§ 31. Maxwell's Law of Molecular Speed Distribution in an Ideal Gas (Grade 10–11 · Uni)
Statistical distribution f(v), most probable speed vₚ, mean ⟨v⟩, and RMS speed vᵣms. Chaos of individual molecular collisions gives rise to a strictly deterministic bell-shaped statistical curve of speed probabilities.
Theory Summary & Scientific Mechanism
Although temperature T fixes the mean kinetic energy of an ideal gas, incessant molecular collisions spread individual molecular speeds across a wide spectrum. In 1859, James Clerk Maxwell derived the equilibrium speed probability density function: f(v) = 4π ≤ft((M)/(2π R T) )^ 3/2 v² exp ≤ft(-(M v²)/(2 R T) ).
The shape of the Maxwell–Boltzmann curve arises from the product of the phase-space shell volume factor 4π v² (which suppresses f(v) → 0 as v → 0) and the Boltzmann exponential factor exp ≤ft(-(Mv²)/(2RT) ) (which suppresses high-energy tails).
Three characteristic speeds obey the invariant ordering vₚ < langle v rangle < vᵣms: 1) most probable speed (peak of f(v)) vₚ = √((2RT)/(M)); 2) mean speed langle v rangle = √((8RT)/(π M)) ≈ 1.128 vₚ; and 3) root-mean-square speed vᵣms = √((3RT)/(M)) ≈ 1.225 vₚ. Raising temperature T or choosing a lighter gas M shifts the peak rightward and lowers its height while preserving total probability ∫₀^∞ f(v) dv = 1.
Everyday Analogy
Air molecules in a room behave like a bustling crowd: a few are nearly stopped after a collision, most cruise around an average speed (sim 500 m/s!), and a few sprinters fly much faster. Heating the gas shifts the entire speed curve upward!
Interactive Question & Derivation
How does the Maxwell speed distribution curve f(v) change when the absolute temperature T of the gas is increased at constant molar mass M?
- A: The peak shifts toward higher speeds (vₚ ∝ √(T)), the curve broadens, and the peak height f(vₚ) decreases so total area remains 1
- B: The curve grows taller at the same speed vₚ without changing width
- C: All molecules acquire the exact same speed vᵣms
Study Hint: When gas temperature T rises (or molar mass M decreases), how does the peak of the Maxwell speed distribution (vₚ = √(2RT/M)) shift?
Proof & Explanation: In Maxwell's distribution f(v) = 4π≤ft((M)/(2π RT) )³/2 v² e⁻(Mv²)/(2RT), all three characteristic speeds scale as √(T/M): most probable vₚ = √(2RT/M), mean langle vrangle = √(8RT/(π M)), and root-mean-square vᵣms = √(3RT/M). Since the normalization integral ∫₀^∞ f(v) dv = 1 is conserved, shifting the peak to higher speeds (vₚ ∝ √(T)) causes the curve to broaden and lower its peak height (f(vₚ) ∝ 1/vₚ), while exponentially increasing the fraction of fast molecules in the high-energy tail!
Key Equations & 3D Simulation Parameters
vₚ : langle vrangle : vᵣms = √(2) : √((8)/(π)) : √(3) ≈ 1 : 1.128 : 1.225
- Gas temperature T: 150 … 900 K (default: 350 K)
- Gas molar mass M (4=He, 28=N₂, 44=CO₂): 4 … 64 g/mol (default: 28 g/mol)
- Visualized molecules N: 40 … 140 pcs (default: 90 pcs)
- Activation threshold vₜhr / vₚ: 0.8 … 2 (default: 1.2)