Isochoric Heating: Pressure Rise at Constant Volume
1 · Predict
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?
- 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.
2 · Set Up
- 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.
3 · Collect Data
| Final temperature T₂ (K) | Final pressure P₂ (kPa) | Heat added Q (J) |
|---|---|---|
| 350 | ||
| 450 | ||
| 550 |
Plot heat added Q (y-axis) against final temperature T₂ (x-axis) for your three trials.
4 · Analyze
- 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.
5 · Extend
- 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.
The Physics Behind This Experiment
Isochoric Heating and Internal Energy
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.