International A Level (IAL)PhysicsEdexcelRevision Notes5. Thermodynamics, Radiation, Oscillations & CosmologyKinetic Theory & Ideal GasesAverage Molecular Kinetic Energy

Average Molecular Kinetic Energy (Edexcel International A Level (IAL) Physics): Revision Note

Exam code: YPH11

Written by: Katie M

Reviewed by: Caroline Carroll

Updated on 5 October 2026

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Average Molecular Kinetic Energy

An important property of molecules in a gas is their average kinetic energy
This can be deduced from the ideal gas equations relating pressure, volume, temperature and speed
Recall the ideal gas equation in terms of number of molecules:

$p V = N k T$

Also, recall the equation linking pressure and mean square speed of the molecules:

$p V = \frac{1}{3} N m {(c_{r m s})}^{2}$

The left-hand side of both equations are equal to pV
This means the right-hand sides of both equations are also equal:

$\frac{1}{3} N m {(c_{r m s})}^{2} = N k T$

N will cancel out on both sides, and multiplying by 3 on both sides also obtains the equation:

$m {(c_{r m s})}^{2} = 3 k T$

Recall the familiar kinetic energy equation from mechanics:

$Kinetic energy = \frac{1}{2} m v^{2}$

Instead of v² for the velocity of one particle, (c_rms)² is the average speed of all molecules

Multiplying both sides of the equation by ½ obtains the average molecular kinetic energy of the molecules of an ideal gas:

$E_{k} = \frac{1}{2} m {(c_{r m s})}^{2} = \frac{3}{2} k T$

Where:
- E_k= kinetic energy of a molecule (J)
- m = mass of one molecule (kg)
- (c_rms)²= mean square speed of a molecule (m² s^-2)
- k = Boltzmann constant
- T = temperature of the gas (K)

Note: this is the average kinetic energy for only one molecule of the gas

To find the total kinetic energy of N molecules of the gas, multiply both sides of the equation by the number of molecules N to obtain:

$E_{k} = \frac{1}{2} N m {(c_{r m s})}^{2} = \frac{3}{2} N k T$

A key feature of this equation is that the mean kinetic energy of an ideal gas molecule is proportional to its thermodynamic temperature

$E_{k} \propto T$

The Boltzmann constant k can be replaced with

$k = \frac{R}{N_{A}}$

Substituting this into the average molecular kinetic energy equation means it can also be written as:

$E_{k} = \frac{1}{2} m {(c_{r m s})}^{2} = \frac{3}{2} k T = \frac{3 R T}{2 N_{A}}$

Worked Example

Helium can be treated as an ideal gas. Helium molecules have a root-mean-square (r.m.s.) speed of 720 m s^-1 at a temperature of 45 °C. Calculate the r.m.s. speed of the molecules at a temperature of 80 °C.

Answer:

Step 1: Write down the equation for the average kinetic energy

$E_{k} = \frac{1}{2} m {(c_{r m s})}^{2} = \frac{3}{2} k T$

Step 2: Determine the relation between and the temperature

Since $m$ and $k$ are constant, ${(c_{r m s})}^{2}$ is directly proportional to $T$
Therefore

$c_{r m s} \propto \sqrt{T}$

Step 3: Change the proportionality into an equation

$c_{r m s} = a \sqrt{T}$

Where $a$ is the constant of proportionality

Step 4: Calculate the constant of proportionality

Since $c_{r m s} = 720 m s^{- 1}$ at a temperature of $45 ° C$ :

$T = 45 ° C + 273.15 = 318.15 K$

$a = \frac{c_{r m s}}{\sqrt{T}} = \frac{720}{\sqrt{318.15}}$

Step 5: Calculate $c_{r m s}$ at $T = 80 ° C$ by substituting the value of $a$

$T = 80 ° C + 273.15 = 353.15 K$

$c_{r m s} = \frac{720}{\sqrt{318.15}} \times \sqrt{353.15}$

$c_{r m s} = 758.57 m s^{- 1} = 760 m s^{- 1} (2 s . f .)$

Examiner Tips and Tricks

You can remember the equation through the rhyme ‘Average KE is three-halves kT’.

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