Thermodynamics
Isochoric heating
Heat at constant V: all heat raises the internal energy.
Isochoric heating — interactive Thermodynamics simulation. Heat at constant V: all heat raises the internal energy. Free browser-based virtual physics lab with live SI measurements and a guided tutorial.
Isochoric heating
Heat at constant V: all heat raises the internal energy.
With rigid walls no work can be done (W = 0), so the first law reduces to Q = ΔU. Every joule of heat increases molecular kinetic energy directly, producing the steepest temperature rise per unit heat among the basic processes.
- W = 0
- Q = ΔU
Investigation brief
Plan the question before you open the lab
The brief mirrors the prerendered page: driving question, competing predictions, variable roles, governing laws, setup, analysis and extension prompts remain visible and in this order.
Driving question
A gas is heated in a rigid, sealed container, so its volume can't change (isochoric). Since the gas can't expand, where does all the added heat go?
Predictions to weigh
- All the heat goes into internal energy, since no work is done at constant volume.
- Most of the heat becomes work even though the container is rigid.
- The heat splits evenly between work and internal energy, just like other processes.
Variable roles
What you set:
- Final temperature T₂ (K)
What you measure:
- Final pressure P₂ (kPa)
- Heat added Q (J)
How the investigation runs
- Open the isochoric-heating preset and press Reset. 1 mol of gas is sealed at 0.0249 m³, starting at 300 K.
- Enable the pressure and heat-added readouts.
- Set the target (final) temperature for each trial, and record the final pressure and the heat added.
Governing equation
Isochoric Heating and Internal Energy — U = f/2·n·R·T
At constant volume, a gas does no work (W_gas = 0), so every joule of added heat goes directly into internal energy: Q = ΔU = (f/2)nRΔT.
What the printable worksheet asks students to work out
- For one trial, compute P₂ = nRT₂/V using n = 1 mol, V = 0.0249 m³, then Q = (f/2)nR(T₂ − T₁) using f = 3, T₁ = 300 K. Compare both to the table.
- Since the gas can't expand (W_gas = 0), the first law gives Q = ΔU exactly. Explain why every joule of added heat shows up entirely as internal energy here, unlike the isobaric-heating experiment.
Where this shows up beyond the lab
- Your P₂-vs-T₂ relationship is Gay-Lussac's Law: at constant volume, pressure is directly proportional to absolute temperature. Explain why this follows from PV = nRT when V is fixed.
- A sealed pressure cooker heats food faster partly because trapped steam can't expand, so pressure (and temperature) rise well above 100°C. Use this experiment's relationship to explain why a sealed rigid container lets temperature climb higher for the same heat input than an open one would.
- AP Physics 2 — Unit 9: Thermodynamics
- IB Physics — B.4 Thermodynamics
- General High School Physics — Heat, temperature & gas laws
- NGSS High School Physics — Thermal energy transfer
- Welcome to Isochoric Heating
- Select the chamber
- Press Play
- Heating at constant V
- Open the Properties panel
- You did it!
Open the interactive simulation to build the scene, press Play, and explore with live measurements and a guided tutorial.