Theme B — The Particulate Nature of Matter
B.3.6 Derivation of the Kinetic Theory of Gases Equation — IB Physics HL Revision Notes & Topical Questions
5 min read 12 parts Lesson 30 of 53 1 quiz
Curriculum Equation Key قوانين الرقم الهيدروجيني والاتزان الأيوني - pH & Acid-Base Equilibrium FormulasFast Exam Solution
pH = -log[H⁺] | pOH = -log[OH⁻] | pH + pOH = 14 | [H⁺][OH⁻] = 10⁻¹⁴
Word Equation & Conditions: الرقم الهيدروجيني = -لوغاريتم تركيز أيون الهيدروجين | pH + pOH = 14
Location & Specifications: الاتزان الأيوني، كيمياء 3 ثانوي، التحصيلي، و AP Chemistry.
Lesson summary
B.3.6 Derivation of the Kinetic Theory of Gases Equation — IB PHYSICS SL notes: Derive p = ⅓ρ⟨c²⟩ from the kinetic model by considering the momentum change at a wall.Aim — to show that the pressure of an ideal gas is p = ⅓ρ⟨c2⟩, starting from Newton's laws and the assumptions of …
- Origin of pressure
- One molecule
- No interaction
- Elastic rebound
- Initial momentum
- Final momentum
Part 1 of 10Setting up the model
Aim
- Aim to show that the pressure of an ideal gas is p = ⅓ρ⟨c2⟩, starting from Newton's laws and the assumptions of the model.
Origin of pressure
- Origin of pressure gas particles collide with the walls and exert forces, and the random motion of many molecules gives an overall pressure.
Cube
- Cube side length L, so V = L3, containing N identical molecules of mass m.
One molecule
- One molecule it moves with speed c, and its components are (vx, vy, vz).
No interaction
- No interaction each molecule is treated alone and the effects are added.
Wall
- Wall consider the wall perpendicular to the x-axis, of area L2.
Part 2 of 10Step 1: change in momentum at a wall
1Elastic rebound
- Elastic rebound no kinetic energy is lost, so the final and initial speeds are equal.
2Initial momentum
- Initial momentum mvx.
3Final momentum
- −mvx.
4Change
- Δp = −mvx − mvx = −2mvx.
5Wall receives
- Wall receives an impulse of +2mvx by conservation of momentum.
- (Newton's third law)
Part 3 of 10Step 2: time between collisions
1Round trip
- Round trip the molecule goes to the opposite wall and back, a distance 2L.
2Time
- Δt = 2L/vx between collisions with the same wall.
Part 4 of 10Step 3: force of one molecule
1Second law
- Second law the average force is momentum change per unit time.
2Equation
- F = 2mvx / (2L/vx) = mvx2/L.
Part 5 of 10Step 4: all the molecules
1Total force
- F = (m/L)(vx12 + vx22 + … + vxN2) = (Nm/L)⟨vx2⟩.
2Pressure
- Pressure p = F/L2 = Nm⟨vx2⟩/L3 = Nm⟨vx2⟩/V.
Part 6 of 10Step 5: three dimensions
1Pythagoras
- Pythagoras c2 = vx2 + vy2 + vz2, and so ⟨c2⟩ = ⟨vx2⟩ + ⟨vy2⟩ + ⟨vz2⟩.
2No preferred direction
- No preferred direction the three averages are equal.
3Result
- ⟨c2⟩ = 3⟨vx2⟩, so ⟨vx2⟩ = ⅓⟨c2⟩.
Part 7 of 10Step 6: final equation
1Substitute
- Substitute pV = ⅓Nm⟨c2⟩.What the symbols mean: p = pressure (Pa) · V = volume (m3) · N = number of molecules · m = mass of one molecule (kg) · <c2> = mean square speed (m2 s−2)
2Density
- Nm/V = ρ, the density of the gas.
3Kinetic theory equation
- Kinetic theory equation p = ⅓ρ⟨c2⟩.
- p is the pressure in Pa, ρ the density in kg m−3 and ⟨c2⟩ the mean square speed in m2 s−2.
Part 8 of 10Commentary
Commentary
Any container
- Any container the result holds for any shape, because pressure is the same on every wall.
Collisions between molecules
- Collisions between molecules they are elastic and between identical molecules, so they exchange momentum without changing the average.
Collision time and gravity
- Collision time and gravity both are ignored.
Mean square speed
- Mean square speed the mean of the squares matters, not the square of the mean, so the rms speed appears.
Link to temperature
- Link to temperature comparing pV = ⅓Nm⟨c2⟩ with pV = NkBT shows the mean kinetic energy is proportional to T.
Part 9 of 10Checking the result
Units
- Units kg m−3 × m2 s−2 = kg m−1 s−2 = N m−2, a pressure.
Behaviour
- Behaviour pressure rises with density and with mean square speed.
Marks
- Marks the stages are the momentum change 2mvx, the time 2L/vx, the force, the pressure and replacing ⟨vx2⟩ by ⅓⟨c2⟩.
Understanding
- Understanding the derivation is not recalled in full, but the physics underlying it and the final result must be understood.
Part 10 of 10More from the specification
More from the specification
Molar gas kinetic equations
- Molar gas kinetic equations quantifying gas substance in moles of colliding atoms provides a statistical means to derive ideal gas pressure.
Constant
- Not changing.
Helium
- Product of the fusion of hydrogen in stars.
Boltzmann constant
- K = R/NA, the gas constant per molecule
Constant volume
- The container is rigid and sealed.
Avogadro constant
- The number of particles per mole, 6.02 × 10²³ mol⁻¹
Thermodynamic temperature
- The absolute temperature in kelvin.
N = nNA
What the symbols mean: N = number of particles · n = amount (mol) · NA = Avogadro constant (6.02 × 1023 mol−1)
Kinetic model assumptions
- Kinetic model assumptions molecules are point-like with negligible volume and exert no forces on each other except during collisions.
Key terms
| Term | Meaning |
|---|---|
| Kinetic theory equation | p = ⅓ρ⟨c2⟩, relating gas pressure to density and mean square speed |
| Mean square speed | Average of the squares of the molecular speeds |
| Impulse | Force multiplied by time, equal to the change in momentum |
| Elastic rebound | Collision in which kinetic energy is conserved |
| Number density | Number of molecules per unit volume |
| Pressure | Force per unit area on the wall |
Avoid these mistakes
1
Using 2mv for the change when the wall is reversed — the molecule's change is −2mvx.
2
Using L as the distance between collisions — it is 2L for a round trip.
3
Forgetting to replace vx2 with one third of c2 — the motion is in three dimensions.
4
Using the square of the mean speed — use the mean of the squares.
5
Saying collisions between molecules change the result — they do not in this model.
Marked practiceQuizzes for this lesson
Last updated:
\n