Topic 9 connects the world you can feel — temperature, heating, boiling — to the world you cannot: billions of molecules colliding. The bridge is kinetic theory.
ΔE = mcΔθ · ΔE = mL · pV = nRT = NkT · pV = ⅓Nm⟨c²⟩ · ½m⟨c²⟩ = (3/2)kTthe last equation is the punchline: temperature IS mean molecular kinetic energy
Absolute temperature is not optional. Every gas equation on this topic requires kelvin. T(K) = θ(°C) + 273. Feed in Celsius and you will not just lose accuracy — you will get physical nonsense (like a gas at negative absolute temperature).
Work through each screen, answer the questions as you go (several are full A-level calculations) and collect ⭐ stars. Press Start when you're ready.
Topic 9 · internal energy
Internal energy and absolute zero
Internal energy is the sum of the randomly distributed kinetic and potential energies of all the molecules in a body.
The kinetic part depends on temperature — raise T and molecules move faster.
The potential part depends on separation — it changes when bonds are broken during melting and boiling.
Absolute zero (0 K = −273.15 °C) is the temperature at which a substance has minimum internal energy. Molecular motion is at its minimum; it is not simply 'nothing moving'.
For an IDEAL gas only, we assume there are no intermolecular forces, so there is no potential energy: the internal energy of an ideal gas is entirely kinetic, and therefore depends only on its absolute temperature.
Topic 9 · heating
Specific heat capacity and specific latent heat
ΔE = mcΔθ · ΔE = mLc in J kg⁻¹ K⁻¹ · L in J kg⁻¹ — note L has no ΔT in it, because there is no temperature change
Specific heat capacity c — the energy needed to raise the temperature of 1 kg by 1 K. Water's is unusually large (4180 J kg⁻¹ K⁻¹), which is why it makes such a good coolant.
Specific latent heat of fusion — energy to melt 1 kg with no temperature change.
Specific latent heat of vaporisation — energy to boil 1 kg with no temperature change. For water it is 2.26 × 10⁶ J kg⁻¹ — over five times the energy needed to heat that kilogram from 0 °C to 100 °C.
Why the temperature stalls during melting: the energy supplied goes into breaking intermolecular bonds — increasing the potential part of the internal energy — not into speeding molecules up. So the mean KE, and hence the temperature, does not change.
Calculate
Your turn — specific heat capacity
1Calculate the energy needed to heat 0.50 kg of water from 20 °C to 100 °C. (c = 4180 J kg⁻¹ K⁻¹.) Give your answer in kJ to 3 significant figures.
kJ
Hint: ΔE = mcΔθ = 0.50 × 4180 × 80 = 167 200 J. Divide by 1000 for kJ.
Calculate
Your turn — latent heat
2Calculate the energy needed to boil away 0.20 kg of water already at 100 °C. (Lv = 2.26 × 10⁶ J kg⁻¹.) Give your answer in kJ to 3 significant figures.
kJ
Hint: ΔE = mL = 0.20 × 2.26 × 10⁶ = 4.52 × 10⁵ J. Divide by 1000.
Quick check
The flat bit of the graph
?A solid is heated steadily. Its temperature–time graph shows a flat section while it melts. Why does the temperature not rise during melting?
Topic 9 · gas laws
The gas laws and the ideal gas equation
Boyle's law (constant T): p ∝ 1/V, so pV = constant.
Charles' law (constant p): V ∝ T.
Pressure law (constant V): p ∝ T.
Combine all three and you get the equation of state for an ideal gas — in two equivalent forms:
pV = nRT · pV = NkTn = moles, R = 8.31 J mol⁻¹ K⁻¹ · N = number of molecules, k = 1.38 × 10⁻²³ J K⁻¹ · and k = R/N_A
32.0 mol of an ideal gas occupies 0.050 m³ at a temperature of 300 K. Calculate its pressure. Give your answer in kPa to 3 significant figures. (R = 8.31 J mol⁻¹ K⁻¹)
kPa
Hint: p = nRT/V = (2.0 × 8.31 × 300) ÷ 0.050 = 4986 ÷ 0.050 Pa. Divide by 1000 for kPa.
Calculate
Your turn — counting molecules
4That same gas has a pressure of 99 720 Pa, a volume of 0.050 m³ and a temperature of 300 K. Use pV = NkT to find the number of molecules N. Give your answer as a multiple of 10²⁴ to 3 significant figures. (k = 1.38 × 10⁻²³ J K⁻¹)
c_rms = √(2.671 × 10⁵) = 517 m s⁻¹ — faster than the speed of sound, which is no coincidence
Calculate
Your turn — rms speed
5Calculate the root-mean-square speed of a nitrogen molecule (m = 4.65 × 10⁻²⁶ kg) at 300 K. Give your answer in m s⁻¹ to 3 significant figures. (k = 1.38 × 10⁻²³ J K⁻¹)
m s⁻¹
Hint: c_rms = √(3kT/m). 3kT = 1.242 × 10⁻²⁰ J. Divide by m = 4.65 × 10⁻²⁶ to get 2.67 × 10⁵, then take the square root.
Quick check
Double the temperature
?The absolute temperature of an ideal gas is doubled. What happens to the mean kinetic energy of a molecule and to the rms speed?
Quick check
Absolute zero
?What is meant by absolute zero?
Quick check
Ideal gas assumptions
?Which of these is NOT an assumption of the kinetic theory of an ideal gas?
Match it
Match the term to its meaning
Tap an item on the left, then its partner on the right.
Term
Meaning
Recap
The big ideas to know
Internal energy = random KE + random PE of the molecules; for an ideal gas it is purely kinetic
Absolute zero: 0 K = −273.15 °C — minimum internal energy; ALWAYS use kelvin in gas equations
Heating: ΔE = mcΔθ (temperature changes) · ΔE = mL (state changes, temperature constant)
Gas laws: Boyle p ∝ 1/V · Charles V ∝ T · pressure law p ∝ T → combined: pV = nRT = NkT
Kinetic theory: pV = ⅓Nm⟨c²⟩; assumptions include point molecules, no intermolecular forces, elastic collisions
The key link: ½m⟨c²⟩ = (3/2)kT — mean molecular KE ∝ absolute temperature
rms speed: c_rms = √(3kT/m) — double T and c_rms rises only by √2
That is Edexcel Topic 9 — from a kettle to the kinetic theory of gases. Press Finish to see your score.
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