rms speed: same energy, different speeds
"rms speed: same energy, different speeds" is a free, interactive Class 11 Physics simulation for NEET and JEE, built on the lesson "Kinetic interpretation of temperature" in Kinetic Theory. The bars show each gas's rms speed; the one you pick is orange. Slide the temperature and watch every bar move together, while the average kinetic energy stays the same for all of them. Temperature fixes the average kinetic energy, (3/2)k_BT, for every gas. Speed then depends on mass: v_rms = √(3k_BT/m), so at one temperature v₁/v₂ = √(M₂/M₁). JEE Main asked 28 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
From the lesson Kinetic interpretation of temperature in Kinetic Theory. Change the values and watch what happens.
In JEE Main
42 shifts analysed
28 questions from Kinetic Theory across the 42 JEE Main shifts Lumi has analysed (2017–2026).
The idea behind it
NCERT §12.4.2
- Rewrite P = ⅓nm⟨v²⟩ with N = nV molecules: PV = ⅔N(½m⟨v²⟩). The bracket is the average translational kinetic energy of one molecule.
- For an ideal gas the internal energy is purely kinetic: E = N × ½m⟨v²⟩, so PV = ⅔E. (E here is the translational part; other degrees of freedom come in §12.5.)
- Compare with PV = Nk_BT: E = (3/2)Nk_BT, i.e. ½m⟨v²⟩ = (3/2)k_BT.
- So a molecule's mean kinetic energy rises in direct proportion to the kelvin temperature, whatever the pressure, the volume or the kind of ideal gas. The Boltzmann constant is the bridge between a measured temperature and a molecular energy.
- Side result: the internal energy of an ideal gas depends only on T, not on P or V. With this reading of temperature, kinetic theory agrees fully with the ideal-gas equation and the gas laws.
- Mixtures: P = ⅓[n₁m₁⟨v₁²⟩ + n₂m₂⟨v₂²⟩ + …]. In equilibrium every species has the same average kinetic energy (3/2)k_BT, so P = (n₁ + n₂ + …)k_BT, which is Dalton's law again.
- Root mean square speed: v_rms = √⟨v²⟩ = √(3k_BT/m).
- Nitrogen at 300 K: m = M/N_A = 28/(6.02 × 10²⁶) = 4.65 × 10⁻²⁶ kg, so ⟨v²⟩ = 3k_BT/m = (516)² m² s⁻² and v_rms = 516 m s⁻¹, comparable to how fast sound travels in air.
- At the same temperature, lighter molecules have larger rms speeds.
Questions
3 answers
- What does the "rms speed: same energy, different speeds" simulation show?
- The bars show each gas's rms speed; the one you pick is orange. Slide the temperature and watch every bar move together, while the average kinetic energy stays the same for all of them.
- Does JEE Main ask questions from Kinetic Theory?
- Yes. JEE Main asked 28 questions from this chapter across the 42 shifts Lumi has analysed (2017–2026).
- Is this simulation free?
- Yes. It is free and runs in the browser on a phone or a laptop. It belongs to the lesson "Kinetic interpretation of temperature" in Kinetic Theory (Class 11 Physics).
Embed this simulation
Free, with a link back
Teaching this idea? Put the simulation on your own page or LMS with this code.