PV = (1/3)Nmv² → PV = nRT. Connect microscopic to macroscopic.

Derive Ideal Gas Equation from Kinetic Theory

Question

Derive the ideal gas equation $PV = nRT$ starting from the kinetic theory expression $PV = \frac{1}{3}Nm\overline{v^2}$ .

Solution — Step by Step

From first principles of kinetic theory, pressure exerted by $N$ molecules of mass $m$ each in a container of volume $V$ is:

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

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

We need to connect $\overline{v^2}$ to temperature. The trick: express the right-hand side in terms of kinetic energy.

PV = \frac{1}{3}Nm\overline{v^2} = \frac{2}{3} \cdot \frac{1}{2}Nm\overline{v^2} = \frac{2}{3} \cdot KE_{total}

So $PV = \frac{2}{3} E_k$ where $E_k$ is the total translational kinetic energy of all molecules.

From kinetic theory, the average translational KE per molecule is directly proportional to absolute temperature:

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

where $k_B = 1.38 \times 10^{-23}$ J/K is Boltzmann’s constant. This is actually the definition of temperature at the molecular level.

Total KE for $N$ molecules:

E_k = N \cdot \frac{3}{2}k_BT

Plug into our Step 2 result:

PV = \frac{2}{3} \cdot N \cdot \frac{3}{2}k_BT = Nk_BT

If we have $n$ moles, then $N = nN_A$ where $N_A = 6.022 \times 10^{23}$ mol $^{-1}$ is Avogadro’s number.

PV = nN_Ak_BT

Since $R = N_Ak_B = 8.314$ J mol $^{-1}$ K $^{-1}$ :

\boxed{PV = nRT}

Why This Works

The derivation reveals something beautiful: temperature is not a vague “hotness” — it is a precise measure of the average translational kinetic energy of molecules. The entire bridge from microscopic chaos to the clean macroscopic law $PV = nRT$ runs through $\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_BT$ .

The factor of $\frac{2}{3}$ appearing and then cancelling with $\frac{3}{2}$ is not an accident. The $3$ comes from three spatial degrees of freedom — the molecule can move in $x$ , $y$ , and $z$ . Pressure along one wall captures only $\frac{1}{3}$ of the total kinetic energy.

The final result $R = N_A k_B$ is worth memorising as a relationship, not just a number. $k_B$ is the gas constant per molecule; $R$ is the gas constant per mole. They differ by exactly Avogadro’s number.

Alternative Method

If the derivation asks you to “verify” rather than “derive”, you can go backwards — start with $PV = nRT$ , substitute $n = N/N_A$ , and $R/N_A = k_B$ to arrive at $PV = Nk_BT$ , then split out the KE form. JEE sometimes words the question this way.

For dimensional verification, check that both sides of $\frac{1}{2}m\overline{v^2} = \frac{3}{2}k_BT$ have units of energy (J = kg·m²·s⁻²). Left side: $\text{kg} \cdot (\text{m/s})^2$ = J ✓. Right side: $k_B$ has units J/K, multiply by K → J ✓.

Common Mistake

Many students write $PV = \frac{1}{3}mv^2$ — dropping $N$ entirely. Remember, $N$ is the total number of molecules in the container. You cannot get pressure from a single molecule. The formula applies to the collective. A related slip: confusing $N$ (number of molecules) with $n$ (number of moles). They differ by a factor of $N_A$ .

Another common error in JEE context: students assume $\overline{v^2} = (\bar{v})^2$ . These are not equal — the mean of squares is always greater than or equal to the square of the mean. $\sqrt{\overline{v^2}}$ is the RMS speed ( $v_{rms}$ ), not the mean speed ( $\bar{v}$ ). The kinetic theory formula uses RMS speed specifically.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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