Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation): Visual walkthrough

Before symbol one, let us name every character in the play.

We also need two "small change" symbols. When I write $d T$ , read it as "a tiny change in $T$ ." The letter $d$ is not a number you multiply — it is a label meaning an infinitesimal nudge. We use tiny nudges because pressure and volume change together and continuously during the process, so we track them one whisker at a time and add up the whiskers at the end (that adding-up is what the integral sign $\int$ will do in Step 6).

Step 1 — Seal the box: $Q = 0$

WHAT. We put the gas in a perfectly insulated cylinder with a piston. No heat leaks through the walls. In symbols the whole process obeys $heat through walls Q = 0.$

WHY. This single condition is the entire difference between adiabatic and everything else. In an Isothermal Process heat pours in to keep $T$ fixed; here we forbid that flow. So whatever energy the gas needs to push its piston, it must find inside itself.

PICTURE. Below, the thick chalk wall carries hatching — that hatching is the universal engineering symbol for "insulated, no heat crosses." The blue wiggly arrows trying to enter are all stopped.

Figure — Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation)

Step 2 — The energy ledger: First Law with no heat

WHAT. Energy is never lost, only moved. The First Law of Thermodynamics is the accountant's equation: $heat in δ Q = change in stored energy d U + work pushed out δ W .$ Setting $δ Q = 0$ (our Step 1) gives $0 = d U + δ W .$

WHY. This is a bookkeeping identity: heat coming in either stays inside (raises $U$ ) or leaves as work $W$ . With no heat coming in, the two terms on the right must cancel — if the gas spends work, its stored energy must drop by exactly that much.

PICTURE. Think of $U$ as money in a piggy bank and $δ W$ as money spent. With no allowance ( $δ Q = 0$ ), spending drains the bank one-to-one.

Step 3 — Name the two terms with gas quantities

WHAT. Replace the abstract $d U$ and $δ W$ with things we can measure. $d U = n C_{V} d T and δ W = P d V .$ Plug both into $0 = d U + δ W$ : $0 = energy the molecules gain/lose n C_{V} d T + work of pushing the piston P d V .$

Here $n$ is the number of moles (how much gas), and $C_{V}$ is the heat capacity at constant volume — how many joules it takes to warm one mole by one degree while volume is held fixed.

WHY each piece?

$d U = n C_{V} d T$ : for an ideal gas the stored energy depends only on temperature. So any change in $U$ is $n C_{V}$ times the temperature change — even while the volume is changing. (This is the subtle point the parent note steel-mans: the " $V$ " in $C_{V}$ tells you how it was measured, not when the formula is allowed.)
$δ W = P d V$ : to push the piston out by a sliver of volume $d V$ , the gas presses with pressure $P$ ; force $\times$ distance works out to $P d V$ . See Work Done in Thermodynamic Processes for that geometry.

PICTURE. The shaded sliver under the piston has area $P d V$ — that thin strip is the work. The little thermometer shows $d T$ , the temperature nudge that goes with it.

Step 4 — Kick out $P$ using the ideal-gas law

WHAT. We have three variables ( $P$ , $V$ , $T$ ) but only want two. The ideal-gas law $P V = n R T$ lets us trade one for the others. Solve it for $P$ : $P = \frac{n R T}{V},$ and substitute into $0 = n C_{V} d T + P d V$ : $0 = n C_{V} d T + this is P \frac{n R T}{V} d V .$

Here $R$ is the universal gas constant (a fixed number that converts moles-and-degrees into energy).

WHY. Three unknowns tangled in one equation cannot be integrated. By eliminating $P$ we are choosing to end up with a relation between $T$ and $V$ first. We could equally eliminate $T$ to head straight for $P V^{γ}$ — same destination, different road.

PICTURE. The ideal-gas law is the bridge connecting the three quantities. We walk across it to remove $P$ from the equation.

Step 5 — Separate the variables and let $γ$ appear

WHAT. Divide every term by $n C_{V} T$ so each variable sits on its own: $\frac{d T}{T} + \frac{R}{C _{V}} \frac{d V}{V} = 0.$ Now the only messy coefficient is $R / C_{V}$ . Using Mayer's Relation $C_{P} - C_{V} = R$ : $\frac{R}{C _{V}} = \frac{C _{P} - C _{V}}{C _{V}} = = γ \frac{C _{P}}{C _{V}} - 1 = γ - 1.$ So the equation becomes $\frac{d T}{T} + (γ - 1) \frac{d V}{V} = 0.$

The symbol $γ = C_{P} / C_{V}$ is the ratio of specific heats, always bigger than $1$ . It secretly encodes how many ways the molecules can store energy — see Degrees of Freedom and $\gamma$.

WHY. "Separating variables" means getting all the $T$ 's on one side of a term and all the $V$ 's on another, so each term can be integrated by itself. And the reason $γ$ shows up at all is Mayer's substitution — that is the exact moment the gas's inner nature ( $γ$ ) enters the derivation.

PICTURE. Watch the tangled coefficient $R / C_{V}$ collapse into the clean $γ - 1$ .

Step 6 — Add up all the whiskers: integrate

WHAT. Each term $\frac{d T}{T}$ and $\frac{d V}{V}$ is a "fractional nudge." Summing fractional nudges gives a logarithm ( $ln$ ), because $ln$ is precisely the running total of "how many percent has this grown so far." Integrating: $ln T + (γ - 1) ln V = const .$

WHY the logarithm and not something else? We need the tool that answers: "if a quantity changes by the fraction $d x / x$ at each step, what is its accumulated change?" The unique function whose tiny change is $d x / x$ is $ln x$ . No other tool fits that shape — that is why $ln$ , not, say, a square root.

Fold it into a product. Use the log rule $ln a + ln b = ln (ab)$ , and $c ln V = ln V^{c}$ : $ln (T V^{γ - 1}) = const ⟹ T V^{γ - 1} = const .$ Taking $ln$ of both sides was the trick; a sum of logs equal to a constant means the product inside is constant.

PICTURE. The area under the curve $1/ T$ from $T_{1}$ to $T_{2}$ is $ln T_{2} - ln T_{1}$ . The figure shows those two shaded areas balancing so their weighted sum stays fixed.

Step 7 — Convert to $P V^{γ}$ and $T^{γ} P^{1 - γ}$

WHAT. We have $T V^{γ - 1} = const$ . To get the pressure form, replace $T$ using $T = \frac{P V}{n R}$ : $\frac{P V}{n R} V^{γ - 1} = const ⟹ P V^{1 + (γ - 1)} = P V^{γ} = const .$ The $V$ powers add: $V^{1} \cdot V^{γ - 1} = V^{γ}$ . The clutter $n R$ is a constant, so it folds into the new constant on the right.

For the third relation, start from $P V^{γ} = const$ and replace $V = \frac{n R T}{P}$ : $P (\frac{n R T}{P})^{γ} = const ⟹ P^{1 - γ} T^{γ} = const .$

WHY. All three relations describe the same curve; they are just re-expressed to match whichever two variables your problem hands you. (Given $V$ -ratio and want $T$ ? Use $T V^{γ - 1}$ . Given $P$ ? Use $P V^{γ}$ .)

PICTURE. One curve, three labels — a triangle of $P$ , $V$ , $T$ with the ideal-gas law converting between any two faces.

Step 8 — Edge cases: don't get caught out

WHAT & WHY. A good picture must survive the corners.

$γ \to 1$ (imaginary "heat-hungry" gas): then $γ - 1 \to 0$ , so $T V^{γ - 1} = T V^{0} = T = const$ and $P V^{γ} \to P V = const$ . The adiabat degenerates into the isotherm. Sanity: $γ = 1$ would mean $C_{P} = C_{V}$ , i.e. adding heat at constant pressure costs nothing extra — a gas that never changes temperature. So the isotherm is the $γ \to 1$ limit. Real gases have $γ > 1$ .
Compression ( $d V < 0$ ): the work term $P d V$ is now negative, so $d U > 0$ : internal energy rises, temperature rises. Compression heats — bicycle pump. Same equation, opposite sign of $d V$ .
Expansion ( $d V > 0$ ): $P d V > 0$ , so $d U < 0$ : expansion cools — rising cloud air.
$V \to 0$ or $T \to 0$ : the logs $ln V$ , $ln T$ blow up to $- \infty$ ; physically the ideal-gas model breaks (real gases liquefy). The relation is a high-temperature, dilute-gas statement — respect its domain.

PICTURE. Two adiabats (steeper, blue) bracketing an isotherm (flatter, pink) on the same $P$ – $V$ axes, with arrows showing "compress $\Rightarrow$ hotter, expand $\Rightarrow$ cooler." The isotherm is the $γ = 1$ degenerate adiabat.

The one-picture summary

Everything above is one chain: seal the box → energy ledger → name the terms → drop $P$ → let $γ$ appear → integrate → rename. The final figure compresses the whole flow, from $Q = 0$ to the three boxed relations.

Recall Feynman retelling — the whole walkthrough in plain words

Picture a thermos with a sliding lid full of air. We seal it so no warmth can sneak in or out (Step 1: $Q = 0$ ). Now the accountant's rule says: any energy the air spends pushing the lid must be stolen from its own inner warmth (Steps 2–3), because there's no heat delivery to refill the piggy bank. We write "spending" as $P d V$ and "warmth lost" as $n C_{V} d T$ , then use the gas law to get rid of the pressure symbol (Step 4). Dividing tidily, the messy factor $R / C_{V}$ turns into the gas's personal number $γ - 1$ (Step 5) — that's where the gas's character enters. Adding up all the tiny fractional changes gives logarithms, and a sum of logs staying constant means a product stays constant: $T V^{γ - 1} = const$ (Step 6). Swap temperature for pressure and you get $P V^{γ} = const$ (Step 7). Squeeze the thermos and the trapped warmth concentrates — it heats; let it swell and the warmth spreads thin — it cools (Step 8). That's the entire story, and $P V^{γ} = const$ is just its rulebook.

Connections

First Law of Thermodynamics — the ledger $δ Q = d U + δ W$ of Step 2.
Isothermal Process — the $γ \to 1$ degenerate case in Step 8.
Mayer's Relation — turns $R / C_{V}$ into $γ - 1$ in Step 5.
Degrees of Freedom and $\gamma$ — what $γ$ secretly counts.
Work Done in Thermodynamic Processes — the $P d V$ sliver of Step 3.
Speed of Sound in Gases — where the adiabatic slope $γ P$ reappears.