Temperature & Molecular Kinetic Energy

The previous topic ended with pV = ⅓Nm⟨v²⟩: pressure as the rate at which N molecules of mass m, with mean-square speed ⟨v²⟩, deliver momentum to the walls. Nothing thermal entered — it was mechanics and averaging. The equation of state, pV = NkT, came from the other direction: from measurements on dilute gases, with T on the ideal-gas absolute scale and k = R/N_A = 1.381 × 10⁻²³ J K⁻¹. A gas has one value of pV, so the two expressions are equal, and cancelling N gives ⅓m⟨v²⟩ = kT. Multiply by 3/2 to make the left side a kinetic energy: ⟨½mv²⟩ = (3/2)kT. Check the dimensions before believing it: k carries J K⁻¹, so kT is a joule, matching the joule on the left. Then read the logic carefully. The thermometer was calibrated first and the molecules were counted second, so this line does not derive temperature; it identifies what a temperature reading had been reporting all along — the mean translational kinetic energy of one molecule, to within the fixed factor 3/2.

The right-hand side contains T and a universal constant, nothing else. Mass cancelled, so at 300 K a helium atom (6.65 × 10⁻²⁷ kg) and a xenon atom (2.18 × 10⁻²⁵ kg, 33 times heavier) both average 6.21 × 10⁻²¹ J; the xenon atom reaches it with a mean-square speed 33 times smaller, an rms speed √33 = 5.7 times smaller. Pressure and volume cancelled, so compressing a gas isothermally to half its volume doubles the rate of wall collisions without changing the energy carried by any of them — that is what isothermal means at the molecular level. Number cancelled, so the result survives in a mixture: nitrogen and oxygen in the same room share one mean kinetic energy, each contributing to the pressure in proportion to its own number, which is Dalton's law arriving from mechanics. One more consequence hides in the 3. A gas at rest has no preferred direction, so ⟨vₓ²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩, and since v² = vₓ² + v_y² + v_z² each component carries ⟨½mvₓ²⟩ = ½kT — at 300 K, 2.07 × 10⁻²¹ J per direction.

The identification fixes one speed only: vᵣₘₛ = √⟨v²⟩ = √(3kT/m). Multiplying numerator and denominator by Avogadro's number turns it into √(3RT/M), with M the molar mass in kg mol⁻¹, which is the form to compute with. Nitrogen at 300 K: √(3 × 8.314 × 300 / 0.0280) = 517 m s⁻¹. Helium: √(3 × 8.314 × 300 / 0.00400) = 1370 m s⁻¹. Both sit above the speed of sound in the same gas at that temperature (347 m s⁻¹ in air, about 1020 m s⁻¹ in helium), as they must, because sound is carried by these very molecules and cannot outrun them. Be careful what vᵣₘₛ is not. It is not the mean speed, because the mean of the squares exceeds the square of the mean whenever the speeds differ: for three molecules at 300, 500 and 700 m s⁻¹ the mean speed is 500 m s⁻¹ but ⟨v²⟩ = 276 670 m² s⁻², so vᵣₘₛ = 526 m s⁻¹. The gap, ⟨v²⟩ − ⟨v⟩² ≥ 0, is the variance of the speeds, and it vanishes only if every molecule moves at the same speed. Temperature is tied to ⟨v²⟩ because kinetic energy is quadratic in speed; the next topic, on the Maxwell-Boltzmann distribution, supplies the mean and most probable speeds, both lower than vᵣₘₛ.

Multiply the mean by the number of molecules and the intensive quantity becomes an extensive one: Uₜᵣₐₙₛ = N × (3/2)kT = (3/2)nRT. One mole of any ideal gas at 300 K stores 1.5 × 8.314 × 300 = 3.74 kJ of translational kinetic energy; ten moles store ten times that at the same temperature. This is where the model can be tested rather than admired. For a monatomic gas — helium, neon, argon — an atom is a point with nothing to spin and nothing to vibrate, so translational energy is the whole internal energy, U = (3/2)nRT, and heating at constant volume must cost C_V = (1/n) dU/dT = (3/2)R = 12.47 J mol⁻¹ K⁻¹, the same for all three. Measured: 12.5, 12.5 and 12.5 J mol⁻¹ K⁻¹. Nitrogen, measured at the same temperature, gives 20.8 J mol⁻¹ K⁻¹ — about (5/2)R — and the excess is the identification marking exactly where its own scope ends. Translation is still (3/2)kT per molecule, but a diatomic molecule also stores energy in rotation, which a thermometer does not see directly and which the equipartition topic will count.

Three qualifications keep the result honest. First, the velocity in ⟨½mv²⟩ is the random one. If the gas as a whole moves at bulk velocity u, write each molecular velocity as u + w with ⟨w⟩ = 0; then ⟨|v|²⟩ = u² + ⟨w²⟩, since the cross term 2u⋅⟨w⟩ averages to zero, and only ⟨½mw²⟩ equals (3/2)kT. Second, the molecules are assumed to interact only in collisions. In a dense gas a molecule spends part of its time in a neighbour's potential well, so U gains a term depending on volume as well as temperature and T no longer measures kinetic energy alone; the real-gas topic quantifies how much. Third, the counting is classical. At room temperature kT = 0.026 eV dwarfs the quantum h²/8mL² that sets the spacing of translational levels in any laboratory box (about 5 × 10⁻¹⁹ eV for helium in a centimetre cube), so the classical average is exact for translation; it also exceeds rotational spacings but not vibrational ones, so the vibration of N₂ stays frozen at 300 K. Correspondence check: (3/2)kT is the limit quantum statistics returns whenever kT is large compared with the level spacing, and for translation it always is.