Step-by-Step Maxwell Boltzmann Distribution Equation and Insights
Consider an ideal system having n particles occupying a volume V, whose total energy is E. Here, the value of E is constant because no energy is being added or taken away from the system. So the total energy of the system is equivalent to the sum of total energies of the individual particles.
Since the system has multiple states, if we consider an ith state of the system, then, the probability at this state will be:
\[P_{i} \alpha E^{-\frac{\epsilon i}{kT}}\]
Where εi is the energy of a system in ith state.
k = Boltzmann constant (= \[1.38 \times 10^{-23} JK^{-1}\])
T =Temperature
Now,
\[P_{i} = \frac{1}{z} E^{-\frac{\epsilon i}{kT}}\]
Here z is the partition function, which is the sum of the energies of all the states in the system.
Maxwell Boltzmann Distribution Derivation
The molecules inside the system travel at varying speeds so two persons named James Maxwell and Ludwig Boltzmann came up with a theory to demonstrate how the speeds of the molecule are distributed for an ideal gas which is Maxwell-Boltzmann distribution theory.
Consider a system having n particles occupying a volume V, whose total energy is E.
So, \[E = \sum_{i=0}^{\infty} n_{i} \epsilon_{i}\]
Where n is the number of particles having energy εi.
The number of particles N is also constant, i.e.,
\[N = \sum_{i=0}^{\infty} n_{i}\]
The number of microstates for the energy levels of molecules of a system can be expressed as:
\[W = \frac{N}{n_{0}! n_{1}! n_{2}}!\] .... (1)
Stirling's equation of eq(1) is,
ln N!. ≅ N ln N - N + 1/2 ln(2πN)...(2)
And, for large N it is:
N!. ≅ N ln N - N
Thus, in W it can be expressed as:
ln W ≅ N ln \[N - \sum_{i} n_{i} ln n_{i} - (N - \sum_{i} n_{i})\]
The term inside the parentheses is zero because, \[N = \sum_{i=0}^{\infty} n_{i}\] is a constant value.
Thus, ln W ≅ \[N ln N - \sum_{i} n_{i} ln n_{i}\] ..(3)
Taking differential of eq(3):
\[d(ln W) = -\sum_{i} (1 + ln n_{i})dn_{i} = - \sum ln n_{i} d n_{i}\] …(4)
Since dni = dN = 0
Now using the Lagrange multipliers here, it is known that:
\[dE = 0 = \sum_{i} n_{i} \epsilon_{i}\] , and
\[dN = 0 = \sum_{i} dn_{i}\] …(5)
Adding (4) to (3), for any constant α and β, following should be true:
\[d(ln W) = - \sum_{i} ln n_{i} d n_{i} - \alpha \sum_{i} d n_{i} - \beta \sum_{i}d n_{i}\]
At maximum, W = 0
0 = - Σ (α + βεi + ln ni )dni
Or, 0 = (α + βεi + ln ni )
ln ni = - α - βεi
Removing the logarithmic, we get:
\[n_{i} =e^{-\alpha} e^{-\beta \epsilon i}\]
Evaluating N using the distribution law here:
\[e^{\alpha} = \frac{N}{e^{-\beta \epsilon i}}\] and z, the partition function = \[\sum_{i} e^{-\beta \epsilon i}\]
Therefore, ni = N/Q e-βεi
Applying zeroth law of thermodynamics, we get
β = 1/kT
Maxwell Boltzmann Distribution Equation Derivation
Maxwell distribution of velocities states that the gaseous molecules inside the system travel at different velocities.
Fraction F(v) = \[4 \pi N(\frac{m}{2 \pi k T})^{3/2} v^{2} e^{-mv^{2/2kT}}\]
The Maxwell distribution of velocities can be derived from Boltzmann’s equation:
\[f (E) = Ae^{-kT}\]
This equation tells us the probability that a molecule will be found with energy E that decreases exponentially with energy; i.e., any molecule is highly unlikely to capture much more than its average part of the total energy available to all the molecules.
(image will be uploaded soon)
If Maxwell-Boltzmann distribution is applied in one dimension of velocity for a molecule in an ideal gas is vz.
Where the factor F = dN/dV
This F states that if we consider a small rectangular inside the above graph, then the probability of the gas molecules traveling at velocity can be found.
Where Fraction F(v) = \[4 \pi N(\frac{m}{2 \pi k T})^{3/2} v^{2} e^{-mv^{2/2kT}}\]
Maxwell Boltzmann Equation Derivation
The probability of finding a molecule at some velocity is one. Then, by using the definite integral form here from minus-infinity to plus-infinity as:
\[\int_{-\infty}^{\infty} e^{-x^{2}} dx = \sqrt{\pi}\] and substituting \[x = \sqrt{\frac{m}{2kTv_{z}}}\]
\[A \sqrt{\frac{m}{2kT} v_{z}} \int_{-\infty}^{\infty} e^{-\frac{m}{2 \pi k T}} dx = \sqrt{\pi}\]
\[F(v_{z}) =\sqrt{\frac{m}{2kT}} v^{2} e^{-mvz^{2/2kT}}\] \[\sqrt{\frac{m}{2kT}} dv_{z} = 1\]
It gives \[A = \sqrt{\frac{m}{2kT}}\]
This normalizes the function to: \[F(v_{z}) =\sqrt{\frac{m}{2kT}} v^{2} e^{-mvz^{2/2kT}}\]
In terms of three dimensions, it becomes:
\[F(v) =4 \pi (\frac{m}{2kT})^{3/2} v^{2} e^{-mvz^{2/2kT}}\]
Maxwell Boltzmann Derivation
Consider the fraction of molecules in a three-dimensional box having the translation energy ε, then, as a function, it will be:
ε = h2 / 8m [nx2 + ny2+ nz2]/[Lx2 + Ly2 + Lz2]
ni2 = nx2 + ny2+ nz2 ,and
εi = n2h2/8mL2
The length of each side of the box is equal.
So, we would consider the density of translational states as a sphere in space.
So the total volume of the sphere is:
\[\epsilon = \frac{4 \pi}{8} \int_{0}^{n} n^{2} dn\]
and dε = π/2 n2 dn
Also n2h2/8mL2 = ½ mu2, and
dn = 2mV1/3/hdu
So, π/2 n2 dn = dNi/N = dε/Q \[e^{-\beta \epsilon i}\]
Substituting in for thermal wavelength, we get:
\[\frac{1}{N} \frac{du}{dN} = 4 \pi u^{2} (m/2 \pi k T)^{3/2} e^{-mu^{2/2kT}}\]
This is the Maxwell-Boltzmann distribution.
FAQs on Maxwell Boltzmann Distribution Derivation Made Easy
1. What does the Maxwell-Boltzmann distribution curve represent in simple terms?
The Maxwell-Boltzmann distribution curve shows how the speeds of particles in a gas are spread out at a certain temperature. The peak of the curve represents the most probable speed — the speed that the largest number of particles have. The curve illustrates that very few particles move extremely slow or extremely fast; most are clustered around an average speed.
2. What are the three main types of molecular speeds explained by this distribution?
The distribution helps us understand three key speeds for gas molecules:
- Most Probable Speed (v_p): The speed at which the highest number of molecules are moving. This corresponds to the peak of the distribution curve.
- Average Speed (v_avg): The arithmetic average of the speeds of all the molecules in the gas.
- Root-Mean-Square (RMS) Speed (v_rms): The square root of the average of the squared speeds, which gives more weight to faster-moving particles.
3. How does changing the temperature of a gas affect the Maxwell-Boltzmann distribution curve?
When you increase the temperature of a gas, the distribution curve flattens and shifts to the right. This indicates two things: first, the average speed of the molecules increases, and second, the range of speeds becomes wider. Essentially, at higher temperatures, more molecules move at higher speeds.
4. What are the fundamental assumptions made when deriving the Maxwell-Boltzmann distribution law?
The derivation relies on the core assumptions of the kinetic theory of gases:
- The gas contains a very large number of identical particles.
- These particles are in constant, random motion.
- Collisions between particles and with the container walls are perfectly elastic.
- The volume of the particles themselves is negligible compared to the container's volume.
- There are no intermolecular forces of attraction or repulsion, except during collisions.
5. Why is this distribution so important for understanding the kinetic theory of gases?
The Maxwell-Boltzmann distribution is crucial because it connects the microscopic world of individual particle speeds to the macroscopic properties we can measure, like pressure and temperature. It provides the mathematical proof for how the collective random motion of countless particles gives rise to the predictable behaviour of a gas as a whole.
6. What are some real-world applications of the Maxwell-Boltzmann distribution?
This distribution is fundamental in many scientific fields. In chemistry, it helps explain reaction rates, as only molecules with sufficient energy (found on the higher-speed end of the curve) can react. It is also used to understand processes like evaporation and is essential in advanced fields like plasma physics and astrophysics to model particle energies.
7. How does Maxwell-Boltzmann statistics differ from quantum statistics like Bose-Einstein or Fermi-Dirac?
Maxwell-Boltzmann statistics describes classical, distinguishable particles, like ideal gas molecules. In contrast, quantum statistics are for indistinguishable particles. Bose-Einstein statistics applies to bosons (e.g., photons), while Fermi-Dirac statistics applies to fermions (e.g., electrons), which obey the Pauli exclusion principle. Maxwell-Boltzmann is an excellent approximation for most gases under normal conditions.
