Adiabatic process calculator

The four basic processes of an ideal gas

An ideal gas changes from one equilibrium state to another in a few standard ways. In an adiabatic process no heat flows in or out (Q = 0), so any work comes out of, or goes into, the internal energy and the temperature changes. In an isothermal process the temperature is held fixed by heat exchange with a reservoir. In an isobaric process the pressure is constant, and in an isochoric process the volume is constant, so the gas does no work. This calculator takes the starting state of the gas and one final value and returns the other two final properties, the work done by the gas, the change in internal energy, the heat and the entropy change.

The formulas

With n moles, the universal gas constant R = 8.314 J/(mol·K) and γ = Cp ÷ Cv, the ideal gas law is pV = nRT. Adiabatic: pV^γ and TV^(γ−1) stay constant, and the work done by the gas is W = (p1V1 − p2V2) ÷ (γ − 1) = n·Cv·(T1 − T2). Isothermal: pV is constant, W = nRT ln(V2 ÷ V1), ΔU = 0 and Q = W. Isobaric: W = p (V2 − V1) and Q = n·Cp·ΔT with Cp = Cv + R. Isochoric: W = 0 and Q = ΔU = n·Cv·ΔT. For every process ΔU = n·Cv·ΔT and Q = ΔU + W, which is the first law written with W as the work done by the gas (OpenStax University Physics, Volume 2, sections 3.2, 3.3, 3.5 and 3.6).

Worked example: squeezing a gasoline-air mixture

These are the numbers of OpenStax Example 3.7, which prints 1.23 × 10⁶ N/m², 600 K (328°C) and −63 J. The gas heats up to about 600 K without any flame, only because the piston did work on it and no heat could escape.

Common mistakes

Using gauge pressure is the most frequent one: a tire gauge reading of 30 psi is about 44.7 psi absolute, and the gas laws need the absolute value. The temperature must be absolute too, so degrees Celsius or Fahrenheit have to be turned into kelvins; the calculator does it when you pick the unit. The next trap is the sign of the work, which depends on whether your book counts the work done by the gas or on it (OpenStax counts the work done by the gas). Last, an isothermal formula is not the right choice for a fast compression: when there is no time for heat to escape, as in a bicycle pump or an engine stroke, the adiabatic formula is closer to the truth.

Limits of this method

The relations pV^γ = constant and the work formulas hold for quasi-static processes, that is, a gas that stays close to equilibrium all the time. A sudden expansion into a vacuum is not one: it has Q = 0 and W = 0, so the temperature of an ideal gas does not change (OpenStax section 3.6), and pV^γ is not conserved. The gas is assumed ideal with constant heat capacities. OpenStax Table 3.3 lists room-temperature values for dilute gases and notes that real gases have somewhat higher heat capacities, which would lower γ a little below the table values; near condensation an ideal gas is a poor model.

What is γ for air and for other gases?

γ is the ratio Cp ÷ Cv of the molar heat capacities at constant pressure and volume. From OpenStax Table 3.3 (dilute gases at room temperature) it is 1.67 for a monatomic gas (Cv = 3/2 R, Cp = 5/2 R), 1.4 for a diatomic gas (Cv = 5/2 R, Cp = 7/2 R) and 1.33 for a polyatomic gas (Cv = 3R, Cp = 4R). Air is mostly nitrogen and oxygen, so it is treated as a diatomic gas with γ = 1.4; OpenStax Example 3.7 treats a gasoline-air mixture the same way. If your problem gives another γ, choose "Other" and type it: the calculator then uses Cv = R ÷ (γ − 1).

How do I find the final temperature and pressure after an adiabatic compression?

For a quasi-static adiabatic process pV^γ and TV^(γ−1) stay constant (OpenStax section 3.6), so T2 = T1 (V1 ÷ V2)^(γ−1) and p2 = p1 (V1 ÷ V2)^γ. Example 3.7: a mixture at 20°C and 1.00 × 10⁵ N/m² is squeezed from 240 to 40 cm³ (a ratio of 6). With γ = 1.4 the pressure becomes 1.23 × 10⁶ N/m² and the temperature 600 K, or 327°C. The pressure rises by 6^1.4 = 12.3, far more than the factor 6 an isothermal squeeze would give, because the gas heats up.

Is the work positive or negative?

OpenStax counts W as the work done by the gas, so W is positive when the gas expands and negative when it is compressed, and the first law reads ΔE_int = Q − W. In Example 3.7 the work is −62.9 J: the minus sign means the piston does work on the mixture, and since no heat flows (Q = 0) that work raises the internal energy by 62.9 J. If your textbook counts the work done on the gas, flip the sign of W and write ΔU = Q + W.

How do I calculate the work in an isothermal expansion?

W = nRT ln(V2 ÷ V1), and since the temperature stays fixed the internal energy of an ideal gas does not change, so Q = W. OpenStax Example 3.5: 1.00 mol of a monatomic gas at 300 K expands until its volume has tripled; W = 1.00 × 8.31 × 300 × ln 3 = 2,740 J (the book writes 2.74 × 10³ J), and the same amount of heat has to be added. Only the ratio V2 ÷ V1 matters, so the "Volume ratio" choice is handy here.

What do isobaric and isochoric processes do?

At constant pressure the work is W = p (V2 − V1) and the heat is Q = n·Cp·ΔT; at constant volume W = 0 and Q = ΔU = n·Cv·ΔT. OpenStax Example 3.4 combines them: an ideal gas goes from (2.10 × 10⁵ N/m², 2.25 × 10⁻³ m³) to (1.05 × 10⁵ N/m², 4.50 × 10⁻³ m³). Along the path that runs at constant pressure first, the gas does 472 J of work; along the path that runs at constant volume first, only 236 J, although both paths join the same two states. Work and heat depend on the path; ΔU does not. Here p1V1 = p2V2, so both end states are at the same temperature and for an ideal gas ΔU = 0: the heat taken in on each path equals its work. (The book takes 515 J for the first path and arrives at 279 J for the second from the difference of the works, a figure that does not fit an ideal gas with these end states.)

Why is an adiabat steeper than an isotherm on a pV diagram?

Along an isotherm dp/dV = −p/V, along an adiabat dp/dV = −γp/V (OpenStax section 3.6). Because γ is greater than 1 the adiabat falls faster. Physically, in an adiabatic expansion the gas cools as it does work, so its pressure drops more than it would at constant temperature; in an adiabatic compression it heats up and the pressure climbs more.

Sources: OpenStax University Physics, Volume 2, sections 3.2 “Work, Heat, and Internal Energy”, 3.3 “First Law of Thermodynamics” (Examples 3.4 and 3.5), 3.5 “Heat Capacities of an Ideal Gas” (Table 3.3) and 3.6 “Adiabatic Processes for an Ideal Gas” (Example 3.7). Related: ideal gas law calculator, combined gas law calculator, Carnot efficiency calculator and entropy change calculator.