Adiabatic Process Calculator
Calculate the final pressure, volume and temperature of an ideal gas after a reversible adiabatic (isentropic) compression or expansion. Enter the initial state, one final quantity and the heat-capacity ratio γ.
Find the Final State of an Ideal Gas After Adiabatic Compression or Expansion
Adiabatic Process
Where:
- \(P_1, V_1, T_1\) — initial pressure, volume and absolute temperature of the gas
- \(P_2, V_2, T_2\) — final pressure, volume and absolute temperature
- \(\gamma = c_p / c_v\) — heat-capacity ratio of the gas (1.67 for monatomic, 1.40 for diatomic gases such as air)
- \(W\) — work done by the gas during the process (J)
These relations describe a reversible adiabatic (isentropic) process of an ideal gas: because no heat is exchanged with the surroundings, pressure, volume and temperature are linked by the constant \(P V^{\gamma}\), and the work term follows from the first law of thermodynamics with \(Q = 0\), so \(\Delta U = -W\).
Adiabatic Process – Calculation Example
Example: 1 m³ of air at 100 kPa and 300 K is compressed adiabatically to half its original volume. For air, γ = 1.4.
\[ P_2 = P_1 \left( \frac{V_1}{V_2} \right)^{\gamma} = 100\,\text{kPa} \times 2^{1.4} \approx 263.9\,\text{kPa} \]
\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1} = 300\,\text{K} \times 2^{0.4} \approx 395.85\,\text{K} \; (\approx 122.7\,\text{°C}) \]
\[ W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1} = \frac{100\,000 \times 1 - 263\,902 \times 0.5}{0.4} \approx -79\,877\,\text{J} \]
Result: the final pressure is ≈ 263.9 kPa (≈ 2.60 atm), the final temperature is ≈ 395.85 K (≈ 122.7 °C), and about 79.9 kJ of work is done on the gas, which is why it warms up.
How to use the calculator
- In the “I know the final” dropdown, choose whether you are given the final volume (and want the final pressure) or the final pressure (and want the final volume).
- Enter the Initial pressure (P₁), Initial volume (V₁) and Initial temperature (T₁), selecting the unit for each field (kPa, bar, atm, psi, mmHg; m³, L, cm³, ft³, in³; K, °C, °F). All inputs are converted to SI units automatically.
- Enter the known final quantity: Final volume (V₂) or Final pressure (P₂), depending on your choice in step 1.
- Enter the Heat-capacity ratio (γ) of the gas — 1.4 for air and other diatomic gases, 1.67 for monatomic gases such as helium or argon, about 1.29 for CO₂.
- Read the results: the final pressure in several units, the final volume in several units, the final temperature in K, °C and °F, and the energy balance (work done, change in internal energy, and the constant \(P V^{\gamma}\) in both states).
What is an adiabatic process?
An adiabatic process is a thermodynamic process in which a gas exchanges no heat with its surroundings (\(Q = 0\)). Energy can still enter or leave the gas — but only in the form of mechanical work. When the gas is compressed, the work done on it increases its internal energy and the temperature rises; when it expands, the gas spends its own internal energy on pushing the surroundings away and cools down. This is why a bicycle pump gets warm when you pump quickly, and why compressed air from an aerosol can feels cold as it expands.
This calculator models the reversible (quasi-static, isentropic) adiabatic process of an ideal gas. In that case the state variables obey \(P V^{\gamma} = \text{constant}\) and \(T V^{\gamma-1} = \text{constant}\). Compared with an isothermal process, the pressure changes more steeply with volume, because temperature changes at the same time: doubling the density of air (\(\gamma = 1.4\)) raises the pressure by a factor \(2^{1.4} \approx 2.64\), not by a factor of 2.
The sign of the work tells you the direction of the process: a negative \(W\) means work was done on the gas (compression, heating), a positive \(W\) means the gas did work on its surroundings (expansion, cooling). Because \(Q = 0\), the first law of thermodynamics reduces to \(\Delta U = -W\): the temperature change you see in the results is a direct consequence of the work exchanged.
Assumptions, limitations and practical tips
- Ideal gas: the relations hold for pressures where the ideal-gas approximation is valid. At very high pressures or near condensation, real-gas effects become important and the results deviate.
- Reversible and truly adiabatic: the calculation assumes the process is slow enough to be quasi-static and fast enough (or well enough insulated) that no heat leaks in or out. Rapid, turbulent compression or expansion generates irreversibilities and gives different final temperatures.
- Constant γ: the heat-capacity ratio is treated as independent of temperature. Over ranges of a few hundred kelvin this is a good approximation for air; for very large temperature swings, tabulated \(\gamma(T)\) values should be used.
- Choose γ carefully: using 1.4 for a monatomic gas (or vice versa) can shift the final temperature by tens of kelvin. Check the gas you are modelling before trusting the result.
- Mass conservation: the calculator assumes a fixed amount of gas. Use the Density calculator if you need to convert between the mass and the amount of gas, and the Pressure calculator for unit checks on the pressures.
Adiabatic compression and expansion are central to engines, compressors, turbines and atmospheric physics — for example, the cooling of rising air parcels is approximated as adiabatic. To place these processes in the wider context of heat engines and their efficiency limits, see the Carnot Efficiency calculator.
Frequently Asked Questions
How do I calculate the final temperature if I only know the pressures?
Use the pressure–temperature form of the adiabatic relation: \( T_2 = T_1 (P_2/P_1)^{(\gamma-1)/\gamma} \). For example, air compressed from 100 kPa and 300 K to 500 kPa with \( \gamma = 1.4 \) reaches \( T_2 = 300 \times 5^{0.4/1.4} \approx 475 \) K. Select “Final pressure (find final volume)” in the calculator to use this path — it also returns the volume for you.
Why does my answer come out wrong when I enter temperature in °C or °F?
It shouldn’t — the calculator converts to kelvin internally — but hand calculations fail without it. The relations \( T V^{\gamma-1} = \text{const} \) only hold for absolute temperature, because ratios like \( T_2/T_1 \) are meaningless on the Celsius or Fahrenheit scale. If you compute manually, always convert first: \( T(\text{K}) = T(^\circ\text{C}) + 273.15 \), then convert back at the end.
What heat-capacity ratio (γ) should I use for my gas?
Use \( \gamma = 1.67 \) for monatomic gases (helium, argon, neon), \( 1.40 \) for diatomic gases near room temperature (air, N₂, O₂, H₂), and roughly 1.29–1.33 for polyatomic gases such as CO₂, methane or steam. The value falls slightly as temperature rises, so for processes spanning many hundreds of kelvin, look up \( \gamma(T) \) tables. Picking the wrong γ can shift the final temperature by tens of kelvin.
What is the difference between adiabatic and isothermal compression?
In an isothermal process the temperature is held constant by heat exchange, so \( PV = \text{const} \). In an adiabatic process no heat is exchanged, the temperature changes, and \( PV^{\gamma} = \text{const} \). Since \( \gamma > 1 \), pressure rises faster during adiabatic compression: halving the volume of air doubles the pressure isothermally but multiplies it by \( 2^{1.4} \approx 2.64 \) adiabatically.
Why is my result different from what happens in a real pump or engine?
Real processes are rarely perfectly reversible and adiabatic. Friction and turbulence make fast compression or expansion irreversible, so the actual final temperature is higher than the isentropic value. Heat also leaks through the walls during slow processes, making them partly isothermal. The calculator gives the ideal isentropic limit; real devices fall between the adiabatic and isothermal cases, and efficiency factors account for the gap.
What does a negative work value mean, and how do I get work from the mass of gas?
The calculator reports work done by the gas, so compression gives negative W — the surroundings did work on the gas. If you prefer the internal-energy route, use \( \Delta U = n\, c_v\, \Delta T \) with \( c_v = R/(\gamma - 1) \) per mole; since \( Q = 0 \), \( W = -\Delta U \). Both methods give identical results for an ideal gas.