Lessons · thermo

Work by an expanding gas: area under the P–V curve

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 3 modeling assumptions (author-asserted)

One mole of an ideal gas at $T = 300$ K expands isothermally from $V_1 = 0.025$ m³ to $V_2 = 0.075$ m³. How much work does it do — and why is the answer a logarithm, not simply pressure times volume change? Watch the work accumulate as the area under the P–V curve.

Isothermal work

W = nRT\ln(V_2/V_1)

(the shaded area)

2740 J

Initial pressure

P_1 = nRT/V_1

99770 Pa

Final pressure

P_2 = nRT/V_2

33260 Pa

If pressure stayed at

P_1

W = P_1\,\Delta V

(the rectangle)

4989 J

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. At constant pressure the area is a rectangle ( $P\,\Delta V$ ); when the pressure varies, it is the area under a curve.

check $\tfrac{dW}{dV} = P$ ; check $W = \int_{V_1}^{V} P\,dV'$ ; recover $W = nRT\ln(V_2/V_1)$ ; collapse the constant-pressure case to $P\,\Delta V$

✓ The work's slope is the pressure: $\tfrac{dW}{dV} = P$ — the area's rate of growth is the curve's height. [structural]
✓ The accumulated work is the area: $W(V) = \int_{V_1}^{V} P\,dV'$ . [structural]
✓ The memorized isothermal result $W = nRT\ln(V_2/V_1)$ is exactly the area at $V_2$ . [structural]
✓ At constant pressure the integral collapses to $W = P\,\Delta V$ — the area is a rectangle (the quadrature). [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “The work done by an expanding gas is just pressure times the change in volume,

W = P\,\Delta V

.”

That is only true at constant pressure (a rectangle). In an isothermal expansion the pressure falls as the gas expands — $P = nRT/V$ — so the work is the area under a curved hyperbola, $nRT\ln(V_2/V_1)$ , which is less than $P_1\,\Delta V$ . Drag the cursor: the shaded area under the curve, not a rectangle, is the work.

Modeling assumptions — author-asserted, disclosed not discharged

Ideal gas: $PV = nRT$ holds exactly (no intermolecular forces, point particles).
Isothermal, reversible, quasi-static expansion — the temperature is held constant by a reservoir.
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):