Kinetic Theory of Gases: Same Energy, Different Speeds

At one temperature every ideal gas has the same average translational kinetic energy per molecule, (3/2)k_BT, whatever its mass. Here k_B is the Boltzmann constant, 1.380649×10⁻²³ J/K, and translational means the energy of the molecule moving as a whole. Lighter molecules therefore move faster. At 300 K all six gases below carry 6.21×10⁻²¹ J per molecule, yet rms speed runs from 412 m/s for CO₂ to 1927 m/s for H₂.

We computed this table in November 2026 from the speed formulas in OpenStax University Physics Volume 2, section 2.4. We used R = 8.314 J/(mol·K), T = 300 K and molar mass in kg/mol, and rounded speeds to the nearest m/s. The energy is ½m·v_rms² per molecule.

You can check our method against a published case. At 293 K the same formulas give N₂ an rms speed of 511 m/s and an energy of 6.07×10⁻²¹ J per molecule, which are the values in the worked example in OpenStax section 2.2.

Three things to read off the table:

• Speed depends on mass. v_rms ∝ 1/√M, so H₂ against O₂ gives √(31.998 ÷ 2.016) = 3.98, which is close to the √16 = 4 you get from rounded masses of 2 and 32.
• The speed ratios never change. The most probable speed v_p, the average speed v_avg and v_rms are in the ratio 1 : √(4/π) : √(3/2) = 1 : 1.128 : 1.225 for every gas at every temperature. So v_avg is 0.921 of v_rms, about 8% below it.
• Energy does not depend on mass at all. Combine the ideal gas law pV = Nk_BT with pV = ⅓Nm⟨v²⟩ (N is the number of molecules, m the mass of one, and ⟨v²⟩ the mean of the squared speeds) and you get ½m⟨v²⟩ = (3/2)k_BT. Mass cancels.

Here rms speed means the square root of the mean of the squared speeds. The most probable speed is the speed at the peak of the Maxwell distribution, which is the spread of molecular speeds in a gas.

The kinetic theory of gases treats a gas as a very large number of tiny molecules in random motion that interact only through elastic collisions. These postulates (also called assumptions) describe an ideal gas, and a real gas follows them closely only when it is dilute and well above its boiling point.

For a board answer, write them like this:

1. A gas contains a very large number of identical molecules in constant, random motion.
2. The size of a molecule is negligible compared with the distance between molecules, so molecules behave as point masses.
3. There are no forces between molecules except during collisions, and gravity has no noticeable effect on their motion.
4. Collisions between molecules, and with the walls, are perfectly elastic (no kinetic energy is lost). A collision lasts a negligible time compared with the time between collisions.
5. Between collisions a molecule moves in a straight line at constant speed, obeying Newton's laws.
6. Molecules are spread uniformly through the container and move equally in all directions.

Postulates 2 and 3 are the ones that break first. At high pressure or low temperature, molecules sit close enough that their size and mutual attraction matter, and the ideal gas results drift away from measured values. Liquids and solids fail the same way, because their molecules are packed almost touching and held by strong forces.

The one third appears because motion is equally likely in all three directions, so a molecule's squared speed splits equally among x, y and z. Only the part along the x direction pushes on a wall facing x. That gives pV = ⅓Nm⟨v²⟩, where N is the number of molecules, m the mass of one, and ⟨v²⟩ the mean square speed.

Here is the argument in order. Take a cube of side L, and a molecule with x-velocity vₓ.

1. Each hit on the wall reverses vₓ, so the momentum change is 2mvₓ.
2. It returns to the same wall after travelling 2L, which takes 2L/vₓ.
3. The average force from that molecule is 2mvₓ ÷ (2L/vₓ) = mvₓ²/L.
4. For N molecules the pressure on a wall of area L² is P = Nm⟨vₓ²⟩/V.
5. Random motion makes ⟨vₓ²⟩ = ⟨v²⟩/3, so P = ⅓(N/V)m⟨v²⟩. Since Nm/V is the density ρ, P = ⅓ρ⟨v²⟩ = ⅓ρv_rms².

A question that comes up at step 3 is why we divide by the time between hits and not by the duration of one collision. The answer is that pressure is the force averaged over time. A wall under gas bombardment feels countless tiny pulses, and you want their steady average, not the size of one pulse. This is why the postulate of negligible collision time is safe to use.

Choose the formula by what the question asks for, then check the units. The table covers the quantities this chapter asks about, with the symbols and units you must use.

Degrees of freedom (f) are the independent ways a molecule can store energy. For board problems use f = 3 for a monatomic gas, which gives γ = 5/3, and f = 5 for a diatomic gas at ordinary temperatures, which gives γ = 7/5. If vibration is included, a diatomic gas has f = 7 and γ = 9/7, and a question will say so. Each degree of freedom carries ½k_BT per molecule, which is the law of equipartition of energy.

Mean free path is the average distance a molecule travels between collisions. As an example of our own working, take N₂ at 300 K and 1 atm (101,325 Pa) with an assumed molecular diameter d = 3.7×10⁻¹⁰ m. Then n = P/(k_BT) = 2.45×10²⁵ per m³ and λ ≈ 6.7×10⁻⁸ m. That is our model, not a measured constant: the answer changes with the diameter you assume.

In our experience, the three that cost the most marks are using Celsius in place of kelvin, putting molar mass in g/mol, and swapping one of the three speeds for another. Each is easy to catch with a one-line check.

1. Celsius for kelvin. Energy and speed are tied to absolute temperature, so convert first with T(K) = t(°C) + 273 (or 273.15). To find how rms speed changes when a gas is heated from 27 °C to 127 °C:

• Correct: v₂/v₁ = √(400 ÷ 300) = 1.15.
• Wrong, using Celsius numbers: √(127 ÷ 27) = 2.17.

The wrong route overstates the change by almost a factor of two.

2. Grams for kilograms. In √(3RT/M), R is in joules, so M must be in kg/mol. Writing 32 for oxygen instead of 0.032 makes M a thousand times too big and your speed √1000 ≈ 31.6 times too small. For O₂ at 300 K the answer should be near 484 m/s, so a result of 15 m/s tells you straight away that the units are wrong.

3. One speed for another. If a question says "root mean square", use √(3RT/M). If it says "average" or "mean" speed, use √(8RT/πM). If it says "most probable", use √(2RT/M). The order is always v_p < v_avg < v_rms, so any answer that breaks that order is wrong.

Two more checks help with ratio questions, and they use the same reasoning:

• Same gas, different temperatures. v_rms ∝ √T, so doubling T multiplies the speed by √2 = 1.41, not by 2.
• Different gases, same rms speed. Argon must be hotter than helium. If argon's rms speed equals helium's at −20 °C (253.15 K), then T_Ar = 253.15 × 39.9 ÷ 4.0 = 2525 K, which is about 2.5×10³ K, using atomic masses of 39.9 atomic mass units (u) for argon and 4.0 u for helium. The heavier atom needs about ten times the temperature to keep up.

For the habit of writing data, formula, units and a sense-check in the same order every time, see our guide to how to solve physics numericals. If a numerical gives you a strange unit, the units and dimensions check catches it quickly.

Our MHT-CET 2027 Long Term Batch, taught by RG Sir (Rahul Jewaani), covers the complete MHT-CET Physics and Chemistry syllabus. It includes live classes, recordings, notes, previous year questions (PYQs) and mock tests. As of November 2026 it costs ₹1,632, listed at ₹1,999. Kinetic theory of gases is part of the Std XII Physics syllabus that MHT-CET is built on.

As of November 2026 the newest syllabus the CET Cell has posted is for A.Y. 2026-27 (CET Cell syllabus page). Check the same page for the A.Y. 2027-28 document when it appears, and compare this chapter's topics against it. Then use the table and the picker above as your drill. Cover the right-hand column, name the formula for each question type, and only then compute.

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