The Ideal Gas Law: Pressure, Volume, and Temperature
1 · Predict
A fixed amount of gas is sealed in a rigid container. If you heat it up, what happens to its pressure?
- Pressure increases proportionally with temperature.
- Pressure decreases as temperature rises.
- Pressure doesn't depend on temperature.
2 · Set Up
- Open the ideal-gas-sandbox preset and press Reset. 1 mol of argon fills a fixed volume of 0.0249 m³.
- Enable the pressure readout on the chamber.
- Set the gas temperature for each trial and record the pressure.
3 · Collect Data
| Temperature T (K) | Internal energy U (J) | Pressure P (kPa) |
|---|---|---|
| 300 | ||
| 450 | ||
| 600 |
Plot pressure P (y-axis) against temperature T (x-axis) for your three trials. Is the line straight through the origin?
4 · Analyze
- For one trial, compute P = nRT/V using n = 1 mol, R = 8.314 J/(mol·K), V = 0.0249 m³. Compare to the table.
- Explain why, at fixed volume and amount of gas, pressure and absolute temperature are directly proportional.
5 · Extend
- The ideal gas law requires temperature in kelvin, not Celsius. Explain why using Celsius (which allows negative and zero values) would break the direct proportionality you found.
- Your internal-energy column also grows with temperature, at fixed volume. Explain, using U = (f/2)nRT, why heating a fixed amount of gas at constant volume increases both its pressure and its internal energy together.
The Physics Behind This Experiment
Ideal Gas Law
For n moles of ideal gas, pressure, volume, and temperature are locked together by P·V = n·R·T. At fixed n and V, pressure rises in direct proportion to absolute temperature.
Internal Energy of an Ideal Gas
An ideal gas's internal energy depends only on its temperature (and degrees of freedom f), not on pressure or volume separately: U = (f/2)nRT.