# Maxwell Speed Distribution > Maxwell speed distribution for JEE Physics: f(v) ∝ v²e^(−mv²/2kT), the 1 : 1.128 : 1.224 ratio of most probable, mean and rms speeds and T-dependence. - Canonical URL: https://prepelephant.com/topics/jee/physics/maxwell-distribution-jee - Exam / course: JEE · Subject: Physics - Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team - First published: 2026-10-02 - Last updated: 2026-10-02 - How to cite: "Maxwell Speed Distribution", PrepElephant, https://prepelephant.com/topics/jee/physics/maxwell-distribution-jee ## Direct answer Molecules in a gas at temperature T do not share one speed: they spread over the Maxwell distribution f(v) = 4πN(m/2πkT)^(3/2) v² e^(−mv²/2kT), a curve that starts at zero (the v² factor), peaks, and dies exponentially. Three characteristic speeds label the curve — most probable v_p = √(2kT/m), mean v = √(8kT/πm) and root-mean-square v_rms = √(3kT/m) — always in the ratio 1 : 1.128 : 1.224, so v_p < v < v_rms in that fixed order. Raising T shifts the peak rightward and flattens it (the area, fixed at N molecules, is conserved); heavier molecules at the same T peak at lower speeds. The distribution underlies why Earth's atmosphere retains oxygen but leaks hydrogen, why reaction rates climb steeply with temperature (only the tail exceeds activation energy), and why pressure fixes only v_rms through PV = ⅓Nm v_rms². ## What you must remember - **Functional form:** f(v) ∝ v² e^(−mv²/2kT); the v² prefactor (from spherical shells in velocity space) forces f(0) = 0, and the exponential kills the high-speed tail. - **The three speeds:** v_p = √(2kT/m), v_mean = √(8kT/πm), v_rms = √(3kT/m); ratio 1 : 1.128 : 1.224 in fixed order v_p < v_mean < v_rms. - **Molar forms:** replace k/m by R/M: v_rms = √(3RT/M), the version needed for numericals; hydrogen at 300 K gives about 1930 m/s, oxygen about 483 m/s. - **Temperature scaling:** each characteristic speed grows as √T, so quadrupling T doubles v_rms while the peak height falls as 1/√T to conserve area. - **Mass dependence:** speeds scale as 1/√m — at the same temperature a lighter gas moves faster and its peak sits lower and further right. - **Area meaning:** the area under f(v) between two speeds counts molecules in that range; total area = N, the molecule count, never changes. - **Pressure connection:** only v_rms is fixed by the ideal-gas relation PV = ⅓Nm v_rms²; v_p and v_mean carry distributional information that pressure alone cannot see. ## Why only rms is fixed Pressure measurements pin down exactly one characteristic speed. Impacts on the wall depend on v², so the average pressure fixes is ⟨v²⟩ — the mean square — and v_rms = √⟨v²⟩ follows; v_p and v_mean need the full distribution to define, which is why Maxwell's 1860 prediction of its shape, confirmed only later by molecular-beam experiments, was a genuine triumph. Run the standard numerical with oxygen at 300 K. v_rms = √(3RT/M) = √(3 × 8.314 × 300/0.032) ≈ 484 m/s. Then read the other two off the ratio ladder: v_p = 484/1.224 ≈ 395 m/s, v_mean = 484 × (1.128/1.224) ≈ 446 m/s. Heat the gas to 1200 K — quadruple T — and every speed doubles: v_rms ≈ 968 m/s, while the peak height halves, the area staying fixed at N. For hydrogen at 300 K, v_rms ≈ 1930 m/s — far below Earth's 11.2 km/s escape speed; the atmosphere keeps its oxygen and loses hydrogen only through rare tail molecules exceeding escape velocity over geological time — why Earth is hydrogen-poor while massive Jupiter kept its own. ## What JEE asks of Maxwell The fixed order v_p < v_mean < v_rms is the most-banked fact: rank-order questions appear almost every year, and the answer follows from the distribution's rightward skew — a long high-speed tail drags the mean above the peak. Second, graph reading: two curves at different temperatures (higher T: flatter, further right) or two gases at one temperature (lighter: further right) — "which curve represents which" is answered by area conservation plus peak movement. Third, the √T scaling: doubling absolute temperature multiplies v_rms by √2, not by 2, and options include both. Fourth, the kinetic energy trap: mean kinetic energy per molecule is always 3/2 kT regardless of mass, so hydrogen and oxygen at the same T share the same energy while hydrogen moves faster — a distinction tested through "which has greater average kinetic energy" phrasing. Main stays at ratio recall and single-formula numericals; Advanced asks for the temperature at which a gas's rms speed doubles, or matches Maxwell curves with effusion (Graham's law) reasoning. ## Frequently asked questions ### What shape does the Maxwell speed distribution have and why? It rises as v² from zero (more velocity-space shells to occupy), peaks at the most probable speed and decays exponentially, since very high speeds are Boltzmann-improbable. ### What is the ratio of most probable, mean and rms speeds? √2 : √(8/π) : √3 in units of √(kT/m), i.e. 1 : 1.128 : 1.224, always in the order v_p < v_mean < v_rms. ### How does the distribution change when temperature rises? The peak shifts to higher speed (all characteristic speeds grow as √T) and the curve flattens, keeping the total area — the number of molecules — constant. ### Why do hydrogen and oxygen at the same temperature have the same average kinetic energy but different speeds? Average translational kinetic energy is 3/2 kT for every molecule at temperature T; mass enters only through speed, so the lighter hydrogen molecules must move faster to carry the same energy. ### Which characteristic speed does gas pressure depend on? The rms speed, via PV = ⅓Nm⟨v²⟩; mean and most probable speeds are invisible to pressure alone because impact momentum goes as v².