The shape of the lesson
By the end of the lesson, learners can
- Collisions make pressure: Pressure is not a property of single molecules; it is the steady drumbeat of countless collisions on the walls.
- From mean-square speed to temperature: Modelling the gas as many small molecules in random motion, kinetic theory gives pV = ⅓Nm⟨c²⟩, where ⟨c²⟩ is the mean of the squares of the speeds and the root-mean-square speed is c_rms = √⟨c²⟩.
- Counting particles: one mole holds N₊ particles, N = nN₊, and k = R / N₊.
- Equation of state: pV = nRT = NkT, with T in kelvin.
- Kinetic theory: pV = ⅓Nm⟨c²⟩, and the mean translational KE per molecule is ½m⟨c²⟩ = ₃⁄₂ kT.
One picture every idea returns to
Treating gas as many molecules in random motion, kinetic theory gives pV = ⅓Nm⟨c²⟩. Comparing with pV = NkT shows the mean translational kinetic energy per molecule is ½m⟨c²⟩ = ₃⁄₂ kT.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, Pressure from molecules.. Kinetic theory explains the large-scale gas laws from the random motion of molecules. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Collisions make pressure | Pressure is not a property of single molecules; it is the steady drumbeat of countless collisions on the walls. Project tab 3 of the lab (Kinetic theory) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: From mean-square speed to temperature | Modelling the gas as many small molecules in random motion, kinetic theory gives pV = ⅓Nm⟨c²⟩, where ⟨c²⟩ is the mean of the squares of the speeds and the root-mean-square speed is c_rms = √⟨c²⟩. | Connect-Extend-Challenge |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (4 questions, 47 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in From mean-square speed to temperature, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Connect-Extend-Challenge
After the new idea, each student writes three lines: how it connects to what they knew, how it extends their thinking, and what still challenges them. Pairs compare, then the challenges become the plenary.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
No examiner trap is recorded for this lesson. Use the quiz key below to spot the distractor most of the class picks.
Kinetic theory of gases: topic worksheet
4 questions, 47 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
Fig. 2.1 represents the molecules of an ideal gas in random motion inside a container.
(a) State three assumptions made in the kinetic theory of an ideal gas. [3]
(b) Kinetic theory gives pV = ⅓Nm⟨c²⟩. Use this together with pV = NkT to show that the mean translational kinetic energy of a molecule is ₃⁄₂ kT. [3]
(c) Calculate the mean translational kinetic energy of a molecule at a temperature of 300 K. [2]
(d) A nitrogen molecule has mass 4.7 × 10⁻²⁶ kg. Calculate its root-mean-square speed at 300 K. [3]
(e) State and explain how the root-mean-square speed changes if the thermodynamic temperature is doubled. [2]
Fig. 8.1 represents the molecules of a gas in rapid random motion.
(a) Write the kinetic theory equation relating pressure to molecular motion. [1]
(b) State three assumptions of the kinetic theory of gases. [3]
(c) A gas has density 1.2 kg m⁻³ at a pressure of 1.0 × 10⁵ Pa. Calculate the root-mean-square speed of its molecules. [3]
(d) State what is meant by the root-mean-square speed. [2]
(e) State and explain how the r.m.s. speed changes if the kelvin temperature is doubled. [2]
This question links the equation of state with the kinetic theory. Part (b) connects pV = NkT with pV = ⅓Nm⟨c²⟩.
(a) Write the expression for the mean translational kinetic energy of a molecule in terms of temperature. [1]
(b) Show that this follows from pV = NkT and pV = ⅓Nm⟨c²⟩. [3]
(c) Calculate the mean translational kinetic energy of a molecule at 300 K. [2]
(d) An oxygen molecule has mass 5.3 × 10⁻²⁶ kg. Calculate its r.m.s. speed at 300 K. [3]
(e) State how the mean translational kinetic energy at a given temperature depends on the type of gas. [2]
A sealed flask of volume 2.0 × 10⁻³ m³ contains helium gas (molar mass 4.0 g mol⁻¹) at 1.5 × 10⁵ Pa and 300 K. This question uses all three lessons of the topic in turn.
(a) Using pV = nRT, calculate the amount of substance. [2]
(b) Calculate the number of helium atoms. [2]
(c) Calculate the mass of helium in the flask. [2]
(d) Calculate the mean translational kinetic energy of a helium atom. [2]
(e) The mass of a helium atom is the molar mass divided by N₊. Calculate the r.m.s. speed of the atoms. [3]
(f) State whether argon atoms (heavier) at the same temperature would have a greater or smaller r.m.s. speed. [1]
Answer key and marking notes
- (a)Any three: a large number of molecules in random motion ✓; molecular volume negligible compared with the container ✓; no intermolecular forces except during collisions, which are perfectly elastic ✓; time of collision negligible compared with time between collisions.
- (b)Equate the two expressions for pV: ⅓Nm⟨c²⟩ = NkT ✓; so ⅓m⟨c²⟩ = kT, giving ½m⟨c²⟩ = ₃⁄₂ kT ✓, which is the mean translational kinetic energy of one molecule ✓
- (c)½m⟨c²⟩ = ₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 300 ✓ = 6.2 × 10⁻²¹ J ✓
- (d)½m c_rms² = 6.2 × 10⁻²¹ J, so c_rms² = 2 × 6.2 × 10⁻²¹ / 4.7 × 10⁻²⁶ ✓ = 2.64 × 10⁵; c_rms = √(2.64 × 10⁵) ✓ = 5.1 × 10² m s⁻¹ ✓
- (e)Since ½m c_rms² = ₃⁄₂ kT, c_rms ∝ √T ✓; doubling T multiplies the r.m.s. speed by √2 (about 1.41) ✓
- (a)pV = ⅓Nm⟨c²⟩ ✓
- (b)Any three: molecules are point-like (negligible volume); no forces between molecules except during collisions; collisions are perfectly elastic; motion is random; collision times are negligible ✓✓✓
- (c)p = ⅓ρ⟨c²⟩, so ⟨c²⟩ = 3p/ρ = 3 × 10⁵/1.2 = 2.5 × 10⁵ ✓✓; c_rms = √(2.5 × 10⁵) = 5.0 × 10² m s⁻¹ ✓
- (d)The square root of the mean of the squares of the molecular speeds ✓✓
- (e)c_rms ∝ √T, so doubling T increases it by a factor of √2 (about 1.4) ✓✓
- (a)½m⟨c²⟩ = ₃⁄₂ kT ✓
- (b)Equating: ⅓Nm⟨c²⟩ = NkT ✓; so m⟨c²⟩ = 3kT ✓; hence ½m⟨c²⟩ = ₃⁄₂ kT ✓
- (c)₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 300 ✓ = 6.2 × 10⁻²¹ J ✓
- (d)½m c² = 6.21 × 10⁻²¹, so c² = 2 × 6.21 × 10⁻²¹ / 5.3 × 10⁻²⁶ ✓✓ = 2.34 × 10⁵; c_rms = 4.8 × 10² m s⁻¹ ✓
- (e)It does not depend on the type of gas; it depends only on the temperature ✓✓
- (a)n = pV/RT = 1.5 × 10⁵ × 2.0 × 10⁻³ / (8.31 × 300) ✓ = 0.12 mol ✓
- (b)N = nN₊ = 0.120 × 6.02 × 10²³ ✓ = 7.2 × 10²² ✓
- (c)mass = nM = 0.120 × 4.0 = 0.48 g = 4.8 × 10⁻⁴ kg ✓✓
- (d)₃⁄₂ kT = 1.5 × 1.38 × 10⁻²³ × 300 ✓ = 6.2 × 10⁻²¹ J ✓
- (e)m = 4.0 × 10⁻³ / 6.02 × 10²³ = 6.6 × 10⁻²⁷ kg ✓; c² = 2 × 6.21 × 10⁻²¹ / 6.64 × 10⁻²⁷ = 1.87 × 10⁶ ✓; c_rms = 1.4 × 10³ m s⁻¹ ✓
- (f)Smaller: the mean kinetic energy is the same, so a larger mass gives a smaller speed ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
A basic assumption of the kinetic theory of an ideal gas is that:
- Aintermolecular forces are negligible except during collisions
- Bmolecules occupy most of the container volume
- Ccollisions lose kinetic energy
- Dall molecules move at the same speed
In the equation pV = ⅓Nm⟨c²⟩, the quantity ⟨c²⟩ is:
- Athe mean of the squares of the molecular speeds
- Bthe square of the mean speed
- Cthe maximum speed
- Dthe speed of sound
The mean translational kinetic energy of a molecule of an ideal gas is:
- A₃⁄₂ kT, proportional to the thermodynamic temperature
- B½ kT
- CkT, independent of temperature
- Dproportional to the pressure only
If the thermodynamic temperature of a gas is increased fourfold, the root-mean-square speed of its molecules:
- Adoubles
- Bquadruples
- Cincreases by √2
- Dis unchanged
Quiz key
Ideal-gas molecules are assumed to have negligible volume and no forces between them except during perfectly elastic collisions.
⟨c²⟩ is the mean-square speed; its square root is the root-mean-square speed c_rms.
Comparing pV = ⅓Nm⟨c²⟩ with pV = NkT gives ½m⟨c²⟩ = ₃⁄₂ kT, proportional to T.
Since ½m c_rms² = ₃⁄₂ kT, c_rms ∝ √T; a fourfold rise in T multiplies c_rms by √4 = 2.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.