Kinetic Theory of Gases Equation

The Root Mean Square Speed

The kinetic theory of gases equation includes the mean square speed of the particles:

$c^{2}$

Where
- c = average speed of the gas particles
- $c^{2}$ has the units m² s⁻²

Since particles travel in all directions in 3D space and velocity is a vector, some particles will have a negative direction and others a positive direction
When there are a large number of particles, the total positive and negative velocity values will cancel out, giving a net zero value overall
In order to find the pressure of the gas, the velocities must be squared
- This is a more useful method, since a negative or positive number squared is always positive
To calculate the average speed of the particles in a gas, take the square root of the mean square speed:

$\sqrt{\bar{c^{2}}}$

This is known as the root-mean-square speed and still has the units of m s⁻¹
- The root-mean-square speed can also have the symbol c_r.m.s.
The mean square speed is not the same as the mean speed

Worked example

A very small group of atoms have the following velocities:

50 m s⁻¹

+80 m s⁻¹

+85 m s⁻¹

−65 m s⁻¹

−90 m s⁻¹

Calculate the mean speed, $\bar{c}$ , mean square speed, $\bar{c^{2}}$ , and r.m.s speed, c_r.m.s, of these atoms.

Step 1: Calculate the mean speed $\bar{c}$ :

- Add all the values and divide by the number of values you have to calculate the mean

$\bar{c}$ = $\frac{50 + 80 + 85 - 65 - 90}{5}$ = +12 m s⁻¹

Step 2: Calculate the mean square speed $\bar{c^{2}}$ :

- Square each value, then add them all and divide by the number of values you have to calculate the mean of the squares

$\bar{c^{2}}$ = $\frac{50^{2} + 80^{2} + {85^{2} + (- 65)}^{2} + {(- 90)}^{2}}{5}$ = +5690 m² s⁻²

- Here the units are also squared, as (m s⁻¹)² = m² s⁻²

Step 3: Calculate the r.m.s speed c_r.m.s

- Find the square root of the mean square speed $\bar{c^{2}}$

c_r.m.s = $\sqrt{\bar{c^{2}}}$ = $\sqrt{5690 m^{2} s^{- 2}}$ = 75.4 m s⁻¹= 75 m s⁻¹(2 s.f.)

Examiner Tip

Make sure you read questions relating to the r.m.s speed carefully! It is easy to get confused between $\bar{c}$ , $\bar{c^{2}}$ and $\sqrt{\bar{c^{2}}}$

The Kinetic Theory of Gases Equation is given by:

$p V = \frac{1}{3} N m \bar{c^{2}}$

Where
- p = pressure (Pa)
- V = volume (m³)
- N = number of molecules
- m = mass of one molecule of gas (kg)
- $\bar{c^{2}}$ = mean square speed of the molecules (m² s⁻²)

The density, ρ of the gas is given by:

$ρ = \frac{mass}{volume} = \frac{N m}{V}$

Rearranging the equation for pressure, p and substituting the density, ρ gives the equation:

$p = \frac{1}{3} ρ \bar{c^{2}}$

Where
- p = pressure (Pa)
- ρ = density (kg m⁻³)
- $\bar{c^{2}}$ = mean square speed of the molecules (m² s⁻²)

Worked example

A gas cylinder contains argon at a pressure of 700 kPa.

The cylinder contains 3.0 × 10²⁵molecules and each molecule has a mass of 4.7 × 10⁻²⁶kg. The r.m.s speed of the molecules is 350 m s⁻¹.

Calculate the volume of the cylinder.

Step 1: List the known quantities

- Number of molecules, N = 3.0 × 10²⁵
- Mass of each molecule, m = 4.7 × 10⁻²⁶kg
- Root mean square speed, c_r.m.s. = 350 m s⁻¹
  - Therefore $\bar{c^{2}}$ = 350²m²s⁻²
- Pressure, P = 700 kPa = 700 × 10³Pa

Step 2: State the equation

$p V = \frac{1}{3} N m \bar{c^{2}}$

Step 3: Rearrange the equation to make volume, V, the subject

$V = \frac{N m \bar{c^{2}}}{3 p}$

Step 3: Substitute the values from the question into the equation and calculate V

$V = \frac{(3.0 \times 10^{25}) \times (4.7 \times 10^{- 26}) \times {(350)}^{2}}{3 \times (700 \times 10^{3})}$

V = 0.08225 m³

Step 4: Write the final answer to the correct number of decimal places

- The volume of the cylinder is 0.082 m³(3 d.p.)

Worked example

An ideal gas has a density of 4.5 kg m⁻³ at a pressure of 9.3 × 10⁵ Pa.

Determine the root mean square speed, c_r.m.s., of the gas atoms at a constant temperature.

Step 1: Write out the known quantities

- Pressure, p = 9.3 × 10⁵ Pa
- Density, ρ = 4.5 kg m⁻³

Step 2: State the equation linking the pressure of an ideal gas with density

$p = \frac{1}{3} ρ \bar{c^{2}}$

Step 3: Rearrange to make $\bar{c^{2}}$ the subject

$\bar{c^{2}} = \frac{3 p}{ρ}$

Step 4: Take square roots of both sides to give an equation for c_r.m.s.

_{$c_{r . m . s .} = \sqrt{\frac{3 p}{ρ}}$}

Step 5: Substitute in known values and calculate c_r.m.s.

_{$c_{r . m . s .} = \sqrt{\frac{3 \times 9.3 \times 10^{5}}{4.5}} = 787.4 = 790 m s^{- 1}$}

Examiner Tip

You are not expected to know the derivation of the kinetic theory equation. However, a common exam question is to use the kinetic model of a gas and Newton's laws of motion to explain how a gas exerts pressure on the walls of its container.

You will also be expected to explain why the mean square speed of the molecules is used in this equation instead of mean velocity. It is important to note that these quantities are not equivalent. The mean velocity would be zero because particles with opposite velocities cancel out.

Kinetic Theory of Gases Equation (OCR A Level Physics)

Revision Note

The Root Mean Square Speed

Worked example

Examiner Tip