What we will do: We will supply heat to a gas in a cylinder through different processes and check how it is split between a rise in temperature and expansion work.
What you see on screen: A cylinder with a movable piston and a heater, a process switch (isochoric, isobaric, isothermal, adiabatic) and a heat slider. Three bars — heat, work and the change in internal energy — are always tied together by the balance ΔU = Q + W. Readouts: process, Q, W, ΔU, temperature and volume.

I will give you a target change in internal energy — pick a process and the energy (slider) to reach it. ΔU = Q + W.
What just happened?
With the piston locked (constant volume) all the supplied heat went into internal energy — the temperature rose, but the gas did no work.
The first law of thermodynamics:
is the energy balance for a gas. The change in internal energy equals the sum of the heat supplied and the work done on the gas:
Energy is not lost — it can enter as heat (Q) or as work (W) and show up as a rise in internal energy (and therefore temperature).
Four processes:
isochoric (constant volume): W = 0, all the heat → ΔU (temperature rises);
isobaric (constant pressure): the gas expands and does work, part of the heat goes into work;
isothermal (constant temperature): ΔU = 0, the supplied heat is converted entirely into work by the gas;
adiabatic (no heat exchange, Q = 0): ΔU = W — compression warms it, expansion cools it.
Why this matters for the exam:
the balance ΔU = Q + W and recognising the processes are the core of thermodynamics. The trick: find what is zero (W for isochoric, ΔU for isothermal, Q for adiabatic) — the rest follows from the balance.
Remember: the internal energy of a gas changes through heat and work: ΔU = Q + W. It is the conservation of energy in its thermal form.
The change in internal energy of a gas equals the sum of the heat supplied and the work done on the gas:
For an ideal gas depends only on temperature — a rise in means a rise in .
Symbols: — change in internal energy, — heat supplied to the gas, — work done on the gas, — pressure, — change in volume, — temperature.
The change in internal energy of a gas equals the sum of the heat supplied and the work done on the gas: ΔU = Q + W.
The gas does no work (W = 0), so all the heat increases the internal energy and the temperature.
A process with no heat exchange (Q = 0) — then ΔU = W: compression heats the gas, while expansion cools it.
Teaching? Embed this experiment on your site.
Free widget — paste it on your school site, blog or notes. It links to the full version.