# Root mean square speed

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u19-root-mean-square-speed/

This sub-topic introduces root mean square (rms) speed, a core statistical quantity in the kinetic theory of ideal gases. You will learn to define, calculate and relate rms speed to gas pressure, density and absolute temperature for CIE A-Level exam questions.

**Prerequisites:** [Kinetic theory assumptions for ideal gases](https://www.owlsprep.com/study/cie-9702-u19-kinetic-theory-assumptions/); [Ideal gas equation](https://www.owlsprep.com/study/cie-9702-u19-ideal-gas-equation/)

## Learning objectives

- Define root mean square (rms) speed for ideal gas molecules
- Relate rms speed to gas pressure, density and absolute temperature
- Calculate rms speed from given gas properties
- Avoid common exam traps when calculating rms speed

## Definition of Root Mean Square Speed

In any sample of ideal gas, individual molecules move randomly at different speeds. We cannot use the speed of a single molecule to describe the behaviour of the whole gas, so we use a statistical average called root mean square speed to represent the typical molecular speed.

**Root mean square speed** — The square root of the mean (average) of the squares of the speeds of all molecules in a gas sample.

*Notation:* $v_{\text{rms}}$

*Example:* For 3 molecules with speeds 200 m/s, 300 m/s, 400 m/s: $v_{\text{rms}} = \sqrt{\frac{200^2 + 300^2 + 400^2}{3}} \approx 311 \, \text{m/s}$

**Worked example:** Four gas molecules have speeds of 100 m/s, 200 m/s, 300 m/s and 400 m/s. Calculate the rms speed of the molecules.

1. First, calculate the square of each speed and sum the squared values:
2. $$100^2 + 200^2 + 300^2 + 400^2 = 10000 + 40000 + 90000 + 160000 = 300000 \, \text{m}^2\text{s}^{-2}$$
3. Find the mean of the squared speeds by dividing by the number of molecules:
4. $$\text{Mean square speed} = \frac{300000}{4} = 75000 \, \text{m}^2\text{s}^{-2}$$
5. Take the square root of the mean square speed to get $v_{\text{rms}}$:
6. $$v_{\text{rms}} = \sqrt{75000} \approx 274 \, \text{m s}^{-1}$$

## rms Speed and Kinetic Gas Equation

From the core assumptions of kinetic theory, we can derive a relationship between the pressure exerted by an ideal gas, its density and the rms speed of its molecules. This is one of the most important equations in ideal gas theory.

**Derivation:** Derive the relation between pressure, density and rms speed

*Starting from:* Kinetic theory assumptions for a cubic container of side $L$ holding $N$ molecules each of mass $m$

1. For one molecule colliding elastically with a container wall, the change in momentum is $\Delta p = 2mv_x$. Time between collisions with the same wall is $\Delta t = 2L/v_x$, so force from the molecule is $F = \Delta p/\Delta t = mv_x^2/L$.
2. For $N$ molecules, total average force is $F = \frac{m}{L}\sum v_{x_i}^2 = \frac{Nm\overline{v_x^2}}{L}$. By symmetry, $\overline{v^2} = 3\overline{v_x^2}$, so $\overline{v_x^2} = v_{\text{rms}}^2/3$.
3. Pressure $p = F/A$, where $A = L^2$ is the area of the wall. Substituting gives $p = \frac{Nmv_{\text{rms}}^2}{3L^3} = \frac{Nm}{3V}v_{\text{rms}}^2$. Density $\rho = Nm/V$, so we get the final relation.

*Conclusion:* p = \frac{1}{3}\rho v_{\text{rms}}^2

**Worked example:** A gas has a density of $1.2 \, \text{kg m}^{-3}$ at a pressure of $1.0 \times 10^5 \, \text{Pa}$. Calculate the rms speed of the gas molecules.

1. Rearrange the kinetic gas equation to isolate $v_{\text{rms}}$:
2. $$v_{\text{rms}} = \sqrt{\frac{3p}{\rho}}$$
3. Substitute the given values:
4. $$v_{\text{rms}} = \sqrt{\frac{3 \times 1.0 \times 10^5}{1.2}} = \sqrt{250000} = 500 \, \text{m s}^{-1}$$

## rms Speed and Absolute Temperature

Combining the kinetic gas equation with the ideal gas equation gives a direct relationship between rms speed, absolute temperature and molar mass of the gas. This relation is used in almost all CIE exam questions on this topic.

Starting from $pV = nRT$ (ideal gas equation) and substituting $p = \frac{1}{3}\frac{Nm}{V}v_{\text{rms}}^2$ gives:

$$v_{\text{rms}} = \sqrt{\frac{3RT}{M}}$$

Where $R$ is the molar gas constant, $T$ is absolute temperature in Kelvin, and $M$ is molar mass of the gas in $\text{kg mol}^{-1}$.

> **info**
>
> rms speed is proportional to $\sqrt{T}$ and inversely proportional to $\sqrt{M}$. This means higher temperature gives higher rms speed, and heavier gases have lower rms speed at the same temperature.

**Worked example:** Calculate the rms speed of oxygen molecules at 27°C. Molar mass of oxygen is $0.032 \, \text{kg mol}^{-1}$, $R = 8.31 \, \text{J mol}^{-1} \text{K}^{-1}$.

1. Convert temperature from Celsius to absolute Kelvin temperature:
2. $$T = 27 + 273 = 300 \, \text{K}$$
3. Substitute values into the rms speed formula:
4. $$v_{\text{rms}} = \sqrt{\frac{3 \times 8.31 \times 300}{0.032}} = \sqrt{233718.75} \approx 480 \, \text{m s}^{-1}$$

## Common pitfalls

- **Wrong:** Using Celsius temperature instead of Kelvin in the $v_{\text{rms}}$ formula
  - Why it fails: The formula $v_{\text{rms}} = \sqrt{3RT/M}$ requires absolute (Kelvin) temperature. Using Celsius gives an incorrect, much lower result.
  - Correct: Always add 273 to any Celsius temperature before substituting into the formula.
- **Wrong:** Calculating the mean speed first, then squaring it
  - Why it fails: rms speed is the root of the mean of the squares, not the square of the mean. This always underestimates $v_{\text{rms}}$.
  - Correct: Square each individual speed first, calculate the mean of the squared values, then take the square root.
- **Wrong:** Using molecular mass (kg per molecule) instead of molar mass (kg per mole)
  - Why it fails: The gas constant $R$ has units of $\text{J mol}^{-1} \text{K}^{-1}$, so M must be molar mass, not mass per molecule. This gives a result ~3 orders of magnitude too large.
  - Correct: If using mass per molecule $m$, use the alternative formula $v_{\text{rms}} = \sqrt{3kT/m}$ where $k$ is Boltzmann constant.
- **Wrong:** Assuming all gases have the same rms speed at the same temperature
  - Why it fails: rms speed depends on molar mass, so heavier gases have slower molecules at the same temperature.
  - Correct: Always compare molar masses when comparing rms speeds of different gases.

## Cheatsheet

| Quantity | Formula |
| --- | --- |
| rms speed from individual speeds | $\sqrt{\frac{\sum v_i^2}{N}}$ |
| rms speed from $p$ and $\rho$ | $\sqrt{\frac{3p}{\rho}}$ |
| rms speed from $T$ and $M$ | $\sqrt{\frac{3RT}{M}}$ |
| rms speed from $T$ and $m$ (molecular mass) | $\sqrt{\frac{3kT}{m}}$ |
| Kinetic gas relation | $p = \frac{1}{3}\rho v_{\text{rms}}^2$ |

## What's next

Root mean square speed is a core statistical quantity that underpins all further study of thermal physics and kinetic theory for CIE A-Level. Understanding how rms speed relates to temperature and pressure allows you to solve problems involving diffusion, internal energy and heat transfer, which appear regularly in both multiple choice and structured questions. This sub-topic also provides the foundation for the key relationship between average molecular kinetic energy and absolute temperature, a core concept that runs through all thermal physics topics on the syllabus.

- [Internal energy of ideal gas](https://www.owlsprep.com/study/cie-9702-u19-internal-energy-of-ideal-gas/)
- [Thermodynamics](https://www.owlsprep.com/study/cie-9702-u20-overview/)
- [Temperature and Thermal Equilibrium](https://www.owlsprep.com/study/cie-9702-u20-temperature-and-thermal-equilibrium/)

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