16.2 Root Mean Square Speed of an Ideal Gas | Statistical Mechanics And Thermodynamics | Physics 12

Why Not Use Average Velocity?

The molecules of an ideal gas move randomly in all directions. Because velocity is a vector, the average velocity of all molecules is zero — positive and negative components cancel. To get a meaningful measure of molecular speed, we use the Root Mean Square (RMS) speed.

$v_{r m s} = ⟨ v^{2} ⟩$

where $⟨ v^{2} ⟩$ is the mean square speed — the average of the squares of all individual molecular speeds.

Derivation of RMS Speed Formulas

From Kinetic Theory Pressure

From the kinetic theory of gases, the pressure exerted by an ideal gas is:

$P = \frac{1}{3} ρ ⟨ v^{2} ⟩$

where $ρ$ is the density of the gas. Rearranging:

$⟨ v^{2} ⟩ = \frac{3 P}{ρ}$

$v_{r m s} = \frac{3 P}{ρ}$

In Terms of Temperature (Boltzmann Constant)

Using the ideal gas law $P V = N k T$ (where $N$ is the number of molecules and $k$ is the Boltzmann constant), and substituting $ρ = N m / V$ :

$v_{r m s} = \frac{3 k T}{m}$

where $m$ is the mass of one molecule and $T$ is the absolute temperature.

Key result: $v_{r m s} \propto T$

In Terms of Molar Mass

Substituting $k = R / N_{A}$ and $m N_{A} = M$ (molar mass):

$v_{r m s} = \frac{3 R T}{M}$

where $R = 8.314 J mol^{- 1} K^{- 1}$ is the universal gas constant and $M$ is the molar mass in kg/mol.

Average Translational Kinetic Energy

The average translational kinetic energy of a single gas molecule is:

$⟨ K E ⟩ = \frac{1}{2} m ⟨ v^{2} ⟩ = \frac{1}{2} m v_{r m s}^{2}$

Substituting $v_{r m s}^{2} = \frac{3 k T}{m}$ :

$⟨ K E ⟩ = \frac{3}{2} k T$

This is one of the most important results of kinetic theory:

The average translational kinetic energy of an ideal gas molecule depends only on the absolute temperature $T$ , and is independent of the type of gas.

For one mole of gas (containing $N_{A}$ molecules), the total translational kinetic energy is:

$K E_{m o l a r} = N_{A} \cdot \frac{3}{2} k T = \frac{3}{2} R T$

Summary Table

Quantity	Formula
RMS speed (density form)	$v_{r m s} = 3 P / ρ$
RMS speed (temperature form)	$v_{r m s} = 3 k T / m$
RMS speed (molar mass form)	$v_{r m s} = 3 R T / M$
Avg. translational KE (per molecule)	$⟨ K E ⟩ = \frac{3}{2} k T$
Avg. translational KE (per mole)	$K E_{m o l a r} = \frac{3}{2} R T$

Worked Example

Calculate the rms speed of nitrogen molecules ( $M = 28 \times 1 0^{- 3}$ kg/mol) at $T = 300$ K.

$v_{r m s} = \frac{3 R T}{M} = \frac{3 \times 8.314 \times 300}{28 \times 1 0 ^{- 3}}$

$v_{r m s} = \frac{7482.6}{0.028} = 267236 \approx 517 m/s$

Why Not Use Average Velocity?

$v_{r m s} = ⟨ v^{2} ⟩$

where $⟨ v^{2} ⟩$ is the mean square speed — the average of the squares of all individual molecular speeds.

Derivation of RMS Speed Formulas

From Kinetic Theory Pressure

From the kinetic theory of gases, the pressure exerted by an ideal gas is:

$P = \frac{1}{3} ρ ⟨ v^{2} ⟩$

where $ρ$ is the density of the gas. Rearranging:

$⟨ v^{2} ⟩ = \frac{3 P}{ρ}$

$v_{r m s} = \frac{3 P}{ρ}$

In Terms of Temperature (Boltzmann Constant)

Using the ideal gas law $P V = N k T$ (where $N$ is the number of molecules and $k$ is the Boltzmann constant), and substituting $ρ = N m / V$ :

$v_{r m s} = \frac{3 k T}{m}$

where $m$ is the mass of one molecule and $T$ is the absolute temperature.

Key result: $v_{r m s} \propto T$

In Terms of Molar Mass

Substituting $k = R / N_{A}$ and $m N_{A} = M$ (molar mass):

$v_{r m s} = \frac{3 R T}{M}$

where $R = 8.314 J mol^{- 1} K^{- 1}$ is the universal gas constant and $M$ is the molar mass in kg/mol.

Average Translational Kinetic Energy

The average translational kinetic energy of a single gas molecule is:

$⟨ K E ⟩ = \frac{1}{2} m ⟨ v^{2} ⟩ = \frac{1}{2} m v_{r m s}^{2}$

Substituting $v_{r m s}^{2} = \frac{3 k T}{m}$ :

$⟨ K E ⟩ = \frac{3}{2} k T$

This is one of the most important results of kinetic theory:

The average translational kinetic energy of an ideal gas molecule depends only on the absolute temperature $T$ , and is independent of the type of gas.

For one mole of gas (containing $N_{A}$ molecules), the total translational kinetic energy is:

$K E_{m o l a r} = N_{A} \cdot \frac{3}{2} k T = \frac{3}{2} R T$

Summary Table

Quantity	Formula
RMS speed (density form)	$v_{r m s} = 3 P / ρ$
RMS speed (temperature form)	$v_{r m s} = 3 k T / m$
RMS speed (molar mass form)	$v_{r m s} = 3 R T / M$
Avg. translational KE (per molecule)	$⟨ K E ⟩ = \frac{3}{2} k T$
Avg. translational KE (per mole)	$K E_{m o l a r} = \frac{3}{2} R T$

Worked Example

Calculate the rms speed of nitrogen molecules ( $M = 28 \times 1 0^{- 3}$ kg/mol) at $T = 300$ K.

$v_{r m s} = \frac{3 R T}{M} = \frac{3 \times 8.314 \times 300}{28 \times 1 0 ^{- 3}}$

$v_{r m s} = \frac{7482.6}{0.028} = 267236 \approx 517 m/s$