Kinetic Theory of Gases — Ideal Gas, Maxwell Distribution, Mean Free Path

From Molecules to Pressure

Kinetic theory connects the microscopic world (random molecular motion) to macroscopic properties (pressure, temperature). A gas is just a huge number of tiny particles moving randomly — and from this simple model, we can derive the ideal gas law, understand temperature at a molecular level, and predict how energy distributes among molecules.

CBSE Class 11 gives 3-5 marks. JEE Main tests 1 question on average, usually on rms speed or degrees of freedom.

graph TD
    A[Kinetic Theory] --> B[Assumptions]
    B --> C[Large number of molecules]
    B --> D[Random motion]
    B --> E[Elastic collisions]
    B --> F[No intermolecular forces]
    A --> G[Key Results]
    G --> H["P = ⅓ρv²rms"]
    G --> I["KE = ³⁄₂kT per molecule"]
    G --> J["vrms = √3kT/m"]
    G --> K[Degrees of freedom → energy]

Essential Formulas

PV = nRT = Nk_BT

$n$ = moles, $R = 8.314$ J/(mol K), $N$ = number of molecules, $k_B = 1.38 \times 10^{-23}$ J/K.

P = \frac{1}{3}\rho v_{rms}^2 = \frac{1}{3}\frac{Nm}{V}v_{rms}^2

Speed	Formula
RMS speed	$v_{rms} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3RT}{M}}$
Average speed	$v_{avg} = \sqrt{\frac{8k_BT}{\pi m}}$
Most probable	$v_{mp} = \sqrt{\frac{2k_BT}{m}}$

Order: $v_{mp} < v_{avg} < v_{rms}$

Energy per molecule $= \frac{f}{2}k_BT$ where $f$ = degrees of freedom.

Gas type	$f$	$C_V$	$C_P$	$\gamma = C_P/C_V$
Monoatomic	3	$\frac{3}{2}R$	$\frac{5}{2}R$	5/3
Diatomic	5	$\frac{5}{2}R$	$\frac{7}{2}R$	7/5
Polyatomic	6+	$\frac{6}{2}R$	$\frac{8}{2}R$	4/3

\lambda = \frac{1}{\sqrt{2}\pi d^2 n}

where $d$ = molecular diameter, $n$ = number density.

Solved Examples

Example 1 (Easy — CBSE)

Find the rms speed of oxygen molecules at 27°C. ( $M = 32 \times 10^{-3}$ kg/mol)

$v_{rms} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3 \times 8.314 \times 300}{32 \times 10^{-3}}} = \sqrt{233,6 \times 10^3/0.032} = \mathbf{483.5 \text{ m/s}}$

Example 2 (Medium — JEE Main)

Two containers have the same gas at temperatures $T_1 = 300$ K and $T_2 = 600$ K. Find the ratio of rms speeds.

$v_{rms} \propto \sqrt{T}$ . Ratio $= \sqrt{300/600} = \sqrt{1/2} = \mathbf{1:\sqrt{2}}$ .

Example 3 (Medium — JEE Main)

A diatomic gas at STP. Find the total internal energy of 1 mole.

$U = \frac{f}{2}nRT = \frac{5}{2} \times 1 \times 8.314 \times 273 = \mathbf{5674 \text{ J}}$

Common Mistakes to Avoid

Mistake 1 — Using Celsius instead of Kelvin. All kinetic theory formulas use absolute temperature (Kelvin). $T(K) = T(°C) + 273$ .

Mistake 2 — Confusing molecular mass with molar mass. $m$ (in $v_{rms} = \sqrt{3k_BT/m}$ ) is mass of ONE molecule. $M$ (in $v_{rms} = \sqrt{3RT/M}$ ) is molar mass in kg/mol.

Mistake 3 — Wrong degrees of freedom at high temperature. At room temperature, diatomic gases have $f = 5$ (3 translational + 2 rotational). Vibrational modes ( $f = 7$ ) activate at higher temperatures.

Practice Questions

Q1. Find the average KE of a gas molecule at 27°C.

$KE = \frac{3}{2}k_BT = 1.5 \times 1.38 \times 10^{-23} \times 300 = 6.21 \times 10^{-21}$ J.

Q2. At what temperature will the rms speed of hydrogen equal that of oxygen at 300 K?

$\sqrt{3RT_H/M_H} = \sqrt{3R \times 300/M_O}$ . $T_H/2 = 300/32$ . $T_H = 18.75$ K.

Q3. If temperature doubles, by what factor does rms speed change?

$v \propto \sqrt{T}$ . Factor $= \sqrt{2} \approx 1.414$ .

Q4. Find $\gamma$ for a mixture of 1 mole He and 1 mole H $_2$ .

$C_V = \frac{1 \times 3R/2 + 1 \times 5R/2}{2} = 2R$ . $C_P = C_V + R = 3R$ . $\gamma = 3/2 = 1.5$ .

FAQs

What is the physical meaning of temperature?

Temperature is a measure of the average kinetic energy of molecules: $\frac{1}{2}mv_{rms}^2 = \frac{3}{2}k_BT$ . Higher temperature means faster molecules.

Why do heavier molecules move slower?

At the same temperature, all gases have the same average KE. Since $KE = \frac{1}{2}mv^2$ , heavier molecules must have lower speed to have the same energy.

What is the Maxwell-Boltzmann distribution?

It describes the spread of molecular speeds in a gas. Most molecules have speeds near $v_{mp}$ , with a tail extending to higher speeds. The distribution broadens and flattens at higher temperatures.

Why does the ideal gas law fail at high pressure and low temperature?

At high pressure, molecules are close together — intermolecular forces matter. At low temperature, molecular volume is significant relative to container volume. Real gas corrections (van der Waals equation) account for these.

Deep Concepts You Must Master

Assumptions of kinetic theory

The kinetic theory model rests on five assumptions. Every derivation and every formula in this chapter follows from these:

A gas consists of a very large number of identical molecules moving randomly in all directions.
The size of molecules is negligible compared to the average distance between them.
Collisions between molecules and with container walls are perfectly elastic — kinetic energy is conserved.
There are no intermolecular forces except during collisions (molecules move in straight lines between collisions).
The time spent in a collision is negligible compared to the time between collisions.

JEE 2023 Shift 2 asked: “Which assumption of kinetic theory breaks down for a real gas at high density?” The answer is assumption 4 — intermolecular forces become significant when molecules are packed close together.

Pressure from molecular motion — the derivation

Consider a cube of side $L$ containing $N$ molecules. Focus on one molecule with velocity components $v_x$ , $v_y$ , $v_z$ .

When the molecule hits the wall perpendicular to the x-axis, it bounces back elastically. Change in momentum per collision = $2mv_x$ . The molecule travels distance $2L$ between consecutive hits on the same wall. Time between hits = $2L/v_x$ . Force from this one molecule = $\frac{2mv_x}{2L/v_x} = \frac{mv_x^2}{L}$ .

Summing over all $N$ molecules and dividing by wall area $L^2$ :

P = \frac{1}{L^3}\sum_{i=1}^{N} mv_{x_i}^2 = \frac{Nm\overline{v_x^2}}{V}

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

P = \frac{1}{3}\frac{Nm}{V}v_{rms}^2 = \frac{1}{3}\rho v_{rms}^2

This is the kinetic theory expression for pressure. From this single result, we derive everything else.

Comparing $PV = \frac{1}{3}Nmv_{rms}^2$ with $PV = Nk_BT$ :

\frac{1}{2}mv_{rms}^2 = \frac{3}{2}k_BT

Temperature IS the average kinetic energy of molecules (up to a constant). This is the deepest result in the chapter.

Law of equipartition of energy

Each quadratic degree of freedom carries energy $\frac{1}{2}k_BT$ per molecule (or $\frac{1}{2}RT$ per mole). A “degree of freedom” is an independent way the molecule can store energy.

Translation: 3 degrees (motion along x, y, z). Every gas has these.
Rotation: Linear molecules have 2 (rotate about two axes perpendicular to the bond). Non-linear molecules have 3.
Vibration: Each vibrational mode has 2 degrees (one kinetic, one potential). Active at high temperatures.

For $f$ degrees of freedom:

C_V = \frac{f}{2}R, \quad C_P = C_V + R = \frac{f+2}{2}R, \quad \gamma = \frac{f+2}{f}

This predicts $\gamma = 5/3$ for monoatomic, $7/5$ for diatomic (at room temperature), and $4/3$ for polyatomic gases.

Students often use $f = 7$ for diatomic gases at room temperature. Vibrational modes activate only above about 1000 K for most diatomic gases. At room temperature, use $f = 5$ (3 translational + 2 rotational).

Mean free path — what it tells us

The mean free path $\lambda$ is the average distance a molecule travels between two successive collisions.

\lambda = \frac{1}{\sqrt{2}\pi d^2 n}

where $d$ is the molecular diameter and $n = N/V$ is the number density. At STP, $\lambda$ for air molecules is about $10^{-7}$ m (0.1 $\mu$ m) — roughly 100 molecular diameters.

Mean free path increases when:

Pressure decreases (fewer molecules per unit volume)
Temperature increases (molecules spread out)

This concept explains why vacuum insulation works — in a good vacuum, $\lambda$ exceeds the container size, so molecules hit walls more than each other, dramatically reducing heat conduction.

More Solved Examples

Example 4 (Hard — JEE Advanced)

The rms speed of hydrogen at 300 K is $v$ . At what temperature will the rms speed of oxygen equal $v$ ?

$v_{rms} = \sqrt{\frac{3RT}{M}}$

For hydrogen at 300 K: $v = \sqrt{\frac{3R \times 300}{2 \times 10^{-3}}}$

For oxygen at temperature $T'$ : $v = \sqrt{\frac{3RT'}{32 \times 10^{-3}}}$

Equating: $\frac{300}{2} = \frac{T'}{32}$

$T' = \frac{300 \times 32}{2} = \mathbf{4800 \text{ K}}$

Oxygen is 16 times heavier than hydrogen, so it needs 16 times the temperature to reach the same rms speed.

Example 5 (Hard — JEE Main)

Find the ratio of average KE per molecule of O $_2$ and N $_2$ at the same temperature.

Average KE per molecule = $\frac{3}{2}k_BT$ — this depends ONLY on temperature, not on the type of gas.

Ratio = $\frac{(3/2)k_BT}{(3/2)k_BT} = \mathbf{1:1}$

This is a common trap question. Students try to use molecular mass, but KE per molecule at the same temperature is always the same — it is speed that differs, not energy.

Additional Practice Questions

Q5. The rms speed of a gas at 300 K is 500 m/s. What is it at 1200 K?

$v \propto \sqrt{T}$ . $v' = 500 \times \sqrt{1200/300} = 500 \times 2 = 1000$ m/s.

Q6. A vessel contains a mixture of 2 moles of He and 1 mole of H $_2$ at temperature $T$ . Find the total internal energy.

He (monoatomic, $f=3$ ): $U_1 = 2 \times \frac{3}{2}RT = 3RT$ . H $_2$ (diatomic, $f=5$ ): $U_2 = 1 \times \frac{5}{2}RT = \frac{5}{2}RT$ . Total: $U = 3RT + \frac{5}{2}RT = \frac{11}{2}RT$ .

Q7. Find the mean free path of nitrogen at STP if the molecular diameter is $3.7 \times 10^{-10}$ m. ( $n = 2.7 \times 10^{25}$ m $^{-3}$ at STP)

$\lambda = \frac{1}{\sqrt{2}\pi d^2 n} = \frac{1}{\sqrt{2} \times 3.14 \times (3.7 \times 10^{-10})^2 \times 2.7 \times 10^{25}}$ $= \frac{1}{1.414 \times 3.14 \times 1.369 \times 10^{-19} \times 2.7 \times 10^{25}}$ $= \frac{1}{1.64 \times 10^{7}} \approx 6.1 \times 10^{-8}$ m $\approx 0.06\,\mu$ m.

Q8. At what temperature does the average kinetic energy of a molecule become $1 \text{ eV}$ ? (1 eV = $1.6 \times 10^{-19}$ J)

$\frac{3}{2}k_BT = 1.6 \times 10^{-19}$ $T = \frac{2 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}} = \frac{3.2 \times 10^{-19}}{4.14 \times 10^{-23}} \approx 7729$ K.

Exam Weightage

CBSE Class 11: 3–5 marks. Expect one derivation (pressure from kinetic theory) and one numerical (rms speed or degrees of freedom). The derivation is a 5-marker in some years.

JEE Main: 1 question per paper on average. Most frequent: rms speed comparison between gases, mixture problems ( $\gamma$ of a mixture), and mean free path. The degrees of freedom table is tested every year.

NEET: Rarely tested directly, but the concept of temperature as average KE helps in thermodynamics questions.

The three speeds always obey $v_{mp} : v_{avg} : v_{rms} = 1 : 1.128 : 1.224$ (or equivalently $\sqrt{2} : \sqrt{8/\pi} : \sqrt{3}$ ). If any problem gives you one speed, you can find the other two immediately.