Physics: Class 11 (Book 2) — Chapter 12: Kinetic Theory

12.1 Introduction & Molecular Nature of Matter

The kinetic theory of gases explains the macroscopic behaviour of gases (such as pressure, temperature, and volume) by analyzing the microscopic motion of rapidly moving atoms or molecules. This theory bridges the gap between mechanics and thermodynamics, providing a molecular interpretation of thermodynamic variables.

Historical Chronicle: The Atomic Hypothesis

  • Ancient Foundations: Acharya Kanada in India (Vaiseshika school, 6th Century B.C.) proposed that matter is composed of eternal, indivisible, and ultimate particles called Paramanu . Democritus in Greece (4th Century B.C.) coined the term "atom" (meaning indivisible) and suggested that different substances differ due to the shape and size of their constituent atoms.
  • Modern Formulation: John Dalton (1808) established the scientific Atomic Theory to explain the laws of definite and multiple proportions. Richard Feynman regarded the discovery that "matter is made of atoms" as the most significant piece of scientific knowledge to pass to future generations.
Key Postulates & Molecular Scales
  • Atomic Size: The typical size of an atom is of the order of 1 \text{ \AA} ($10^{-10}\text{ m}$).
  • Interatomic Distances: In solids and liquids, atoms are tightly packed with a spacing of about 2 \text{ \AA} . In gases, the interatomic distances are much larger, typically in tens of angstroms ( 20 \text{ \AA} to 30 \text{ \AA} ).
  • Intermolecular Forces: Molecules experience long-range attraction at larger separations and short-range repulsion when squeezed close together. In gases, because molecules are far apart, these forces are negligible except during collisions.
12.2 Behaviour of Gases & The Ideal Gas Equation

An ideal gas is a theoretical gas model whose molecules have negligible volume and no mutual interactions. Real gases approach ideal gas behaviour at low pressures and high temperatures .

The Equation of State

The state of an ideal gas is governed by the relation:

$$PV = \mu R T$$

Where:

  • $P$ is the pressure of the gas.
  • $V$ is the volume.
  • $T$ is the absolute temperature in Kelvin.
  • $\mu$ is the number of moles, defined as $\mu = \frac{M}{M_0} = \frac{N}{N_A}$.
  • $R$ is the universal gas constant ($R = 8.314\text{ J mol}^{-1}\text{K}^{-1}$).

Alternative forms of the ideal gas equation:

$$PV = k_B N T \quad \text{or} \quad P = k_B n T$$

Where $k_B = \frac{R}{N_A} = 1.38 \times 10^{-23}\text{ J K}^{-1}$ is the Boltzmann constant , $N$ is the total number of molecules, and $n = \frac{N}{V}$ is the number density.

Gas Laws Derived from the Ideal Gas Equation
  • Boyle's Law: For a fixed mass of gas at constant temperature, the volume is inversely proportional to pressure ($PV = \text{constant}$).
  • Charles's Law: For a fixed mass of gas at constant pressure, the volume is directly proportional to absolute temperature ($V \propto T$).
  • Dalton's Law of Partial Pressures: The total pressure exerted by a mixture of non-interacting ideal gases is equal to the sum of the partial pressures of individual gases: $$P = P_1 + P_2 + \dots = (\mu_1 + \mu_2 + \dots)\frac{RT}{V}$$

Mathematical Problem Statement: Estimate the ratio of the molecular volume to the total volume occupied by water vapour at $100^\circ\text{C}$ and $1\text{ atm}$ pressure. Given: Density of liquid water = $1000\text{ kg m}^{-3}$, Density of water vapour = $0.6\text{ kg m}^{-3}$.


Step-by-Step Algorithmic Solution Trace:

  1. The molecular volume is approximately equal to the volume occupied by the same mass of water in its liquid state, where molecules are closely packed.
  2. Let $M$ be a given mass of water. The volume in the liquid state is $V_{\text{liquid}} = \frac{M}{\rho_{\text{water}}}$.
  3. The volume in the vapour state is $V_{\text{vapour}} = \frac{M}{\rho_{\text{vapour}}}$.
  4. The ratio of molecular volume to total volume is: $$\frac{V_{\text{liquid}}}{V_{\text{vapour}}} = \frac{\rho_{\text{vapour}}}{\rho_{\text{water}}} = \frac{0.6}{1000} = 6 \times 10^{-4}$$

Final Evaluation Output: The ratio of molecular volume to total volume is $6 \times 10^{-4}$.

12.3 Kinetic Theory of an Ideal Gas: Pressure & Temperature
Derivation of Pressure Exerted by a Gas

Consider a gas enclosed in a cube of side $l$ (volume $V = l^3$). Let a molecule of mass $m$ move with velocity $(v_x, v_y, v_z)$.

  1. When the molecule collides elastically with the wall parallel to the $yz$-plane, its $x$-component of velocity reverses from $v_x$ to $-v_x$, while $v_y$ and $v_z$ remain unchanged.
  2. The change in momentum of the molecule is: $$\Delta p_x = -m v_x - m v_x = -2 m v_x$$
  3. By conservation of momentum, the momentum imparted to the wall in a single collision is $2 m v_x$.
  4. In a small time interval $\Delta t$, the molecules that can hit the wall of area $A = l^2$ must be within a distance $v_x \Delta t$. The volume of this region is $A v_x \Delta t$.
  5. On average, only half of the molecules in this volume are moving towards the wall. Thus, the number of collisions in time $\Delta t$ is: $$N_{\text{collisions}} = \frac{1}{2} A v_x \Delta t n$$ where $n$ is the number density of molecules.
  6. The total momentum transferred to the wall in time $\Delta t$ is: $$Q = (2 m v_x) \left(\frac{1}{2} n A v_x \Delta t\right) = n m A v_x^2 \Delta t$$
  7. The force exerted on the wall is the rate of momentum transfer: $$F = \frac{Q}{\Delta t} = n m A v_x^2$$
  8. The pressure $P$ is force per unit area: $$P = \frac{F}{A} = n m v_x^2$$
  9. Since molecules have a distribution of velocities, we replace $v_x^2$ with its average value $\bar{v_x^2}$: $$P = n m \bar{v_x^2}$$
  10. Assuming the gas is isotropic, there is no preferred direction of motion: $$\bar{v_x^2} = \bar{v_y^2} = \bar{v_z^2} = \frac{1}{3} \bar{v^2}$$ where $\bar{v^2}$ is the mean square speed.
  11. Substituting this gives the fundamental pressure equation: $$P = \frac{1}{3} n m \bar{v^2} = \frac{1}{3} \rho \bar{v^2}$$ where $\rho = n m$ is the mass density of the gas.
Kinetic Interpretation of Temperature

The pressure equation can be rewritten as:

$$PV = \frac{1}{3} N m \bar{v^2} = \frac{2}{3} N \left(\frac{1}{2} m \bar{v^2}\right)$$

Comparing this with the ideal gas equation $PV = N k_B T$, we get:

$$\frac{1}{2} m \bar{v^2} = \frac{3}{2} k_B T$$

This shows that the average translational kinetic energy of a molecule is directly proportional to the absolute temperature of the gas and is independent of its pressure, volume, or nature.

The total internal energy $E$ of a mole of ideal gas (purely translational) is:

$$E = \frac{3}{2} R T$$
Root Mean Square (rms) Speed

The root mean square speed $v_{\text{rms}}$ is defined as:

$$v_{\text{rms}} = \sqrt{\bar{v^2}} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M_0}}$$

Mathematical Problem Statement: A flask contains argon and chlorine in the ratio of 2:1 by mass at $27^\circ\text{C}$. Find the ratio of (i) average kinetic energy per molecule, and (ii) root mean square speed $v_{\text{rms}}$ of the molecules of the two gases. (Atomic mass of $\text{Ar} = 39.9\text{ u}$, Molecular mass of $\text{Cl}_2 = 70.9\text{ u}$).


Step-by-Step Algorithmic Solution Trace:

  1. The average kinetic energy per molecule of any ideal gas depends only on temperature: $E_{\text{avg}} = \frac{3}{2} k_B T$. Since both gases are at the same temperature ($27^\circ\text{C}$), the ratio of their average kinetic energies is $1:1$.
  2. The rms speed is given by $v_{\text{rms}} = \sqrt{\frac{3 R T}{M_0}}$.
  3. The ratio of rms speeds is inversely proportional to the square root of their molecular masses: $$\frac{v_{\text{rms, Ar}}}{v_{\text{rms, Cl}}} = \sqrt{\frac{M_{\text{Cl}}}{M_{\text{Ar}}}} = \sqrt{\frac{70.9}{39.9}} \approx 1.33$$

Final Evaluation Output: (i) Ratio of average kinetic energy = $1:1$, (ii) Ratio of rms speeds = $1.33$.

12.4 Law of Equipartition of Energy

The law of equipartition of energy states that for any dynamic system in thermal equilibrium, the total energy is distributed equally among all its degrees of freedom, and the average energy associated with each quadratic degree of freedom is $\frac{1}{2} k_B T$.

Degrees of Freedom ($f$)

The number of independent coordinates required to completely specify the position and configuration of a system is called its degrees of freedom .

  • Monatomic Gas: Has only 3 translational degrees of freedom ($f = 3$). Example: Helium, Argon.
  • Diatomic Gas (Rigid Rotator): Has 3 translational and 2 rotational degrees of freedom ($f = 5$). Example: $\text{O}_2$, $\text{N}_2$, $\text{CO}$.
  • Diatomic Gas (Non-Rigid): At high temperatures, vibrational modes are excited. A single vibrational mode contributes 2 quadratic terms (kinetic energy and potential energy), adding 2 degrees of freedom ($f = 7$).
12.5 Specific Heat Capacities of Gases & Solids

Using the law of equipartition of energy, we can theoretically predict the molar specific heat capacities at constant volume ($C_v$) and constant pressure ($C_p$), and their ratio ($\gamma = \frac{C_p}{C_v}$).

Theoretical Predictions for Gases
Gas Type Degrees of Freedom ($f$) Internal Energy ($U$) $C_v$ ($\text{J mol}^{-1}\text{K}^{-1}$) $C_p$ ($\text{J mol}^{-1}\text{K}^{-1}$) $\gamma = \frac{C_p}{C_v}$
Monatomic 3 $\frac{3}{2}RT$ $\frac{3}{2}R \approx 12.5$ $\frac{5}{2}R \approx 20.8$ $\frac{5}{3} \approx 1.67$
Diatomic (Rigid) 5 $\frac{5}{2}RT$ $\frac{5}{2}R \approx 20.8$ $\frac{7}{2}R \approx 29.1$ $\frac{7}{5} = 1.40$
Diatomic (Vibrational) 7 $\frac{7}{2}RT$ $\frac{7}{2}R \approx 29.1$ $\frac{9}{2}R \approx 37.4$ $\frac{9}{7} \approx 1.29$
Polyatomic $3 + 3 + f$ $(3+f)RT$ $(3+f)R$ $(4+f)R$ $\frac{4+f}{3+f}$
Specific Heat Capacity of Solids

In a solid, each atom is held in a lattice and vibrates about its mean position in 3 dimensions. Each 1D oscillation has 2 quadratic energy terms (kinetic and potential), so a 3D oscillation has 6 quadratic terms.

  • Average energy per atom = $3 k_B T$.
  • Total internal energy of 1 mole of solid ($N = N_A$): $$U = 3 R T$$
  • Molar specific heat capacity of solids (Dulong-Petit Law): $$C = \frac{dU}{dT} = 3 R \approx 24.9\text{ J mol}^{-1}\text{K}^{-1}$$

Mathematical Problem Statement: A cylinder of fixed capacity $44.8\text{ litres}$ contains helium gas at standard temperature and pressure (STP). What is the amount of heat needed to raise the temperature of the gas in the cylinder by $15.0^\circ\text{C}$? (Given $R = 8.31\text{ J mol}^{-1}\text{K}^{-1}$).


Step-by-Step Algorithmic Solution Trace:

  1. At STP, 1 mole of any ideal gas occupies $22.4\text{ litres}$. Thus, the number of moles of helium is: $$\mu = \frac{44.8}{22.4} = 2\text{ moles}$$
  2. Since the volume of the cylinder is fixed, the process is isochoric (constant volume). The heat required is given by: $$Q = \mu C_v \Delta T$$
  3. Helium is a monatomic gas, so its molar specific heat at constant volume is $C_v = \frac{3}{2} R$.
  4. Substitute the values: $$Q = 2 \times \left(\frac{3}{2} R\right) \times 15.0 = 45 R$$
  5. Calculate the final value: $$Q = 45 \times 8.31 = 374\text{ J}$$

Final Evaluation Output: The amount of heat needed is $374\text{ J}$.

12.6 Mean Free Path

The mean free path ($l$) is the average distance a gas molecule travels between two successive collisions.

Derivation of Mean Free Path Formula
  1. Assume molecules are spheres of diameter $d$. Let a single molecule move with average speed $\langle v \rangle$ while all other molecules are temporarily assumed to be at rest.
  2. In time $\Delta t$, the moving molecule sweeps out a collision cylinder of volume: $$V_{\text{cylinder}} = \pi d^2 \langle v \rangle \Delta t$$
  3. If $n$ is the number density of molecules, the number of collisions in time $\Delta t$ is: $$N_{\text{collisions}} = n \pi d^2 \langle v \rangle \Delta t$$
  4. The mean free path $l$ is the total distance traveled divided by the number of collisions: $$l = \frac{\langle v \rangle \Delta t}{n \pi d^2 \langle v \rangle \Delta t} = \frac{1}{n \pi d^2}$$
  5. When the motion of all molecules is taken into account using Maxwell's velocity distribution, the relative velocity $\langle v_r \rangle = \sqrt{2}\langle v \rangle$ is used, yielding the exact formula: $$l = \frac{1}{\sqrt{2} n \pi d^2}$$

Mathematical Problem Statement: Estimate the mean free path for a water molecule in water vapour at $373\text{ K}$. Use the molecular diameter $d = 2 \times 10^{-10}\text{ m}$ and the number density at STP adjusted for $373\text{ K}$.


Step-by-Step Algorithmic Solution Trace:

  1. The number density $n$ is inversely proportional to absolute temperature. At STP ($273\text{ K}$), $n_0 \approx 2.7 \times 10^{25}\text{ m}^{-3}$.
  2. At $373\text{ K}$, the adjusted number density is: $$n = 2.7 \times 10^{25} \times \frac{273}{373} \approx 2 \times 10^{25}\text{ m}^{-3}$$
  3. Using the exact mean free path formula: $$l = \frac{1}{\sqrt{2} n \pi d^2}$$
  4. Substitute the values ($d = 2 \times 10^{-10}\text{ m}$): $$l = \frac{1}{1.414 \times (2 \times 10^{25}) \times 3.1416 \times (2 \times 10^{-10})^2}$$ $$l = \frac{1}{1.414 \times 2 \times 10^{25} \times 3.1416 \times 4 \times 10^{-20}} \approx 4 \times 10^{-7}\text{ m}$$

Final Evaluation Output: The estimated mean free path is $4 \times 10^{-7}\text{ m}$.