Kinetic Interpretation of Temperature

Temperature is one of the most familiar physical quantities, but its microscopic meaning only emerges from kinetic theory: temperature measures the average translational kinetic energy of a molecule.

Concept

Starting from the pressure result $PV = \tfrac{1}{3} N m \langle v^2\rangle$ and the ideal gas law $PV = N k_B T$, we equate:

$\frac{1}{3} N m \langle v^2\rangle = N k_B T$

Solving:

$\boxed{\tfrac{1}{2} m \langle v^2\rangle = \tfrac{3}{2} k_B T}$

The left side is the average translational kinetic energy per molecule, $\langle KE\rangle$. So

$\langle KE\rangle_{\text{trans}} = \tfrac{3}{2} k_B T$

Each translational degree of freedom contributes $\tfrac{1}{2} k_B T$ to the average energy.

RMS speed

From $\tfrac{1}{2} m \langle v^2\rangle = \tfrac{3}{2} k_B T$:

$v_{\text{rms}} = \sqrt{\langle v^2\rangle} = \sqrt{\frac{3 k_B T}{m}} = \sqrt{\frac{3 R T}{M}}$

where $M$ is the molar mass.

Implications

• Temperature is absolute: as $T \to 0$, $\langle v^2\rangle \to 0$.
• At the same temperature, all gases have the same average translational KE per molecule. Heavier molecules therefore have smaller mean speeds.
• Temperature has nothing to do with rotational or vibrational energy in this strict translational sense (those add separately via equipartition).

Derivation Recap

1. Kinetic theory: $PV = \tfrac{1}{3} N m \langle v^2\rangle$.
2. Ideal gas law: $PV = N k_B T$.
3. Equate: $\tfrac{1}{3} m \langle v^2\rangle = k_B T \Rightarrow \tfrac{1}{2} m \langle v^2\rangle = \tfrac{3}{2} k_B T$.

Worked Example

Q: Compare the RMS speed of hydrogen ($M_{H_2} = 2$ g/mol) and oxygen ($M_{O_2} = 32$ g/mol) molecules at the same temperature.

A: Since $\tfrac{1}{2} m v_{\text{rms}}^2 = \tfrac{3}{2} k_B T$ at the same $T$, both gases have the same average translational KE, but

$\frac{v_{\text{rms}}(H_2)}{v_{\text{rms}}(O_2)} = \sqrt{\frac{M_{O_2}}{M_{H_2}}} = \sqrt{\frac{32}{2}} = 4$

Hydrogen molecules move four times faster (RMS) than oxygen molecules at the same temperature. This is also why H$_2$ leaks faster and escapes Earth's atmosphere more readily.

Q2: At what temperature is the RMS speed of nitrogen ($M = 28$ g/mol) molecules equal to 500 m/s?

A: $v_{\text{rms}}^2 = 3 R T / M \Rightarrow T = M v_{\text{rms}}^2 / (3 R) = (0.028)(500)^2 / (3 \times 8.314) \approx 281$ K.

Common Confusions

• "Heat" and "temperature" are different. Two objects can be at the same temperature with very different heat contents (depends on mass and specific heat).
• The formula $\tfrac{3}{2} k_B T$ refers only to translational kinetic energy. A diatomic molecule has additional rotational energy at higher temperatures.
• $T$ must be in kelvin — the equation makes no sense at "negative temperature" in this kinetic interpretation.

Key Takeaways

• $\langle KE\rangle_{\text{trans}} = \tfrac{3}{2} k_B T$ per molecule, independent of mass.
• $v_{\text{rms}} = \sqrt{3 k_B T / m} = \sqrt{3 R T / M}$.
• Temperature is a statistical measure: it has meaning only for a large collection of particles, not a single molecule.