Welcome to Lecture 108 of our comprehensive JEE Physics Master Series! This landmark lecture concludes Unit 9: Kinetic Theory of Gases with an in-depth study of molecular transport foundations: the Mean Free Path, Collision Frequency, and Avogadro’s Number. In previous lectures, we treated molecules as point masses obeying classical kinematics. Here, we introduce finite molecular dimensions ($d$), derive the collision cross-section ($\sigma = \pi d^2$), establish Clausius and Maxwell’s relative-velocity formulation for mean free path ($\lambda = \frac{1}{\sqrt{2} n \pi d^2}$), explore its temperature and pressure proportionalities, analyze collision frequency ($Z_1$) and relaxation time ($\tau$), investigate classical experimental methods for determining Avogadro’s Number ($N_A$), and master rigorous JEE Advanced problem patterns.
Detailed Concept Card: Mean Free Path & Molecular Collisions
1. Physical Picture & Collision Cross-Section
Gas molecules are modeled as rigid hard spheres of effective diameter $d$. A collision occurs whenever the center-to-center distance between two molecules becomes less than or equal to $d$:
- Collision Cross-Section Area ($\sigma$): The effective circular target area centered on a molecule within which another molecule’s center cannot enter without colliding: $$\mathbf{\sigma = \pi d^2}$$
- Collision Volume Swept per Unit Time ($\mathcal{V}$): If a test molecule moves with average relative velocity $v_{\text{rel}}$, the cylindrical volume swept out in time $\Delta t$ is: $$\mathcal{V} = \sigma \langle v_{\text{rel}} \rangle \Delta t = \pi d^2 \langle v_{\text{rel}} \rangle \Delta t$$
2. Derivation of the Mean Free Path ($\lambda$)
The mean free path ($\lambda$) is defined as the average distance traversed by a molecule between two successive collisions:
$$\lambda = \frac{\text{Total Distance Traversed in time } \Delta t}{\text{Total Number of Collisions in time } \Delta t} = \frac{\bar{v} \Delta t}{N_{\text{coll}}}$$
A. Simplified Model (Stationary Target Molecules)
If all other molecules are assumed frozen in space while a single molecule moves at speed $\bar{v}$, the number of collisions equals the number of molecules inside the swept cylinder of volume $\pi d^2 \bar{v} \Delta t$:
$$N_{\text{coll, simple}} = n \cdot (\pi d^2 \bar{v} \Delta t)$$
where $n = N/V$ is the molecular number density. Hence:
$$\lambda_{\text{simple}} = \frac{\bar{v} \Delta t}{n \pi d^2 \bar{v} \Delta t} = \frac{1}{n \pi d^2}$$
B. Maxwell-Clausius Relative Velocity Correction
In thermal equilibrium, target molecules also move randomly in all directions according to the Maxwell-Boltzmann distribution. The average relative velocity between two identical particles moving isotropically is:
$$\langle v_{\text{rel}} \rangle = \sqrt{\langle |\vec{v}_1 – \vec{v}_2|^2 \rangle} = \sqrt{\langle v_1^2 + v_2^2 – 2 \vec{v}_1 \cdot \vec{v}_2 \rangle} = \sqrt{\bar{v}^2 + \bar{v}^2 – 0} = \mathbf{\sqrt{2} \, \bar{v}}$$
The actual number of collisions in time $\Delta t$ is therefore scaled up by $\sqrt{2}$:
$$N_{\text{coll}} = n (\pi d^2 \langle v_{\text{rel}} \rangle \Delta t) = \sqrt{2} n \pi d^2 \bar{v} \Delta t$$
Substituting into the mean free path definition yields the exact Maxwell formula:
$$\mathbf{\lambda = \frac{1}{\sqrt{2} n \pi d^2}}$$
3. Parametric Proportionalities of $\lambda$
Using the ideal gas equation of state $P = n k_B T \implies n = \frac{P}{k_B T}$, the mean free path can be expressed in terms of macroscopic state variables:
$$\mathbf{\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}}$$
Expressed in terms of gas mass density $\rho = m n = \frac{M P}{R T}$ (where $m$ is the mass of a single molecule):
$$\mathbf{\lambda = \frac{m}{\sqrt{2} \pi d^2 \rho}}$$
Core Thermodynamic Proportionalities for JEE:
1. Isochoric Process ($V = \text{const}, \rho = \text{const}, n = \text{const}$): $\lambda = \text{constant}$! The mean free path is strictly independent of temperature and pressure in a rigid closed vessel.
2. Isobaric Process ($P = \text{const}$): $\lambda \propto T \propto V$. Heating at constant pressure causes the gas to expand, reducing density and increasing free path.
3. Isothermal Process ($T = \text{const}$): $\lambda \propto \frac{1}{P} \propto V$. Compressing a gas isothermally reduces the clearance between molecules linearly.
4. Adiabatic Process ($P V^\gamma = \text{const}$): Since $\lambda \propto V$, $\lambda \propto P^{-1/\gamma}$. As the gas expands adiabatically, $\lambda$ increases.
4. Collision Frequency ($Z_1$) & Relaxation Time ($\tau$)
- Mean Collision Time (Relaxation Time, $\tau$): The average time between consecutive collisions of a molecule: $$\tau = \frac{\lambda}{\bar{v}} = \frac{1}{\sqrt{2} n \pi d^2 \bar{v}} = \frac{k_B T}{\sqrt{2} \pi d^2 P \bar{v}}$$
- Collision Frequency of a Single Molecule ($Z_1$): Number of collisions experienced by one molecule per unit time: $$Z_1 = \frac{1}{\tau} = \mathbf{\sqrt{2} n \pi d^2 \bar{v}}$$ Since average speed $\bar{v} = \sqrt{\frac{8 k_B T}{\pi m}} \propto \sqrt{T}$: – At constant volume ($n = \text{const}$): $Z_1 \propto \bar{v} \propto \mathbf{\sqrt{T}}$. – At constant pressure ($n \propto 1/T$): $Z_1 \propto \frac{1}{T} \sqrt{T} = \mathbf{\frac{1}{\sqrt{T}}}$.
- Total Collision Rate per Unit Volume ($Z_{11}$): The total number of binary collisions occurring per unit volume per second (divided by $2$ to prevent double-counting): $$Z_{11} = \frac{1}{2} n Z_1 = \mathbf{\frac{1}{\sqrt{2}} n^2 \pi d^2 \bar{v}}$$
5. Avogadro’s Number ($N_A$) & Methods of Determination
Avogadro’s number ($N_A \approx 6.022 \times 10^{23}\text{ mol}^{-1}$) represents the fundamental bridge between macroscopic laboratory quantities and individual atomic masses ($R = N_A k_B$). Historically, its exact determination validated the atomic hypothesis:
- Perrin’s Sedimentation Equilibrium: Microscopic colloidal particles suspended in liquid undergo Brownian motion. In thermal equilibrium under gravity, their vertical concentration obeys the Boltzmann barometric formula: $$n(z) = n_0 \exp\left(-\frac{m_{\text{eff}} g z}{k_B T}\right) = n_0 \exp\left(-\frac{N_A m_{\text{eff}} g z}{R T}\right)$$ where $m_{\text{eff}} = V_{\text{particle}}(\rho – \rho_{\text{fluid}})$ is the buoyant effective mass. By measuring $n(z)$ under a microscope, Jean Perrin determined $N_A$ (Nobel Prize 1926).
- Electrolysis (Faraday’s Law): The Faraday constant $F \approx 96485\text{ C/mol}$ is the total charge carried by one mole of univalent ions: $$N_A = \frac{F}{e}$$ where $e = 1.602 \times 10^{-19}\text{ C}$ is the elementary electronic charge.
- X-Ray Crystallography: For an elemental crystalline solid of molar mass $M$, density $\rho$, and cubic unit cell edge length $a$ containing $n_{\text{cell}}$ atoms per unit cell: $$N_A = \frac{n_{\text{cell}} M}{\rho a^3}$$
Four Graded Solved Examples
Example 1 (JEE Main Level: Microscopic Collision Dynamics of Nitrogen)
Problem Statement: Nitrogen gas ($\text{N}_2$, molar mass $M = 28.0\text{ g/mol}$) is maintained at standard room temperature $T = 300\text{ K}$ and atmospheric pressure $P = 1.013 \times 10^5\text{ Pa}$. The effective collision diameter of a nitrogen molecule is $d = 0.36\text{ nm} = 3.6 \times 10^{-10}\text{ m}$. Given $k_B = 1.38 \times 10^{-23}\text{ J/K}$ and $R = 8.314\text{ J/(mol}\cdot\text{K)}$:
1. Calculate the molecular number density $n$ in molecules per $\text{m}^3$.
2. Calculate the mean free path $\lambda$ of the nitrogen molecules in nanometers.
3. Calculate the average molecular speed $\bar{v}$.
4. Find the mean time between collisions ($\tau$) and the collision frequency ($Z_1$).
Solution:
Step 1: Molecular Number Density ($n$).
From the ideal gas law:
$$n = \frac{P}{k_B T} = \frac{1.013 \times 10^5\text{ Pa}}{(1.38 \times 10^{-23}\text{ J/K})(300\text{ K})} = \frac{101300}{4.14 \times 10^{-21}} \approx \mathbf{2.447 \times 10^{25}\text{ molecules/m}^3}$$
Step 2: Mean Free Path ($\lambda$).
The collision cross-section is:
$$\sigma = \pi d^2 = \pi (3.6 \times 10^{-10}\text{ m})^2 \approx 4.0715 \times 10^{-19}\text{ m}^2$$
Using the Maxwell formula:
$$\lambda = \frac{1}{\sqrt{2} n \pi d^2} = \frac{1}{\sqrt{2} (2.447 \times 10^{25}\text{ m}^{-3})(4.0715 \times 10^{-19}\text{ m}^2)}$$
$$\lambda = \frac{1}{1.4142 \times 9.963 \times 10^6} = \frac{1}{1.409 \times 10^7} \approx 7.098 \times 10^{-8}\text{ m} \approx \mathbf{71.0\text{ nm}}$$
Step 3: Average Molecular Speed ($\bar{v}$).
$$\bar{v} = \sqrt{\frac{8 R T}{\pi M}} = \sqrt{\frac{8 \times 8.314 \times 300}{\pi \times 0.0280\text{ kg/mol}}} = \sqrt{\frac{19953.6}{0.087965}} \approx \sqrt{226836} \approx \mathbf{476.3\text{ m/s}}$$
Step 4: Relaxation Time & Collision Frequency.
$$\tau = \frac{\lambda}{\bar{v}} = \frac{7.098 \times 10^{-8}\text{ m}}{476.3\text{ m/s}} \approx \mathbf{1.49 \times 10^{-10}\text{ s}}$$
$$Z_1 = \frac{1}{\tau} \approx \mathbf{6.71 \times 10^9\text{ collisions/second}} \quad (\approx 6.71\text{ GHz})$$
Takeaway: A nitrogen molecule under ambient conditions undergoes nearly $7\text{ billion}$ collisions every second, traveling roughly $200$ times its own molecular diameter between consecutive collisions!
Example 2 (Mathematical Formulation: Evolution of $\lambda$ Across Thermodynamic Paths)
Problem Statement: A fixed mass of an ideal gas initially has a mean free path $\lambda_0$ at state $(P_0, V_0, T_0)$. It is taken to a final volume $V_f = 2V_0$ along four different reversible paths:
1. Path A: Isobaric expansion ($P = P_0$).
2. Path B: Isothermal expansion ($T = T_0$).
3. Path C: Adiabatic expansion ($P V^\gamma = \text{constant}$, with $\gamma = 5/3$).
4. Path D: The gas is instead held in a rigid container of volume $V_0$ and heated isochorically until its temperature doubles to $2T_0$.
Determine the final mean free path ($\lambda_f$) in terms of $\lambda_0$ for each of the four processes.
Solution:
Fundamental Spatial Insight:
Recall that the number density of a closed system containing $N$ molecules is $n = N/V$. Therefore:
$$\lambda = \frac{1}{\sqrt{2} n \pi d^2} = \frac{1}{\sqrt{2} \pi d^2} \left(\frac{V}{N}\right)$$
Since $N$ and $d$ are constants for a fixed quantity of gas, the mean free path $\lambda$ is strictly directly proportional to the volume of the container ($V$):
$$\mathbf{\lambda \propto V}$$
Now evaluate each process:
– Path A (Isobaric Expansion): $V_f = 2V_0 \implies \mathbf{\lambda_f = 2 \lambda_0}$.
– Path B (Isothermal Expansion): $V_f = 2V_0 \implies \mathbf{\lambda_f = 2 \lambda_0}$.
– Path C (Adiabatic Expansion): $V_f = 2V_0 \implies \mathbf{\lambda_f = 2 \lambda_0}$.
– Path D (Isochoric Heating): The volume is fixed ($V_f = V_0$), so number density $n = N/V_0$ remains strictly constant:
$$\mathbf{\lambda_f = \lambda_0}$$
Conclusion: In any process where volume doubles, $\lambda$ doubles, regardless of whether the path is isobaric, isothermal, or adiabatic! In a rigid closed vessel, heating or cooling changes molecular speed and collision frequency, but leaves the mean free path completely unaltered.
Example 3 (JEE Advanced Level: Perrin’s Sedimentation Equilibrium & Avogadro’s Number)
Problem Statement: In Jean Perrin’s classic sedimentation experiment, spherical gamboge particles of density $\rho = 1.20 \times 10^3\text{ kg/m}^3$ and radius $r = 2.0 \times 10^{-7}\text{ m}$ are suspended in pure water of density $\rho_0 = 1.00 \times 10^3\text{ kg/m}^3$ at uniform temperature $T = 300\text{ K}$. Take $g = 9.80\text{ m/s}^2$ and $R = 8.314\text{ J/(mol}\cdot\text{K)}$.
1. Formulate the sedimentation equilibrium relation $n(z)$ in terms of effective buoyant mass $m_{\text{eff}}$.
2. Calculate the effective mass $m_{\text{eff}}$ of a single colloidal particle.
3. If the particle concentration decreases by a factor of $e$ over a vertical height separation of $h = 6.0 \times 10^{-5}\text{ m}$ ($60\ \mu\text{m}$), determine the experimental value of Boltzmann’s constant $k_B$ and Avogadro’s number $N_A$.
Solution:
Step 1: Sedimentation Equilibrium Formulation.
At thermal equilibrium, colloidal particles suspended in liquid behave as giant gas molecules possessing mean thermal kinetic energy $\frac{3}{2} k_B T$. In a gravitational field, balancing the downward effective gravity against the upward osmotic diffusion pressure yields the barometric distribution:
$$n(z) = n_0 \exp\left(-\frac{m_{\text{eff}} g z}{k_B T}\right)$$
Step 2: Effective Buoyant Mass ($m_{\text{eff}}$).
Volume of one spherical particle:
$$V_p = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (2.0 \times 10^{-7}\text{ m})^3 = \frac{32 \pi}{3} \times 10^{-21}\text{ m}^3 \approx 3.351 \times 10^{-20}\text{ m}^3$$
Density difference: $\Delta \rho = \rho – \rho_0 = (1.20 – 1.00) \times 10^3\text{ kg/m}^3 = 200\text{ kg/m}^3$.
$$m_{\text{eff}} = V_p \Delta \rho = (3.351 \times 10^{-20}\text{ m}^3)(200\text{ kg/m}^3) \approx \mathbf{6.702 \times 10^{-18}\text{ kg}}$$
Step 3: Determining $k_B$ and $N_A$.
When $z = h$, the concentration drops by $1/e$:
$$\frac{m_{\text{eff}} g h}{k_B T} = 1 \implies k_B = \frac{m_{\text{eff}} g h}{T}$$
$$m_{\text{eff}} g h = (6.702 \times 10^{-18}\text{ kg}) \times (9.80\text{ m/s}^2) \times (6.0 \times 10^{-5}\text{ m}) \approx 3.941 \times 10^{-21}\text{ J}$$
$$k_B = \frac{3.941 \times 10^{-21}\text{ J}}{300\text{ K}} \approx \mathbf{1.314 \times 10^{-23}\text{ J/K}}$$
Using the macroscopic gas constant $R = 8.314\text{ J/(mol}\cdot\text{K)}$:
$$N_A = \frac{R}{k_B} = \frac{8.314\text{ J/(mol}\cdot\text{K)}}{1.314 \times 10^{-23}\text{ J/K}} \approx \mathbf{6.33 \times 10^{23}\text{ mol}^{-1}}$$
Historical Note: This remarkable experiment by Perrin was among the first direct empirical proofs of the physical reality of molecules and atoms!
Example 4 (JEE Advanced Level: High-Vacuum Knudsen Regime Transition)
Problem Statement: A spherical glass chamber of radius $R_0 = 5.0\text{ cm} = 0.050\text{ m}$ contains Nitrogen gas ($\text{N}_2$, $d = 3.6 \times 10^{-10}\text{ m}$) at temperature $T = 300\text{ K}$.
1. Calculate the critical chamber pressure $P_c$ below which the mean free path $\lambda$ exceeds the internal diameter of the chamber ($2 R_0 = 0.10\text{ m}$).
2. Describe the physical changes in molecular transport when $P < P_c$ (the Knudsen regime).
Solution:
Step 1: Calculate Critical Pressure ($P_c$).
Set $\lambda = 2 R_0 = 0.10\text{ m}$:
$$\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P_c} \implies P_c = \frac{k_B T}{\sqrt{2} \pi d^2 \lambda}$$
$$k_B T = (1.38 \times 10^{-23}\text{ J/K})(300\text{ K}) = 4.14 \times 10^{-21}\text{ J}$$
$$\sqrt{2} \pi d^2 = 1.4142 \times \pi \times (3.6 \times 10^{-10}\text{ m})^2 \approx 5.758 \times 10^{-19}\text{ m}^2$$
$$P_c = \frac{4.14 \times 10^{-21}\text{ J}}{(5.758 \times 10^{-19}\text{ m}^2)(0.10\text{ m})} = \frac{4.14 \times 10^{-21}}{5.758 \times 10^{-20}} \approx \mathbf{0.0719\text{ Pa} \approx 7.2 \times 10^{-2}\text{ Pa}}$$
In vacuum units: $P_c \approx 5.4 \times 10^{-4}\text{ torr}$ (or $\approx 7.1 \times 10^{-7}\text{ atm}$).
Step 2: Physics of the Knudsen Regime.
When $P < P_c$, the Knudsen number $Kn = \frac{\lambda}{L} > 1$ (where $L = 2R_0$ is the characteristic system dimension):
– Intermolecular collisions become extremely rare compared to collisions with the container walls.
– Molecules fly freely along straight-line ballistic trajectories from wall to wall.
– Standard continuum fluid mechanics breaks down: continuum viscosity and thermal conductivity relations cease to apply, and heat transfer is governed by free-molecular transport directly between opposite boundaries.
Practice Worksheet: 10 JEE Problems
Questions 1 to 5: JEE Main Level (Single Correct)
Q1. For a fixed mass of an ideal gas kept at constant temperature, if the pressure is doubled ($P \to 2P$), the mean free path $\lambda$ of its molecules will:
(A) Double ($2\lambda$)
(B) Quadruple ($4\lambda$)
(C) Halve ($\lambda / 2$)
(D) Remain unchanged
Q2. An ideal gas is enclosed in a rigid, sealed container of fixed volume. When the gas is heated so that its absolute temperature increases from $300\text{ K}$ to $600\text{ K}$, the mean free path of the gas molecules:
(A) Increases by a factor of $2$
(B) Increases by a factor of $\sqrt{2}$
(C) Decreases by a factor of $2$
(D) Remains strictly unchanged
Q3. The mean free path ($\lambda$) of gas molecules modeled as hard spheres of diameter $d$ is proportional to:
(A) $d$
(B) $d^{-1}$
(C) $d^2$
(D) $d^{-2}$
Q4. If the average molecular speed of a gas is $\bar{v}$ and its mean free path is $\lambda$, the collision frequency ($Z_1$) experienced by an individual molecule is given by:
(A) $\lambda / \bar{v}$
(B) $\bar{v} / \lambda$
(C) $\sqrt{\bar{v} / \lambda}$
(D) $\bar{v} \cdot \lambda$
Q5. Avogadro’s number $N_A$ is fundamentally related to the universal gas constant $R$ and Boltzmann’s constant $k_B$ through:
(A) $N_A = \frac{R}{k_B}$
(B) $N_A = \frac{k_B}{R}$
(C) $N_A = R \cdot k_B$
(D) $N_A = \frac{R^2}{k_B}$
Questions 6 to 8: JEE Advanced Level (One or More Than One Correct)
Q6. Which of the following statements regarding the mean free path ($\lambda$) and collision frequency ($Z_1$) of ideal gas molecules is/are correct?
(A) In a closed container of fixed volume, the mean free path $\lambda$ is strictly independent of temperature.
(B) In a closed container of fixed volume, the collision frequency of a single molecule scales as $Z_1 \propto \sqrt{T}$.
(C) In an isobaric expansion ($P = \text{constant}$), the mean free path scales as $\lambda \propto T$.
(D) In an isobaric expansion ($P = \text{constant}$), the collision frequency of a single molecule scales as $Z_1 \propto \frac{1}{\sqrt{T}}$.
Q7. An ideal gas undergoes a reversible adiabatic expansion governed by $P V^\gamma = \text{constant}$. Which of the following statements is/are correct regarding the mean free path $\lambda$ and molecular collision dynamics?
(A) The mean free path is directly proportional to volume ($\lambda \propto V$).
(B) The mean free path is related to pressure by $\lambda \propto P^{-1/\gamma}$.
(C) As the gas expands adiabatically, its mean free path increases.
(D) As the gas expands adiabatically, its molecular collision frequency $Z_1$ decreases.
Q8. Which of the following independent experimental methodologies can be utilized to determine Avogadro’s number $N_A$?
(A) Electrolysis by measuring the Faraday constant $F$ and the electronic charge $e$ ($N_A = F / e$).
(B) Perrin’s sedimentation equilibrium of colloidal suspensions in a gravitational field.
(C) X-ray diffraction measurement of lattice constants and density in crystalline solids ($N_A = \frac{n_{\text{cell}} M}{\rho a^3}$).
(D) Observation of mean square Brownian displacement of colloidal particles over time via Einstein’s diffusion relation.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A rigid, closed container holds an ideal gas at temperature $T_1 = 300\text{ K}$. If the gas is heated isochorically until its temperature reaches $T_2 = 1200\text{ K}$, the collision frequency ($Z_1$) of an individual molecule increases by a factor of $n$. Determine the integer value of $n$.
Q10. An ideal gas of molecular diameter $d$ has a mean free path $\lambda_1$ at initial temperature $T_1 = 300\text{ K}$ and pressure $P_1 = 100\text{ kPa}$. The gas is heated to temperature $T_2 = 450\text{ K}$ and compressed to a final pressure $P_2 = 600\text{ kPa}$. If the final mean free path is expressed as $\lambda_2 = \frac{\lambda_1}{m}$, find the integer value of $m$.
Detailed Solutions & Answer Key
Answer Key
| Question | Correct Option / Answer | Topic / Concept |
|---|---|---|
| Q1 | (C) $\lambda / 2$ | Isothermal Pressure Dependence of $\lambda$ |
| Q2 | (D) Remains strictly unchanged | Isochoric Invariance of Mean Free Path |
| Q3 | (D) $d^{-2}$ | Molecular Diameter Scaling of $\lambda$ |
| Q4 | (B) $\bar{v} / \lambda$ | Collision Frequency Definition |
| Q5 | (A) $R / k_B$ | Avogadro’s Number & Gas Constants |
| Q6 | (A), (B), (C), (D) | Comprehensive Thermodynamic Scaling of $\lambda$ and $Z_1$ |
| Q7 | (A), (B), (C), (D) | Adiabatic Dynamics of Mean Free Path & Collisions |
| Q8 | (A), (B), (C), (D) | Experimental Determination Methods of $N_A$ |
| Q9 | 2 | Collision Frequency Ratio at Constant Volume ($\sqrt{1200/300} = 2$) |
| Q10 | 4 | Combined $T$ and $P$ Scaling of Mean Free Path ($m = 4$) |
Complete Step-by-Step Solutions
Solution 1:
From $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P}$: at constant temperature $T$, $\lambda \propto \frac{1}{P}$.
When pressure is doubled ($P \to 2P$), the mean free path is halved ($\lambda \to \lambda / 2$).
Correct Option: (C)
Solution 2:
In a sealed container of fixed volume $V$, the number density $n = N/V$ is constant. Since $\lambda = \frac{1}{\sqrt{2} n \pi d^2}$, the mean free path depends strictly on the spatial density of molecules and is completely independent of temperature. Therefore, $\lambda$ remains strictly unchanged.
Correct Option: (D)
Solution 3:
The collision cross-section is proportional to $d^2$. The mean free path is inversely proportional to cross-section: $\lambda = \frac{1}{\sqrt{2} n \pi d^2} \propto d^{-2}$.
Correct Option: (D)
Solution 4:
The relaxation time (average time between collisions) is $\tau = \lambda / \bar{v}$. The collision frequency is the reciprocal: $Z_1 = 1 / \tau = \bar{v} / \lambda$.
Correct Option: (B)
Solution 5:
By definition of Boltzmann’s constant, $k_B = R / N_A \implies N_A = R / k_B$.
Correct Option: (A)
Solution 6:
(A) Correct: $n = N/V = \text{const} \implies \lambda = \frac{1}{\sqrt{2} n \pi d^2} = \text{const}$.
(B) Correct: $Z_1 = \bar{v} / \lambda$. Since $\lambda$ is constant and $\bar{v} \propto \sqrt{T}$, $Z_1 \propto \sqrt{T}$.
(C) Correct: For $P = \text{const}$, $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} \propto T$.
(D) Correct: For $P = \text{const}$, $Z_1 = \frac{\bar{v}}{\lambda} \propto \frac{\sqrt{T}}{T} = \frac{1}{\sqrt{T}}$.
Correct Options: (A), (B), (C), (D)
Solution 7:
(A) Correct: $\lambda = \frac{V}{\sqrt{2} N \pi d^2} \propto V$.
(B) Correct: For $P V^\gamma = \text{const} \implies V \propto P^{-1/\gamma}$, so $\lambda \propto P^{-1/\gamma}$.
(C) Correct: Expansion means $V$ increases, so $\lambda$ increases.
(D) Correct: During adiabatic expansion, the gas cools ($T$ drops) and volume expands ($\lambda$ increases). Both factors reduce collision frequency: $Z_1 = \bar{v} / \lambda \downarrow$.
Correct Options: (A), (B), (C), (D)
Solution 8:
All four methods are historically verified and conceptually fundamental techniques to determine Avogadro’s number $N_A$ across chemistry, thermodynamics, crystallography, and statistical physics.
Correct Options: (A), (B), (C), (D)
Solution 9:
In a rigid closed container, the volume is constant, so the mean free path $\lambda$ remains unchanged.
The collision frequency is:
$$Z_1 = \frac{\bar{v}}{\lambda} \propto \bar{v} \propto \sqrt{T}$$
$$\frac{Z_2}{Z_1} = \sqrt{\frac{T_2}{T_1}} = \sqrt{\frac{1200\text{ K}}{300\text{ K}}} = \sqrt{4} = \mathbf{2}$$
Thus, $n = 2$.
Answer: 2
Solution 10:
The mean free path is given by $\lambda = \frac{k_B T}{\sqrt{2} \pi d^2 P} \propto \frac{T}{P}$.
$$\frac{\lambda_2}{\lambda_1} = \left(\frac{T_2}{T_1}\right) \left(\frac{P_1}{P_2}\right) = \left(\frac{450\text{ K}}{300\text{ K}}\right) \left(\frac{100\text{ kPa}}{600\text{ kPa}}\right)$$
$$\frac{\lambda_2}{\lambda_1} = \left(\frac{3}{2}\right) \left(\frac{1}{6}\right) = \frac{3}{12} = \frac{1}{4}$$
Therefore, $\lambda_2 = \frac{\lambda_1}{4}$, which implies $m = 4$.
Answer: 4