Kinetic Theory: Basic Postulates and Assumptions

Understand the fundamental assumptions of kinetic theory of gases for JEE preparation

8 min read

Kinetic Theory: Basic Postulates and Assumptions

Real-Life Hook

Why does the smell of perfume spread throughout a room? The answer lies in kinetic theory! Perfume molecules are in constant random motion, colliding with air molecules billions of times per second. This chaotic motion, called Brownian motion (visible when you see dust particles dancing in sunlight), is direct evidence of the kinetic theory. The same principle explains why helium balloons deflate faster than regular balloons - smaller He atoms slip through microscopic pores more easily due to their higher average speeds!

Core Concepts

What is Kinetic Theory?

Kinetic theory explains the macroscopic properties of gases (pressure, temperature, volume) in terms of the microscopic behavior of individual molecules. It bridges the gap between Newton’s laws at the molecular level and thermodynamic laws at the bulk level.

Fundamental Postulates of Kinetic Theory

The Five Key Assumptions

  1. Gas consists of a large number of molecules: Typically ~10²³ molecules in a mole
  2. Molecules are in random motion: They move in all directions with varying speeds
  3. Molecular size is negligible: Volume of molecules « Volume of container
  4. Collisions are perfectly elastic: No kinetic energy is lost during collisions
  5. No intermolecular forces: Molecules don’t attract or repel each other (except during collision)

Let’s examine each assumption in detail:

1. Large Number of Molecules

  • Allows us to use statistical methods
  • Individual molecular behavior averages out
  • Macroscopic properties become predictable and stable

Mathematical consequence: We can use probability distributions to describe molecular speeds

2. Random Motion

  • Molecules move in straight lines between collisions
  • All directions are equally probable
  • No preferred direction (isotropy)

Consequence: Average velocity of all molecules = 0, but average speed ≠ 0

3. Negligible Molecular Size

Volume Comparison

$$V_{molecules} << V_{container}$$

Typically: $\frac{V_{molecules}}{V_{container}} \approx 10^{-3}$ at STP

Physical meaning: Most of the gas is empty space!

Why this matters: At high pressures, molecules are compressed, and this assumption breaks down → real gas behavior

4. Perfectly Elastic Collisions

Elastic Collision Condition

$$KE_{before} = KE_{after}$$

No energy is converted to heat, sound, or deformation

Implications:

  • Total kinetic energy of the gas remains constant (at constant T)
  • Individual molecular speeds change, but total energy is conserved
  • Momentum is conserved in each collision

Why this matters: This assumption allows us to relate temperature to average kinetic energy

5. No Intermolecular Forces

  • Molecules behave as non-interacting point particles
  • Potential energy between molecules = 0
  • Only kinetic energy matters

Exception: Very brief repulsive force during actual collision

Breakdown: At low temperatures and high pressures, Van der Waals forces become significant → real gas behavior

Pressure from Kinetic Theory

Derivation of Pressure

Consider a cubical container of side $L$ with $N$ molecules of mass $m$:

Pressure Formula from Kinetic Theory

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

Where:

  • $N$ = number of molecules
  • $m$ = mass of each molecule
  • $\overline{v^2}$ = mean square speed
  • $V$ = volume of container

Alternative form:

$$PV = \frac{1}{3}Nm\overline{v^2}$$

Connecting to Ideal Gas Equation

From kinetic theory: $PV = \frac{1}{3}Nm\overline{v^2}$

From thermodynamics: $PV = NkT$

Equating both:

$$\frac{1}{3}Nm\overline{v^2} = NkT$$

Temperature and Kinetic Energy

$$\frac{1}{2}m\overline{v^2} = \frac{3}{2}kT$$

Average kinetic energy per molecule: $\overline{KE} = \frac{3}{2}kT$

Key Insight: Temperature is a direct measure of average molecular kinetic energy!

Interactive Demo: Kinetic Theory in Motion

Visualize how molecular motion relates to temperature and pressure in real-time.

For One Mole of Gas

Average kinetic energy per mole:

$$\overline{KE}_{mole} = \frac{3}{2}RT$$

Where $R = N_A k = 8.314$ J/(mol·K)

Memory Tricks

“LRSEF” for Five Assumptions

  • Large number of molecules
  • Random motion
  • Small size (negligible)
  • Elastic collisions
  • Force-free (no intermolecular forces)

“KET = 3/2 kT” Mnemonic

Kinetic Energy per molecule at Temperature T = (3/2)kT

Remember the 1/3 Factor

The factor 1/3 comes from three dimensions (x, y, z). In one dimension, pressure would be $P = \frac{Nm\overline{v^2}}{V}$, but in 3D it’s distributed equally: $P = \frac{1}{3}\frac{Nm\overline{v^2}}{V}$

Common Unit Conversion Mistakes

Boltzmann Constant (k)

  • k = 1.38 × 10⁻²³ J/K (most common form)
  • Never confuse with $k$ (spring constant) or $K$ (Kelvin)

Mass of Molecules

Common Mistake: Using molar mass instead of molecular mass

  • Molecular mass: $m = \frac{M}{N_A}$ (in kg)
  • Example: For O₂, $M = 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol}$
  • $m = \frac{32 \times 10^{-3}}{6.022 \times 10^{23}} = 5.31 \times 10^{-26}$ kg

Energy Units

  • Always use Joules (J) for energy
  • Temperature must be in Kelvin (K)
  • $1 \text{ eV} = 1.6 \times 10^{-19}$ J (sometimes used in atomic physics)

Important Formulas

Kinetic Energy Relations

Average KE per molecule

$$\overline{KE} = \frac{3}{2}kT$$

Average KE per mole

$$\overline{KE}_{mole} = \frac{3}{2}RT$$

Total KE of N molecules

$$KE_{total} = \frac{3}{2}NkT = \frac{3}{2}nRT$$

Root Mean Square Speed

RMS Speed

$$v_{rms} = \sqrt{\overline{v^2}} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3RT}{M}}$$

Where M = molar mass (kg/mol)

3-Level Practice Problems

Level 1: JEE Main Basics

Problem 1: Calculate the average kinetic energy of one molecule of an ideal gas at 27°C. (k = 1.38 × 10⁻²³ J/K)

Solution

Temperature: $T = 27 + 273 = 300$ K

Average KE per molecule:

$$\overline{KE} = \frac{3}{2}kT = \frac{3}{2} \times 1.38 \times 10^{-23} \times 300$$ $$\overline{KE} = 6.21 \times 10^{-21} \text{ J}$$

Answer: $6.21 \times 10^{-21}$ J

Problem 2: The rms speed of oxygen molecules at a certain temperature is 400 m/s. What is the rms speed of hydrogen molecules at the same temperature? (Molar mass: O₂ = 32 g/mol, H₂ = 2 g/mol)

Solution

Since $v_{rms} \propto \frac{1}{\sqrt{M}}$ at constant temperature:

$$\frac{v_{rms(H_2)}}{v_{rms(O_2)}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$$ $$v_{rms(H_2)} = 4 \times 400 = 1600 \text{ m/s}$$

Answer: 1600 m/s

Level 2: JEE Main Advanced

Problem 3: At what temperature will the average kinetic energy of a gas molecule be exactly equal to 1 eV? (1 eV = 1.6 × 10⁻¹⁹ J, k = 1.38 × 10⁻²³ J/K)

Solution

Given: $\overline{KE} = 1 \text{ eV} = 1.6 \times 10^{-19}$ J

Using: $\overline{KE} = \frac{3}{2}kT$

$$T = \frac{2 \times \overline{KE}}{3k} = \frac{2 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}}$$ $$T = \frac{3.2 \times 10^{-19}}{4.14 \times 10^{-23}} = 7729 \text{ K} \approx 7730 \text{ K}$$

Answer: 7730 K (or about 7457°C)

Problem 4: A vessel contains 2 moles of nitrogen gas at 27°C. Calculate the total kinetic energy of the gas molecules. (R = 8.314 J/(mol·K))

Solution

Given: $n = 2$ mol, $T = 300$ K

Total kinetic energy:

$$KE_{total} = \frac{3}{2}nRT = \frac{3}{2} \times 2 \times 8.314 \times 300$$ $$KE_{total} = 7482.6 \text{ J} \approx 7.48 \text{ kJ}$$

Answer: 7.48 kJ

Level 3: JEE Advanced

Problem 5: A container has two compartments separated by a partition. One compartment has N molecules of He at temperature T₁ and the other has 2N molecules of Ar at temperature T₂. If T₁ = 2T₂, compare the total kinetic energies and pressures in both compartments (assuming equal volumes).

Solution

Kinetic Energy Comparison:

For He: $KE_1 = \frac{3}{2}NkT_1 = \frac{3}{2}Nk(2T_2) = 3NkT_2$

For Ar: $KE_2 = \frac{3}{2}(2N)kT_2 = 3NkT_2$

$$\frac{KE_1}{KE_2} = \frac{3NkT_2}{3NkT_2} = 1$$

Pressure Comparison:

Using $PV = NkT$:

For He: $P_1 = \frac{NkT_1}{V} = \frac{2NkT_2}{V}$

For Ar: $P_2 = \frac{2NkT_2}{V}$

$$\frac{P_1}{P_2} = \frac{2NkT_2/V}{2NkT_2/V} = 1$$

Answer: Both kinetic energies are equal; both pressures are equal.

Key Insight: Total KE depends on nT product, and pressure depends on NT/V, regardless of the type of gas!

Problem 6: Show that the pressure exerted by an ideal gas is equal to two-thirds of the kinetic energy per unit volume.

Solution

From kinetic theory:

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

We know that average kinetic energy per molecule:

$$\overline{KE} = \frac{1}{2}m\overline{v^2}$$

Therefore:

$$m\overline{v^2} = 2\overline{KE}$$

Substituting:

$$P = \frac{1}{3}\frac{N \times 2\overline{KE}}{V} = \frac{2}{3}\frac{N\overline{KE}}{V}$$

Since $N\overline{KE}$ = total kinetic energy and $\frac{N\overline{KE}}{V}$ = kinetic energy per unit volume:

$$P = \frac{2}{3} \times (\text{KE per unit volume})$$

Proved

This is a fundamental result connecting pressure to kinetic energy density!

When Kinetic Theory Breaks Down

Real Gas Behavior

Kinetic theory assumptions fail when:

  1. High Pressure: Molecular volume becomes significant

    • Van der Waals correction: $(P + \frac{an^2}{V^2})(V - nb) = nRT$

  2. Low Temperature: Intermolecular forces become important

    • Gases liquefy when attractive forces dominate

  3. Near Critical Point: Distinct liquid and gas phases merge

Deviations from Ideal Behavior

Compressibility factor: $Z = \frac{PV}{nRT}$

  • $Z = 1$ for ideal gas
  • $Z > 1$ for real gas (repulsive forces dominate)
  • $Z < 1$ for real gas (attractive forces dominate)

Advanced Topics

Key Takeaways

  1. Temperature = Average KE: $\overline{KE} = \frac{3}{2}kT$ is the most important relation
  2. Pressure from motion: Gas pressure arises from molecular collisions with walls
  3. 1/3 factor: Comes from equal distribution of KE in three dimensions
  4. Elastic collisions: Essential for energy conservation at constant temperature
  5. Large N required: Statistical laws need ~10²³ molecules to be valid
  6. Ideal gas assumptions: Break down at high P and low T

Exam Tips

  • JEE Main: Focus on calculating average KE, understanding assumptions, and basic pressure derivations
  • JEE Advanced: Expect conceptual questions on when assumptions break down, derivations, and multi-part problems
  • Common trap: Confusing mass of molecule (m) with molar mass (M)
  • Quick check: Average KE at room temperature (300 K) ≈ 6 × 10⁻²¹ J
  • Conceptual clarity: Understand WHY each assumption is needed, not just WHAT it is

Last updated: February 18, 2025