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Entropy Change of an Ideal Gas Calculator

Enter the moles, heat capacity, and the two temperatures and volumes to find the entropy change.

How it is calculated

For an ideal gas with constant Cv the tool adds two terms: ΔS = n·Cv·ln(T2 ÷ T1) + n·R·ln(V2 ÷ V1), with R = 8.314 J/(mol·K). Here n is moles, Cv the molar heat capacity at constant volume in J/(mol·K), T1 and T2 the initial and final temperatures in kelvin, and V1 and V2 the initial and final volumes. The answer is in J/K, and the temperature and volume contributions are shown separately. All temperatures and volumes must be greater than zero. Only the ratios matter, so volumes can be in any unit if both use the same one. Temperatures must be absolute: using °C breaks the ratio. Cv is assumed constant over the temperature range, and a negative result simply means the gas lost entropy through cooling or compression.

Worked example

Take 2 mol of a diatomic gas with Cv = 20.785 J/(mol·K), heated from 300 K to 450 K while expanding from 10 m³ to 25 m³. Temperature term: 2 × 20.785 × ln(450 ÷ 300) = 41.57 × 0.4055 = 16.8552 J/K. Volume term: 2 × 8.314 × ln(25 ÷ 10) = 16.628 × 0.9163 = 15.2361 J/K. Total: 32.0913 J/K, so the gas gains entropy, because both heating and expansion raise it.

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Common mistakes

  • Entering temperatures in °C or °F. The ratio T2/T1 needs kelvin: 27 °C is 300 K.
  • Using Cp instead of Cv. The formula needs the heat capacity at constant volume; for an ideal gas Cp = Cv + R.
  • Entering the temperature and volume values for the same state in swapped fields, which flips the sign of the result.
  • Mixing volume units between V1 and V2, for example litres and cubic metres. Only a consistent unit gives the correct ratio.
  • Applying the formula to a real gas near condensation or to a large temperature range where Cv is not constant.

About this calculator

Entropy is the thermodynamic measure of a system's disorder, and for an ideal gas undergoing any process, the change in entropy can be split cleanly into a temperature-dependent term and a volume-dependent term: ΔS = nCv·ln(T2/T1) + nR·ln(V2/V1), where n is the moles of gas, Cv is the molar heat capacity at constant volume, and R is the universal gas constant. Heating a gas or letting it expand both increase its entropy, and this formula quantifies exactly how much.

This calculator takes the moles, heat capacity, and initial/final temperatures and volumes and computes the total entropy change. It's a standard calculation in physical chemistry and chemical engineering thermodynamics courses, used to analyze whether a process is reversible, to evaluate the efficiency of heat engines and refrigeration cycles, and to check compliance with the second law of thermodynamics, which requires total entropy to never decrease in an isolated system.

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Comments (1)

  • Editorial team

    Try the free expansion of one mole at constant temperature. Set T1 equal to T2 and double the volume: the temperature term reads zero and the volume term is R times ln 2, about 5.76 J/K, because entropy is a state function and depends only on the end states. Use Cv of 12.47 for a monatomic gas. Which process would you like a worked example for?

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