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R.m.s. speed of gas particles

Derive the r.m.s. speed of gas particles from kinetic temperature and apply it with consistent units.

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  • University Physics Year 1
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Kinetic energy depends on squared speed. The root-mean-square speed is therefore the speed that reproduces the mean translational kinetic energy, rather than the speed of an average molecule. You will connect it to temperature, molecular mass and pressure, and decide which average a question actually needs.

At a glance

  • Prerequisites: kinetic temperature relation, ideal gas constants.
  • Outcomes: compute vᵣₘₛ from T and molecular mass.
  • Key result: vᵣₘₛ = square root of (3kBT/m) = square root of (3RT/M).
  • Common trap: using molar mass in grams instead of kilograms.

Define the energy-based speed average

For molecules of mass m in a dilute classical ideal gas at equilibrium, use speeds relative to the gas’s mean motion. Here the gas has no bulk flow:

(1/2)m⟨v²⟩ = (3/2)kBT.

Define

vᵣₘₛ≡ square root of (⟨v²⟩).

Molar mass M = NA m is the mass per mole, rather than the mass of this particular sample. Pair m with kB, or M with R; the result does not depend on how many moles the sample contains.

Physical interpretation:

  • vᵣₘₛ is the speed that matches the mean kinetic energy, because kinetic energy depends on v².
  • It is not the same as the mean speed v bar = ⟨v⟩; those become distinct once you use the full Maxwell-Boltzmann speed distribution.

Connect temperature, mass and rms speed

From kinetic temperature:

⟨v²⟩ = 3kBT/m.

Take square root:

vᵣₘₛ = square root of (3kBT/m).

Using R = NAkB and M = NA m:

vᵣₘₛ = square root of (3RT/M).

Units check:

[RT/M] = (J mol⁻¹)/(kg mol⁻¹) = J/kg = m²/s²,

so the square root gives m s⁻¹.

For a single species at equilibrium with no bulk flow, pressure gives a second route when mass density ρ is known. With ρ = Nm/V, the kinetic pressure formula gives P = (1/3)ρ⟨v²⟩, vᵣₘₛ = square root of (3P/ρ). This uses mass density in kg m⁻³, not number density or molar mass. The ideal-gas equation also gives T = PM/(ρ R) for a known pure gas. Pressure alone does not fix speed: density or temperature and particle mass is also needed.

Physical sanity check:

  • vᵣₘₛ ∝ square root of T: heating increases molecular speeds sublinearly.
  • vᵣₘₛ ∝ 1/square root of M: lighter gases move faster at same T.

Limit checks:

  • As T → 0 (classical prediction), vᵣₘₛ → 0.
  • At fixed T, M → ∞ gives vᵣₘₛ → 0 (very heavy molecules move slowly).
Pitfalls (RMS speed)
  • Use T in kelvin and M in kg mol⁻¹ with SI R. Substituting the numerical value in grams makes the speed too small by square root of 1000.
  • RMS speed comes from ⟨v²⟩. The most probable speed and mean speed describe different features of the distribution.

Comparing nitrogen and helium speeds

Worked example 1

Molecular mass and rms speed

Problem

Find rms speeds for ideal nitrogen and helium at 300 K. Use M_N₂ = 28.0 × 10⁻³ kg mol⁻¹, M_He = 4.0 × 10⁻³ kg mol⁻¹ and R = 8.314 J mol⁻¹K⁻¹.
Show full solution

  1. Calculate nitrogen's rms speed

    Method

    Use vᵣₘₛ = square root of (3RT/M) with mass in kilograms per mole.

    Reason

    The formula equates mean translational kinetic energy to 3kBT/2; molar units pair M with R.

    Working

    Answer

    vᵣₘₛ = square root of (3(8.314)(300)/(28.0 × 10⁻³)) = 5.17 × 10² m s⁻¹.

  2. Calculate helium's speed and compare

    Working

    Answer

    For helium, vᵣₘₛ = square root of (3(8.314)(300)/(4.0 × 10⁻³)):

    v_(rms,He) ≈ 1.37 × 10³ m s⁻¹,

    Nitrogen has vᵣₘₛ ≈ 517 m s⁻¹; helium has about 1.4 × 10³ m s⁻¹. The ratio is square root of (28.0/4.0) = square root of 7 ≈ 2.65: lighter particles move faster at the same mean translational energy.

Practice set

Check your understanding 1

Find the rms speed of ideal oxygen at 400 K with M = 32 × 10⁻³ kg mol⁻¹. Use R = 8.314 J mol⁻¹K⁻¹.

Show hint
Use 3RT/M for mean squared speed, then take the square root.
Show answer
vᵣₘₛ = square root of (3(8.314)(400)/(32 × 10⁻³)) ≈ 558 m s⁻¹, about 5.6 × 10² m s⁻¹. The quantity inside the root has units m²s⁻².

Check your understanding 2

For the same ideal gas, temperature rises from 300 K to 1200 K. By what factor does rms speed change?

Show hint

Use a ratio to cancel the gas constants and molar mass.

Show answer

vf/vᵢ = square root of (Tf/Tᵢ) = square root of (1200/300) = 2. The mean squared speed quadruples, while rms speed doubles.

Check your understanding 3

Two ideal gases have the same temperature. Gas A has nine times the molar mass of gas B. Find v_((rms, A))/v_((rms, B)).

Show hint

Mass appears in the denominator inside a square root.

Show answer

v_((rms, A))/v_((rms, B)) = square root of (MB/MA) = square root of (1/9) = 1/3. The heavier gas has the same mean translational energy but a lower speed.

Recover a speed from bulk measurements

Guided practice 2

Find speed from pressure and density

About 8 min

Problem

A pure dilute ideal gas has pressure 1.20 × 10⁵ Pa, mass density 1.50 kg m⁻³ and molar mass 0.0280 kg mol⁻¹. It is at equilibrium with no bulk flow. Use R = 8.314 J mol⁻¹ K⁻¹. Find temperature, mean squared speed and rms speed. Explain why these data are sufficient without the vessel volume or sample amount.

Work through the model, then compare your reasoning

Unit: K
Unit: m²/s²
Unit: m/s

Hints

Hint 1

Check the model

Separate a mass per unit volume from a mass per mole. Which ratio in the pressure equation removes sample size?
Show solution step by step

  1. Connect the two density descriptions

    Method

    Use P = ρ RT/M and the kinetic pressure relation.

    Reason

    The amount per volume is ρ/M, so volume and total amount need not be known.

    Working

    Answer

    T = PM/(ρ R) ≈ 269 K and ⟨v²⟩ = 3P/ρ = 2.40 × 10⁵ m² s⁻².

  2. Convert the second moment to speed

    Method

    Take the square root and check the independent temperature route.

    Reason

    A squared-speed result is not itself a speed.

    Working

    Answer

    vᵣₘₛ ≈ 490 m s⁻¹. Substitution into square root of (3RT/M) gives the same result. The units of P/ρ are m² s⁻². Pressure and mass density each contain the same molecule-number-to-volume factor. Dividing one by the other cancels sample size.

Choose an average for a mixture

For an ideal mixture in isotropic equilibrium with no bulk flow, add each species’ pressure contribution. If Nⱼ particles of species j each have mass mⱼ, then P = 1/3V∑ⱼNⱼmⱼ⟨vⱼ²⟩, ρ = 1/V∑ⱼNⱼmⱼ. Consequently, 3P/ρ = (∑ⱼNⱼmⱼ⟨vⱼ²⟩)/∑ⱼNⱼmⱼ, a mass-weighted mean square. An rms speed across the particle population instead uses v_(rms,population) = square root of ((∑ⱼNⱼ⟨vⱼ²⟩)/∑ⱼNⱼ), with number weights. The two averages agree for a single molecular mass, but need not agree for a mixture. State which population and weighting the measurement or question requires.

Check your understanding 4

An ideal mixture contains equal numbers of helium and neon atoms at 300 K, with no bulk flow. Their molar masses are 0.00400 and 0.0200 kg mol⁻¹. Use R = 8.314 J mol⁻¹ K⁻¹. Find the number-weighted mean squared speed and rms speed of the combined particle population. Also evaluate square root of (3P/ρ) for the mixture’s total pressure and mass density, using its composition and ideal-gas state. A solution first averages the two molar masses, then inserts that average into square root of (3RT/M). Decide whether this gives the requested population average, explain which weighting the bulk pressure/density result uses, and compare the species’ mean translational energies.

Unit: m²/s²
Unit: m/s
Unit: m/s
Show answer

The combined mean square is the number-weighted mean of species mean squares: ⟨v²⟩ = (1/2)(3RT/M_He + 3RT/M_Ne) = 1.12239 × 10⁶ m² s⁻². Taking its square root gives about 1.06 × 10³ m s⁻¹. This number-weighted average uses inverse masses, not masses. Taking the arithmetic mean of the two species’ rms speeds before squaring would also give a different quantity, about 990 m s⁻¹.

The mean molar mass of this equinumerous mixture is M bar = 0.0120 kg mol⁻¹. Its ideal-gas state gives P/ρ = RT/M bar, so square root of (3P/ρ) = square root of (3RT/M bar) ≈ 790 m s⁻¹. That is a meaningful mass-weighted rms, but it does not answer the number-weighted population question. Helium and neon carry mass weights 1/6 and 5/6, rather than the number weights 1/2 and 1/2.

The two species have equal mean translational energy 3kBT/2 despite different speed distributions. Check that your explanation identifies the quantity being averaged, its weights and why the bulk measurement supplies a different average.

What to keep

For a single molecular mass, rms speed reproduces mean translational kinetic energy. Pair molecular mass with kB, molar mass with R, and mass density with pressure. Average squared speeds before taking the root, with the weighting the population requires.

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