Thermodynamic Processes Calculator: Isothermal, Adiabatic, Isobaric and Isochoric

Work, heat and internal energy for the standard gas processes, with the pressure-volume graph.

Calculated from standard textbook formulas, with every step shown. The formulas are fixed, so the working does not need checking: what matters is the numbers you type in. Use these calculators only if you are competent to choose the inputs (loads, material strength, dimensions, supports) and check your units and assumptions. For learning and checking your own working: design and sign-off are for a qualified engineer. Drafting Central does not store the numbers you type.

How this calculator works

Choose the process and enter the starting pressure, volume and temperature, and the final volume (or the final pressure for a constant-volume process). The calculator works out the mass of gas from the ideal gas law and then the final state, the work done by the gas, the heat added, and the changes in internal energy and enthalpy.

The work is the area under the process line on the pressure-volume graph, and the first law (heat added minus work done equals the change in internal energy) is checked at the end. For an isothermal process all the heat becomes work; for an adiabatic one there is no heat and the gas cools as it does work; for an isobaric one the heat goes into both; for an isochoric one there is no work.

The cycle option takes a rectangle of two constant-pressure and two constant-volume steps, and finds the net work (the enclosed area), the heat in and out, the efficiency and the Carnot limit. Use it for compressed-air receivers, cylinders, simple engine cycles and coursework.

Formulas used

  • W = P_1 V_1 \ln\frac{V_2}{V_1}\ (\text{isothermal})
  • P V^{\gamma} = \text{constant},\ W = \frac{P_1V_1 - P_2V_2}{\gamma - 1}
  • W = P\Delta V\ (\text{isobaric})
  • Q = \Delta U + W

Worked example

With the default numbers in the calculator above:

  1. Mass of gas (from the ideal gas law): 0.034826 kg
  2. Specific heats: cv = 717.50, cp = 1004.50
  3. The process: process equation
  4. Final state: 150.00 kPa, 20.00 L, 27.00 °C
  5. Work done by the gas: 2079.44 J (2.0794 kJ)
  6. Heat added to the gas: 2079.44 J (2.0794 kJ)
  7. Change in internal energy: 0.00 J
  8. Change in enthalpy: 0.00 J

Change the numbers in the calculator to see the full working, every step, for your own case.

A second worked example

A different case, with these numbers: Which process? Adiabatic (no heat): given the final volume.

  1. Mass of gas (from the ideal gas law): 0.034826 kg
  2. Specific heats: cv = 717.50, cp = 1004.50
  3. The process: process equation
  4. Final state: 113.68 kPa, 20.00 L, -45.68 °C
  5. Work done by the gas: 1816.06 J (1.8161 kJ)
  6. Heat added to the gas: 0.00 J (0 kJ)
  7. Change in internal energy: -1816.06 J
  8. Change in enthalpy: -2542.49 J

Common questions

What is the difference between isothermal and adiabatic?

In an isothermal process the temperature is held constant, so heat flows in or out. In an adiabatic process no heat flows, so the temperature changes. Fast compression, with no time for heat to escape, is close to adiabatic.

Why is the work in the cycle the area?

On a pressure-volume graph the work is the area under each line. Going round a loop, the work out on the top lines minus the work in on the bottom lines is the area inside.

Does it work for steam or refrigerants?

No. It is for an ideal gas with constant specific heats. Use the property tables for steam and refrigerants.

Related calculators

Got a question about this? Ask an engineering question and get a plain-language answer with references.

Further reading: Cengel and Boles, Thermodynamics: An Engineering Approach (ideal gas processes); Moran and Shapiro, Fundamentals of Engineering Thermodynamics.

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