JEE Main Physics Kinetic Theory of Gases 2027: RMS Speed, Degrees of Freedom & Equipartition
From PV = nRT to equipartition of energy: RMS speed, degrees of freedom, specific heats and mean free path, with solved examples.
Edurack
September 28, 2026
Temperature is not a mysterious property. It is a measure of how fast molecules jiggle. Kinetic Theory of Gases turns that idea into precise formulas, and JEE Main questions here are mostly clean substitutions once you know the pattern.
Temperature is average molecular kinetic energy in disguise.
Chapter at a Glance
| Snapshot | Detail |
|---|---|
| NTA unit | Unit 9 of 20 — Kinetic Theory of Gases |
| Priority (trend-based) | Moderate |
| Typical question style | Formula-substitution MCQs on speeds, energy and specific heats |
| Best first step | Remember three speeds and the degrees-of-freedom table |
Priority reflects past-paper trends, not an official NTA weightage.
What the NTA Syllabus Covers
- Equation of state of a perfect gas, work done on compressing a gas
- Assumptions of kinetic theory, concept of pressure, kinetic interpretation of temperature
- RMS speed of molecules, degrees of freedom, law of equipartition and specific heats of gases
- Mean free path, Avogadro's number
Master These Topics
1. Pressure, Temperature and Three Speeds
The ideal gas law is PV = nRT. Kinetic theory links pressure to molecular motion: P = ⅓ρ v²_rms. The average translational kinetic energy per molecule is (3/2)kT, which depends only on temperature.
Three characteristic speeds:
- RMS speed:
v_rms = √(3RT/M) - Average speed:
v_avg = √(8RT/πM) - Most probable speed:
v_mp = √(2RT/M)
Their ratio is v_mp : v_avg : v_rms = 1 : 1.128 : 1.224.
Worked example: RMS speed of oxygen (M = 0.032 kg/mol) at 300 K: √(3 × 8.31 × 300 / 0.032) ≈ 483 m/s. At the same temperature, hydrogen (M = 0.002) is faster by √(32/2) = 4 times.
2. Degrees of Freedom and Equipartition
The law of equipartition says each degree of freedom carries ½kT of energy per molecule. With f degrees of freedom:
- Internal energy per mole:
U = (f/2) RT Cv = (f/2) R,Cp = (f/2 + 1) Rγ = 1 + 2/f
| Gas type | f | Cv | γ |
|---|---|---|---|
| Monatomic | 3 | 3R/2 | 5/3 |
| Diatomic (rigid) | 5 | 5R/2 | 7/5 |
Worked example: Internal energy of 2 mol of a diatomic gas at 300 K = (5/2) × 2 × 8.31 × 300 ≈ 12,465 J.
Trap: Internal energy of an ideal gas depends only on temperature, not on pressure or volume. Vibrational modes are considered only if the question says so.
3. Mean Free Path
The average distance between collisions is λ = kT / (√2 π d² P), where d is the molecular diameter. It increases with temperature and decreases with pressure.
Common Traps to Avoid
- Using the wrong speed formula. Read whether RMS, average or most probable is asked.
- Forgetting to convert molar mass to kg/mol.
- Assuming internal energy depends on volume for an ideal gas.
- Using γ = 5/3 for a diatomic gas. Use 7/5 unless told otherwise.
60-Second Revision Sheet
v_rms = √(3RT/M),v_avg = √(8RT/πM),v_mp = √(2RT/M)- Average KE per molecule
(3/2)kT Cv = fR/2,γ = 1 + 2/f- Mean free path
λ = kT/(√2 π d² P)
Your Study Plan
- Day 1: gas laws and pressure derivation basics.
- Day 2: the three speeds and their ratios.
- Day 3: degrees of freedom, Cv, Cp, γ for mixtures too.
- Day 4: mean free path and mixed timed set.
Practice Kinetic Theory of Gases Questions Free → (opens in a new tab)
Continue Your Physics Journey
- Previous chapter: Thermodynamics
- Next chapter: Oscillations & Waves
- All 20 JEE Main Physics chapters
- Complete JEE Main Syllabus 2027 guide
Frequently Asked Questions
What is the difference between RMS and average speed?
RMS speed is the square root of the mean of squared speeds, and is slightly larger than the plain average speed. Their ratio is about 1.086.
How many degrees of freedom does a diatomic gas have?
Five at ordinary temperatures: three translational and two rotational. Vibration adds more only if the problem states it.