understand that the root-mean-square speed cr.m.s. is given by c<>

Kinetic Theory of Gases

Objective

To understand that the root‑mean‑square speed of gas molecules is given by

$$c_{\text{rms}} = \sqrt{\dfrac{3k_{\mathrm B}T}{m}} = \sqrt{\dfrac{3RT}{M}}$$

Key Assumptions of the Kinetic Theory

• The gas consists of a large number of tiny particles (atoms or molecules) in constant random motion.
• The particles are point masses; their own volume is negligible compared with the volume of the container.
• No intermolecular forces act except during perfectly elastic collisions.
• Collisions between particles and with the walls of the container are perfectly elastic.
• The time between collisions is much larger than the duration of a collision.

Derivation of the Root‑Mean‑Square Speed

Consider a cubic container of side length $L$ containing $N$ molecules, each of mass $m$. The pressure exerted on a wall arises from the change in momentum of molecules colliding with that wall.

1. For a single molecule moving with velocity components $(v_x, v_y, v_z)$, the momentum change on striking a wall perpendicular to the $x$‑axis is $2mv_x$.
2. The time between successive collisions of this molecule with the same wall is $\Delta t = \dfrac{2L}{|v_x|}$.
3. The average force contributed by this molecule is $$F = \frac{2mv_x}{\Delta t} = \frac{mv_x^2}{L}.$$
4. Summing over all $N$ molecules and using the fact that the motion is isotropic ($\langle v_x^2\rangle = \langle v_y^2\rangle = \langle v_z^2\rangle$), the total pressure is $$p = \frac{1}{3}\,\frac{Nm\langle v^2\rangle}{V},$$ where $V=L^3$ and $\langle v^2\rangle = \langle v_x^2+v_y^2+v_z^2\rangle$.
5. Combining this result with the ideal‑gas equation $pV = Nk_{\mathrm B}T$ gives $$\frac{1}{3}Nm\langle v^2\rangle = Nk_{\mathrm B}T,$$ leading to $$\langle v^2\rangle = \frac{3k_{\mathrm B}T}{m}.$$
6. The root‑mean‑square (r.m.s.) speed is defined as $$c_{\text{rms}} = \sqrt{\langle v^2\rangle} = \sqrt{\frac{3k_{\mathrm B}T}{m}}.$$
7. Using the molar gas constant $R = N_{\mathrm A}k_{\mathrm B}$ and the molar mass $M = N_{\mathrm A}m$, the expression can be written in macroscopic form: $$c_{\text{rms}} = \sqrt{\frac{3RT}{M}}.$$

Physical Significance

• The r.m.s. speed is a statistical measure of the average speed of molecules in a gas at temperature $T$.
• It increases with temperature and decreases with molecular mass.
• It is directly related to the kinetic energy per molecule: $\frac{1}{2}m c_{\text{rms}}^{2}= \frac{3}{2}k_{\mathrm B}T$.

Example Calculation

Find the r.m.s. speed of nitrogen ($\mathrm{N_2}$) molecules at $300\ \text{K}$.

1. Molar mass of $\mathrm{N_2}$: $M = 28.02\ \text{g mol}^{-1}=2.802\times10^{-2}\ \text{kg mol}^{-1}$.
2. Use $R = 8.314\ \text{J mol}^{-1}\text{K}^{-1}$.
3. Apply the formula: $$c_{\text{rms}} = \sqrt{\frac{3RT}{M}} = \sqrt{\frac{3(8.314)(300)}{2.802\times10^{-2}}} \approx 517\ \text{m s}^{-1}.$$

Typical r.m.s. Speeds of Common Gases at 300 K

Gas	Molar Mass $M$ (kg mol⁻¹)	r.m.s. Speed $c_{\text{rms}}$ (m s⁻¹)
Helium (He)	4.00 × 10⁻³	\overline{1} 300
Hydrogen (H₂)	2.02 × 10⁻³	\overline{1} 950
Nitrogen (N₂)	2.80 × 10⁻²	\overline{517}
Oxygen (O₂)	3.20 × 10⁻²	\overline{480}
Carbon Dioxide (CO₂)	4.44 × 10⁻²	\overline{426}

Key Points to Remember

• The r.m.s. speed depends only on temperature and molecular mass; pressure and volume do not appear explicitly in the final expression.
• Higher temperature → higher kinetic energy → higher $c_{\text{rms}}$.
• Lighter molecules move faster than heavier ones at the same temperature.
• The r.m.s. speed is useful for estimating rates of diffusion, effusion, and the speed distribution described by the Maxwell‑Boltzmann law.