Work by gases and p-V diagrams

Put a gas in a cylinder closed by a piston of area A and let the piston slide out by dx. The gas presses on the piston with force F = pA, so it does work δW = F dx = pA dx. But A dx is exactly the extra volume the gas now occupies, dV, so δW = p dV and the apparatus has dropped out of the statement. Integrating along the process gives W = ∫ p dV from V₁ to V₂. A number makes the equivalence concrete: a piston of area 8.0 × 10⁻³ m², with the gas held at 1.5 × 10⁵ Pa, carries a force of 1.2 × 10³ N, and moving it 0.020 m costs 24 J of work; the same 24 J follows from p ΔV = 1.5 × 10⁵ Pa × 1.6 × 10⁻⁴ m³. The second route never mentions the piston, which is why it works just as well for a balloon, a bubble or a cylinder of any cross-section.

The p in ∫ p dV is the gas's own pressure at the moving boundary, and giving the whole gas one pressure assumes the change is quasi-static: slow enough for the pressure to even out, so every intermediate state is a point on the diagram rather than a smear. Ram the piston in fast and the gas piles up against it, so the boundary pressure runs above the bulk value and the compression costs more than the drawn curve suggests. The sharpest case is free expansion. Open a valve between a gas at 1.5 × 10⁵ Pa and an evacuated flask and the volume may double, but the gas pushes on nothing that moves, so W = 0 even though ΔV is large and the pressure is nowhere zero. Free expansion has no p-V curve at all, only two dots, because in between the gas is not in a single equilibrium state to plot.

Once a change is quasi-static it draws a curve on a p-V diagram, and ∫ p dV is the area between that curve and the volume axis. Two consequences follow. Geometry can replace calculus whenever the curve is simple: a straight line from (2.0 × 10⁻³ m³, 300 kPa) to (6.0 × 10⁻³ m³, 100 kPa) is a trapezium of mean height ½(300 + 100) kPa = 200 kPa across a width of 4.0 × 10⁻³ m³, so W = 2.00 × 10⁵ Pa × 4.0 × 10⁻³ m³ = 8.0 × 10² J. And the sign lives in dV, never in p: gas pressure is always positive, so each slice p dV takes the sign of dV. Sweep to the right and the gas does positive work on its surroundings; sweep to the left and the integral is negative, meaning the surroundings did the work. A vertical leg has dV = 0 at every point and contributes exactly nothing, however violently the pressure changes along it.

A rectangle is only available while p is constant. Otherwise you need p as a function of V along the actual path, and then you integrate. Take a cylinder whose loading makes the measured curve fit pV² = C. At V₁ = 1.50 × 10⁻³ m³ the pressure is 2.40 × 10⁵ Pa, so C = p₁V₁² = 0.540 Pa m⁶. Let it expand to V₂ = 4.50 × 10⁻³ m³, where the pressure has fallen to C/V₂² = 2.67 × 10⁴ Pa. Then W = ∫ C V⁻² dV = C(1/V₁ − 1/V₂) = 0.540 Pa m⁶ × (666.7 − 222.2) m⁻³ = 240 J. Set that against what the shortcuts would have claimed. Using the starting pressure throughout gives p₁ ΔV = 720 J, three times too large; the straight chord joining the two end states gives 400 J. The curve p = C/V² sags below its own chord, and only the integral knows by how much.

Take a gas from state A (1.0 × 10⁻³ m³, 300 kPa) to state C (4.0 × 10⁻³ m³, 100 kPa) by two routes. Route 1 expands it at a steady 300 kPa out to 4.0 × 10⁻³ m³, then cools it at fixed volume down to 100 kPa: W = 3.00 × 10⁵ Pa × 3.0 × 10⁻³ m³ + 0 = 900 J. Route 2 cools it first at fixed 1.0 × 10⁻³ m³ down to 100 kPa, then expands at that pressure: W = 0 + 1.00 × 10⁵ Pa × 3.0 × 10⁻³ m³ = 300 J. Same two states, three times the work. Now add the first law. ΔU is fixed by the end states alone, and for an ideal gas U follows pV, which is 300 J at A and 400 J at C, so a monatomic sample has ΔU = 1.5 × 100 J = 150 J on either route. Then Q = ΔU + W is 1050 J on route 1 and 450 J on route 2. Heat is path-dependent for precisely the reason work is: the 600 J of extra work route 1 delivers is 600 J of extra heat it must first absorb.

Run route 1 outwards and route 2 back and the gas returns to A, having drawn a closed rectangle. Every leg still contributes its own ∫ p dV, but the two horizontal legs are now swept in opposite directions, so their areas subtract and what survives is the area enclosed: (300 − 100) kPa × 3.0 × 10⁻³ m³ = 6.0 × 10² J. Direction sets the sign. Go round clockwise — expand along the high-pressure leg, return along the low — and Wₙₑₜ = ∮ p dV is positive, so the gas delivers net work each time round; go anticlockwise and Wₙₑₜ is negative, so the surroundings pay. Because U comes back to its starting value, ΔU = 0 over any cycle, and the first law then forces Qₙₑₜ = Wₙₑₜ exactly. The enclosed area is therefore both the net work per cycle and the net heat absorbed per cycle — and a loop that encloses no area does nothing at all.