Kinetic Theory

CBSE Class 11 Physics • Chapter 12

Chapter Overview

This chapter covers the fundamental concepts of Kinetic Theory. Ensure you understand the definitions and derivations thoroughly.

12.1 Assumptions of Kinetic Theory

12.2 Pressure of an Ideal Gas

$v_x$ Area A

Figure 12.1: Molecule colliding with wall of container

Consider a cube of side $L$. A molecule with velocity $(v_x, v_y, v_z)$ hits the wall perpendicular to x-axis.

Momentum change per collision: $\Delta p = -mv_x - (mv_x) = -2mv_x$. Momentum imparted to wall = $+2mv_x$.

Time between collisions: $\Delta t = 2L/v_x$.

Force $F = \frac{\Delta p}{\Delta t} = \frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}$.

Total Force (sum over N molecules): $F_{net} = \frac{m}{L} \sum v_{x}^2$.

Pressure $P = F/A = F/L^2 = \frac{m}{L^3} \sum v_{x}^2 = \frac{m}{V} (N \overline{v_x^2})$.

Since random motion: $\overline{v_x^2} = \overline{v_y^2} = \overline{v_z^2} = \frac{1}{3} \overline{v^2}$.

$$ P = \frac{1}{3} \frac{Nm}{V} \overline{v^2} = \frac{1}{3} \rho \overline{v^2} $$

where $\overline{v^2}$ is Mean Square Speed.

Kinetic Interpretation of Temperature

From pressure eq: $PV = \frac{1}{3} Nm \overline{v^2}$. Also $PV = nRT = (N/N_A) RT = N k_B T$.

$\frac{1}{3} Nm \overline{v^2} = N k_B T \Rightarrow \frac{1}{2} m \overline{v^2} = \frac{3}{2} k_B T$.

Average Kinetic Energy of a molecule is proportional to Absolute Temperature.

$$ E = \frac{3}{2} k_B T $$
Practice Problem 1

Q: Calculation the rms speed of Oxygen molecules at $27^\circ C$. ($M = 32g$)

Ans: $v_{rms} = \sqrt{\frac{3RT}{M}}$. $T = 300K$, $M = 32 \times 10^{-3} kg/mol$.
$v_{rms} = \sqrt{\frac{3 \times 8.31 \times 300}{32 \times 10^{-3}}} = \sqrt{\frac{7479}{0.032}} = \sqrt{233718} \approx \mathbf{483 \text{ m/s}}$.

12.3 Degrees of Freedom (f)

The number of independent terms in the expression for internal energy of a molecule.

12.4 Law of Equipartition of Energy

Statement: In thermal equilibrium, the total energy of a dynamic system is equally distributed amongst its degrees of freedom, and the energy associated with each degree of freedom is $\frac{1}{2} k_B T$.

Total Internal Energy $U = \frac{f}{2} N k_B T = \frac{f}{2} n R T$.

12.5 Specific Heat Capacity

Mayer's Formula: $C_p - C_v = R$.

Ratio of Specific Heats $\gamma = C_p / C_v = 1 + \frac{2}{f}$.

12.6 Mean Free Path ($\lambda$)

$\lambda_1$ $\lambda_2$ $\lambda_3$ $\lambda_4$ Collision Cylinder ($\pi d^2 v \Delta t$)

Figure 12.2: Random path and Mean Free Path

Average distance a molecule travels between two successive collisions.

$$ \lambda = \frac{1}{\sqrt{2} n \pi d^2} $$
Practice Problem 2 (Mean Free Path)

Q: Estimate the mean free path for a water molecule in water vapour at $373 K$. ($d \approx 2 \times 10^{-10} m$, $n \approx 3 \times 10^{25} m^{-3}$ for steam at 1 atm).

Ans: $\lambda = \frac{1}{1.414 \times 3 \times 10^{25} \times 3.14 \times (2 \times 10^{-10})^2}$.
$\lambda = \frac{1}{1.414 \times 3 \times 3.14 \times 4 \times 10^{-20} \times 10^{25}}$.
$\lambda \approx \frac{1}{53.3 \times 10^5} \approx \mathbf{1.9 \times 10^{-7} m}$. (Comparable to inter-molecular distance).

Important Question Bank

Q3. Conceptual

Question: Why does the temperature of a gas rise when it is compressed rapidly?

Answer: Rapid compression is an **Adiabatic** process ($\Delta Q = 0$). By First Law, $\Delta W$ is negative (work done on gas), so $\Delta U = -\Delta W$ is positive. Internal energy increases, so Temperature ($T \propto U$) increases.

Study Tip: Master the derivations in this chapter as they are frequently asked in exams. Practice numericals based on the formulas.