Derive RMS speed of gas molecules from kinetic theory

Question

Using the kinetic theory of gases, derive the expression for the RMS speed of gas molecules.

(NCERT Class 11, Chapter 13)

Solution — Step by Step

From kinetic theory, the pressure exerted by an ideal gas is:

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

where $\rho$ is the gas density and $\overline{v^2}$ is the mean square speed.

This comes from averaging the momentum transfer of molecules hitting the walls of the container.

For $N$ molecules in volume $V$ with molecular mass $m$ :

$\rho = \frac{Nm}{V}$ , so:

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

From the ideal gas law: $PV = NkT$ (where $k$ is Boltzmann’s constant).

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

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

The RMS speed is:

v_{\text{rms}} = \sqrt{\overline{v^2}} = \sqrt{\frac{3kT}{m}}

For one mole ( $N = N_A$ , $Nk = R$ , $Nm = M$ = molar mass):

\boxed{v_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3kT}{m}}}

Why This Works

The kinetic theory connects microscopic molecular motion to macroscopic quantities like pressure and temperature. The RMS speed is the square root of the average of squared speeds — it represents a “typical” speed that accounts for the energy of molecular motion.

Key proportionalities:

$v_{\text{rms}} \propto \sqrt{T}$ — hotter gas, faster molecules
$v_{\text{rms}} \propto 1/\sqrt{M}$ — lighter molecules move faster

At room temperature ( $300\,\text{K}$ ), $v_{\text{rms}}$ for nitrogen is about $517\,\text{m/s}$ and for hydrogen is about $1920\,\text{m/s}$ . Hydrogen is 14 times lighter, so it’s $\sqrt{14} \approx 3.7$ times faster.

Alternative Method — From kinetic energy

The average KE of a molecule is $\frac{3}{2}kT$ (from equipartition theorem, 3 translational degrees of freedom).

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

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

v_{\text{rms}} = \sqrt{\frac{3kT}{m}}

This is faster and more direct if you already know the equipartition result.

Three important speeds in kinetic theory: $v_{\text{rms}} = \sqrt{3RT/M}$ , $v_{\text{avg}} = \sqrt{8RT/(\pi M)}$ , $v_{\text{mp}} = \sqrt{2RT/M}$ . Their ratio is $v_{\text{mp}} : v_{\text{avg}} : v_{\text{rms}} = 1 : 1.128 : 1.224$ . JEE frequently asks you to rank these or use the correct one in a given context.

Common Mistake

Students often confuse $m$ (mass of one molecule) with $M$ (molar mass). The formula $v_{\text{rms}} = \sqrt{3kT/m}$ uses molecular mass $m$ with Boltzmann constant $k$ . The formula $v_{\text{rms}} = \sqrt{3RT/M}$ uses molar mass $M$ with gas constant $R$ . Mixing them — using $R$ with $m$ or $k$ with $M$ — gives answers off by a factor of $\sqrt{N_A} \approx 7.76 \times 10^{11}$ . Always check: if using $R$ , use molar mass; if using $k$ , use molecular mass.

Question

Solution — Step by Step

Why This Works

Alternative Method — From kinetic energy

Common Mistake

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