Chemistry · Lesson

Kinetic molecular ideas and deviations from ideal gases

Use particle motion to interpret temperature and gas behaviour. Separate a distribution of speeds from one summary speed, then recognize the competing effects that make real gases depart from the ideal model.

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By the end, you can

Distinguish mean translational energy from a single common particle speed.
Calculate and interpret root-mean-square speed.
Use the dimensionless compressibility factor to describe departure from ideality.

Particles do not all move at one speed

A gas at a fixed temperature contains particles with a range of speeds. Collisions exchange energy and change directions. An equilibrium speed distribution describes the population; temperature does not assign one identical speed to every molecule.

The classical three-dimensional ideal model gives mean translational kinetic energy per mole as (3/2)RT. Translational means motion of the molecule as a whole. Rotational, vibrational and other contributions are separate, so this formula is not the complete internal energy of every molecular gas.

Root-mean-square speed summarizes the distribution

The root-mean-square speed, uᵣₘₛ, is the square root of the mean squared speed. In the classical ideal model, uᵣₘₛ = √(3RT/M), with M the molar mass. When R is in J mol⁻¹ K⁻¹, insert M in kg mol⁻¹ to obtain speed in m s⁻¹.

For an illustrative nitrogen calculation with M = 0.0280 kg mol⁻¹, the model gives about 517 m s⁻¹ at 300 K and 731 m s⁻¹ at 600 K. Doubling absolute temperature doubles mean translational energy but multiplies rms speed by √2. The square root makes the speed curve grow more slowly than a straight proportional relation.

Rms speed grows with the square root of temperature

Rms speed / m s⁻¹

Absolute temperature T / K

Selected calculated rms speeds
— Ideal-model square-root relation

Underlying data

Absolute temperature T / K	Rms speed / m s⁻¹
150	366 · Selected calculated rms speeds
300	517 · Selected calculated rms speeds
600	731 · Selected calculated rms speeds

Original ideal-model calculation for M = 0.0280 kg mol⁻¹. Circular markers give selected rounded speeds; the dashed curve uses √(3RT/M). This is translational rms speed, not measured velocities or the arithmetic mean speed.

Equal temperature does not imply equal speed

At equal temperature, gases in this model have the same mean translational energy per molecule. A larger mass requires a smaller speed to give the same kinetic energy. Consequently, rms speed varies as 1/√M at fixed T.

Rms speed is not the arithmetic mean speed or the most probable speed. It is also not an individual molecule’s permanent speed. Use the name of the summary consistently when comparing a calculation with an experiment or a distribution.

Real size and interactions change the prediction

An ideal model neglects particle volume and intermolecular interactions. In a real gas, attractive interactions can reduce pressure relative to the ideal prediction at the same V, n and T. At high density, finite occupied volume and short-range repulsive effects can instead raise it. These effects compete.

The compressibility factor Z = PV/(nRT) is dimensionless. Z = 1 is the ideal value; real gases can have Z below or above 1. A value near 1 at one state may reflect small effects or their partial cancellation. It does not establish that the gas is ideal at every pressure or temperature.

Worked examples

Keep mass units and the speed summary explicit

Estimate nitrogen’s ideal-model rms speed at 300 K using the illustrative M = 0.0280 kg mol⁻¹. Treat 300 K as stated to three significant figures.

Use uᵣₘₛ = √(3RT/M).
Substitute √[(3 × 8.314462618… × 300)/0.0280].
The units reduce to √(J/kg) = m s⁻¹.
The unrounded model result is 516.962846… m s⁻¹.

uᵣₘₛ ≈ 517 m s⁻¹. Individual particles have a distribution of speeds, and this summary is not the mean speed.

Try it yourself

If absolute temperature doubles, by what factor does ideal-model rms speed change?
Show the answer and reasoning
√2, about 1.414.
The speed depends on the square root of T, while mean translational energy depends directly on T.
At the same T, which has larger ideal-model rms speed: M or 4M?
Show the answer and reasoning
The gas with molar mass M.
Its rms speed is twice that of the gas with 4M.
What unit does Z = PV/(nRT) have?
Show the answer and reasoning
No unit.
The numerator and denominator have the same energy units, so their ratio is dimensionless.
Can a real gas have Z greater than 1 as well as less than 1?
Show the answer and reasoning
Yes.
Finite-volume and repulsive effects compete with attractions; the balance depends on the state.

What to carry forward

A temperature describes a distribution, not one common particle speed.
Use kg mol⁻¹ in the SI rms-speed formula.
Distinguish translational energy from all internal energy.
Z measures departure from ideality at the stated conditions.

References

Teaching resources and reference material supporting this lesson’s explanations and examples.