Isothermal Compression: Work Done at Constant Temperature
1 · Predict
A gas is compressed to a smaller volume while its temperature is held constant (isothermal). Is more work required to compress it further?
- Compressing to a smaller final volume requires more work (a more negative Wgas).
- Compressing to a smaller final volume requires less work.
- The work needed is the same no matter the final volume.
2 · Set Up
- Open the isothermal-compression preset and press Reset. 1 mol of gas starts at 0.0499 m³, 300 K.
- Enable the work-done-on-gas readout.
- Set the target (final) volume for each trial and record the work done by the gas.
3 · Collect Data
| Final volume V₂ (m³) | Work done by gas W_gas (J) |
|---|---|
| 0.0249426 | |
| 0.016628399999999998 | |
| 0.0124713 |
Plot W_gas (y-axis) against ln(V₂/V₁) (x-axis) for your three trials, using V₁ = 0.0499 m³.
4 · Analyze
- For one trial, compute W_gas = nRT·ln(V₂/V₁) using n = 1 mol, R = 8.314 J/(mol·K), T = 300 K. Compare to the table.
- W_gas is negative in every trial. Explain what a negative value means physically for a gas being compressed, and why compressing to a smaller V₂ makes W_gas more negative.
5 · Extend
- Since ΔU = 0 for any isothermal process (temperature doesn't change), the first law says Q = W_gas here too. Is heat flowing into or out of the gas during this compression? Explain using your sign convention.
- To keep the gas at constant temperature while compressing it, real experiments need a heat reservoir in contact with the cylinder. Explain why compression would heat the gas up if that reservoir weren't there.
The Physics Behind This Experiment
First Law of Thermodynamics
Internal energy change equals heat added minus work done by the gas: ΔU = Q − W. For an isothermal ideal-gas process, ΔU = 0, so all the work done on the gas leaves as heat.