Audited 31 Jul 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

Isentropic Flow Calculator

Find how much an ideal gas cools as it expands, or heats as it is compressed, with no losses: T₂/T₁ from the absolute pressure ratio and heat-capacity ratio γ.

Isentropic Flow Calculator

Pa
Pa
Isentropic temperature ratio T2/T1
0.8203
The final absolute temperature divided by the initial one, T₂/T₁, for a loss-free adiabatic change; it has no units.
Model scope
Calorically perfect ideal-gas, reversible adiabatic ratio with constant γ; shocks, friction, heat transfer, real-gas effects, variable heat capacity, stagnation/static distinctions, and mass flow are outside it.

Background.

When a gas expands without exchanging heat and without friction, its temperature falls by an amount fixed entirely by how far the pressure drops. Halve the pressure of air this way and its absolute temperature falls by about 18%: air at 300 K comes out near 246 K, well below freezing. Run the process the other way, compressing instead of expanding, and the same law explains why a compressor outlet runs hot and why a bicycle pump warms in your hand.

Such a process is called isentropic, because the entropy of the gas stays constant. It is an idealisation, but a central one. Nozzles, diffusers, turbines and compressors are all designed against it, and the gap between a real machine and the isentropic version of the same pressure change is what engineers call isentropic efficiency. Friction, turbulence and shock waves all add entropy, so a real compression ends hotter than the ideal one and a real expansion ends less cold.

The relation used here, T₂/T₁ = (p₂/p₁)^((γ−1)/γ), holds for an ideal gas whose heat capacities do not change with temperature. γ is the ratio of the gas's heat capacity at constant pressure to that at constant volume, and it sets how strongly temperature follows pressure. For air near room temperature it is about 1.4, giving an exponent of 2/7. For a monatomic gas such as helium or argon it is 5/3, giving 2/5, so helium cools more for the same pressure drop.

The answer is a ratio, not a temperature. Multiply it by the starting temperature in kelvin, never in degrees Celsius. The pressures must be absolute, since a gauge reading leaves out the atmosphere's 101,325 Pa and changes the ratio. The same physics, applied to a rising parcel of dry air, gives the atmosphere's cooling of about 9.8 °C per kilometre of ascent.

What is isentropic flow calculator?

The isentropic temperature ratio is the factor by which the absolute temperature of an ideal gas changes when its pressure changes reversibly and adiabatically, meaning with no heat exchanged and no losses. It has no units. A value below 1 means the gas cools as it expands; a value above 1 means it heats as it is compressed. It depends only on the pressure ratio p₂/p₁ and on the heat-capacity ratio γ = c_p/c_v, not on the size of the equipment or the speed of the flow.

How to use this calculator.

  1. Enter the starting and final pressures as absolute pressures in pascals. Add 101,325 Pa to gauge readings, and multiply values in bar by 100,000.
  2. Enter γ for the gas: about 1.4 for air at ordinary temperatures, or 5/3 (about 1.667) for monatomic gases such as helium and argon. It must be greater than 1.
  3. Read the temperature ratio and multiply it by the starting temperature in kelvin to get the ideal final temperature. Subtract 273.15 to convert back to °C.
  4. For a real compressor, turbine or nozzle, treat the answer as the loss-free limit: the actual outlet will be warmer than this, whether the gas was compressed or expanded.

The formula.

T₂/T₁ = (p₂/p₁)^((γ-1)/γ)

T₂/T₁ = (p₂/p₁)^((γ − 1)/γ). Subscripts 1 and 2 mark the start and end of the process, p is absolute pressure and T is absolute temperature. The exponent comes from combining the ideal-gas law with the isentropic relation p·v^γ = constant, which follows from the first law of thermodynamics for a reversible process with no heat transfer and constant heat capacities. γ is the ratio of specific heats, c_p/c_v. Only the ratio of the two pressures matters, so any unit works as long as both pressures share it; the page asks for pascals. The exponent is small, 0.2857 for air, so temperature responds far less than proportionally: compressing air tenfold raises its absolute temperature by a factor of about 1.93, not 10. The page requires positive pressures and γ above 1, and it leaves out shocks, friction, heat transfer and real-gas behaviour.

A worked example.

Example

Air (γ = 1.4) expands isentropically from 200,000 Pa to 100,000 Pa absolute, roughly 2 bar down to 1 bar. The pressure ratio is 100,000 ÷ 200,000 = 0.5. The exponent is (1.4 − 1) ÷ 1.4 = 0.4 ÷ 1.4 = 0.285714. Raising 0.5 to that power gives a temperature ratio of 0.8203. If the air started at 300 K (about 27 °C), it ends at 300 × 0.820335 = 246.1 K, about −27.0 °C. Helium making the same expansion, with γ = 5/3 and an exponent of 0.4, reaches a ratio of 0.757858 and cools to 227.4 K. Compressing the air the other way, from 1 bar back to 2, gives the reciprocal ratio of 1.219.

final Pressure Pa100,000
initial Pressure Pa200,000
heat Capacity Ratio1.4

Frequently asked questions.

Why must I enter absolute pressure instead of gauge pressure?
The formula depends on the ratio of the pressures, and a gauge reading drops the atmosphere. Air going from 100 kPa gauge to 0 gauge is really 201,325 Pa to 101,325 Pa absolute, a ratio of 0.5033 and a temperature ratio of 0.8219. The gauge figures cannot even be entered, since 0 is not allowed.
Can I multiply the ratio by a temperature in degrees Celsius?
No. The ratio applies to absolute temperature. Air at 27 °C halving its pressure ends at about −27 °C, but 27 × 0.8203 would wrongly suggest 22 °C. Convert to kelvin by adding 273.15, apply the ratio, then convert back.
What value of γ should I use for my gas?
About 1.4 for air and other diatomic gases such as nitrogen and oxygen near room temperature, and 5/3 for monatomic gases like helium and argon. Gases with more complex molecules have lower values, and for air γ itself falls as the temperature climbs, which this constant-γ model does not capture.
Why does a real compressor run hotter than this result?
Friction and turbulence turn some of the work into extra heat. Engineers compare the ideal and actual temperature rises using isentropic efficiency. A tenfold compression of air from 300 K has an ideal rise of about 279 K; a machine that is 80% efficient by that measure would rise about 349 K, to roughly 649 K.
Does the formula work for steam or refrigerants?
Not reliably. It assumes an ideal gas with constant heat capacities, while steam near saturation and refrigerants depart from ideal-gas behaviour and may condense or boil during the process. For those, isentropic states are read from property tables or software using the entropy directly.
How much does the density change in the same expansion?
For an isentropic ideal gas, density follows ρ₂/ρ₁ = (p₂/p₁)^(1/γ). Halving the pressure of air gives 0.5^(1/1.4) = 0.6095, so the gas ends with about 61% of its original density. The page does not report this, but it follows from the same assumptions.

How this page was produced

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Quanta Calculator
Primary sources
3 cited below
Method
T₂/T₁ = (p₂/p₁)^((γ-1)/γ)
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