Kinetic Theory and RMS Speed

Kinetic theory derives macroscopic gas variables ($P$, $T$) from molecular motion under the assumptions of point particles, elastic collisions, and no intermolecular forces.

Concept

For an ideal gas of $N$ molecules of mass $m$ in volume $V$ with mean square speed $\langle v^2\rangle$: $PV = \tfrac{1}{3}Nm\langle v^2\rangle$

Comparing with $PV = nRT = Nk_BT$ gives the average translational KE per molecule: $\tfrac{1}{2}m\langle v^2\rangle = \tfrac{3}{2}k_B T$

Three characteristic speeds:

• RMS: $v_{rms} = \sqrt{3k_BT/m} = \sqrt{3RT/M}$
• Average: $\bar v = \sqrt{8k_BT/(\pi m)}$
• Most probable: $v_p = \sqrt{2k_BT/m}$

Ratio: $v_p : \bar v : v_{rms} = \sqrt 2 : \sqrt{8/\pi} : \sqrt 3 \approx 1 : 1.13 : 1.22$.

Mean free path: $\lambda = 1/(\sqrt 2\,n\pi d^2)$ where $n$ is number density, $d$ molecular diameter.

Derivation

Consider one molecule moving along $x$ between two walls separated by $L$. Time between collisions with one wall: $2L/v_x$. Momentum change per hit: $2mv_x$. Force on wall: $mv_x^2/L$. Pressure from $N$ molecules: $P = (N/V)m\langle v_x^2\rangle$.

By isotropy $\langle v_x^2\rangle = \langle v^2\rangle/3$, so: $P = \frac{Nm\langle v^2\rangle}{3V}$

— the fundamental kinetic-theory result.

JEE Worked Example

Problem: Compute $v_{rms}$ of $\text{N}_2$ at 300 K. ($M = 28$ g/mol.)

Solution: $v_{rms}=\sqrt{\frac{3RT}{M}}=\sqrt{\frac{3\times 8.314\times 300}{0.028}}=\sqrt{2.67\times 10^5}\approx 517\;\text{m/s}$

Traps

• Use $M$ in kg/mol with $R$, or use $m$ per molecule with $k_B$.
• $v_{rms}$ is not the average speed; never confuse with $\bar v$.
• Pressure depends on KE per unit volume, not per molecule alone.
• Temperature is a measure of average translational KE only — independent of mass.
• Mean free path increases as density falls; at low pressure, $\lambda$ can exceed container size (Knudsen regime).

Key Takeaways

• $PV = \tfrac13 Nm\langle v^2\rangle$ and $\tfrac12 m\langle v^2\rangle = \tfrac32 k_BT$.
• $v_{rms} = \sqrt{3RT/M}$.
• Lighter gases move faster at the same $T$.
• Mean free path: $\lambda = 1/(\sqrt 2\,n\pi d^2)$.