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 \$(vx, vy, vz)\$, the momentum change on striking a wall perpendicular to the \$x\$‑axis is \$2mvx\$.
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{2mvx}{\Delta t} = \frac{mvx^2}{L}.\$

4. Summing over all \$N\$ molecules and using the fact that the motion is isotropic (\$\langle vx^2\rangle = \langle vy^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 vx^2+vy^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

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.