Lessons · thermo

Adiabatic work: the same area, a steeper curve, a cooler gas

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 diatomic gas ( $\gamma = 1.4$ ) expands from $V_1 = 1$ L at $P_1 = 100$ kPa to $V_2 = 3$ L with no heat exchanged ( $PV^\gamma =$ const). How much work does it do, and what happens to its temperature? The work is again the area under the $P$ – $V$ curve — but the adiabat falls off faster than an isotherm, so the gas does less work and cools. Drag the cursor out along the adiabat to watch the work accumulate; raise the initial pressure $P_1$ to lift the whole curve while its shape (and the cooling ratio) stays fixed by $\gamma$ .

Adiabatic work

W = \dfrac{P_1V_1 - P_2V_2}{\gamma-1}

(the shaded area)

88.9 J

Final pressure

P_2 = P_1(V_1/V_2)^\gamma

21480 Pa

If pressure stayed at

P_1

W = P_1\,\Delta V

(the rectangle)

200 J

Temperature ratio

T_2/T_1 = (V_1/V_2)^{\gamma-1}

— the gas cools

0.6444

Work is the area under the pressure–volume curve

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

As before, the work done by the gas is the integral of pressure over volume — the area under the $P$ – $V$ curve. What changes from the isotherm is only the shape of $P(V)$ .

check $\tfrac{dW}{dV} = P$ ; check $W = \int_{V_1}^{V} P\,dV'$ ; recover $W = (P_1V_1 - P_2V_2)/(\gamma-1)$ ; 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: $W(V) = \int_{V_1}^{V} P\,dV'$ . [simplify]
✓ The memorized adiabatic work $W = \dfrac{P_1V_1 - P_2V_2}{\gamma-1}$ is exactly the area at $V_2$ . [simplify]
✓ 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: “An expanding gas always does the same work for a given change in volume.”

Work is the area under the $P$ – $V$ curve, and the path sets the curve. An adiabat ( $P\propto V^{-\gamma}$ ) falls faster than an isotherm ( $P\propto V^{-1}$ ), so for the same expansion it encloses less area and does less work. And because no heat flows in, that work is paid out of the gas's internal energy — so the gas cools, $T_2/T_1 = (V_1/V_2)^{\gamma-1}$ . Same volume change, different area, different work.

Modeling assumptions — author-asserted, disclosed not discharged

Quasi-static, reversible adiabatic process: no heat crosses the boundary ( $Q = 0$ ), so $PV^\gamma =$ const holds at every step.
Ideal gas with constant $\gamma$ over the temperature range (no vibrational modes switching on).

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):