Temperature & Molecular Kinetic Energy

The kinetic model of the previous topic delivered pV = (1/3)Nm⟨v²⟩ from collisions with the walls alone, with no thermometer anywhere in the derivation. Measurement delivers pV = NkT, with no molecules anywhere in it. They describe the same gas, so set them equal: (1/3)Nm⟨v²⟩ = NkT. The N cancels — the result is per molecule and says nothing about how many there are — and multiplying through by 3/2 gives ⟨½mv²⟩ = (3/2)kT. Read what that claims. On the left is a mechanical average over some 10²³ molecules; on the right is a thermometer reading times a constant. Neither side mentions volume, pressure, or which gas it is. At 300 K the number is (3/2)(1.381 × 10⁻²³)(300) = 6.21 × 10⁻²¹ J per molecule, the same for helium and for carbon dioxide. Absolute temperature is not a second quantity riding alongside the motion; for an ideal gas it is that motion, measured in kelvin.

vᵣₘₛ is defined by the route you take to it: square every speed, average the squares, then take the root. The order matters. Four molecules at 300, 400, 500 and 600 m s⁻¹ have a mean speed of 450 m s⁻¹ but an rms speed of √(2.150 × 10⁵) = 464 m s⁻¹, because squaring gives the fast ones more weight. Only ⟨v²⟩ appears in the pressure and in the energy, so vᵣₘₛ is the speed the thermodynamics actually uses, and it always sits a little above the arithmetic mean of the same set. Rearranging ⟨½mv²⟩ = (3/2)kT gives vᵣₘₛ = √(3kT/m); multiplying numerator and denominator inside the root by Avogadro's number turns it into the laboratory form vᵣₘₛ = √(3RT/M). For nitrogen at 300 K, with M = 2.80 × 10⁻² kg mol⁻¹, that is √(3 × 8.314 × 300 / 0.0280) = 517 m s⁻¹, comfortably supersonic. How the speeds spread around this value is the next topic's business.

Put helium and nitrogen in one flask and wait. Collisions keep passing energy back and forth until the exchange balances, and the balance point is equal average translational kinetic energy per molecule — not equal speed. That is the microscopic reading of the zeroth law, and the reason a single thermometer can serve both gases at once. If ½m⟨v²⟩ is shared while m is not, the speeds must be unequal in exactly the compensating ratio: vᵣₘₛ(He)/vᵣₘₛ(N₂) = √(M(N₂)/M(He)) = √(28.0/4.00) = 2.65, which at 300 K means 1.37 × 10³ m s⁻¹ against 517 m s⁻¹. This is measurable, not decorative. Lighter molecules strike a small hole more often and effuse through it faster in the same √M ratio, which is why helium seeps out of a balloon in a day while the nitrogen around it stays put.

Where does the 3 in (3/2)kT come from? No direction is special, so ⟨vₓ²⟩ = ⟨v_y²⟩ = ⟨v_z²⟩ = ⟨v²⟩/3, and each direction carries ½m⟨vₓ²⟩ = ½kT. Three independent directions, three halves of kT. Equipartition generalises exactly that statement: in classical equilibrium, every term in a molecule's energy that is quadratic in a coordinate or a momentum averages ½kT. A rigid diatomic can also spin about the two axes perpendicular to its bond, adding ½I₁ω₁² and ½I₂ω₂², so f = 5 and ⟨E⟩ = (5/2)kT; spinning about the bond axis has so small a moment of inertia that its quantum is enormous, and it does not count. A vibrating bond is worth two terms, the kinetic ½μv² and the elastic ½κx², so switching vibration on takes a diatomic from f = 5 to f = 7. Choosing f is a modelling decision about the molecule, not a fit to data.

Add (f/2)kT over N molecules and use Nk = nR: U = (f/2)nRT. Two things in that result deserve a pause. Volume does not appear. An ideal gas has no intermolecular potential energy by assumption, so separating the molecules costs nothing and U cannot depend on V; a real gas, whose molecules attract, cools slightly when it expands into a vacuum, and the size of that cooling measures how far from ideal it is. Second, f is testable. Differentiating gives the molar heat capacity at constant volume, C_V = (1/n) dU/dT = (f/2)R, predicting 12.47 J mol⁻¹ K⁻¹ for a monatomic gas and 20.79 J mol⁻¹ K⁻¹ for a rigid diatomic. Argon measures 12.5 and nitrogen 20.8 near room temperature. One mole of argon at 300 K therefore holds (3/2)(8.314)(300) = 3.74 kJ, all of it translational, while a mole of nitrogen holds 6.24 kJ of which only 60% is.

If every quadratic term always took ½kT, each gas would have a fixed C_V. It does not. Nitrogen holds 20.8 J mol⁻¹ K⁻¹ from about 200 K to 600 K, then climbs towards the f = 7 value of 29.1 as it gets hot; hydrogen falls from 20.8 towards 12.5 below about 80 K, losing its rotations. Quantum mechanics sets the rule. A mode's energy comes in steps of size ΔE, and a collision can only excite it if the energy on offer is comparable, so the mode is active when kT ≫ ΔE and frozen out when kT ≪ ΔE. Define θ = ΔE/k. For N₂ the vibrational quantum is ΔE = 4.69 × 10⁻²⁰ J, so θ_vib = 3.40 × 10³ K, and at 295 K the ratio ΔE/kT is 11.5: collisions essentially never reach the first vibrational level. Rotation is cheap by comparison, θᵣₒₜ ≈ 2.9 K for N₂ and 85 K for H₂, which is why f = 5 works for air and only hydrogen shows the shutdown in a laboratory.