Work done in each process — derivation: Visual walkthrough

This page is the picture-companion to the parent derivation. Every symbol is earned before use.

Step 1 — What "work" even means for a pushing gas

WHAT. A gas is trapped in a cylinder sealed by a piston (a sliding lid). The gas presses outward; if the piston slides a little, the gas has done work.

WHY start here. In physics, work = force $\times$ distance moved in the direction of the force. Symbol $W$ . It measures "how much pushing effort actually got converted into motion." We must start from force because that is the only definition of work that is always true — everything else we build from it.

PICTURE. Look at the cylinder below. The gas fills the left; the piston is the amber bar. The white arrow is the force $F$ the gas exerts on the piston. If the piston does not move, distance $= 0$ , so $W = 0$ — pushing a wall that won't budge does no work.

Figure — Work done in each process — derivation

Step 2 — Turn force into pressure (why pressure is the natural variable)

WHAT. Replace the force $F$ by something the gas actually "feels everywhere": pressure.

WHY pressure and not force? Force depends on how big the piston is; a wider piston feels more total force from the same gas. Pressure $P$ = force spread over area, $P = F / A$ , so it is a property of the gas alone, independent of piston size. That is why every gas law is written in $P$ , not $F$ . Rearranging gives the force the gas produces:

F = pressure of gas P \times piston area A

$P$ ::: force per unit area (pascals). The intrinsic "push strength" of the gas.
$A$ ::: the flat area of the piston the gas pushes on.
$P A$ ::: total force, because pressure acts over the whole face.

PICTURE. The face of the piston is shaded; the many small cyan arrows are pressure acting on every patch. Add them all up over area $A$ and you recover the single big force arrow $F = P A$ .

Step 3 — One tiny push: $d W = P d V$

WHAT. Let the piston slide out by a tiny distance we call $d x$ (the " $d$ " just means "a very small amount of"). The tiny work done is

d W = F d x = (P A) d x = P = d V (A d x) = P d V

$d x$ ::: a sliver of distance the piston moves — small enough that $P$ hasn't changed yet.
$A d x$ ::: area $\times$ sliver of length = a thin slab of volume swept out. We rename it $d V$ .
$d V$ ::: the tiny extra volume the gas gained.
$d W = P d V$ ::: the tiny work for one tiny expansion.

WHY do it in tiny steps at all? Because $P$ is not constant while the gas expands — as the gas spreads out it pushes more weakly. If we tried $W = F \times (whole distance)$ in one shot, we'd be using the wrong $P$ for most of the trip. Cutting the journey into slivers so thin that $P$ is effectively constant inside each sliver is exactly what a tiny " $d$ " buys us.

PICTURE. The amber slab is the swept volume $d V$ ; its width is $d x$ , its area is $A$ . Its "cost in work" is $P$ times its width — that is the height-times-width of one thin strip.

Step 4 — Add the tiny pushes = the integral = area under the curve

WHAT. The full expansion from volume $V_{1}$ to $V_{2}$ is millions of these slivers laid side by side. "Add up infinitely many infinitely thin pieces" has a name and a symbol: the integral $\int$ .

W = \int_{V_{1}}^{V_{2}} P d V

$\int$ ::: a stretched "S" for Sum — it adds every sliver $P d V$ .
$V_{1}$ (bottom) ::: where the piston starts.
$V_{2}$ (top) ::: where the piston stops.
$P$ inside ::: the height of each strip; it may change strip to strip.

WHY an integral and not a multiplication? Because a plain multiplication $P Δ V$ only works if the height $P$ never changes. The integral is precisely the tool that handles a changing height. If $P$ happens to be constant, the integral collapses back to a rectangle — no contradiction.

WHAT IT LOOKS LIKE. Plot $P$ up, $V$ across — a $P$ – $V$ diagram (see P-V Diagrams). Each strip is a thin rectangle of height $P$ , width $d V$ . Stacking them fills the region under the curve. So:

W = area under the P - V curve

Step 5 — Case A: Isobaric (flat curve → a rectangle)

WHAT. "Iso-baric" = same pressure. $P$ is constant, so pull it out of the sum:

W = \int_{V_{1}}^{V_{2}} P d V = P \int_{V_{1}}^{V_{2}} d V = height P width (V_{2} - V_{1})

WHY it's the easy one. Constant height means the area under the curve is just a plain rectangle: height $P$ , width $Δ V = V_{2} - V_{1}$ . No summing tricks needed.

PICTURE. A horizontal line at height $P$ ; the shaded area beneath it is a clean rectangle.

Using the ideal-gas law $P V = n R T$ , this also equals $W = n R (T_{2} - T_{1})$ .

Step 6 — Case B: Isochoric (no width → zero area)

WHAT. "Iso-choric" = same volume. The piston is pinned; $V$ never changes so $d V = 0$ .

W = \int_{V_{1}}^{V_{1}} P d V = 0

WHY zero. No distance travelled means no width for any strip. On the diagram the "path" is a vertical line, and a vertical line has zero area beneath it — you cannot make a rectangle with zero width, no matter how tall.

PICTURE. A vertical segment at fixed $V$ ; the shaded region collapses to a line — nothing to fill.

Step 7 — Case C: Isothermal (a curved roof → a logarithm)

WHAT. "Iso-thermal" = same temperature $T$ . Now $P$ genuinely changes. From $P V = n R T$ we solve for the height at each volume:

P = \frac{n R T}{V}

Feed this changing height into the sum:

W = \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = n R T \int_{V_{1}}^{V_{2}} \frac{d V}{V} = n R T [ln V]_{V_{1}}^{V_{2}}

W = n R T ln \frac{V _{2}}{V _{1}}

$\frac{n R T}{V}$ ::: the height of the curve — it sinks as $V$ grows (weaker push).
$\int \frac{d V}{V}$ ::: the one integral whose answer is a logarithm, $ln$ .
$ln (V_{2} / V_{1})$ ::: "how many times over did the volume grow, measured on a multiply-scale."

WHY a logarithm appears. The strips get shorter as the curve dives (a hyperbola $P \propto 1/ V$ ). Adding up the area under a $1/ V$ curve is the unique job that produces $ln$ . That's why the tool here is the logarithm and not a power — the shape demanded it. See Isothermal Process and Boyle's Law.

PICTURE. A downward-curving hyperbola; the shaded area under it is fatter near the start (tall strips) and thin at the end (short strips).

Step 8 — Case D: Adiabatic (a steeper curve → a power law)

WHAT. "A-diabatic" = no heat crosses the wall, $Q = 0$ . Here the gas obeys $P V^{γ} = K$ (a constant), where $γ$ (gamma) is a number bigger than 1 set by the gas type. Solve for height: $P = K V^{- γ}$ .

W = \int_{V_{1}}^{V_{2}} K V^{- γ} d V = K [\frac{V ^{1 - γ}}{1 - γ}]_{V_{1}}^{V_{2}} = \frac{K}{1 - γ} (V_{2}^{1 - γ} - V_{1}^{1 - γ})

Because $K = P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ}$ , the ugly powers snap shut into pressures:

W = \frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1} = \frac{n R ( T _{1} - T _{2} )}{γ - 1}

$K V^{- γ}$ ::: the height — it drops faster than the isothermal $1/ V$ , because $γ > 1$ steepens the fall.
$\int V^{- γ} d V = \frac{V ^{1 - γ}}{1 - γ}$ ::: the power rule for integrating; used here because the height is a power of $V$ , not $1/ V$ .
$γ - 1$ in the denominator ::: what's left after $1 - γ$ flips sign to keep expansion positive.

WHY a power law, not a log. An adiabatic curve is $P \propto V^{- γ}$ — a power of $V$ . Integrating a power (other than $- 1$ ) always gives another power via the power rule. Only the special exponent $- 1$ gives a log; adiabatic uses $- γ \neq = - 1$ , so we get a power answer. Source of $P V^{γ} = K$ : Adiabatic Process and PV^gamma.

PICTURE. Two curves from the same start: the flatter cyan is isothermal, the steeper amber is adiabatic. The adiabatic dives faster, so it encloses less area — less work for the same volume change.

Step 9 — Case E: Cyclic (loop → enclosed area)

WHAT. A cycle returns to its start, tracing a closed loop. The net work is the area enclosed by the loop (written $\oint$ , the circle-integral):

W_{net} = \oint P d V = area inside the loop

WHY it comes out to the enclosed area. Going right along the top (expansion) adds a big positive area; coming back left along the bottom (compression) subtracts a smaller negative area. What survives is the difference — the region trapped between the two paths.

Clockwise loop ::: top path is the higher-pressure one ⇒ $W_{net} > 0$ (engine).
Anticlockwise loop ::: $W_{net} < 0$ (refrigerator).

PICTURE. A closed loop; the outgoing top arc (amber) and returning bottom arc (cyan) enclose a shaded region — that shaded region is the net work. This is the heart of the Carnot Cycle.

The one-picture summary

All four processes on one $P$ – $V$ diagram, sharing the same start point. Read the shape → read the work:

Flat line = isobaric = rectangle = $P Δ V$ .
Vertical line = isochoric = zero width = $0$ .
Gentle curve = isothermal = area under $1/ V$ = $n R T ln (V_{2} / V_{1})$ .
Steep curve = adiabatic = area under $V^{- γ}$ = $\frac{P _{1} V _{1} - P _{2} V _{2}}{γ - 1}$ .

Recall Feynman retelling — the whole walkthrough in plain words

A gas pushes a lid. Work is push strength × how far the lid moved. We swapped "force" for "pressure" because pressure belongs to the gas no matter how big the lid is. But the gas gets weaker as it spreads, so we can't push once and multiply — we chop the trip into slivers so thin the strength is steady inside each, find the little work $P d V$ for each sliver, and stack them all up. Stacking = the integral = the area under the pressure-versus-volume curve. Now the four cases are just four shapes: a flat roof gives a plain rectangle (isobaric); a wall gives no width so no work (isochoric); a gently sagging roof gives a logarithm (isothermal, because the shape is $1/ V$ ); a steeper sagging roof gives a power answer (adiabatic, because heat is trapped and the curve is $V^{- γ}$ ); and a closed loop just traps an area between its two arcs (cyclic). One idea — area under the curve — wearing four costumes.

Recall

What single equation is every work formula secretly? ::: $W = \int_{V_{1}}^{V_{2}} P d V$ — area under the P–V curve. Why does isochoric work vanish? ::: The path is a vertical line — zero width, zero area, piston doesn't move. Why isothermal gives a log but adiabatic gives a power? ::: Isothermal height is $1/ V$ (integral of $1/ V$ is $ln$ ); adiabatic height is $V^{- γ}$ with $γ \neq = 1$ (power rule gives a power). What does the net cyclic work equal geometrically? ::: The area enclosed by the loop; clockwise is positive.

Connections

Parent: Work done in each process
First Law of Thermodynamics — $Δ U = Q - W$ consumes these $W$ values
P-V Diagrams — work = area under the curve
Adiabatic Process and PV^gamma — origin of $P V^{γ} = K$
Isothermal Process and Boyle's Law — origin of the $1/ V$ curve
Internal Energy and Cv — adiabatic $W = - Δ U$
Carnot Cycle — the cyclic enclosed-area idea