Relate molecular kinetic energy to temperature
Compare the kinetic-theory equation with the ideal-gas equation to derive and use the average translational kinetic energy per molecule.
A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 15. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.
Before you begin
- Use pressure, volume, temperature in kelvins, energy and scientific notation.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the average translational kinetic energy of one ideal-gas molecule at temperature T?
What you will learn
- Compare pV = ⅓Nm⟨c²⟩ with pV = NkT
- Derive average translational kinetic energy 3/2 kT
- Calculate the average translational kinetic energy per molecule
Equate the two descriptions of pV
The kinetic-theory relation is pV = ⅓Nm⟨c²⟩. The ideal-gas equation is pV = NkT. Equating them and cancelling N gives ⅓m⟨c²⟩ = kT.
Multiplying by 3/2 shows that ½m⟨c²⟩ = 3/2 kT. The left side is the average translational kinetic energy of one molecule, so its average value is 3/2 kT.
Average kinetic energy depends on absolute temperature
Use the thermodynamic temperature in kelvins and Boltzmann constant k in J K⁻¹. The result 3/2 kT is in joules per molecule.
At a higher temperature, the average translational kinetic energy is proportionally higher. This is an average across a distribution of molecular energies; individual molecules do not all have exactly this kinetic energy.
Pause and explain
If thermodynamic temperature doubles, what happens to average translational kinetic energy?
Worked example
Calculate the average translational kinetic energy per molecule at 300 K using k = 1.38 × 10⁻²³ J K⁻¹.
Show the worked solution
- Use Ē = 3/2 kT.
- Substitute 1.38 × 10⁻²³ × 300.
- Give joules per molecule.
Answer Ē = 6.21 × 10⁻²¹ J per molecule.
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Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Calculate the average translational kinetic energy per molecule at 600 K using k = 1.38 × 10⁻²³ J K⁻¹.
Give me a hint
Use 3/2 kT with T in kelvins.
Compare my reasoning
- Ē = 3/2 kT.
- Ē = 1.5 × 1.38 × 10⁻²³ × 600.
- Give joules per molecule.
Ē = 1.24 × 10⁻²⁰ J per molecule.
Look for these in your work
- Equation used
- Temperature in kelvins
- Energy value and unit stated
A molecule's average translational kinetic energy is 8.28 × 10⁻²¹ J. Estimate T using k = 1.38 × 10⁻²³ J K⁻¹.
Give me a hint
Rearrange 3/2 kT = Ē to make T the subject.
Compare my reasoning
- T = 2Ē/(3k).
- T = 2 × 8.28 × 10⁻²¹/(3 × 1.38 × 10⁻²³).
- Give kelvins.
T = 400 K.
Look for these in your work
- Equation rearranged
- Powers of ten handled
- Thermodynamic unit stated
Explain why 3/2 kT gives an average rather than the exact kinetic energy of every molecule in a gas.
Give me a hint
Connect the formula to a statistical average over random molecular motion.
Compare my reasoning
- Gas molecules have a distribution of speeds.
- The kinetic-theory quantity ⟨c²⟩ is averaged over many molecules.
- Therefore 3/2 kT gives average translational kinetic energy, not the value for each individual molecule.
Molecular speeds vary, so 3/2 kT is the ensemble average; individual molecules have different kinetic energies.
Look for these in your work
- Distribution recognised
- Mean-square connection identified
- Average distinguished from individual
Common mistakes
- Using Celsius instead of kelvin
- Forgetting that 3/2 multiplies kT
- Claiming every molecule has exactly the average kinetic energy
What can you explain now?
What is the average translational kinetic energy of one ideal-gas molecule at temperature T?
Compare with the explanation
3/2 kT
This follows by comparing both equations for pV.
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Make it stick
How does comparing the two gas equations lead to average molecular translational kinetic energy?
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