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DSE Physics — Gas Laws and Kinetic Theory: Lesson Prep Guide (Tutor Edition)
Students can quote pV = nRT but substitute Celsius, or assume compression speeds molecules up. This lesson folds three laws into one equation, then uses kinetic theory to explain where pressure comes from.
Objectives — Students will be able to: (1) use p1V1 = p2V2 to show that 2 x 10^5 Pa in 300 cm3 becomes 5 x 10^5 Pa at 120 cm3; (2) use the constant-volume relation to take a tyre from 2.2 x 10^5 Pa at 300 K to 2.42 x 10^5 Pa at 330 K; (3) find 0.963 mol in 24 dm3 at 1.0 x 10^5 Pa and 300 K, then compute mean kinetic energy of 6.21e-21 J and an rms speed near 517 m/s for nitrogen.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Pressure | Definition and unit of pressure | Force over area, the pascal (N/m2) |
| Absolute temperature | How do you convert Celsius? | Add 273, so 0 C is 273 K |
| The mole | How many particles in a mole? | 6.02 x 10^23 |
| Kinetic energy | The kinetic energy formula | Half m v squared |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0–8 min | Unit warm-up | Convert three Celsius temperatures to kelvin | Convert and justify the change |
| 8–20 min | Compress at constant T | Syringe demonstration recording p and V | Reach 5 x 10^5 Pa |
| 20–32 min | Heat at constant V | Substitute the tyre data into p/T | Reach 2.42 x 10^5 Pa |
| 32–44 min | One equation | Derive pV = nRT and compute n and molar volume | 0.963 mol and 24.9 L per mole |
| 44–52 min | Question drill | One calculation, one graph, one explanation, one design item | Work independently for eight minutes |
| 52–60 min | Exit check | Mark three written items on the spot | Complete independently |
03 Teaching script (read it out)
Scene 1: Three laws, one equation
At constant temperature p1V1 = p2V2: 2 x 10^5 times 300 equals p2 times 120, giving 5 x 10^5 Pa. At constant volume p/T is fixed: 2.2 x 10^5 over 300 equals p2 over 330, giving 2.42 x 10^5 Pa. Combining them gives pV = nRT: 1.0 x 10^5 times 0.024 divided by (8.31 times 300) is 0.963 mol, so one mole occupies 24.9 L at 300 K, more than the 22.4 L at 0 C. Students should say that comparison themselves.
Ask students: what nonsense appears if Celsius is used instead of kelvin?
Scene 2: Pressure is molecules hitting the wall
Molecules move constantly and each collision changes momentum, so the wall feels a force. Hotter gas means higher mean kinetic energy, harder and more frequent collisions, higher pressure. At 300 K the mean kinetic energy is 1.5 times 1.381 x 10^-23 times 300, which is 6.21e-21 J; nitrogen with molar mass 0.028 kg gives an rms speed of the square root of 3RT/M, about 517 m/s. Students must see that rms is the square root of the mean square, not the mean speed, and that at one temperature all gases share the same mean kinetic energy, so heavier molecules are slower.
Ask students: at the same temperature, which moves faster, hydrogen or nitrogen, and why?
Scene 3: Graphs and models check each other
pV constant is a hyperbola on a p-V plot and a straight line through the origin on p against 1/V, with slope nRT. A line that misses the origin usually means cubic centimetres were never converted to cubic metres, or the temperature was not constant. If the gas warms during compression, pressure sits above the isotherm, which is one reason real gases depart from the ideal model.
Ask students: what does the slope of the p against 1/V graph represent?
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | New pressure after isothermal compression | Convert units, write p1V1 = p2V2, solve and check the order of magnitude | 5 x 10^5 Pa |
| Graph | Replot p-V data as p against 1/V and find the slope | Compute 1/V point by point, draw a best-fit line, read the slope as nRT | Slope about 8.0 J |
| Explanation | Use kinetic theory to explain heating at constant volume | Temperature up, kinetic energy up, speed up, collision rate and momentum change up | Pressure is proportional to absolute temperature at constant volume |
| Experimental design | Verify Boyle law with a syringe | Trapped fixed mass, step pressure, read volume, keep temperature steady and move slowly | p times V should stay near 60000000 Pa cm3 |
Sanity check (teach this): Sanity check: add 273 to every temperature; keep volumes in one unit; the product must survive compression (2 x 10^5 times 300 equals 5 x 10^5 times 120, both 60000000); an rms speed of 517 m/s is right, while 500 km/s signals a unit error.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| 2 x 10^5 Pa in 300 cm3 compressed to 120 cm3: new pressure? | 5 x 10^5 Pa | Check the student states temperature is constant |
| How many moles in 24 dm3 at 1.0 x 10^5 Pa and 300 K? | 0.963 mol | Check the volume is converted to cubic metres |
| What is the rms speed of nitrogen at 300 K? | About 517 m/s | Check kelvin and a molar mass in kilograms |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Substituting Celsius | p1/T1 with T1 equal to 27 | The equation needs kelvin: 27 C is 300 K, not 27. |
| Mixing units | Volume in cm3 with R as 8.31 | R as 8.31 pairs with cubic metres and pascals; divide cm3 by 10^6. |
| Thinking compression speeds molecules up | Squeezed gas means faster molecules | At constant temperature kinetic energy is unchanged; collision frequency and density rise. |
| Confusing rms with mean speed | rms speed is the average speed | rms squares first, averages, then roots, so it sits slightly above the mean. |
07 Homework (paper, section and question numbers)
- Secondary Physics chapter index — Read the index and write the condition for each of the three laws
- Heat capacity and latent heat (S4 chapter) — Redo practice question 3 there and compare heating with compressing as ways to change internal energy
- First law of thermodynamics (S5 chapter) — Use that sign convention to explain the sign of W in isothermal compression
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| 2.0 x 10^5 Pa in 300 cm3 compressed to 120 cm3 at constant temperature: new pressure? | 5 x 10^5 Pa | If temperature is ignored: reteach Boyle law |
| Why must temperature be in kelvin? | The law scales from absolute zero; Celsius breaks the ratio | If “because it is the convention”: reteach absolute temperature |
| At one temperature, which has the larger rms speed, nitrogen or hydrogen? | Hydrogen, because its molar mass is smaller at equal kinetic energy | If nitrogen: reteach the kinetic energy and mass relation |
09 Tutor FAQ
Which year level is this for?
S5; S6 revision of thermal physics and ideal-gas assumptions can reuse the numbers.
Is 60 minutes enough?
Enough for three acts and one drill; spend five minutes on unit conversion and move the graph question to homework.
Is there practice attached?
The homework section links to the site chapter pages, where answers appear under independent practice.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.