Library / Thermodynamics / THE-11
Thermodynamics
Isothermal expansion work
Work done by a gas expanding slowly at constant temperature.
- Ref
- THE-11
- Rev
- v1
- Field
- Thermodynamics
- Result
- W [J]
n is the amount of gas in moles, T the constant temperature in kelvin, and V₁ and V₂ the starting and finishing volumes in the same unit. W is positive for expansion (V₂ > V₁), when the gas does work, and negative for compression. The gas must be heated or cooled along the way to hold T constant.
Work done by a gas expanding slowly at constant temperature.
Calculator
- Heat absorbed from the surroundings (ΔU = 0 for an ideal gas)1,729 J
- Final pressure nRT/V₂1,247 Pa
What drives the result
Each variable is moved across its typical range while the others stay at the values above. The longest bar is the key driver.
- Doubling the volume of 1 mol at 300 K does 0.693 × nRT, about 1.73 kJ, and the gas must take in the same amount of heat to keep its temperature from falling.
- Each doubling of the volume does the same work again (nRT ln 2), so expanding from 1 to 4 m³ does twice as much as 1 to 2 m³.
Chart data as tables
Ranked variables
| Variable | Typical range | Lowest result | Highest result | Swing (J) | Local step | Change in result |
|---|---|---|---|---|---|---|
| n Amount of gas | 0.001 to 1,000 mol | 1.729 | 1,729,000 | 1,729,000 | +10% | +10 % |
| V2 Final volume | 0.001 to 100 m³ | −17,230 | 11,490 | 28,720 | +10% | +13.8 % |
| V1 Initial volume | 0.001 to 100 m³ | −9,758 | 18,960 | 28,720 | +10% | −13.8 % |
| T Temperature | 100 to 3,000 K | 576.3 | 17,290 | 16,710 | +10% | +10 % |
Sweep of Final volume
| Final volume (m³) | Work done (J) |
|---|---|
| 0.001 | −17,230 |
| 0.004217 | −13,640 |
| 0.01778 | −10,050 |
| 0.07499 | −6,461 |
| 0.3162 | −2,872 |
| 1.334 | 717.9 |
| 5.623 | 4,308 |
| 23.71 | 7,897 |
| 100 | 11,490 |
Sweep one variable
Variables and how each one contributes
Work done
What the equation gives you. Everything else feeds it.
Amount of gas
The amount of gas, in moles. Directly proportional: twice the gas does twice the work for the same volume ratio.
Temperature
The constant temperature of the process, in kelvin. Directly proportional: a hotter gas pushes harder at each volume, so it does more work.
Initial volume
The volume at the start. Raising V₁ towards V₂ shrinks the expansion ratio and lowers the work. A 10 % rise in V₁ lowers W by about 14 %.
Final volume
The volume at the end. Acts through the logarithm, so it has diminishing returns: a 10 % rise in V₂ raises W by about 14 %. A further doubling adds the same fixed amount again.
Worked example with the values above
- Volume ratio: V₂/V₁ = 2 / 1 = 2; ln = 0.69315.
- nRT = 1 × 8.314 × 300 = 2,494 J.
- W = 2,494 × 0.69315 = 1,729 J.
Assumptions and limits
- An ideal gas, so that pV = nRT.
- A slow, reversible process at constant temperature, which can be approached with good heat exchange.
Common mistakes
- Using log₁₀ instead of the natural logarithm.
- Entering T in °C.
Statistics and uncertainty
W ∝ ln(V₂/V₁), so a 1 % error in the volume ratio gives about a 1.4 % error in the work at a ratio of 2, and the error grows as the ratio approaches 1, where W is small. Measure T in kelvin: a 3 °C error on 27 °C is only 1 %.
Where it is used
- Maximum work from a slow gas expansion, as in a piston or bellows.
- Work needed to compress gas slowly with cooling, as a bound on compressor power.
Maths you need
- Natural logarithm: ln of the volume ratio
- Ideal gas law: pV = nRT