Derive PV = (1/3)nmv squared from kinetic theory

Question

Derive the expression $PV = \frac{1}{3}nm\bar{v^2}$ from kinetic theory of gases. Define each term in the expression.

Solution — Step by Step

The kinetic theory models a gas as a collection of a very large number of molecules with these assumptions:

Molecules are identical, rigid point particles (negligible size)
Molecules move randomly in all directions with varying speeds
All collisions (molecule-molecule and molecule-wall) are perfectly elastic
No intermolecular forces except during collisions
Average spacing between molecules >> molecular diameter

Under these assumptions, we can calculate the pressure exerted by gas molecules colliding with the container walls.

Consider a cubical container of side length $l$ , volume $V = l^3$ . The gas contains $N$ molecules, each of mass $m$ .

Consider one molecule moving with velocity component $v_x$ in the x-direction (toward the right wall).

When it hits the right wall (elastic collision), $v_x$ reverses: $v_x \to -v_x$ .

Change in momentum of the molecule = $m(-v_x) - m(v_x) = -2mv_x$

Impulse on the wall = $+2mv_x$ (by Newton’s third law)

After hitting the right wall, the molecule travels to the left wall (distance $l$ ), bounces, and returns.

Time between successive collisions with the right wall = $\frac{2l}{v_x}$

Force from this one molecule = $\frac{\text{Impulse}}{\text{Time}} = \frac{2mv_x}{2l/v_x} = \frac{mv_x^2}{l}$

For $N$ molecules with x-velocity components $v_{x1}, v_{x2}, \ldots, v_{xN}$ :

Total force on right wall = $\frac{m}{l}(v_{x1}^2 + v_{x2}^2 + \cdots + v_{xN}^2) = \frac{mN\bar{v_x^2}}{l}$

where $\bar{v_x^2} = \frac{1}{N}\sum v_{xi}^2$ is the mean square of x-velocity components.

Pressure = $\frac{\text{Force}}{\text{Area}} = \frac{mN\bar{v_x^2}/l}{l^2} = \frac{mN\bar{v_x^2}}{l^3} = \frac{mN\bar{v_x^2}}{V}$

For random molecular motion (no preferred direction), the average kinetic energy is equally distributed among all three directions:

\bar{v_x^2} = \bar{v_y^2} = \bar{v_z^2}

Since total speed $v^2 = v_x^2 + v_y^2 + v_z^2$ :

\overline{v^2} = \bar{v_x^2} + \bar{v_y^2} + \bar{v_z^2} = 3\bar{v_x^2}

Therefore: $\bar{v_x^2} = \frac{\overline{v^2}}{3}$

Substituting back:

P = \frac{mN \cdot \overline{v^2}/3}{V}

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

Where $n = N/V$ is the number density, this can also be written as $P = \frac{1}{3}nm\overline{v^2}$ .

Why This Works

The derivation rests on two key ideas: (1) Newton’s laws applied to molecular collisions give the impulse per collision, and (2) the equipartition of kinetic energy among the three spatial directions connects $\bar{v_x^2}$ to $\overline{v^2}$ .

The result $PV = \frac{1}{3}Nm\overline{v^2}$ is a bridge between microscopic molecular properties (mass $m$ , speed $v$ ) and macroscopic observables ( $P$ , $V$ ). Combined with $PV = nRT$ (ideal gas law), it gives:

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

The average kinetic energy of a molecule is directly proportional to its absolute temperature — this is one of the most profound results in physics. It tells us that temperature is a measure of the average translational kinetic energy of molecules.

Alternative Method

The derivation can be extended to give the root mean square speed ( $v_{rms}$ ):

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

where $M$ is the molar mass. At higher temperatures or lower molar mass, $v_{rms}$ is larger — molecules move faster. At room temperature (298 K), nitrogen molecules ( $M = 28$ g/mol) have $v_{rms} \approx 515$ m/s.

For JEE Main and CBSE Class 11, this derivation appears as a 5-mark question. The three key steps are: (1) momentum change per collision = $2mv_x$ , (2) time between collisions = $2l/v_x$ , (3) the isotropy argument $\bar{v_x^2} = \overline{v^2}/3$ . The isotropy step is often under-explained but is crucial — explicitly state “since molecular motion is isotropic (no preferred direction), $\bar{v_x^2} = \bar{v_y^2} = \bar{v_z^2}$ .”

Common Mistake

Students often skip the step connecting $\bar{v_x^2}$ to $\overline{v^2}$ , jumping directly from the expression with $v_x^2$ to the final formula with $v^2$ . This is the crucial isotropy argument — without it, you’d have $PV = Nm\bar{v_x^2}$ (not the correct factor of $1/3$ ). The factor of 3 in the denominator comes specifically from distributing kinetic energy equally among x, y, and z directions. Another mistake: using density ( $\rho = nm$ ) incorrectly. In the formula $P = \frac{1}{3}\rho\overline{v^2}$ , $\rho$ is the mass density (kg/m³), while in $PV = \frac{1}{3}Nm\overline{v^2}$ , $N$ is the total number of molecules.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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