Ideal Gases and Processes
The ideal gas law, specific heats, and the isothermal, isobaric, isochoric, adiabatic and polytropic processes with their work.
The ideal gas law
At moderate pressures and temperatures well above their condensation point, gases such as air, nitrogen, oxygen and hydrogen obey the ideal gas law very closely:
p V = m R T p v = R T R = Ru ÷ M
For 1 kg of air at 100 kPa and 300 K the volume is V = 0.287 × 300 ÷ 100 = 0.861 m³, a density of 1.161 kg/m³.
Specific heats
The specific heat at constant volume cv gives the change in internal energy, Δu = cv ΔT; at constant pressure cp gives the change in enthalpy, Δh = cp ΔT. They are related by cp − cv = R and their ratio is k = cp ÷ cv.
| Gas | R (kJ/kg·K) | cp (kJ/kg·K) | cv (kJ/kg·K) | k |
|---|---|---|---|---|
| Air | 0.287 | 1.005 | 0.718 | 1.40 |
| Nitrogen | 0.2968 | 1.039 | 0.743 | 1.40 |
| Oxygen | 0.2598 | 0.918 | 0.658 | 1.40 |
| Carbon dioxide | 0.1889 | 0.846 | 0.657 | 1.29 |
| Helium | 2.077 | 5.193 | 3.116 | 1.67 |
| Hydrogen | 4.124 | 14.31 | 10.18 | 1.41 |
| Methane | 0.5182 | 2.254 | 1.736 | 1.30 |
Values are near 300 K; cp rises with temperature (air is 1.005 at 300 K, 1.14 at 1000 K). Monatomic gases have k ≈ 1.67, diatomic k ≈ 1.4, polyatomic lower.
The five processes
| Process | Held constant | Relation | Work done by the gas |
|---|---|---|---|
| Isothermal | T | p V = constant | m R T ln(V₂ ÷ V₁) |
| Isobaric | p | V ÷ T = constant | p (V₂ − V₁) |
| Isochoric | V | p ÷ T = constant | 0 |
| Adiabatic (isentropic) | Q = 0 (and reversible) | p Vk = constant, T Vk−1 = constant | m R (T₁ − T₂) ÷ (k − 1) |
| Polytropic | general | p Vn = constant | m R (T₁ − T₂) ÷ (n − 1) |
The polytropic process covers the rest with an exponent n: n = 0 is isobaric, n = 1 isothermal, n = k adiabatic and n = ∞ isochoric. A real compressor lies between 1 and k, typically n = 1.2 to 1.35 for an air-cooled machine.
One kilogram compressed three ways
Worked example: compress 1 kg of air to half its volume
Air starts at 100 kPa and 300 K (0.861 m³) and is compressed to 0.431 m³. Work is entered here as work done on the gas, so the numbers are positive.
| Process | Final pressure | Final temperature | Work on the gas (kJ) |
|---|---|---|---|
| Isothermal (n = 1) | 200 kPa | 300 K | 59.7 |
| Polytropic (n = 1.25) | 238 kPa | 356.8 K | 65.2 |
| Adiabatic (n = 1.4) | 264 kPa | 395.8 K | 68.8 |
- The isothermal route needs 59.7 kJ: the least, because heat flows out as fast as the gas heats.
- The adiabatic route needs 68.8 kJ (15 percent more) and ends 96 K hotter.
- The heat removed in the isothermal case equals the work, 59.7 kJ, because the internal energy of an ideal gas depends on temperature alone.
Worked example: heating the same air at constant volume and constant pressure
- Constant volume, 300 K to 600 K: Q = m cv ΔT = 0.718 × 300 = 215.4 kJ; no work; the pressure doubles to 200 kPa.
- Constant pressure: Q = m cp ΔT = 1.005 × 300 = 301.5 kJ, of which 86.1 kJ (29 percent) leaves as expansion work, p ΔV = m R ΔT.
Staged compression and intercooling
Compression work grows with the final temperature, so a large pressure ratio is split into stages with cooling between them. For two equal stages of ratio √r the work is least; the optimum intermediate pressure is pi = √(p₁ p₂).
Worked example: 100 kPa to 900 kPa
- One stage (isentropic): T₂ = 562 K (289 °C), w = cp T₁ (90.2857 − 1) = 263.2 kJ/kg.
- Two stages with ideal intercooling back to 300 K, each of ratio 3: w = 2 cp T₁ (30.2857 − 1) = 222.2 kJ/kg, a saving of 15.6 percent and a far cooler discharge.
This is why industrial air compressors above about 7 bar use two or more stages, and why the discharge temperature on the datasheet matters. See Compressed Air and Gas Flow for the flow side and the Compressed Air Consumption Calculator for the demand.
When is a gas not ideal?
The ideal gas law fails near the saturation point, at high pressure and at low temperature. The compressibility factor Z = p v ÷ (R T) measures the departure: Z = 1 is ideal. Air at 10 MPa and 300 K has Z ≈ 0.99, but steam near saturation, refrigerant vapor and CO₂ near its critical point deviate strongly. For those, use property tables or software (see Steam, Water and Property Tables and Refrigeration and Heat Pumps). A good rule: treat a gas as ideal when it is above about twice its critical temperature or below a few bar.
Related
- Previous: Properties, State and the First Law. Next: The Second Law, Entropy and Carnot.
- Compressed Air Consumption Calculator.
- Pressure converter.
- Thermodynamics and Steam: the whole guide.
FAQ
Why must I use kelvin?
The gas law is proportional to absolute temperature. Using °C would give nonsense, for example a gas at 0 °C would have no volume.
How hot does air get when compressed?
For a single stage with ratio r, T₂ = T₁ r(k−1)/k. A ratio of 9 takes 300 K to 562 K, hot enough to burn skin and to ignite oil vapor. Real discharge is hotter again because of inefficiency.
What is the difference between isothermal and adiabatic?
Isothermal keeps the temperature constant by removing heat as fast as work is added (slow compression with cooling). Adiabatic allows no heat transfer (fast compression). Real machines are in between.
Can I use the ideal gas law for steam?
Only for low-pressure superheated steam, and then only roughly. Use the steam tables otherwise.
References
- Çengel, Y. A. and Boles, M. A., Thermodynamics: An Engineering Approach, McGraw-Hill (ideal gas, processes, compressors).
- Moran, M. J., et al., Fundamentals of Engineering Thermodynamics, Wiley.
- CAGI, Compressed Air and Gas Handbook, Compressed Air and Gas Institute.
Last reviewed: October 2026. Gas properties are typical room-temperature values.