Isothermal, isobaric, isochoric, and adiabatic processes

Write the first law once, with its convention stated: ΔU = Q − W, where Q is heat into the gas and W = ∫p dV is work done by the gas. Two of the three terms depend on the path; only ΔU is a state function. For an ideal gas that state function is unusually simple, because Joule's expansion showed U depends on temperature alone: ΔU = nC_VΔT on every path, isobaric and adiabatic included, not just at the constant volume where C_V was defined. So the working routine for any named process is the same three lines. Find the end temperatures and write ΔU = nC_VΔT. Evaluate W from the path, an integral the constraint usually makes trivial. Then Q = ΔU + W is the remainder, and its sign tells you which way heat flowed. Every named process below is this routine with one term made easy.

Hold T fixed by keeping the gas in thermal contact with a large reservoir and moving the piston slowly enough that the gas never leaves that temperature. For an ideal gas ΔU = 0 at once, and p = nRT/V along the whole path, so W = ∫nRT dV/V = nRT ln(V₂/V₁). The logarithm is the signature of a pressure that falls as 1/V: doubling the volume costs the same work whatever volume you start from. Take 2.00 mol at 300 K expanding from 10.0 L to 20.0 L: W = 2.00 × 8.314 × 300 × ln 2 = 3.46 kJ done by the gas, and since ΔU = 0 the reservoir supplied exactly Q = +3.46 kJ. Compress instead and both signs flip: the same 3.46 kJ of work done on the gas leaves as heat. Isothermal is not the process with no heat. It is the process in which heat and work are equal, and it is slow by definition.

At constant volume the piston is clamped, ∫p dV = 0, and the heat is the whole story: Q_V = ΔU = nC_VΔT. This is the measurement that defines C_V. At constant pressure the gas must push the surroundings back as it warms, W = pΔV, and for an ideal gas pΔV = nRΔT, so the heat needed is Qₚ = nC_VΔT + nRΔT = nCₚΔT with Cₚ = C_V + R. Numbers make the split visible. Warm 1.00 mol of nitrogen (C_V = 5R/2 = 20.8 J mol⁻¹ K⁻¹) by 50.0 K at constant pressure: Qₚ = 29.1 × 50.0 = 1.45 kJ enters, of which ΔU = 20.8 × 50.0 = 1.04 kJ stays as internal energy and W = 8.314 × 50.0 = 416 J leaves as work on the piston. Clamp the piston and the same 50.0 K rise needs only 1.04 kJ. The extra 416 J is not lost, it is the work you can watch the piston doing, and it is the whole reason Cₚ exceeds C_V.

Insulate the cylinder, or move the piston fast compared with the time heat needs to flow, and Q = 0. The first law then reads dU = −p dV for every infinitesimal step, and with dU = nC_V dT and p = nRT/V that is nC_V dT = −nRT dV/V. Separate the variables: dT/T = −(R/C_V) dV/V. The ratio R/C_V is (Cₚ − C_V)/C_V = γ − 1, so integrating from state 1 to state 2 gives ln(T₂/T₁) = −(γ − 1) ln(V₂/V₁), or TV^(γ−1) = constant. Substitute T = pV/nR and pV^γ = constant follows. Compare slopes at a shared point: on the isotherm dp/dV = −p/V, on the adiabat dp/dV = −γp/V, steeper by γ, because compressing without letting heat escape raises T as well as squeezing V, and both push p up. A limiting case checks the sign: γ → 1 means C_V → ∞, a gas that absorbs unlimited energy without warming, and the adiabat collapses onto the isotherm.

With p = K V^(−γ) along the path, W = ∫p dV = K(V₂^(1−γ) − V₁^(1−γ))/(1 − γ) = (p₁V₁ − p₂V₂)/(γ − 1). Since Q = 0 this must also equal −ΔU = nC_V(T₁ − T₂), and it does, because p₁V₁ − p₂V₂ = nR(T₁ − T₂) and R/(γ − 1) = C_V. Try air (γ = 1.40) at 1.00 × 10⁵ Pa and 300 K in 1.00 L, compressed reversibly to 0.100 L: T₂ = 300 × 10^0.40 = 754 K and p₂ = 10^1.40 × 10⁵ Pa = 25.1 × 10⁵ Pa. The work is (100 J − 251 J)/0.40 = −378 J done by the gas, so 378 J was done on it, and every joule sits in the gas as internal energy. Check by the other route: n = p₁V₁/RT₁ = 0.0401 mol, C_V = 20.8 J mol⁻¹ K⁻¹, ΔT = 454 K, nC_VΔT = 378 J. That 754 K is why a diesel engine needs no spark plug, and why adiabatic compression always costs more work than isothermal compression to the same volume.

The derivation borrowed three things, and pV^γ = constant is only as good as all three. First, an ideal gas: U = U(T) and pV = nRT both entered, so near condensation or at high density the curve is wrong even if Q = 0 exactly. Second, a reversible path: the gas had a single pressure p at every instant, which requires the piston stroke to take far longer than the sound-crossing time of the cylinder while still being short compared with the time heat needs to leak out. Adiabatic is not the same as pV^γ: a free expansion into vacuum has Q = 0 and W = 0, so ΔU = 0 and the ideal gas ends on its original isotherm, not on the adiabat. Third, a constant γ: pulling γ − 1 outside the integral needs C_V independent of T. For nitrogen C_V rises from 5R/2 towards 7R/2 as vibration wakes up above roughly 1000 K, so γ drifts from 1.40 towards 1.29, and a compression that reaches 750 K already wants ∫C_V(T) dT/T in place of the power law.