Thermal physics: gases, systems and energy transfer
Key idea: Connect thermodynamic temperature and molecular motion to ideal-gas behaviour, internal energy, equilibrium, work, the thermodynamic laws and thermal-property energy balances.
Before you start: Energy & Fields objective chainQuantities & Measurement objective chain
By the end, you can
- Use the absolute thermodynamic scale and ideal-gas equations with particle and mole quantities.
- Apply the kinetic model to derive gas pressure and relate temperature to mean translational kinetic energy.
- Distinguish internal energy, temperature and heating, and explain thermal equilibrium.
- Apply work sign conventions and the zeroth and first laws without changing convention mid-solution.
- Use specific heat capacity and specific latent heat in thermal energy balances.
Starting-point self-check
1. Check your starting point
Attempt all six groups without notes and mark the first scale, particle-count, collision, energy-store or sign decision you cannot justify. Use the recorded topic diagnostic above when you want scoring and a personalised repair plan.
Kinetic model, pressure derivation and molecular energy 12(e)–(g)
Question 1
State the ideal-gas assumptions and show the final pressure and energy relations that follow from molecular collisions.
Check the model response
Particles are numerous point masses in random motion; their own volume and intermolecular forces are negligible except during brief elastic collisions, and Newtonian mechanics applies. A one-dimensional wall-collision argument plus ⟨cₓ²⟩ = ⅓⟨c²⟩ gives pV = ⅓Nm⟨c²⟩. Comparing with pV = NkT gives ½m⟨c²⟩ = 3kT/2.
repair
2. Repair the common breaks
Use only the correction matching an error, then retry the corresponding diagnostic.
Kinetic model, pressure derivation and molecular energy 12(e)–(g)
Check this idea
Misconception: Random molecular velocities cancel, so gas pressure is zero.
Repair: Opposite velocities cancel in the mean velocity, but pressure depends on mean squared components, which are positive.
Check this idea
Misconception: Mean molecular speed is directly proportional to T.
Repair: Mean translational kinetic energy and mean squared speed are proportional to T; rms speed is proportional to √T.
worked example
3. Follow six worked models
Follow how each solution fixes the scale, gas amount, collision axis, system boundary or work convention before calculating.
Kinetic model, pressure derivation and molecular energy 12(e)–(g)
Model 1
Derive the pressure contribution of molecules moving between opposite walls of a cube, then obtain the three-dimensional result.
Check the model response
For one molecule, a wall collision changes x-momentum by 2mcₓ and successive hits on that wall are 2L/cₓ apart, giving mean force mcₓ²/L. Summing and dividing by wall area L² gives pV = Nm⟨cₓ²⟩. Random isotropic motion gives equal component means and ⟨cₓ²⟩ = ⅓⟨c²⟩, hence pV = ⅓Nm⟨c²⟩.
guided practice
4. Guided practice
Use each hint only to choose the governing definition, equation or sign convention.
Kinetic model, pressure derivation and molecular energy 12(e)–(g)
Question 1
The thermodynamic temperature of a monatomic ideal gas quadruples. State the factors for mean translational kinetic energy and rms speed.
Hint: Separate a squared speed from speed.
Check the model response
Mean translational kinetic energy quadruples because it is 3kT/2. Since ½m⟨c²⟩ ∝ T, rms speed √⟨c²⟩ doubles.
independent practice
5. Independent practice
Solve without repair notes and state every idealisation, system boundary and sign convention used.
Kinetic model, pressure derivation and molecular energy 12(e)–(g)
Question 1
Starting from a molecule's elastic wall collision, explain why gas pressure depends on mean squared speed rather than mean speed.
Check the model response
Each impulse is proportional to cₓ and collision frequency is also proportional to cₓ, so force is proportional to cₓ². Summing yields pV = Nm⟨cₓ²⟩ = ⅓Nm⟨c²⟩; opposite velocities do not cancel when squared.
Practice exit check
6. Practice assessment
Use this as extra closed-book practice, then complete the separate recorded assessment in your plan.
Kinetic model, pressure derivation and molecular energy 12(e)–(g)
Question 1
State the kinetic-theory assumptions and derive pV = ⅓Nm⟨c²⟩ far enough to identify the isotropy step.
Check the model response
Brief elastic wall collisions give pV = Nm⟨cₓ²⟩. Random isotropic motion makes ⟨cₓ²⟩ = ⟨cᵧ²⟩ = ⟨c_z²⟩ and their sum ⟨c²⟩, so each is one third. The model assumes point particles, negligible forces between collisions and random Newtonian motion.
Re-test practice
7. Delayed re-test practice
Return after at least three days and solve these fresh contexts without reopening earlier responses. The recorded plan enforces the delay and uses a separate re-test family for selected-response skill-group evidence.
Kinetic model, pressure derivation and molecular energy 12(e)–(g)
Question 1
At 300 K, find mean translational kinetic energy per molecule using k = 1.38 × 10⁻²³ J K⁻¹.
Check the model response
Mean energy = 3kT/2 = 1.5(1.38 × 10⁻²³)(300) = 6.21 × 10⁻²¹ J.