Visual walkthrough — Thermodynamic processes — isothermal (T const), isochoric (V const), isobaric (P const), adiabatic (Q = 0)

We lean on the First Law of Thermodynamics, the Ideal Gas Law, Heat Capacities Cv and Cp, and Work done by gas — PV diagrams. Everything traces back to the parent: the topic note.

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Step 1 — What "adiabatic" even means (a wall heat cannot cross)

WHAT. We put gas in a cylinder wrapped in perfect insulation. A piston caps it.

WHY. Setting $Q=0$ is the single constraint that makes this process special. Every other process lets heat sneak in or out; here the gas is on its own — whatever work it does must come from its own energy.

WHAT IT LOOKS LIKE. In the figure, the orange hatched wall blocks the wavy red heat arrows: they bounce back. Only the piston can move.

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Step 2 — The First Law, drawn as an energy bucket

WHAT. The First Law of Thermodynamics says energy is just bookkeeping: $\Delta U=Q-W$. In tiny-step form, $dU=dQ-dW$.

WHY. We need a relation that connects the gas's temperature change to the work it does, because that link is what will bend the curve. The First Law is the only law that ties $Q$, $W$, and $U$ together.

WHAT IT LOOKS LIKE. Picture a bucket labelled $U$. Heat pours in from the top ($dQ$), work leaks out the side ($dW$). With $dQ=0$, the top is capped — the bucket can only drop as work leaks out.

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Step 3 — Naming the two little pieces: $dU$ and $dW$

WHAT. We replace the vague $dU$ and $dW$ with concrete formulas.

• Internal energy of an ideal gas: $dU=n{C}_{V}dT$. This holds in every process (it is the parent's second "mistake" — $C_V$ is just the dial from temperature to internal energy).
• Work when the piston sweeps a tiny volume $dV$: $dW=PdV$ (from Work done by gas — PV diagrams — force $\times$ distance repackaged as pressure $\times$ swept volume).

WHY. To combine them into one equation we must speak the same language — both must become expressions in the state variables $P$, $V$, $T$.

WHAT IT LOOKS LIKE. The figure shows the piston nudged out by the swept sliver of volume $dV$ (blue). The gas cooled a touch: $dT<0$, drawn as the thermometer dropping.

Substitute both into $dU=-dW$: nC_V\,dT = -\,P\,dV \tag{$\ast$}

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Step 4 — The problem: three variables, we want only two

WHAT. Equation $(\ast )$ mixes $T$, $P$, and $V$. But a curve on a $P$–$V$ diagram only knows about $P$ and $V$. We must evict $T$.

WHY. The Ideal Gas Law $PV=nRT$ is the bridge. It ties $T$ to $P$ and $V$, so we can rewrite any $dT$ using $dP$ and $dV$ instead.

WHAT IT LOOKS LIKE. Think of $T$ as a shadow of the point $(P,V)$: the picture shows a point moving on the $P$–$V$ plane, and $T$ is a colour that follows it. We want the equation in the plane, so we trade the colour for coordinates.

So $nRdT=PdV+VdP$, giving $ndT=\frac{PdV+VdP}{R}$.

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Step 5 — Fuse the two equations and let $\gamma$ appear

WHAT. Put the $T$-expression from Step 4 into $(\ast )$.

WHY. This is the moment the two physical laws (energy and gas law) become one equation with only $P$ and $V$ — exactly what a curve needs.

Starting from $(\ast )$: $C_V(ndT)=-PdV$. Replace $ndT$: $C_V\cdot \frac{PdV+VdP}{R}=-PdV$

Multiply through by $R/C_V$: $PdV+VdP=-\frac{R}{C_V}PdV$

Now Heat Capacities Cv and Cp hands us Mayer's relation $C_P=C_V+R$, so $\frac{R}{C_V}=\frac{C_P-C_V}{C_V}=\gamma -1$, where $\gamma \equiv \frac{C_P}{C_V}$.

WHAT IT LOOKS LIKE. The figure shows the messy three-variable relation collapsing, with $\gamma$ dropping in as the single knob that controls steepness.

Substitute $\gamma -1$: $PdV+VdP=-(\gamma -1)PdV$

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Step 6 — Tidy into a clean separated form

WHAT. Move everything to one side and group the $dV$ terms.

WHY. We want the pattern $\text{(stuff)}\frac{dV}{V}+\frac{dP}{P}=0$, because that pattern is instantly integrable — each piece is the tiny change of a logarithm.

$PdV+VdP+(\gamma -1)PdV=0$ $\gamma PdV+VdP=0$

Divide by $PV$ (both positive — a gas always has real pressure and volume): $\gamma \frac{dV}{V}+\frac{dP}{P}=0$

• $\gamma \frac{dV}{V}$ :: the fractional change in volume, weighted by $\gamma$.
• $\frac{dP}{P}$ :: the fractional change in pressure.
• $=0$ :: they must cancel — when volume grows, pressure must fall to keep the balance.

WHAT IT LOOKS LIKE. The figure shows a seesaw: pushing $V$ up by a fraction forces $P$ down by $\gamma$ times that fraction. This is why the adiabat drops faster than the isotherm (where the weight would be just $1$, not $\gamma$).

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Step 7 — Integrate: logs turn into the boxed law

WHAT. Integrate both terms.

$\int \gamma \frac{dV}{V}+\int \frac{dP}{P}=\int 0$ $\gamma \ln V+\ln P=\text{const}$

Use log rules ($a\ln x=\ln x^a$ and $\ln x+\ln y=\ln xy$): $\ln \left(P{V}^{\gamma }\right)=\text{const}⟹\boxed{P{V}^{\gamma }=\text{const}}$

WHY. Exponentiating removes the log, leaving the pure product $P{V}^{\gamma }$ frozen along the whole path.

WHAT IT LOOKS LIKE. The figure plots the finished adiabat (orange) against an isotherm (blue) from the same start point — the adiabat plunges below it, exactly the steepness $\gamma$ bought us in Step 6.

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Step 8 — Edge cases: what if $\gamma \to 1$, or the gas just sits still?

Every derivation must survive its own extremes.

• $\gamma \to 1$ (isothermal limit). If $\gamma =1$, the boxed law becomes $P{V}^1=\text{const}$ — the plain hyperbola of the isothermal process. Physically $\gamma =1$ would mean $R=C_V\to \infty$, i.e. infinite heat capacity: the gas soaks up energy without changing temperature. So the adiabat continuously deforms into the isotherm as $\gamma \to 1$. The two curves are cousins, not strangers.
• $dV=0$ (piston jammed). Then $dW=PdV=0$, and $(\ast )$ gives $dT=0$: no work, no temperature change — the adiabatic and isochoric conditions coincide at a single frozen point. No contradiction, just a still frame.
• Compression ($dV<0$). The balance $\gamma \frac{dV}{V}+\frac{dP}{P}=0$ now forces $dP>0$ and $(\ast )$ gives $dT>0$: pushing in heats the gas with nowhere for heat to escape. This is the diesel-ignition case flagged in the parent's sign mistake.

WHAT IT LOOKS LIKE. The figure overlays three adiabats for $\gamma =1$ (=isotherm), $\gamma =1.4$, $\gamma =1.67$ from one point — steeper as $\gamma$ climbs — plus an arrow marking the compression direction where $T$ rises.

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The one-picture summary

The whole story on one canvas: insulated cylinder ($Q=0$) → energy bucket with capped lid ($dU=-dW$) → name the pieces ($n{C}_{V}dT=-PdV$) → evict $T$ with $PV=nRT$ → $\gamma$ appears → separate → integrate the logs → $P{V}^{\gamma }=\text{const}$, steeper than the isotherm.

Recall Feynman retelling (say it to a friend)

Wrap a gas cylinder in a perfect blanket so no heat can get in or out. Now let the gas push the piston. Where does the pushing energy come from? It has nowhere to borrow from — no heat is allowed in — so it spends its own internal jiggling energy, and cools as it expands.

To find the exact curve: energy says "temperature drop equals work done" ($n{C}_{V}dT=-PdV$). But a $P$–$V$ picture doesn't know about temperature, so we use $PV=nRT$ to trade the temperature term for pressure-and-volume terms. When we do, the ratio $\gamma =C_P/C_V$ pops out as the natural weight. The equation tidies into "a $\gamma$-weighted fractional change in volume must cancel a fractional change in pressure." Summing those tiny fractional changes is exactly what a logarithm does, and un-logging leaves the clean, frozen quantity $P{V}^{\gamma }$. Because $\gamma >1$, volume changes are amplified, so the adiabat dives steeper than the plain constant-temperature hyperbola. And if you dial $\gamma$ down to $1$, the adiabat morphs smoothly back into that hyperbola.

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Quick self-check

Fill the reveals before peeking.