Pick a topic, read the steps, then try them right in the sandbox below.
Class 11 — Kinetic theory: where pressure actually comes from
Concept: gas pressure isn't magic — it's the real, measurable result of
countless particles colliding with the container walls. More particles or higher
temperature (faster particles) means more, harder collisions, which means more
pressure.
- Switch to Kinetic theory gas mode.
- Note the starting Pressure (measured) value.
- Raise Temperature T and watch the particles visibly speed up — Pressure
(measured) should rise too.
- Reset the temperature, then instead raise Particles N — pressure rises
again, this time because more particles are hitting the walls, not because they're
faster.
- Finally, drag the Piston (volume) control to shrink the box — pressure
rises sharply as the same particles now hit the walls far more often.
Takeaway: pressure isn't one single "thing" — it's the combined effect of
how many particles there are, how fast they're moving, and how much space they have.
Class 11 — The four PV processes
Concept: the same gas can be compressed or expanded in different ways —
keeping temperature fixed (isothermal), pressure fixed (isobaric), volume fixed
(isochoric), or with no heat exchange at all (adiabatic) — and each one does a
different amount of work, shown as the area under the curve on the PV diagram.
- Switch to PV diagram mode.
- Select isothermal and drag Volume V to expand the gas. Note the
Work done W=∫PdV value.
- Reset, select adiabatic instead, and expand the volume by the same
amount. Compare the work done — an adiabatic expansion does less work for the same
volume change, because the gas cools as it expands instead of drawing in heat to
stay at constant temperature.
- Try isochoric (volume fixed) — confirm the work done stays at zero no
matter what you do to Temperature, since W = ∫PdV needs the volume to actually
change.
Takeaway: "work done by a gas" always means the area under its path on a PV
diagram — and that area depends entirely on which of the four processes it follows.
Class 11 — The Carnot engine and the refrigerator
Concept: a heat engine takes heat Qh from a hot reservoir, turns part of it into work W and rejects the rest, Qc, to a cold one. The most efficient
engine possible between two temperatures is the Carnot engine, with η = 1 − Tc ÷ Th.
- Open Heat engines & Carnot cycle: monatomic gas, Th = 500 K, Tc = 300 K, expansion ratio 2. The gas takes in Qh = nRTh ln 2 = 2881 J
on the hot isotherm and rejects Qc = nRTc ln 2 = 1729 J on the cold one. The work is 2881 − 1729 = 1153 J, the area inside the loop.
- The efficiency is W ÷ Qh = 40.0%, exactly 1 − 300 ÷ 500. Change the expansion ratio: the work changes but the efficiency does not.
- Raise Th to 600 K: η = 1 − 300 ÷ 600 = 50%. Lower Tc to 250 K at Th = 500 K: also 50%. Only the two temperatures matter.
- Tick “run it backwards”: it becomes a refrigerator. It takes 1729 J from the cold side using 1153 J of work and gives 2881 J to the hot side. The coefficient of performance
is Tc ÷ (Th − Tc) = 300 ÷ 200 = 1.50.
- On the left graph the dot always lies on or below the Carnot curve: no engine can do better (the second law).
Takeaway: η = 1 − Tc ÷ Th and Qc ÷ Qh = Tc ÷ Th; a refrigerator has COP = Tc ÷ (Th − Tc).
Class 11 — The Otto cycle and other cycles
Concept: for any cycle the work done is the area enclosed on the PV diagram, and the efficiency is W ÷ Qin, where Qin is all the heat absorbed.
- Choose the Otto cycle (petrol engine). The simulation switches to a diatomic gas (air, γ = 1.4). With compression ratio r = 8, intake 300 K and a peak of 1800 K:
T2 = 300 × 80.4 = 689 K, Qin = nCv(1800 − 689) = 23 088 J, T4 = 783 K and Qout = 10 049 J.
- The efficiency is 1 − r1−γ = 1 − 8−0.4 = 56.5%. Raise r to 10: 60.2%. A higher compression ratio always helps.
- Choose the rectangle cycle with a monatomic gas, lowest temperature 300 K, pressure ratio 2.5 and volume ratio 2. The work is the area of the rectangle, (2.5 − 1) × 105 × 24.94 L = 3741 J.
- Heat is absorbed twice: nCv(TA − TD) = 5612 J by the isochoric heating and nCp(TB − TA) = 15 589 J by the isobaric expansion, so Qin = 21 201 J and η = 17.6%,
far below the Carnot value 1 − 300 ÷ 1500 = 80% for the same extreme temperatures.
Takeaway: W = enclosed area; η = W ÷ (all heat absorbed). Otto: η = 1 − r1−γ.
Class 11 — Specific heat, latent heat and the heating curve
Concept: heat that changes the temperature obeys Q = mcΔT; heat that changes the state (melting, boiling) at constant temperature obeys Q = mL.
- Open Heating curve & latent heat: 0.2 kg of ice at −20 °C with a 1000 W heater. Warming the ice to 0 °C takes 0.2 × 2100 × 20 = 8.4 kJ.
- Melting takes mLf = 0.2 × 334 000 = 66.8 kJ with the temperature stuck at 0 °C. Warming the water from 0 to 100 °C needs 0.2 × 4186 × 100 = 83.7 kJ.
- Boiling takes mLv = 0.2 × 2 260 000 = 452 kJ, more than five times the heat needed to heat the water to 100 °C (5.4 times), and another 20.1 kJ heats the steam to 150 °C.
The total is 631 kJ, which takes 631 s at 1000 W.
- Double the mass: every stage doubles. Double the power: the times halve but the energies do not change.
- This is why steam at 100 °C burns far worse than water at 100 °C: condensing releases the latent heat as well.
Takeaway: Q = mcΔT for a temperature change and Q = mL for a change of state, with no temperature change.
Class 11 — The principle of calorimetry (mixtures)
Concept: in an insulated container, heat lost by the hot body = heat gained by the cold body.
- Open Mixing (calorimetry) with 0.5 kg of water at 80 °C and 0.5 kg of copper at 20 °C. Tf = (m1c1T1 + m2c2T2) ÷ (m1c1 + m2c2) = (167 440 + 3850) ÷ (2093 + 192.5) = 74.95 °C.
- The result is close to the water's temperature because the water's heat capacity (2093 J/K) is about ten times the copper's (192.5 J/K). The heat that moves is 0.5 × 385 × 54.95 = 10.58 kJ.
- On the left graph the final temperature is where the two lines cross.
- Now take body 1 as ice: 0.2 kg at −10 °C, and body 2 as 0.5 kg of water at 40 °C. Heat available from the water = 83.7 kJ; warming the ice needs 4.2 kJ and melting it 66.8 kJ, so all of it melts and
Tf = (83.7 − 4.2 − 66.8) kJ ÷ (0.7 × 4186) = 4.34 °C.
- Try 0.5 kg of ice with the same water: there is not enough heat to melt all the ice and the mixture stays at 0 °C.
Takeaway: heat lost = heat gained; with ice, check first whether there is enough heat to warm and melt it.
Class 11 — Conduction and thermal resistance
Concept: the heat current through a rod is H = kAΔT ÷ L. Thermal resistances add in series, and heat currents add in parallel.
- Open Conduction, One rod of copper (k = 385), A = 10 cm2, L = 20 cm, from 100 °C to 0 °C: H = 385 × 0.001 × 100 ÷ 0.20 = 192.5 W, and R = L ÷ kA = 0.519 K/W.
- Double the length: the current halves. Double the area: it doubles.
- Choose two rods end to end: steel (k = 50) then aluminium (k = 205), both 20 cm and 10 cm2. R1 = 4.00 K/W, R2 = 0.976 K/W, so H = 100 ÷ 4.976 = 20.1 W.
The interface temperature is 100 − 20.1 × 4.00 = 19.6 °C: nearly all the drop is across the steel, the poorer conductor.
- Choose side by side with copper and wood: the currents add (192.5 W + 0.06 W), so the copper does almost all the work.
- Try air (k = 0.026) or polystyrene foam as a material: they are excellent insulators.
Takeaway: H = kAΔT ÷ L; series: R = R1 + R2; parallel: H = H1 + H2.
Class 11 — Radiation: Stefan's and Wien's laws
Concept: every body radiates a power P = εσAT4, and the wavelength at which it radiates most is λmax = b ÷ T.
- Open Radiation with T = 3000 K (a lamp filament), A = 1 m2, ε = 1: P = 5.67 × 10−8 × 30004 = 4.59 MW and λmax = 2.898 × 10−3 ÷ 3000 = 966 nm, in the infrared. Most of a filament's output is not light.
- Change T to 6000 K: the power is 16 times larger (73.5 MW) and the peak moves to 483 nm, in the visible range. The Sun (5778 K) peaks at 502 nm.
- Set T = 300 K (a person or a room): P = 459 W per square metre and the peak is at 9.66 µm, in the infrared; the net power is zero when the surroundings are also at 300 K.
- Lower ε to 0.5: the power halves. A shiny surface radiates (and absorbs) less than a dull black one.
Takeaway: P = εσAT4; λmaxT = 2.898 × 10−3 m·K; net power = εσA(T4 − Ts4).
Class 11 — Newton's law of cooling
Concept: the rate of cooling is proportional to the excess temperature: dT/dt = −k(T − Ts), so the excess dies away exponentially.
- Open Newton's law of cooling with a cup at 90 °C in a 20 °C room, k = 0.05 per minute. T(t) = 20 + 70e−0.05t.
- After 10 minutes the cup is at 62.5 °C, after 20 minutes at 45.8 °C: it cools 27.5 °C in the first 10 minutes but only 16.7 °C in the next 10.
- It reaches 40 °C after t = (1 ÷ 0.05) ln(70 ÷ 20) = 25.1 min. The excess temperature halves every ln 2 ÷ k = 13.9 minutes.
- The average-temperature method: (90 − 62.5) ÷ 10 = k × 56.2, giving k ≈ 0.049, close to the exact 0.05.
- Double k (a wider cup): everything happens twice as fast. The graph of ln(T − Ts) against time is a straight line with slope −k.
Takeaway: T = Ts + (T0 − Ts)e−kt; the room temperature is the limit the object approaches but never passes.
Class 11 — Thermal expansion and thermal stress
Concept: a rod lengthens by ΔL = αL0ΔT. If it cannot expand, it presses on its supports with a stress YαΔT.
- Open Thermal expansion, a free steel rod, L0 = 1.00 m, ΔT = 60 K: ΔL = 12 × 10−6 × 1 × 60 = 0.72 mm. An area grows by 2αΔT = 0.144% and a volume by 3αΔT = 0.216%.
- Compare materials: aluminium (23 × 10−6) expands almost twice as much as steel; invar (1.2 × 10−6) hardly at all.
- Choose a rod held between rigid walls, A = 4 cm2: the stress is YαΔT = 200 × 109 × 12 × 10−6 × 60 = 144 MPa and the force 57.6 kN, whatever the length.
- Choose the bimetallic strip with steel (α = 12) and brass (α = 19), thickness 2 mm, length 10 cm, ΔT = 60 K: R = 2h ÷ 3(α2 − α1)ΔT = 3.17 m, and the end moves L2 ÷ 2R = 1.57 mm. The strip bends towards the steel, which expands less.
Takeaway: ΔL = αL0ΔT, β = 2α, γ = 3α; thermal stress = YαΔT.
Class 11 — Molecular speeds and degrees of freedom
Concept: the molecules of a gas have a spread of speeds (the Maxwell distribution). Their rms speed is √(3RT ÷ M), and the heat capacity depends on the degrees of freedom f.
- Open Molecular speeds with nitrogen at 300 K: vp = 422 m/s, v̄ = 476 m/s, vrms = 517 m/s, in the ratio 1 : 1.128 : 1.225.
- Choose helium: vrms = 1367 m/s. Hydrogen: 1927 m/s. Oxygen: 484 m/s. Hydrogen is √(28.01 ÷ 2.016) = 3.73 times as fast as nitrogen, though all have the same average kinetic energy (3/2)kT.
- Raise T from 300 K to 1200 K for nitrogen: the speeds double (to 1034 m/s), not quadruple.
- Shaded area: the fraction of nitrogen molecules between 300 and 600 m/s is 54.2%. Move the sliders and read the fraction.
- Degrees of freedom: helium f = 3 (Cv = 12.47, γ = 1.667), nitrogen f = 5 (Cv = 20.79, γ = 1.40), water vapour f = 6 (γ = 1.33).
Takeaway: vp : v̄ : vrms = √2 : √(8/π) : √3; Cv = (f/2)R, γ = 1 + 2/f.