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Kinetic temperature

Derive gas pressure from molecular motion and connect kinetic temperature to average translational kinetic energy.

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  • University Physics Year 1
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A gas can exert pressure even when its average velocity is zero. Molecules arriving at a wall transfer momentum, and faster molecules both hit harder and arrive more often. Here you will connect that microscopic motion to pressure and temperature, keeping random motion separate from motion of the gas as a whole.

At a glance

  • Prerequisites: Newton’s second law, momentum, averages, ideal gas law.
  • Outcomes: derive P = (Nm⟨v²⟩)/3V and ⟨Eₜᵣₐₙₛ⟩ = (3/2)kB T.
  • Key result: temperature measures average translational kinetic energy.
  • Common trap: forgetting isotropy, so using ⟨vₓ²⟩ = ⟨v²⟩ (incorrect).

Define thermal motion and the wall model

Take N molecules (mass m each) in a cubic box of side d and volume V = d³. The box and gas are at rest on average. Use a dilute, non-relativistic classical gas in thermal equilibrium; the velocities below are random velocities in that rest frame.

  • Molecular velocities are random: no preferred direction.
  • Molecule-wall collisions are elastic.
  • For the pressure derivation, ignore molecule-molecule collisions first.

Physical interpretation (what the symbols mean):

  • ⟨v²⟩ is an average of squared speed over many molecules (or over time for one molecule in equilibrium). It is not (⟨v⟩)².
  • The relation (1/2)m⟨v²⟩ = (3/2)kBT is the kinetic definition of temperature for an ideal gas: temperature measures the typical translational kinetic energy per molecule.
  • The factor 3 comes from three translational components (x,y,z) under isotropy.

Sign convention:

  • Pressure on wall is positive outward normal force per area.

Derive pressure from repeated impulses

For one molecule with x-component vₓ hitting the wall normal to x:

Δ pₓ = (-mvₓ) - (mvₓ) = -2mvₓ.

For the displayed momentum change, the molecule approaches the positive-x wall with vₓ > 0. More generally, the impulse magnitude delivered per hit is 2m|vₓ|. In this simple collision-free flight argument, the return time is 2d/|vₓ|, so the average force contribution is:

Fᵢ = 2m|v_(x,i)|/(2d/|v_(x,i)|) = (m v_(x,i)²)/d.

Summing over all molecules:

F = m/d∑ᵢ₌₁^N v_(x,i)² = (Nm⟨vₓ²⟩)/d.

Pressure P = F/A = F/d² gives

P = (Nm⟨vₓ²⟩)/d³ = (Nm⟨vₓ²⟩)/V.

Isotropy gives

⟨vₓ²⟩ = ⟨vy²⟩ = ⟨vz²⟩ = 1/3⟨v²⟩,

therefore

P = (Nm⟨v²⟩)/3V = (2/3)(N/V)((1/2)m⟨v²⟩).

Combine with ideal gas law PV = NkBT:

(1/2)m⟨v²⟩ = (3/2)kB T.

If the gas has local mean flow velocity u, use the fluctuations c = v-u for its kinetic temperature: (1/2)m⟨c²⟩ = (3/2)kBT. For isotropic local equilibrium, the thermal pressure is P = (N/V)m⟨cₓ²⟩. The extra laboratory kinetic energy associated with u is bulk motion, not a temperature rise. A directed beam or anisotropic distribution need not have one equilibrium temperature.

Limit checks (where this picture needs care):

  • Requires an ideal gas in equilibrium with isotropic velocities. In directed flows or near shocks, different components can have different “temperatures”.
  • At very low T (quantum regime) or very high densities (strong interactions), the simple classical link between T and translational kinetic energy needs modification.
Pitfalls (kinetic temperature)
  • Isotropy gives ⟨vₓ²⟩ = ⟨v²⟩/3, rather than the whole mean squared speed.
  • The average velocity can be zero while mean squared speed is large. Opposite velocities cancel in the vector average; their squared contributions add.
  • In a flowing gas, subtract the local mean velocity before using random kinetic energy to find temperature.

Units check:

[(1/2)m⟨v²⟩] = kg · m²/s² = J, [kBT] = J.

Sign/sanity check:

  • P ∝ ⟨v²⟩, so faster random motion means higher pressure.
  • Squared velocity ensures positive pressure contribution even if vₓ < 0 before collision.

Translational energy at room temperature

Worked example 1

Energy per molecule and per mole

Problem

For a classical monatomic ideal gas at 300 K, find the mean translational kinetic energy per molecule and per mole. Use kB = 1.38 × 10⁻²³ J K⁻¹ and R = 8.314 J mol⁻¹K⁻¹.
Show full solution

  1. Evaluate the molecular average

    Method

    Use ⟨Eₜᵣₐₙₛ⟩ = (3/2)kBT.

    Reason

    Three independent translational components contribute to the average, irrespective of molecular mass.

    Working

    Answer

    ⟨Eₜᵣₐₙₛ⟩ = (3/2)kB T = (3/2)(1.38 × 10⁻²³)(300) ≈ 6.21 × 10⁻²¹ J.

  2. Convert to a molar amount

    Working

    Answer

    The molecular average is 6.21 × 10⁻²¹ J. Multiplying by NA replaces kB with R, giving

    (3/2)RT ≈ (3/2)(8.314)(300) = 3.74 × 10³ J mol⁻¹.

Practice set

Check your understanding 1

Show that ⟨vₓ²⟩ = ⟨v²⟩/3 for isotropic molecular motion.

Show hint

Average v² = vₓ² + vy² + vz², then use symmetry.

Show answer

Isotropy gives equal mean squares in all three directions. Hence ⟨v²⟩ = 3⟨vₓ²⟩, so each component contributes one-third, not the whole mean squared speed.

Check your understanding 2

An ideal gas’s absolute temperature doubles at fixed N and V. What happens to its pressure?

Show hint

Keep the fixed factors in P = NkBT/V unchanged.

Show answer

Pressure doubles because P ∝ T at fixed N,V. The mean translational energy also doubles.

Check your understanding 3

Find the mean translational kinetic energy per molecule of a classical ideal gas at 500 K. Use kB = 1.38 × 10⁻²³ J K⁻¹.

Show hint
Use the average for all three translational components.
Show answer
⟨Eₜᵣₐₙₛ⟩ = (3/2)(1.38 × 10⁻²³)(500) = 1.035 × 10⁻²⁰ J ≈ 1.04 × 10⁻²⁰ J per molecule.

Relate random motion to pressure

Guided practice 2

Connect three components to one temperature

About 8 min

Problem

A dilute gas at equilibrium has zero mean velocity and rms velocity component 300 m s⁻¹ in each of the three directions. Each molecule has mass m = 4.65 × 10⁻²⁶ kg, and number density is nN = 2.00 × 10²⁵ m⁻³. Use kB = 1.38 × 10⁻²³ J K⁻¹. Find mean squared speed, thermal pressure and temperature. Explain why zero mean velocity does not make any of these vanish.

Work through the model, then compare your reasoning

Unit: m²/s²
Unit: Pa
Unit: K

Hints

Hint 1

Check the model

Which average is zero, and which averages enter energy and momentum transfer?
Show solution step by step

  1. Combine the components

    Method

    Average the squared speed, rather than the velocity vector.

    Reason

    Isotropy gives equal component mean squares.

    Working

    Answer

    ⟨v²⟩ = 3(300)² = 2.70 × 10⁵ m² s⁻². Pressure is P = nNm⟨vₓ²⟩ = 8.37 × 10⁴ Pa; using the whole squared speed without the factor one-third would triple it.

  2. Identify the thermal scale

    Method

    Use one component or the full translational energy consistently.

    Reason

    A zero first moment does not imply a zero second moment.

    Working

    Answer

    T = m⟨vₓ²⟩/kB ≈ 303 K. Check P = nNkBT and ⟨Eₜᵣₐₙₛ⟩ = 3kBT/2. Opposite velocities cancel in the vector average, while their squared contributions add.

Separate flow from thermal motion

Check your understanding 4

A locally equilibrated gas flows uniformly with mean laboratory velocity ⟨v⟩ = (200,0,0) m s⁻¹. Laboratory measurements give ⟨vₓ²⟩ = 1.30 × 10⁵ and ⟨vy²⟩ = ⟨vz²⟩ = 9.00 × 10⁴, all in m² s⁻². Use the same molecular mass and kB as above. Find the x-component fluctuation mean square, the thermal mean squared speed and temperature. A calculation uses all the laboratory kinetic energy as thermal energy. Explain the error and whether the fluctuation data are consistent with isotropy.

Unit: m²/s²
Unit: m²/s²
Unit: K
Show answer

Subtract the mean motion before averaging energy: ⟨cₓ²⟩ = ⟨vₓ²⟩-⟨vₓ⟩² = 90000 m² s⁻². The other mean components vanish, so all three fluctuation mean squares are equal and ⟨c²⟩ = 270000 m² s⁻². Thus T = m⟨c²⟩/(3kB) ≈ 303 K. Using 310000 instead would include the bulk-flow energy and give about 348 K. Equal fluctuation variances support this isotropic local model; the laboratory second moments themselves need not be equal.

What to keep

Pressure depends on the normal random-velocity second moment. For isotropic classical equilibrium, each direction contributes one-third of the mean squared thermal speed. Remove mean flow before interpreting kinetic energy as temperature.

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