Lessons · thermo

Constant-pressure work: the area is a rectangle

Regime 3 — an algebra-only domain, but the calculus underpinning is clean and worth seeing: the work is the area under the curve, ∫P dV. SymPy proves that area is exactly the memorized result — and that the constant-pressure case collapses to the rectangle.

Regime 3 · algebra-only Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A gas expands at constant pressure $P = 100$ kPa from $V_1 = 1$ L to $V_2 = 3$ L (a piston under a fixed load). How much work does it do? Because the pressure never changes, the $P$ – $V$ curve is a horizontal line and the work is the rectangular area under it: $W = P\,\Delta V$ . This is the simplest integral in the course — a constant integrand — so the memorized algebra formula and the calculus area are literally the same. Drag $P$ to scale the rectangle; sweep the cursor to watch the work accumulate as a straight line.

Isobaric work

W = P\,\Delta V

(the shaded rectangle)

200 J

Pressure

P

(constant — the height of the rectangle)

100000 Pa

Volume change

\Delta V = V_2 - V_1

(the width)

0.002 m³

Work is the area under the pressure–volume curve

W = \int_{V_1}^{V_2} P\,dV

Work done by the gas is the integral of pressure over volume — the area under the $P$ – $V$ curve, for any process. The shape of that area depends on how $P$ varies as the gas expands.

check $\tfrac{dW}{dV} = P$ ; check $W = \int_{V_1}^{V} P\,dV'$ ; collapse the constant-integrand integral to the rectangle $P\,\Delta V$

✓ The work's slope is the pressure: $\tfrac{dW}{dV} = P$ — the area's rate of growth is the (constant) height. [structural]
✓ The accumulated work is the area: $W(V) = \int_{V_1}^{V} P\,dV'$ . [structural]
✓ A constant integrand makes the integral a rectangle: $\int_{V_1}^{V_2} P\,dV = P\,\Delta V$ — the algebra formula is the integral already evaluated. [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “Every expansion from

V_1

V_2

does the same work, since the volume change is the same.”

No — work is the area under the $P$ – $V$ curve, and the path sets the area. At constant pressure the area is the full rectangle $P\,\Delta V$ . If instead the gas expands isothermally or adiabatically, the pressure falls as it expands, so the area under the curve is smaller and the work is less. Same $\Delta V$ , lower average pressure, less work. The isobaric rectangle is the largest of the three for a given starting pressure.

Modeling assumptions — author-asserted, disclosed not discharged

The pressure is held constant throughout the expansion (an isobaric process — e.g. a piston free to move under a fixed external load).
Work is the area under the P–V curve, $W = \int P\,dV$ , with no friction or dissipation.

The P–V graph, fully annotated

A static rendering (Matplotlib): the shaded area under P is the accumulated integral W, and the slope of W is P. The interactive version with a draggable cursor is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):