Isothermal & Adiabatic Processes
Slow and in contact, or fast and insulated: the two limits of a gas process.
Lesson at a glanceShow the key points
- Isothermal: T constant, ΔU = 0, Q = W = nRT ln(V_f/V_i); C infinite; slow process in contact with a reservoir.
- Adiabatic: Q = 0, W = −ΔU; C = 0; expansion cools, compression heats.
- Reversible adiabatic: pV^γ = constant (27.9); T^γ/p^(γ−1) = constant (27.10); TV^(γ−1) = constant (27.11).
- Adiabatic work W = (p₁V₁ − p₂V₂)/(γ − 1) (27.12).
Squeeze a gas slowly in a metal cylinder and it stays at room temperature. Squeeze it fast in a bicycle pump and the pump gets hot. These are the two limits: isothermal and adiabatic.
T constant. Ideal gas: ΔU = 0, so heat in = work done, W = nRT ln(V_f/V_i). Boyle’s law holds. Heat capacity ΔQ/(nΔT) is infinite. Achieved by working slowly in contact with a large reservoir (e.g. a metal cylinder in air).
No heat exchanged; heat capacity zero. Work done by the gas comes from its internal energy: expanding, it cools; compressed, it warms. Achieved with insulation or by acting quickly.
pV^γ is constant in a reversible adiabatic process
With dQ = 0 the first law gives 0 = dU + p dV = nCᵥdT + p dV. From pV = nRT, n dT = (p dV + V dp)/R. Substituting and using Cᵥ + R = Cₚ:
Integrating between the initial and final states:
Dry air at 20°C and 8.0 atm is suddenly released to 1.0 atm. Treating it as reversible and adiabatic with γ = 1.40, find the final temperature.
Air at 300 K (γ = 1.4) is compressed quickly to half its volume. Estimate its temperature.
Work in an adiabatic process
With pV^γ = K, W = ∫K V^(−γ) dV between V₁ and V₂:
- 1Isothermal: T fixed, W = nRT ln(V_f/V_i) = Q; heat capacity infinite.
- 2Adiabatic: Q = 0, W = −ΔU; heat capacity zero; expansion cools, compression heats.
- 3pV^γ, T^γ/p^(γ−1) and TV^(γ−1) are constant in a reversible adiabatic process.
- 4Adiabatic work W = (p₁V₁ − p₂V₂)/(γ − 1).
Check yourself
Keep exploring
Hand-picked videos, simulations and readings that explain this idea from a different angle.