where β =
The MaxwellBoltzmann distribution law for the particles in the states is
After using the values
We get
The number of particles dN(E) having energies in the range from E to E +dE is dN(E) = f (E)g(E)dE where f (E) is distribution function and g(E)dE is number of level (quantum state) in the range of E to E + dE
The number of particles dn(E) having energies in the range from E to E + dE in three dimensional space
is fraction of particles that have energy between from E to E + dE is
which is popularly known as probability to find gas have energy between from E to E + dE , where is identified probability density.
This is known as the MaxwellBoltzmann energy distribution law for an ideal gas, where λ = is defined as the thermal wavelength.
For the MaxwellBoltzmann energy distribution law, average energy (E) of the particles is
Hence, the average of a particle is per degree of freedom, for three degree of freedom it is
Mean square of energy (E^{2}) =
Root mean Energy
Most probable Energy
Probability density then The most probable energy is given by
MaxwellBoltzmann distribution law is applicable for ideal gas, where molecules have no vibrational or rotational energies.
In the equilibrium state of the molecules, molecules have completed their random motion and probability that a molecule has a given velocity component is independent of other two components.
In the given figure dv is volume element in velocity space for a molecule at velocity
We need to calculate number of molecules simultaneously having component in the range and , which is equation of sphere and dv_{x}dv_{y}dv_{z }can be replaced by . There is an assumption in MaxwellBoltzmann distribution law that probability that a molecule selected at random has velocities in a given range is a purely function of the magnitude of velocity and the width of the interval.
We need to calculate number of molecules simultaneously having component in the range and which is equation of sphere and _{ }can be replaced by dv_{x}dv_{y}dv_{z} .
where
Average Speed
=
Mean square Speed
=
Root Mean Square speed
Most probable speed
Average velocity of
= 0
= 0
Similarly,
Example 15: For Maxwellian gas, find the
As, = ⇒
Example 16: If v_{x} and are v_{y} are x and y component of velocity then find the average value of
Similarly, (v_{y}) = 0 and
Therefore,
=
Example 17: (a) Write down expression of Maxwell distribution function for speed in two dimensional in equilibrium temperature T
(b) Find Average speed for two dimensions in two dimensional at equilibrium temperature T
(c) Find RMS speed for two dimensions in two dimensional at equilibrium temperature T
(d) Write down expression of Maxwell distribution function for energy in two dimensional at equilibrium temperature T
(e) Find average energy for twodimensional system at equilibrium temperature T.
(a)
(b)
(c) , ⇒ ; 0 < E < ∞
(d)
Example 18: Using the Maxwell distribution function, calculate the mean velocity
projection v_{x} and the mean value of the modulus of this projection , if the mass of each molecule is equal to m and the gas temperature is T.
We know that Mean Velocity
= 0
Mean speed,
=
Example 19: If N number of distinguishable particle is kept into one dimensional box of length L. What is average energy at temperature T .
for two dimensional system g(E)dE = and distribution Function is given by
(E) =
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