Welcome, future physicists! In this detailed session, we're going to dive deep into two fundamental concepts in the Kinetic Theory of Gases:
Mean Free Path and
Degrees of Freedom. These ideas are crucial for understanding the microscopic behavior of gases and lay the groundwork for many advanced topics in thermodynamics. Let's start with the journey from the basics.
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1. Mean Free Path (λ)
Imagine a gas inside a container. According to the Kinetic Theory, the gas molecules are in constant, random motion, colliding with each other and with the walls of the container. While we often idealize molecules as point masses for simplicity in some contexts, for understanding collisions and their consequences, we must consider them to have a finite size. If they have finite size, they *will* collide.
Now, picture a single molecule hurtling through this chaotic environment. It travels some distance, then collides with another molecule, changes direction, travels another distance, collides again, and so on. The path between two successive collisions is called a
free path. Since the motion is random and the number of molecules is enormous, these free paths will vary in length. The
mean free path (denoted by
λ) is simply the average distance a molecule travels between two successive collisions.
Conceptual Significance:
* It tells us how "crowded" a gas is. A shorter mean free path means more frequent collisions, indicating a denser gas.
* It's vital for understanding transport phenomena like diffusion, viscosity, and thermal conductivity in gases.
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1.1 Derivation of the Mean Free Path Formula
Let's derive an expression for the mean free path. We'll make some simplifying assumptions initially and then refine them.
Assumption 1 (Simplified Model):
Consider a single gas molecule of diameter 'd' moving with an average speed 'v', while all other molecules are stationary.
For a collision to occur, the center of our moving molecule must come within a distance 'd' from the center of any other stationary molecule. This means our moving molecule effectively sweeps out a cylindrical volume of radius 'd' (and cross-sectional area
πd²) as it moves.
In a time interval Δt, this molecule travels a distance
vΔt. The volume swept out is
V_swept = (πd²) (vΔt).
If 'n' is the number density of the gas (number of molecules per unit volume,
n = N/V), then the number of collisions in time Δt would be the number of stationary molecules within this swept volume:
Number of collisions = n * V_swept = n (πd² vΔt)
The collision frequency (Z), which is the number of collisions per unit time, would be:
Z = n πd² v
The mean free path (λ) is the average distance traveled divided by the average number of collisions in that distance (or distance traveled in unit time divided by collisions in unit time):
λ = v / Z = v / (n πd² v) = 1 / (n πd²)
Assumption 2 (More Realistic Model - Accounting for Relative Motion):
The previous derivation assumed only one molecule is moving, and all others are stationary. This isn't true; all molecules are moving randomly. When we account for the relative velocity between colliding molecules, the effective collision cross-section changes.
Considering the average relative speed (
v_rel) between molecules moving randomly, it can be shown that
v_rel ≈ √2 * v, where 'v' is the average speed of a single molecule.
Incorporating this factor, the refined formula for the mean free path becomes:
λ = 1 / (√2 n π d²)
Where:
*
λ = mean free path
*
n = number density of gas molecules (
N/V), in
molecules/m³
*
d = diameter of a gas molecule, in meters (assuming spherical molecules)
JEE Focus: This formula is extremely important. Understand its derivation helps in remembering it and its dependencies.
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1.2 Factors Affecting Mean Free Path
From the formula
λ = 1 / (√2 n π d²) , we can analyze the factors:
1.
Number Density (n): As 'n' increases (more molecules per unit volume), molecules are closer together, leading to more frequent collisions and a
smaller mean free path.
λ ∝ 1/n.
2.
Molecular Diameter (d): Larger molecules present a bigger target for collisions. Therefore, a larger molecular diameter leads to a
smaller mean free path.
λ ∝ 1/d².
*
Important Note: The dependence on
d² means molecular size has a significant impact!
3.
Temperature (T) and Pressure (P): We can express 'n' in terms of P and T using the ideal gas law for a single molecule:
PV = NkT, so
n = N/V = P/(kT), where 'k' is Boltzmann's constant.
Substituting this into the mean free path formula:
λ = 1 / (√2 (P/kT) π d²) = kT / (√2 π d² P)
*
Temperature (T): At constant pressure, as temperature increases, molecules move faster, but the density 'n' decreases. The decrease in density dominates, so higher temperature leads to a
larger mean free path.
λ ∝ T (at constant P).
*
Pressure (P): At constant temperature, as pressure increases, the gas becomes denser ('n' increases), leading to more frequent collisions and a
smaller mean free path.
λ ∝ 1/P (at constant T).
Summary Table:
Parameter |
Effect on Mean Free Path (λ) |
Relationship |
|---|
Number Density (n) |
Inverse relation (↑ n, ↓ λ) |
λ ∝ 1/n |
Molecular Diameter (d) |
Inverse square relation (↑ d, ↓ λ) |
λ ∝ 1/d² |
Temperature (T) |
Direct relation (↑ T, ↑ λ) at constant Pressure |
λ ∝ T (at constant P) |
Pressure (P) |
Inverse relation (↑ P, ↓ λ) at constant Temperature |
λ ∝ 1/P (at constant T) |
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1.3 Collision Frequency (Z)
The
collision frequency is the average number of collisions a single molecule undergoes per unit time. We derived it during the mean free path derivation (simplified model):
Z = n π d² v (simplified)
Using the refined mean free path and average speed:
Z = v_avg / λ
where
v_avg is the average speed of the molecules.
Substituting λ:
Z = v_avg * (√2 n π d²)
Since
v_avg ∝ √T (from Maxwell-Boltzmann distribution,
v_avg = √(8kT/πm)) and
n ∝ P/T, we can see:
Z ∝ (√T) * (P/T) = P/√T.
So, collision frequency increases with pressure and decreases with temperature (at constant P, the effect of T on n dominates over the effect of T on v_avg, leading to lower collision frequency).
Example 1: Mean Free Path Calculation
Consider an ideal gas at STP (Standard Temperature and Pressure: T = 273.15 K, P = 1 atm = 1.013 x 10⁵ Pa). Assume the molecular diameter 'd' is 3 x 10⁻¹⁰ m. Calculate the mean free path.
(Boltzmann constant
k = 1.38 x 10⁻²³ J/K)
Step-by-step Solution:
1.
Identify Given Values:
P = 1.013 x 10⁵ Pa
T = 273.15 K
d = 3 x 10⁻¹⁰ m
k = 1.38 x 10⁻²³ J/K
2.
Choose the Appropriate Formula:
Since P and T are given, use
λ = kT / (√2 π d² P)
3.
Substitute and Calculate:
π ≈ 3.14159
d² = (3 x 10⁻¹⁰ m)² = 9 x 10⁻²⁰ m²
√2 ≈ 1.414
λ = (1.38 x 10⁻²³ J/K * 273.15 K) / (1.414 * 3.14159 * 9 x 10⁻²⁰ m² * 1.013 x 10⁵ Pa)
λ ≈ (3.769 x 10⁻²¹) / (4.045 x 10⁻¹³)
λ ≈ 9.31 x 10⁻⁸ m
4.
Result: The mean free path is approximately
9.31 x 10⁻⁸ meters (or 93.1 nm). This is about 300 times the molecular diameter, indicating that molecules travel quite far, relatively speaking, between collisions.
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2. Degrees of Freedom (f)
The concept of
degrees of freedom is fundamental to understanding how a molecule can store energy. It's not just about kinetic energy of translation; molecules can also rotate and vibrate.
A
degree of freedom of a dynamical system is defined as the total number of independent ways in which the system can store energy or the total number of independent coordinates required to completely specify the position and orientation of a molecule in space.
Each degree of freedom is associated with a specific type of motion (translational, rotational, or vibrational) and, according to the
Law of Equipartition of Energy (which we'll discuss briefly here and in more detail in later sections), contributes
(1/2)kT of energy to the internal energy of the system, where 'k' is Boltzmann's constant and 'T' is the absolute temperature.
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2.1 Types of Degrees of Freedom
1.
Translational Degrees of Freedom:
These correspond to the independent ways a molecule can move as a whole in space. Since space is 3-dimensional, any molecule (whether monoatomic, diatomic, or polyatomic) can move along the x, y, and z axes.
* Therefore, every molecule has
3 translational degrees of freedom.
* Energy associated with each:
(1/2)mvₓ², (1/2)mvᵧ², (1/2)mv_z².
2.
Rotational Degrees of Freedom:
These correspond to the independent ways a molecule can rotate about its center of mass.
*
Monoatomic Gases (e.g., He, Ne, Ar): These molecules are essentially point masses or perfect spheres. Their moment of inertia about any axis passing through their center is extremely small, making the rotational kinetic energy negligible.
* Hence, monoatomic gases have
0 rotational degrees of freedom. (In some advanced texts, it's considered 3 but with negligible energy contribution, so effectively 0).
*
Diatomic Gases (e.g., H₂, O₂, N₂, CO): These molecules can be visualized as two point masses joined by a rigid rod. They can rotate about two axes perpendicular to the line joining the atoms. Rotation about the internuclear axis is generally negligible because the mass is concentrated along this axis, giving a very small moment of inertia.
* Hence, diatomic gases have
2 rotational degrees of freedom.
* Energy associated with each:
(1/2)Iω₁² and
(1/2)Iω₂².
*
Polyatomic Gases:
*
Linear Polyatomic (e.g., CO₂, C₂H₂): Similar to diatomic molecules, they have 2 rotational degrees of freedom, as rotation about the molecular axis is negligible.
*
Non-linear Polyatomic (e.g., H₂O, NH₃, CH₄): These molecules can rotate about three mutually perpendicular axes.
* Hence, non-linear polyatomic gases have
3 rotational degrees of freedom.
* Energy associated with each:
(1/2)Iₓωₓ², (1/2)Iᵧωᵧ², (1/2)I_zω_z².
3.
Vibrational Degrees of Freedom:
These correspond to the ways atoms within a molecule can oscillate relative to each other. Vibrational motion involves both kinetic and potential energy. Each vibrational mode contributes
two degrees of freedom (one for kinetic energy and one for potential energy).
*
Monoatomic Gases: No bonds, no vibrational motion.
0 vibrational degrees of freedom.
*
Diatomic Gases: Can vibrate along the internuclear axis (stretching and compressing the bond). This is one vibrational mode.
* Each vibrational mode counts as
2 degrees of freedom. So, a diatomic molecule has
2 vibrational degrees of freedom.
*
Polyatomic Gases: Can have multiple vibrational modes (e.g., symmetric stretch, asymmetric stretch, bending modes). The number of vibrational modes for a non-linear molecule with N atoms is
3N - 6. For a linear molecule, it is
3N - 5.
* Each mode contributes
2 degrees of freedom.
JEE Focus - Important Consideration for Vibrational Modes:
Vibrational modes are generally "activated" only at
high temperatures. At moderate (room) temperatures, the energy available is often insufficient to excite these vibrational modes significantly. Unless specified otherwise, for JEE problems, it's usually assumed that vibrational degrees of freedom are
not excited at room temperature. When a problem mentions "high temperature," then you must include vibrational degrees of freedom.
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2.2 Total Degrees of Freedom (f) for Different Gases
Let's summarize the total degrees of freedom (f) for different types of gases:
Gas Type |
Translational |
Rotational |
Vibrational (at moderate T) |
Total (f) at moderate T |
Total (f) at high T |
|---|
Monoatomic (e.g., He, Ar) |
3 |
0 |
0 |
3 |
3 |
Diatomic (e.g., H₂, O₂, N₂) |
3 |
2 |
0 |
5 |
7 (3+2+2) |
Polyatomic (Non-linear) (e.g., H₂O, NH₃) |
3 |
3 |
0 |
6 |
6 + (2 * (3N-6)) |
Polyatomic (Linear) (e.g., CO₂, C₂H₂) |
3 |
2 |
0 |
5 |
5 + (2 * (3N-5)) |
Example 2: Degrees of Freedom
Determine the total degrees of freedom for the following molecules at room temperature and at high temperature, where applicable:
a) Neon (Ne)
b) Carbon Monoxide (CO)
c) Methane (CH₄)
d) Carbon Dioxide (CO₂)
Step-by-step Solution:
a)
Neon (Ne):
* Type: Monoatomic gas.
* Translational: 3
* Rotational: 0 (negligible moment of inertia)
* Vibrational: 0 (no bonds)
*
Total (f) at room temperature = 3
*
Total (f) at high temperature = 3 (vibrational modes are not possible)
b)
Carbon Monoxide (CO):
* Type: Diatomic gas.
* Translational: 3
* Rotational: 2 (about two axes perpendicular to internuclear axis)
* Vibrational: 0 (at room temperature, not excited)
*
Total (f) at room temperature = 3 + 2 + 0 = 5
* Vibrational (at high temperature): 1 mode, contributing 2 degrees of freedom.
*
Total (f) at high temperature = 3 + 2 + 2 = 7
c)
Methane (CH₄):
* Type: Polyatomic, non-linear gas (Carbon in center, four Hydrogens tetrahedrally arranged). (N=5 atoms)
* Translational: 3
* Rotational: 3 (can rotate about x, y, z axes)
* Vibrational: 0 (at room temperature, not excited)
*
Total (f) at room temperature = 3 + 3 + 0 = 6
* Vibrational (at high temperature): Number of vibrational modes =
3N - 6 = 3(5) - 6 = 15 - 6 = 9 modes. Each mode contributes 2 degrees of freedom. So,
2 * 9 = 18 vibrational degrees of freedom.
*
Total (f) at high temperature = 3 + 3 + 18 = 24 (Note: This value is rarely asked for JEE unless specifically stated, focusing mostly on simpler molecules or up to diatomic high-T cases).
d)
Carbon Dioxide (CO₂):
* Type: Polyatomic, linear gas (O=C=O). (N=3 atoms)
* Translational: 3
* Rotational: 2 (like diatomic, rotation about O-C-O axis is negligible)
* Vibrational: 0 (at room temperature, not excited)
*
Total (f) at room temperature = 3 + 2 + 0 = 5
* Vibrational (at high temperature): Number of vibrational modes =
3N - 5 = 3(3) - 5 = 9 - 5 = 4 modes. So,
2 * 4 = 8 vibrational degrees of freedom.
*
Total (f) at high temperature = 3 + 2 + 8 = 13
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3. Interrelation and JEE Significance
Mean free path and degrees of freedom, while distinct concepts, are both crucial for understanding the macroscopic properties of gases from a microscopic perspective.
* The
mean free path helps us quantify the collision rate and, therefore, transport properties like viscosity, thermal conductivity, and diffusion. It's a measure of how "spread out" the gas molecules are and how often they interact.
*
Degrees of freedom are essential for determining the internal energy of a gas, its specific heat capacities (Cv and Cp), and thus the adiabatic index (γ). They quantify how much energy a molecule can store in its various motions.
Both concepts are frequently tested in JEE, often in combination with other topics like the ideal gas law, specific heats, and thermodynamic processes. A solid understanding of these foundational elements is key to mastering Kinetic Theory of Gases. Remember to pay close attention to the conditions (temperature, pressure) given in problems, especially when deciding whether to include vibrational degrees of freedom!