Mastering Molecular Speed Distributions, Mean Free Path, Maxwell-Boltzmann Distribution, and Energy Distributions
Statistical Mechanics
Maxwell-Boltzmann Distribution
Mean Free Path
Molecular Speeds
Energy Distribution
Reading Time: 30 min
📜 Historical Background
Statistical mechanics developed in the 19th century to bridge the gap between microscopic molecular behavior and macroscopic thermodynamics:
- James Clerk Maxwell (1860): First derived the distribution of molecular speeds
- Ludwig Boltzmann (1870s): Developed statistical interpretation of entropy
- Josiah Willard Gibbs (1902): Formulated the foundations of statistical mechanics
- Albert Einstein (1905): Applied statistical mechanics to Brownian motion
These developments connected the microscopic world of atoms and molecules to the macroscopic properties we observe.
Introduction to Statistical Mechanics
🔬 What is Statistical Mechanics?
Statistical mechanics is the branch of physics that uses probability theory to study the thermodynamic behavior of systems composed of a large number of particles. It bridges the microscopic world of atoms and molecules with the macroscopic world of thermodynamics.
While thermodynamics deals with relationships between macroscopic properties, statistical mechanics explains these relationships in terms of the statistical behavior of microscopic constituents.
📝 From Averages to Distributions
In the previous chapter (Kinetic Theory), we determined the average translational kinetic energy of gas molecules. However, knowing the average tells us nothing about how individual molecular speeds are distributed around this average.
Statistical mechanics provides the tools to understand these distributions and use them to compute macroscopic properties of collections of molecules.
Statistical Distributions and Mean Values
📊 Statistical Distributions
When dealing with a large number of objects (say N) moving with different speeds, we can organize the data using a histogram. Each rectangular area in the histogram has:
- Width equal to the size of the sorting interval
- Height equal to the number of observations or relative frequency in that interval
Calculating Average Speed
🧮 Mean Speed Calculation
Step 1: Total Number of Observations
\[ N = \sum_i n(v_i) \]
where \( n(v_i) \) is the number of observations in interval i
Step 2: Average Speed Formula
\[ \bar{v} = \frac{\sum_i v_i n(v_i)}{\sum_i n(v_i)} \]
\[ = \frac{\sum_i v_i n(v_i)}{N} \]
Step 3: Using Relative Frequency
\[ f(v_i) = \frac{n(v_i)}{N} \]
\[ \bar{v} = \sum_i v_i f(v_i) \]
Sample Problem: Molecular Speed Calculation
The speeds of a group of 10 molecules are 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 km/s. Find (a) the average speed and (b) the root mean square speed.
Given: \( v_i = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \) km/s
N = 10 molecules
(a) Average Speed:
\[ \bar{v} = \frac{2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11}{10} \]
\[ = \frac{65}{10} \]
\[ = 6.5 \, \text{km/s} \]
(b) Root Mean Square Speed:
\[ \bar{v^2} = \frac{2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 + 11^2}{10} \]
\[ = \frac{4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121}{10} \]
\[ = \frac{505}{10} \]
\[ = 50.5 \, \text{km}^2/\text{s}^2 \]
\[ v_{rms} = \sqrt{\bar{v^2}} = \sqrt{50.5} \]
\[ = 7.11 \, \text{km/s} \]
Sample Problem: Car Speed Distribution
Sample Problem 1: Car Speed Distribution
Based on observation of 1205 cars, the speed distribution is as follows:
| Interval |
Speed Range (mph) |
Number of Cars |
| 1 | 0-5 | 23 |
| 2 | 5-10 | 41 |
| 3 | 10-15 | 54 |
| 4 | 15-20 | 95 |
| 5 | 20-25 | 123 |
| 6 | 25-30 | 142 |
| 7 | 30-35 | 177 |
| 8 | 35-40 | 186 |
| 9 | 40-45 | 170 |
| 10 | 45-50 | 122 |
| 11 | 50-55 | 50 |
| 12 | 55-60 | 15 |
| 13 | 60-65 | 7 |
Find the mean value of speed for this distribution, taking the speed at the middle of each interval as representing the entire interval.
Given: Total Cars N = 1205
We calculate the mean speed using:
\[ \bar{v} = \frac{\sum v_i n(v_i)}{N} \]
Calculations:
\[ \sum v_i n(v_i) = (2.5 \times 23) + (7.5 \times 41) + (12.5 \times 54) + (17.5 \times 95) + (22.5 \times 123) + (27.5 \times 142) + (32.5 \times 177) + (37.5 \times 186) + (42.5 \times 170) + (47.5 \times 122) + (52.5 \times 50) + (57.5 \times 15) + (62.5 \times 7) \]
\[ = 57.5 + 307.5 + 675 + 1662.5 + 2767.5 + 3905 + 5752.5 + 6975 + 7225 + 5795 + 2625 + 862.5 + 437.5 \]
\[ = 42446.5 \]
Mean speed:
\[ \bar{v} = \frac{42446.5}{1205} \]
\[ = 35.2 \, \text{mph} \]
Mean Free Path
🔬 Mean Free Path Definition
The mean free path (λ) is the average distance a molecule travels between successive collisions with other molecules. This concept is crucial for understanding transport phenomena in gases.
Exponential Decay Model
🧮 Exponential Decay Derivation
Step 1: Probability of No Collision
Let P(x) be the probability that a molecule travels distance x without collision.
\[ P(x + dx) = P(x) \times (\text{probability of no collision in dx}) \]
Step 2: Collision Probability
Probability of collision in distance dx is dx/λ, where λ is mean free path.
\[ P(x + dx) = P(x) \left(1 - \frac{dx}{\lambda}\right) \]
Step 3: Differential Equation
\[ \frac{dP}{dx} = \lim_{dx \to 0} \frac{P(x + dx) - P(x)}{dx} \]
\[ = \frac{P(x)(1 - dx/\lambda) - P(x)}{dx} \]
\[ = -\frac{P(x)}{\lambda} \]
Step 4: Solution
\[ \frac{dP}{P} = -\frac{dx}{\lambda} \]
\[ \ln P = -\frac{x}{\lambda} + C \]
\[ P(x) = P(0) e^{-x/\lambda} \]
Step 5: Initial Condition
\[ P(0) = 1 \]
\[ P(x) = e^{-x/\lambda} \]
Microscopic Calculation
🧮 Mean Free Path Calculation
Step 1: Collision Cross Section
Consider a molecule of diameter d moving with speed v. It will collide with any molecule whose center lies within a cylinder of radius d.
\[ \text{Collision cross section} = \pi d^2 \]
Step 2: Number of Collisions
In time t, the molecule sweeps out a volume:
\[ V = \pi d^2 v t \]
Number of collisions in time t:
\[ N_{\text{coll}} = (\pi d^2 v t) n \]
where n is the number density of molecules
Step 3: Mean Free Path
\[ \lambda = \frac{\text{total distance}}{\text{number of collisions}} \]
\[ = \frac{v t}{\pi d^2 v t n} \]
\[ = \frac{1}{\pi d^2 n} \]
Step 4: More Accurate Calculation
Considering relative motion between molecules:
\[ \lambda = \frac{1}{\sqrt{2} \pi d^2 n} \]
Step 5: Using Ideal Gas Law
\[ n = \frac{N}{V} = \frac{P}{k_B T} \]
\[ \lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} \]
Sample Problems
Sample Problem 2: Mean Free Path Calculation
Calculate the mean free path of nitrogen molecules at 1.0 atm pressure and 25°C temperature. The diameter of nitrogen molecule is 3.15 × 10⁻¹⁰ m.
Given:
\[ P = 1.0 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa} \]
\[ T = 25^\circ C = 298 \, K \]
\[ d = 3.15 \times 10^{-10} \, \text{m} \]
\[ k_B = 1.38 \times 10^{-23} \, \text{J/K} \]
Using the formula:
\[ \lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} \]
\[ = \frac{(1.38 \times 10^{-23}) \times 298}{\sqrt{2} \times \pi \times (3.15 \times 10^{-10})^2 \times (1.013 \times 10^5)} \]
\[ = \frac{4.1124 \times 10^{-21}}{1.414 \times 3.1416 \times 9.9225 \times 10^{-20} \times 1.013 \times 10^5} \]
\[ = \frac{4.1124 \times 10^{-21}}{4.45 \times 10^{-14}} \]
\[ = 9.24 \times 10^{-8} \, \text{m} \]
\[ = 92.4 \, \text{nm} \]
Sample Problem 3: Probability Calculation
For nitrogen molecules at STP, the mean free path is 8.0 × 10⁻⁸ m. What fraction of molecules travel 10 cm without collision?
Given:
\[ \lambda = 8.0 \times 10^{-8} \, \text{m} \]
\[ x = 10 \, \text{cm} = 0.1 \, \text{m} \]
Using exponential decay formula:
\[ P(x) = e^{-x/\lambda} \]
\[ = e^{-0.1/(8.0 \times 10^{-8})} \]
\[ = e^{-1.25 \times 10^6} \]
\[ \approx 0 \]
This extremely small probability shows that at STP, molecules undergo frequent collisions.
Maxwell-Boltzmann Speed Distribution
🔬 Maxwell-Boltzmann Distribution
The Maxwell-Boltzmann distribution describes the probability distribution of molecular speeds in a gas at thermal equilibrium. For an ideal gas in three dimensions:
\[ f(v) dv = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{-mv^2/(2k_B T)} dv \]
where f(v)dv gives the fraction of molecules with speeds between v and v+dv.
📈 Maxwell-Boltzmann Speed Distribution
[Graph: Maxwell-Boltzmann distribution showing molecular speed distribution at different temperatures]
The graph shows how molecular speeds are distributed in a gas. As temperature increases, the distribution shifts to higher speeds and becomes broader.
Derivation of Average Speed
🧮 Average Speed Calculation
Step 1: Definition of Average Speed
\[ \bar{v} = \int_0^\infty v f(v) dv \]
Step 2: Substitute Distribution Function
\[ \bar{v} = \int_0^\infty v \left[ 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{-mv^2/(2k_B T)} \right] dv \]
\[ = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} \int_0^\infty v^3 e^{-mv^2/(2k_B T)} dv \]
Step 3: Use Standard Integral
\[ \int_0^\infty v^3 e^{-av^2} dv = \frac{1}{2a^2} \]
\[ \text{where } a = \frac{m}{2k_B T} \]
Step 4: Evaluate Integral
\[ \int_0^\infty v^3 e^{-mv^2/(2k_B T)} dv = \frac{1}{2(m/(2k_B T))^2} \]
\[ = \frac{1}{2(m^2/(4k_B^2 T^2))} \]
\[ = \frac{2k_B^2 T^2}{m^2} \]
Step 5: Final Expression
\[ \bar{v} = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} \times \frac{2k_B^2 T^2}{m^2} \]
\[ = 4\pi \left(\frac{m^{3/2}}{(2\pi k_B T)^{3/2}}\right) \times \frac{2k_B^2 T^2}{m^2} \]
\[ = \frac{8\pi k_B^2 T^2}{m^{1/2} (2\pi k_B T)^{3/2}} \]
\[ = \frac{8\pi k_B^2 T^2}{m^{1/2} \times 2^{3/2} \pi^{3/2} k_B^{3/2} T^{3/2}} \]
\[ = \frac{8}{2^{3/2} \pi^{1/2}} \times \frac{k_B^{1/2} T^{1/2}}{m^{1/2}} \]
\[ = \sqrt{\frac{8}{\pi}} \times \sqrt{\frac{k_B T}{m}} \]
\[ \bar{v} = \sqrt{\frac{8k_B T}{\pi m}} \]
Most Probable Speed
🧮 Most Probable Speed Calculation
Step 1: Condition for Maximum
The most probable speed occurs where df(v)/dv = 0
\[ f(v) = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{-mv^2/(2k_B T)} \]
Step 2: Differentiate
\[ \frac{df}{dv} = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} \left[ 2v e^{-mv^2/(2k_B T)} + v^2 \left(-\frac{mv}{k_B T}\right) e^{-mv^2/(2k_B T)} \right] \]
\[ = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} e^{-mv^2/(2k_B T)} \left[ 2v - \frac{mv^3}{k_B T} \right] \]
Step 3: Set Derivative to Zero
\[ 2v - \frac{mv^3}{k_B T} = 0 \]
\[ 2 = \frac{mv^2}{k_B T} \]
\[ v^2 = \frac{2k_B T}{m} \]
\[ v_p = \sqrt{\frac{2k_B T}{m}} \]
Root Mean Square Speed
🧮 RMS Speed Calculation
Step 1: Definition of Mean Square Speed
\[ \bar{v^2} = \int_0^\infty v^2 f(v) dv \]
Step 2: Substitute Distribution Function
\[ \bar{v^2} = \int_0^\infty v^2 \left[ 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{-mv^2/(2k_B T)} \right] dv \]
\[ = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} \int_0^\infty v^4 e^{-mv^2/(2k_B T)} dv \]
Step 3: Use Standard Integral
\[ \int_0^\infty v^4 e^{-av^2} dv = \frac{3\sqrt{\pi}}{8a^{5/2}} \]
\[ \text{where } a = \frac{m}{2k_B T} \]
Step 4: Evaluate Integral
\[ \int_0^\infty v^4 e^{-mv^2/(2k_B T)} dv = \frac{3\sqrt{\pi}}{8(m/(2k_B T))^{5/2}} \]
\[ = \frac{3\sqrt{\pi}}{8} \times \left(\frac{2k_B T}{m}\right)^{5/2} \]
Step 5: Final Expression
\[ \bar{v^2} = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} \times \frac{3\sqrt{\pi}}{8} \times \left(\frac{2k_B T}{m}\right)^{5/2} \]
\[ = 4\pi \times \frac{m^{3/2}}{(2\pi k_B T)^{3/2}} \times \frac{3\sqrt{\pi}}{8} \times \frac{(2k_B T)^{5/2}}{m^{5/2}} \]
\[ = \frac{3}{2} \times \frac{2k_B T}{m} \]
\[ \bar{v^2} = \frac{3k_B T}{m} \]
\[ v_{rms} = \sqrt{\bar{v^2}} = \sqrt{\frac{3k_B T}{m}} \]
📊 Comparison of Molecular Speeds
| Speed Type |
Formula |
Value Relative to vp |
| Most Probable Speed (vp) |
\(\sqrt{\frac{2k_B T}{m}}\) |
1.000 vp |
| Average Speed (\(\bar{v}\)) |
\(\sqrt{\frac{8k_B T}{\pi m}}\) |
1.128 vp |
| Root Mean Square Speed (vrms) |
\(\sqrt{\frac{3k_B T}{m}}\) |
1.225 vp |
Sample Problem: Oxygen Molecules
Sample Problem 4: Molecular Speeds of Oxygen
Calculate the most probable speed, average speed, and root mean square speed for oxygen molecules at 300 K. The molar mass of oxygen is 32 g/mol.
Given:
\[ T = 300 \, K \]
\[ M = 32 \, \text{g/mol} = 0.032 \, \text{kg/mol} \]
\[ R = 8.314 \, \text{J/mol·K} \]
\[ m = \frac{M}{N_A} = \frac{0.032}{6.022 \times 10^{23}} = 5.314 \times 10^{-26} \, \text{kg} \]
Most Probable Speed:
\[ v_p = \sqrt{\frac{2k_B T}{m}} = \sqrt{\frac{2RT}{M}} \]
\[ = \sqrt{\frac{2 \times 8.314 \times 300}{0.032}} \]
\[ = \sqrt{\frac{4988.4}{0.032}} \]
\[ = \sqrt{155887.5} \]
\[ = 394.8 \, \text{m/s} \]
Average Speed:
\[ \bar{v} = \sqrt{\frac{8RT}{\pi M}} \]
\[ = \sqrt{\frac{8 \times 8.314 \times 300}{\pi \times 0.032}} \]
\[ = \sqrt{\frac{19953.6}{0.10053}} \]
\[ = \sqrt{198457.5} \]
\[ = 445.5 \, \text{m/s} \]
Root Mean Square Speed:
\[ v_{rms} = \sqrt{\frac{3RT}{M}} \]
\[ = \sqrt{\frac{3 \times 8.314 \times 300}{0.032}} \]
\[ = \sqrt{\frac{7482.6}{0.032}} \]
\[ = \sqrt{233831.25} \]
\[ = 483.6 \, \text{m/s} \]
Distribution of Energies
🔬 Energy Distribution Function
The distribution of molecular energies can be derived from the speed distribution. For an ideal gas, the fraction of molecules with energy between E and E+dE is:
\[ f(E) dE = \frac{2}{\sqrt{\pi}} \frac{1}{(k_B T)^{3/2}} \sqrt{E} e^{-E/(k_B T)} dE \]
where E = ½mv² is the kinetic energy of a molecule.
Derivation of Energy Distribution
🧮 Energy Distribution Derivation
Step 1: Relationship Between Speed and Energy
\[ E = \frac{1}{2} m v^2 \]
\[ v = \sqrt{\frac{2E}{m}} \]
\[ dv = \frac{dE}{\sqrt{2mE}} \]
Step 2: Transform Speed Distribution
\[ f(v) dv = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} v^2 e^{-mv^2/(2k_B T)} dv \]
\[ = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} \left(\frac{2E}{m}\right) e^{-E/(k_B T)} \frac{dE}{\sqrt{2mE}} \]
Step 3: Simplify Expression
\[ f(E) dE = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} \frac{2E}{m} e^{-E/(k_B T)} \frac{dE}{\sqrt{2mE}} \]
\[ = 4\pi \left(\frac{m}{2\pi k_B T}\right)^{3/2} \frac{2\sqrt{E}}{\sqrt{2m}} e^{-E/(k_B T)} dE \]
\[ = \frac{4\pi}{(2\pi k_B T)^{3/2}} \sqrt{\frac{2E}{m}} e^{-E/(k_B T)} dE \]
Step 4: Final Form
\[ f(E) dE = \frac{2}{\sqrt{\pi}} \frac{1}{(k_B T)^{3/2}} \sqrt{E} e^{-E/(k_B T)} dE \]
Internal Energy of Ideal Gas
🧮 Internal Energy Calculation
Step 1: Definition of Internal Energy
\[ U = N \bar{E} \]
where \(\bar{E}\) is the average energy per molecule
Step 2: Average Energy Calculation
\[ \bar{E} = \int_0^\infty E f(E) dE \]
\[ = \int_0^\infty E \left[ \frac{2}{\sqrt{\pi}} \frac{1}{(k_B T)^{3/2}} \sqrt{E} e^{-E/(k_B T)} \right] dE \]
\[ = \frac{2}{\sqrt{\pi}} \frac{1}{(k_B T)^{3/2}} \int_0^\infty E^{3/2} e^{-E/(k_B T)} dE \]
Step 3: Use Gamma Function
\[ \int_0^\infty x^{n} e^{-ax} dx = \frac{\Gamma(n+1)}{a^{n+1}} \]
\[ \text{Let } x = E, n = 3/2, a = 1/(k_B T) \]
\[ \int_0^\infty E^{3/2} e^{-E/(k_B T)} dE = \frac{\Gamma(5/2)}{(1/(k_B T))^{5/2}} \]
\[ = \frac{(3\sqrt{\pi}/4)}{(k_B T)^{-5/2}} \]
\[ = \frac{3\sqrt{\pi}}{4} (k_B T)^{5/2} \]
Step 4: Final Expression
\[ \bar{E} = \frac{2}{\sqrt{\pi}} \frac{1}{(k_B T)^{3/2}} \times \frac{3\sqrt{\pi}}{4} (k_B T)^{5/2} \]
\[ = \frac{3}{2} k_B T \]
\[ U = N \bar{E} = \frac{3}{2} N k_B T = \frac{3}{2} n R T \]
Mean and Most Probable Energy
🧮 Mean and Most Probable Energy
Mean Energy
\[ \bar{E} = \frac{3}{2} k_B T \]
As derived above from the energy distribution
Most Probable Energy
The most probable energy occurs where df(E)/dE = 0
\[ f(E) = \frac{2}{\sqrt{\pi}} \frac{1}{(k_B T)^{3/2}} \sqrt{E} e^{-E/(k_B T)} \]
\[ \frac{df}{dE} \propto \frac{1}{2\sqrt{E}} e^{-E/(k_B T)} - \frac{\sqrt{E}}{k_B T} e^{-E/(k_B T)} \]
\[ = e^{-E/(k_B T)} \left( \frac{1}{2\sqrt{E}} - \frac{\sqrt{E}}{k_B T} \right) \]
Set Derivative to Zero
\[ \frac{1}{2\sqrt{E}} - \frac{\sqrt{E}}{k_B T} = 0 \]
\[ \frac{1}{2\sqrt{E}} = \frac{\sqrt{E}}{k_B T} \]
\[ \frac{1}{2} = \frac{E}{k_B T} \]
\[ E_p = \frac{1}{2} k_B T \]
📊 Comparison of Molecular Energies
| Energy Type |
Formula |
Value Relative to Ep |
| Most Probable Energy (Ep) |
\(\frac{1}{2} k_B T\) |
1.000 Ep |
| Average Energy (\(\bar{E}\)) |
\(\frac{3}{2} k_B T\) |
3.000 Ep |
Frequently Asked Questions
❓ Why are there three different measures of molecular speed?
The three different measures of molecular speed serve different purposes:
- Most Probable Speed (vp): The speed at which the distribution function reaches its maximum. This is the speed most likely to be observed for any individual molecule.
- Average Speed (\(\bar{v}\)): The arithmetic mean of all molecular speeds. Useful for calculating transport properties like diffusion.
- Root Mean Square Speed (vrms): Related to the average kinetic energy. Most relevant for pressure and temperature calculations.
Their relationship is: vp < \(\bar{v}\) < vrms
❓ How does temperature affect the Maxwell-Boltzmann distribution?
Temperature has a significant effect on the Maxwell-Boltzmann distribution:
- Peak Shifts: As temperature increases, the peak of the distribution shifts to higher speeds
- Distribution Broadens: The distribution becomes wider and flatter at higher temperatures
- Area Under Curve: The total area under the curve remains constant (normalization condition)
- High-Speed Tail: The fraction of molecules with very high speeds increases with temperature
This explains why chemical reaction rates typically increase with temperature - more molecules have sufficient energy to overcome activation barriers.
❓ What is the physical significance of the mean free path?
The mean free path is crucial for understanding various physical phenomena:
- Transport Properties: Determines rates of diffusion, thermal conductivity, and viscosity in gases
- Vacuum Physics: At very low pressures, mean free path can exceed container dimensions
- Atmospheric Science: Explains why the sky is blue (Rayleigh scattering)
- Engineering Applications: Important in vacuum technology, semiconductor manufacturing, and aerosol science
At standard temperature and pressure, the mean free path for air molecules is about 68 nm, while at an altitude of 100 km, it's about 10 cm.
Read More HRK Physics Notes
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