Formula used
For a reversible isothermal process of an ideal gas, the work done by the gas is:
- n is the amount of gas in moles.
- T is absolute temperature in kelvin.
- R is the molar gas constant.
- The logarithm is natural: ln(x).
Note: This calculator assumes ideal-gas behavior and reversibility. For irreversible paths, the work depends on the external pressure history.
How to use this calculator
- Select an input mode: volume ratio (V1, V2) or pressure ratio (P1, P2).
- Enter the gas amount n and the temperature T with units.
- Fill the matching pair (volumes or pressures) and choose units.
- Pick a sign convention and an output unit (J or kJ).
- Click Calculate Work to view results above the form.
- Use the download buttons to export CSV or PDF reports.
Example data table
Example for 1 mol ideal gas at 300 K expanding from 10 L to 20 L.
| n (mol) | T (K) | V1 (L) | V2 (L) | ln(V2/V1) | Wby (J) |
|---|---|---|---|---|---|
| 1.0 | 300 | 10 | 20 | 0.6931 | ≈ 1729 |
Your computed value may differ slightly due to constants, rounding, or chosen precision.
Isothermal work in practice
1) Why isothermal paths matter
In an isothermal process the temperature stays constant while the system exchanges heat with its surroundings. For an ideal gas, internal energy depends only on temperature, so ΔU is zero when the temperature does not change. That makes the work term the key energy transfer to quantify during slow expansions and compressions.
2) Reversible work gives the benchmark
This calculator uses the reversible expression, which represents the maximum work obtainable from an expansion and the minimum work required for a compression at the same endpoints. Reversibility implies quasi‑static behavior with negligible dissipative losses and a well-defined pressure along the path.
3) Core equation and typical magnitudes
The work by the gas is Wby = nRT ln(V2/V1). The logarithm grows slowly: doubling volume gives ln(2)=0.693. For n=1 mol and T=300 K, doubling volume produces about 1.73×103 J. Raising temperature to 600 K doubles the work for the same ratio.
4) Using pressure mode correctly
Because PV is constant for an ideal gas at fixed temperature, V2/V1 equals P1/P2. Pressure inputs are convenient when you measure initial and final pressures directly, such as in piston-cylinder experiments or controlled gas handling lines.
5) Sign conventions and interpretation
Thermodynamics texts differ on the sign of work. This tool lets you choose either convention. With “work by the gas is positive,” an expansion typically yields positive work. With “work on the gas is positive,” the same expansion becomes negative. Always report the chosen convention with results.
6) Unit handling and reporting
Inputs can be entered in common pressure, volume, and temperature units, then converted internally to SI. Output can be displayed in joules or kilojoules. Significant figures only affect presentation; the underlying calculation uses full floating‑point arithmetic for consistency.
7) Assumptions, limits, and sources of error
Ideal-gas behavior is most accurate at low pressure and moderate temperature. Real gases deviate near phase changes or at high pressure, where compressibility factors differ from one. If the path is irreversible, the work depends on external pressure versus time, so endpoint ratios alone are insufficient.
8) Quick validation checks
For expansions, V2 > V1 should give ln(V2/V1) > 0 and positive Wby. For compressions, the log term is negative and Wby is negative. If pressure mode is used, verify that P2 < P1 corresponds to expansion.
FAQs
1) What kind of work does this compute?
It computes reversible boundary work for an ideal gas during an isothermal change between two states, using the logarithmic relation based on either volumes or pressures.
2) Why does the formula use a natural logarithm?
For a reversible isothermal ideal gas, pressure varies as P = nRT/V. Integrating ∫(nRT/V)dV produces ln(V), so the result depends on ln(V2/V1).
3) Can I use this for irreversible expansions?
Not directly. Irreversible work depends on the external pressure history. This tool gives the reversible benchmark, which is useful for limits, comparisons, and efficiency discussions.
4) What if I only know one pressure and one volume?
You need a ratio of volumes or a ratio of pressures for the isothermal work expression. If you have mixed data, use the ideal-gas relation PV = nRT at the same temperature to infer the missing endpoint.
5) How do I interpret a negative result?
Under the “work by the gas is positive” convention, negative work typically indicates compression. Under the “work on the gas is positive” convention, expansions often appear negative. Always note the chosen convention.
6) Does changing units change the physics?
No. Units only change numerical representation. The calculator converts inputs to SI internally, then reports the final work in your selected output unit.
7) What conditions make ideal-gas results less accurate?
High pressures, very low temperatures, and regions near condensation increase non‑ideal behavior. In those cases, real-gas models using compressibility factors or an equation of state can improve accuracy.