Gas work, p-V paths, and path dependence

Let the gas fill a cylinder closed by a piston of area A. When the piston moves out by dx, the gas pushes it with force pA and does work dW = pA dx = p dV, because A dx is the volume swept. Sum the steps and W = ∫p dV from V₁ to V₂. The dimensions check at once: Pa × m³ = (N m⁻²)(m³) = N m = J. The sign is carried by dV alone. An expansion has dV > 0 and W > 0, the gas doing work on its surroundings; a compression has dV < 0 and W < 0, meaning work is done on the gas, and in the convention ΔU = Q − W that negative W raises U. On a p-V diagram the integral is the area between the path and the V axis, counted positive when the path runs to the right and negative when it runs to the left. Nothing in the derivation used an ideal gas or a particular process. It used only that the gas had a pressure p at the piston face, and that is the assumption to watch.

Take a gas from state 1, p₁ = 2.0 × 10⁵ Pa and V₁ = 1.0 × 10⁻³ m³, to state 2, p₂ = 1.0 × 10⁵ Pa and V₂ = 2.0 × 10⁻³ m³. Route A: expand at constant p₁ to V₂, then drop the pressure at fixed volume. The isochoric leg has dV = 0 and contributes nothing, so W_A = p₁(V₂ − V₁) = 2.0 × 10⁵ × 1.0 × 10⁻³ = 200 J. Route B: drop the pressure first at V₁, then expand at p₂: W_B = 1.0 × 10⁵ × 1.0 × 10⁻³ = 100 J. Same end states, and the gas hands over twice as much energy along A as along B. The difference, 100 J, is the area of the rectangle the two routes enclose, (p₁ − p₂)(V₂ − V₁), and it is what a cycle running out along A and back along B would deliver per loop. Since p₁V₁ = p₂V₂, the two states also share an isotherm, and for an ideal gas the curved route p = p₁V₁/V gives a third answer between the other two: 200 J × ln 2 = 139 J.

For an ideal gas at fixed T, substitute p = nRT/V into the integral: W = nRT ∫dV/V = nRT ln(V₂/V₁), and since nRT = p₁V₁ = p₂V₂ on the isotherm, W = p₁V₁ ln(V₂/V₁) as well. The argument is a ratio, dimensionless as a logarithm's must be. Now check the limits. For a tiny change V₂ = V₁(1 + ε), ln(1 + ε) ≈ ε and W ≈ p₁V₁ε = p₁ΔV: every quasi-static path looks isobaric over a short enough step, which is why the rectangle is the building block of all of them. For a large expansion the logarithm is slow. One mole at 300 K doubling its volume does nRT ln 2 = 8.314 × 300 × 0.6931 = 1729 J, and doubling again adds another 1729 J rather than 3458 J, because the pressure has halved and each further litre is bought at a lower price. The result sits between the rectangles, p₁ΔV above and p₂ΔV below, which for the first doubling are 2494 J and 1247 J. For a compression V₂ < V₁ the logarithm is negative and the same formula returns the work done on the gas with the sign already right.

Internal energy has a value at every state, so its change between two states is fixed before any path is named: dU is an exact differential and ∮dU = 0. Work is different, and the calculus says so directly. A differential M dV + N dp on the p-V plane is exact only if ∂M/∂p = ∂N/∂V. For work M = p and N = 0, so ∂M/∂p = 1 while ∂N/∂V = 0. The test fails, there is no function W(p, V) whose differences give the work, and the loop integral ∮p dV is the enclosed area rather than zero. That is why the increment is written đW or δW rather than dW: it is a small quantity of work transferred, not a small change in something the gas possesses. The practical consequence is a rule of grammar. 'The work done by the gas between states 1 and 2' is not a complete sentence until the path is stated, whereas 'the change in internal energy between states 1 and 2' is. Heat inherits the same status through the first law, Q = ΔU + W: a fixed ΔU plus a path-dependent W is path-dependent.

The integrand in ∫p dV is the pressure of the gas, which is defined only when the gas is in internal equilibrium. Move the piston and the molecules next to it are compressed first; the news travels through the gas at the speed of sound, cₛ ≈ 340 m s⁻¹ in air near room temperature, and only after several crossings of the cylinder length L is the pressure uniform again. So 'slow' has a number: the process time must be long compared with L/cₛ. For L = 0.20 m that is 0.6 ms. A stroke of that full length lasting 1 s spans some 1700 crossing times; the piston moves at 0.2 m s⁻¹, the pressure across the cylinder differs by roughly that speed over the sound speed, under one part in a thousand, and the process can be drawn as a curve of equilibrium points. The same stroke completed in 1 ms cannot. Two more conditions travel with the word: the piston must be frictionless, or part of the force at the face is spent on the wall instead of the gas, and any heat exchange must be slow enough that T is defined too. Meet them and the path is reversible in the sense the section uses later: run backwards, it retraces the same curve.

Release a latched piston and the gas rushes out; the pressure at the piston face collapses below the pressure deep in the cylinder, and there is no single p to integrate. The end points are still equilibrium states and can be marked on the diagram, but no curve joins them. What the surroundings receive is set on their side of the boundary: W = ∫pₑₓₜ dV, with pₑₓₜ the external pressure the piston pushes against. Hold 1 mol at 300 K in contact with a bath and let it double its volume from V₁ = nRT/p₁ = 0.02494 m³ (p₁ = 1.00 × 10⁵ Pa) against a constant pₑₓₜ equal to the final pressure of 0.50 × 10⁵ Pa: W = 0.50 × 10⁵ × 0.02494 = 1247 J, against 1729 J for the quasi-static isotherm between the same states. Push it back suddenly against pₑₓₜ = p₁ = 1.00 × 10⁵ Pa and the work done on the gas is 1.00 × 10⁵ × 0.02494 = 2494 J, against 1729 J the slow way. Expand into a vacuum and pₑₓₜ = 0: the gas doubles its volume, its temperature is unchanged, and W = 0. The pattern is general and is the seed of the Carnot bound: of all routes between two states that exchange heat only with the one bath at T, the quasi-static one extracts the most work on expansion and costs the least on compression. The qualifier matters, since routes that trade heat at other temperatures, like the isobaric-then-isochoric rectangle of the worked comparison, can exceed the isotherm.