BORING EDUCATION — LIVE EXPERIMENTProduct is live. Experiments are running. Feedback is being collected. Improvements are underway.Welcome to Boring Education. We're glad you're here while we're figuring out how to make learning better.
BORINGEDUCATION

13.3Free

Pressure of a gas: from one bouncing molecule to PV = ⅓Nm⟨v2⟩

Your guide: Sir HamzaBelieves every formula has a story, and units never lie.

The problem

Throw a tennis ball at a wall and catch it on the rebound, again and again. Your wall feels a push each time. Now imagine billions of balls every second, too small to see. That steady push is gas pressure.

This section turns that picture into an equation. It links what we measure (pressure and volume) to what we cannot see (the mass and speed of molecules). We start with one molecule in a cube of side L.

For N molecules of mass m in volume V: PV = ⅓Nm⟨v2⟩, where ⟨v2⟩ is the mean-square speed. With density ρ = Nm/V, this becomes P = ⅓ρ⟨v2⟩.

Step 1 / 7

Notes, short questions and MCQs

Read the full notes: key terms, model answers and MCQs with answers

The problem

Throw a tennis ball at a wall and catch it on the rebound, again and again. Your wall feels a push each time. Now imagine billions of balls every second, too small to see. That steady push is gas pressure.

This section turns that picture into an equation. It links what we measure (pressure and volume) to what we cannot see (the mass and speed of molecules). We start with one molecule in a cube of side L.

For N molecules of mass m in volume V: PV = ⅓Nm⟨v2⟩, where ⟨v2⟩ is the mean-square speed. With density ρ = Nm/V, this becomes P = ⅓ρ⟨v2⟩.

Key terms

Momentum change
A molecule moving at vₓ bounces elastically off the wall and comes back at −vₓ. Its momentum changes from mvₓ to −mvₓ.
Time between hits
Before it hits the same wall again, the molecule must cross the box and come back: a distance 2L at speed vₓ.
Force and pressure of one molecule
Force is the rate of change of momentum: 2mvₓ ÷ (2L/vₓ) = mvₓ2/L. Dividing by the wall's area L2 gives the pressure from one molecule.
All molecules, all directions
Adding N molecules gives P = (Nm/V)⟨vₓ2⟩. Motion is random, so the average square speed is shared equally by x, y and z: ⟨vₓ2⟩ = ⅓⟨v2⟩.

Short questions with model answers

  1. Q1. Derive the force exerted on a wall by one molecule of mass m moving with velocity vₓ in a cube of side L.

    • Δp = −mvₓ − (mvₓ) = −2mvₓ
    • Δt = 2L ÷ vₓ
    • F = 2mvₓ ÷ (2L/vₓ) = mvₓ2 ÷ L

    F = mvₓ2 ÷ L

  2. Q2. Starting from P = (Nm/V)⟨vₓ2⟩, show that PV = ⅓Nm⟨v2⟩ and P = ⅓ρ⟨v2⟩.

    • ⟨v2⟩ = ⟨vₓ2⟩ + ⟨v_y2⟩ + ⟨v_z2⟩ = 3⟨vₓ2⟩
    • P = ⅓(Nm/V)⟨v2⟩ → PV = ⅓Nm⟨v2⟩
    • ρ = Nm ÷ V → P = ⅓ρ⟨v2⟩

    PV = ⅓Nm⟨v2⟩ · P = ⅓ρ⟨v2⟩

  3. Q3. Numerical 13.2: find the pressure of oxygen at 0 °C if its density is 1.44 kg m−3 and its rms speed is 456.4 m s−1.

    • P = ⅓ρ v2ᵣₘₛ
    • P = ⅓ × 1.44 × (456.4)2 = ⅓ × 1.44 × 208,301
    • P = 99,985 Pa ≈ 1.0 × 105 N m−2

    P ≈ 1.0 × 105 N m−2

Common mistakes

  • ✗ Writing Δp = 0 because the molecule's speed is the same before and after the hit.

    ✓ Momentum is a vector. The speed is the same but the direction reverses, so Δp = −2mvₓ.

  • ✗ Using the average velocity of the molecules in the formula.

    ✓ The average velocity is zero, because +v and −v cancel. The formula needs the mean-square speed ⟨v2⟩, or the rms speed squared.

  • ✗ Forgetting the ⅓ and writing PV = Nm⟨v2⟩.

    ✓ Only the x-part of the motion pushes on the wall, and on average it carries one-third of ⟨v2⟩. So PV = ⅓Nm⟨v2⟩.

MCQs

  1. 1. A molecule of mass m hits a wall with velocity vₓ and rebounds elastically. Its change in momentum is:

    1. (a) 0
    2. (b) mvₓ
    3. (c) −2mvₓ
    4. (d) 2L/vₓ
    Show answer

    (c) −mvₓ − mvₓ = −2mvₓ: the direction reverses.

  2. 2. The time between two collisions of the molecule with the same wall is:

    1. (a) L/vₓ
    2. (b) 2L/vₓ
    3. (c) vₓ/2L
    4. (d) L2/vₓ
    Show answer

    (b) It must travel 2L (there and back) at speed vₓ.

  3. 3. In a closed room a gas has pressure 3.2 × 105 N m−2 and density 6 kg m−3. The effective (rms) speed of its particles is:

    1. (a) 400 m s−1
    2. (b) 300 m s−1
    3. (c) 250 m s−1
    4. (d) 330 m s−1
    Show answer

    (a) v = √(3P/ρ) = √(3 × 3.2 × 105 ÷ 6) = √(1.6 × 105) = 400 m s−1.

  4. 4. If the density of a gas is doubled, with everything else unchanged, its pressure:

    1. (a) halves
    2. (b) stays the same
    3. (c) doubles
    4. (d) becomes four times
    Show answer

    (c) P = ⅓ρ⟨v2⟩: with ⟨v2⟩ fixed, pressure is directly proportional to density.

  5. 5. Why is ⟨vₓ2⟩ = ⅓⟨v2⟩?

    1. (a) because the box is a cube
    2. (b) because motion is random, so x, y and z share it equally
    3. (c) because only one-third of the molecules move
    4. (d) because collisions are elastic
    Show answer

    (b) No direction is preferred, so ⟨vₓ2⟩ = ⟨v_y2⟩ = ⟨v_z2⟩, and together they make ⟨v2⟩.

Quick revision

  • One molecule: Δp = −2mvₓ, Δt = 2L/vₓ, F = mvₓ2/L, P = mvₓ2/L3.
  • All molecules, random directions: ⟨vₓ2⟩ = ⅓⟨v2⟩, so PV = ⅓Nm⟨v2⟩ and P = ⅓ρ⟨v2⟩.
  • Doubling the speed quadruples the pressure: twice the push per hit, twice as many hits.