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Isothermal Work (Ideal Gas)

Calculates the work done by an ideal gas expanding or compressing at constant temperature between two volumes.

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Isothermal Work CalculatorW = n · R · T · ln(V₂ / V₁)

W = n · R · T · ln( V₂ / V₁ )
W = work (J)  ·  n = moles  ·  R = gas constant (J/mol·K)  ·  T = temperature (K)  ·  V₁, V₂ = initial & final volumes (m³)
⟹ SolveW, n, R, T, V₁, V₂
J
mol
J/mol·K
K
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W
W: n: R: T: V₁: V₂:
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W = n · R · T · ln(V₂ / V₁)  ·  Isothermal process (constant temperature) for an ideal gas

Variables

SymbolQuantityUnit
WWork done by the gasJ
nMoles of gasmol
RUniversal gas constant8.314 J/mol.K
TAbsolute temperature (constant)K
V1, V2Initial and final volumesm3

What it means

For an ideal gas undergoing a reversible isothermal process (constant temperature), the work done by the gas is given by W = nRT ln(V₂/V₁), where n is the number of moles, R is the universal gas constant, T is the absolute temperature, and V₁ and V₂ are the initial and final volumes. This expression is obtained by integrating PdV with P = nRT/V. Since the temperature is constant, the internal energy change is zero (ΔU = 0) for an ideal gas, so all heat added goes into work. This process is important in many engineering applications, such as compressors, turbines, and refrigeration cycles. The natural logarithm indicates that more work is required to compress a gas at high volume ratios. The isothermal work is the maximum work that can be extracted from a gas between two volumes at a given temperature. In practice, isothermal processes are approximated by slow processes with heat exchange to maintain constant temperature.

Worked example

Isothermal Work – Two Examples

Real‑World
Scenario 1 – Expansion: 1 mol of gas expands from 0.01 m³ to 0.02 m³ at 300 K. Find work done (isothermal).
ParameterValue
n1 mol
T300 K
V10.01 m³
V20.02 m³
1W = nRT ln(V2/V1) = 1×8.314×300×ln(2) = 2494×0.693 ≈ 1729 J
ResultW ≈ 1729 J✓ work output
Scenario 2 – Compression: 2 mol at 350 K compressed from 0.04 m³ to 0.02 m³. Find work (on gas).
ParameterValue
n2 mol
T350 K
V10.04 m³
V20.02 m³
1W = 2×8.314×350×ln(0.02/0.04) = 5819×ln(0.5) = 5819×(−0.693) ≈ −4034 J (work done on gas)
ResultW ≈ −4034 J✓ input
Key insight: For isothermal process, work depends on the logarithm of volume ratio.

Common mistakes

  • Temperature in Kelvin: T must be absolute (Kelvin).
  • Natural log: Use ln(V₂/V₁), not log₁₀.
  • Isothermal condition: This assumes constant temperature throughout the process – not always true.
  • Ideal gas: Only for ideal gases; real gases deviate.
  • Work sign: If expansion (V₂ > V₁), W is positive (work done by the gas).

Applications

Isothermal work for an ideal gas is the work done during a constant‑temperature expansion or compression, expressed as nRT ln(V₂/V₁). This process is encountered in many practical applications, such as in isothermal compressors, gas storage, and certain chemical reactions. In the petroleum industry, isothermal compression is used in natural gas pipelines to transport gas efficiently. In pneumatic systems, isothermal expansion can be approximated in slow processes. The formula also plays a role in the analysis of heat engines and refrigeration cycles that involve isothermal heat transfer. By understanding isothermal work, engineers can design processes that minimise energy losses and improve the efficiency of gas handling systems.

  • Design of isothermal compressors and expanders
  • Natural gas pipeline compression and storage
  • Thermodynamic cycle analysis (Carnot, Stirling)
  • Biological and chemical processes at constant temperature
  • Pneumatic system design and analysis