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

Before anything: two housekeeping notes, then the toolbox.

Definition The gas law and reversibility — the ground rules for this whole page

Every example here assumes an ideal gas, whose three state variables always obey the ideal gas law $P V = n R T,$ where $P$ = pressure, $V$ = volume, $T$ = absolute temperature (in kelvin), $n$ = number of moles, $R = 8.314 J mol^{- 1} K^{- 1}$ = the universal gas constant. We lean on this law constantly (e.g. to swap $P \leftrightarrow T$ ).

Every example also assumes the process is reversible (quasi-static): it happens slowly enough that $P, V, T$ have a single well-defined value at every instant, so the gas is always "on the curve." This is exactly the condition under which the parent note's relations $P V^{γ} = const$ , $T V^{γ - 1} = const$ , $T^{γ} P^{1 - γ} = const$ were derived — a fast, turbulent (irreversible) $Q = 0$ process still has $Δ U = - W_{by}$ , but those three curve-equations no longer describe it.

Definition Sign convention for work, and the

$δ$ vs $d$ notation We use the physicist's "work done BY the gas" convention, written $W_{by}$ . The first law reads $δ Q = d U + δ W_{by}, δ W_{by} = P d V .$ Why $δ Q, δ W$ but $d U$ ? Internal energy $U$ is a state function — the gas has a definite amount of it at each $(P, V, T)$ point, so a change $d U$ is exact: it depends only on start and end states, and $\int d U = U_{2} - U_{1}$ no matter the path. Heat $Q$ and work $W$ are not stored quantities — they are amounts that flow during a path. Two different paths between the same two states can exchange different heat and work. To flag "this is a path-dependent sliver, not the change of a stored quantity," we write $δ$ (an inexact differential) instead of $d$ . That is the entire meaning of the symbol: $d$ = "change of something the gas owns," $δ$ = "a bit that crossed the boundary along the way."

With this: when the gas expands ( $d V > 0$ ) it does positive work. For an ideal gas the internal energy depends only on temperature, giving $Δ U = n C_{V} (T_{2} - T_{1})$ for any process (the " $U = U (T)$ only" fact from the parent note — the $V$ -subscript on $C_{V}$ just says how it was measured, not when it applies). In an adiabatic process $Q = 0$ , so $0 = Δ U + W_{by}$ , i.e. $W_{by} = - Δ U = n C_{V} (T_{1} - T_{2})$ .

The scenario matrix

Every adiabatic problem lives in one of these cells. Each worked example below is tagged with the cell(s) it covers.

#	Cell class	What makes it distinct	Example
A	Compression, find $T$	$V$ shrinks → $T$ must rise	Ex 1
B	Expansion, find $P$	$V$ grows → $P$ must fall	Ex 2
C	Given $T$ & $P$ , find the other $T$ / $P$	uses the $T P$ relation	Ex 3
D	Work / energy	no $V$ -ratio given, use $Δ U$	Ex 4
E	Degenerate $γ \to 1$	adiabat collapses onto isotherm	Ex 5
F	Limiting behaviour (huge compression $V \to 0$ )	how fast does $T$ or $P$ blow up?	Ex 6
G	Real-world word problem	rising air parcel cooling (clouds)	Ex 7
H	Exam twist — mixed process	adiabat then isotherm, or slope comparison	Ex 8
I	Inverse: find a $V$ -ratio	given $T$ (or $P$ ) ratio, solve for $V_{2} / V_{1}$	Ex 9

Worked examples

Worked example Ex 1 — Cell A: compression, find temperature

Air ( $γ = 1.4$ ) starts at $T_{1} = 300$ K and is adiabatically compressed to one-third its volume, $V_{2} = V_{1} /3$ . Find $T_{2}$ .

Forecast: volume shrank → gas should get hotter, so expect $T_{2} > 300$ K. Guess a number before reading on.

Pick the tool. We know a volume ratio and want a temperature. That is exactly the job of $T V^{γ - 1} = const$ . Why this step? No pressure is mentioned, so avoid any relation containing $P$ — that would drag in an unknown.
Write the two states equal. $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \Rightarrow T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} .$ Why this step? "const" means the value is the same at state 1 and state 2 — so set the two products equal and solve.
Plug in. $\frac{V _{1}}{V _{2}} = 3$ , $γ - 1 = 0.4$ : $T_{2} = 300 \times 3^{0.4} = 300 \times e^{0.4 l n 3} = 300 \times e^{0.4394} = 300 \times 1.5518 \approx 466 K .$ Why this step? We rewrite $3^{0.4}$ as $e^{0.4 l n 3}$ because a fractional power is only computable through the exponential/log — this is the concrete arithmetic behind the symbol.

Verify: $466 > 300$ ✔ — compression heated it, matching the forecast. Units: $K \times (dimensionless ratio) = K$ ✔.

Worked example Ex 2 — Cell B: expansion, find pressure

A monatomic gas ( $γ = 5/3$ ) at $P_{1} = 8$ atm expands adiabatically to quadruple its volume, $V_{2} = 4 V_{1}$ . Find $P_{2}$ .

Forecast: volume grew fourfold → pressure must drop a lot. Expect $P_{2} ≪ 8$ atm.

Pick the tool. Given $P$ and a volume ratio → use $P V^{γ} = const$ . Why this step? We want pressure and only have volumes; the $P V$ form contains exactly those two.
Equate states. $P_{1} V_{1}^{γ} = P_{2} V_{2}^{γ} \Rightarrow P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} .$ Why this step? The quantity $P V^{γ}$ is the same number at states 1 and 2, so we equate the two and divide by $V_{2}^{γ}$ to isolate $P_{2}$ .
Plug in. $\frac{V _{1}}{V _{2}} = \frac{1}{4}$ , $γ = \frac{5}{3} = 1.6667$ : $P_{2} = 8 \times (0.25)^{1.6667} = 8 \times e^{1.6667 l n 0.25} = 8 \times e^{- 2.3105} = 8 \times 0.09921 \approx 0.794 atm .$ Why this step? The ratio $V_{1} / V_{2} = 1/4 < 1$ raised to a positive power stays below 1, which is what guarantees $P_{2} < P_{1}$ — a built-in check on the direction.

Verify: $0.79 < 8$ ✔ — expansion dropped the pressure sharply. Note it fell faster than the $1/4$ an isotherm would give ( $8/4 = 2$ atm), because the extra $γ$ -power steepens the drop — consistent with the Isothermal Process contrast in the parent note.

Worked example Ex 3 — Cell C: given a pressure ratio, find temperature

A diatomic gas ( $γ = 1.4$ ) at $T_{1} = 350$ K is compressed adiabatically until its pressure doubles, $P_{2} = 2 P_{1}$ . Find $T_{2}$ .

Forecast: compressed (pressure up) → temperature up.

Pick the tool. We have a pressure ratio and want temperature — no volume in sight. Use $T^{γ} P^{1 - γ} = const$ . Why this step? Choosing $T V^{γ - 1}$ would force us to first find the volume ratio; the $T P$ relation skips that.
Rearrange for the $T$ ratio. Writing the relation at both states: $T_{1}^{γ} P_{1}^{1 - γ} = T_{2}^{γ} P_{2}^{1 - γ} \Rightarrow (\frac{T _{2}}{T _{1}})^{γ} = (\frac{P _{2}}{P _{1}})^{γ - 1} \Rightarrow \frac{T _{2}}{T _{1}} = (\frac{P _{2}}{P _{1}})^{\frac{γ - 1}{γ}} .$ Why this step? Isolate the ratio you want by moving $P$ to the other side; the exponent becomes $(γ - 1) / γ$ .
Plug in. $\frac{γ - 1}{γ} = \frac{0.4}{1.4} = 0.2857$ : $T_{2} = 350 \times 2^{0.2857} = 350 \times e^{0.2857 l n 2} = 350 \times e^{0.1980} = 350 \times 1.2189 \approx 427 K .$ Why this step? The exponent $(γ - 1) / γ = 0.2857$ is positive and less than 1, so doubling the pressure raises $T$ only modestly (by $\approx 22%$ ) — the physical expectation for a gentle compression.

Verify: $427 > 350$ ✔ — pressure rose, so temperature rose. Cross-check via the ideal-gas law would need volumes, but the sign already confirms physics.

Worked example Ex 4 — Cell D: work done during expansion

Two moles ( $n = 2$ ) of a diatomic gas ( $C_{V} = \frac{5}{2} R$ ) expand adiabatically, cooling from $T_{1} = 600$ K to $T_{2} = 400$ K. Find the work $W_{by}$ done by the gas.

Forecast: the gas expanded (did work) → $W_{by} > 0$ . That work must equal the lost internal energy.

Use energy conservation, not $P d V$ . For an ideal gas $Δ U = n C_{V} (T_{2} - T_{1})$ (internal energy depends on $T$ alone). Since $Q = 0$ , the first law $0 = Δ U + W_{by}$ gives $W_{by} = - Δ U = - n C_{V} (T_{2} - T_{1}) = n C_{V} (T_{1} - T_{2}) .$ Why this step? We are not given $P$ or $V$ , only temperatures — integrating $\int P d V$ would be harder; energy bookkeeping is direct. The minus sign flips $(T_{2} - T_{1})$ into $(T_{1} - T_{2})$ , which is positive here since $T_{1} > T_{2}$ .
Plug in. $C_{V} = \frac{5}{2} R = \frac{5}{2} (8.314) = 20.785 J mol^{- 1} K^{- 1}$ : $W_{by} = 2 \times 20.785 \times (600 - 400) = 2 \times 20.785 \times 200 = 8314 J \approx 8.31 kJ .$ Why this step? We substitute $C_{V} = \frac{5}{2} R$ (diatomic value from Degrees of Freedom and $\gamma$) so the whole answer is expressed in known SI constants, giving joules.

Verify: positive ✔ — expansion does positive work. Units: $mol \cdot \frac{J}{mol K} \cdot K = J$ ✔. See Work Done in Thermodynamic Processes for the $\int P d V$ route that gives the same number.

Worked example Ex 5 — Cell E: the degenerate case

$γ \to 1$ What does the adiabatic relation become if $γ = 1$ ? Physically, why is this the boundary case?

Forecast: $γ = 1$ would mean $C_{P} = C_{V}$ , i.e. Mayer's relation $C_{P} - C_{V} = R$ demands $R = 0$ — impossible for a real gas, but instructive as a limit.

Put $γ = 1$ into $P V^{γ}$ . $P V^{1} = const \Rightarrow P V = const .$ Why this step? Just substitute the value; the exponent collapses to 1.
Recognise what $P V = const$ is. By the ideal gas law $P V = n R T$ , so $P V = const$ means $T = const$ — an isothermal process (see Isothermal Process). Why this step? We interpret the algebra physically to see which real process the degenerate case mimics.
Check the $T V$ form too. $T V^{γ - 1} = T V^{0} = T = const$ ✔ — consistent, $T$ is fixed. Why this step? A good limit must be self-consistent across all three tools; if the $T V$ form disagreed with the $P V$ form we would suspect an error. Here both say " $T$ constant," confirming the collapse onto the isotherm.

Verify: As $γ \to 1^{+}$ the adiabat's slope $- γ P / V \to - P / V$ , which is exactly the isotherm's slope. So the adiabat continuously merges into the isotherm as $γ \to 1$ — the boundary is real, only that no gas actually sits there.

Worked example Ex 6 — Cell F: limiting behaviour, extreme compression

A monatomic gas ( $γ = 5/3$ ) is compressed adiabatically so its volume shrinks by a factor of $1000$ ( $V_{2} = V_{1} /1000$ ). Starting at $T_{1} = 300$ K, find $T_{2}$ and comment on how fast $T$ diverges as $V \to 0$ .

Forecast: volume down a thousandfold → temperature soars. Because the exponent $γ - 1 = \frac{2}{3} < 1$ , it soars but slower than the volume factor itself.

Use the $T V$ relation. $T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 300 \times 100 0^{2/3}$ . Why this step? Volume ratio → temperature, so $T V^{γ - 1}$ .
Evaluate. $100 0^{2/3} = (1 0^{3})^{2/3} = 1 0^{2} = 100$ : $T_{2} = 300 \times 100 = 30000 K .$ Why this step? We rewrite $1000$ as $1 0^{3}$ so the fractional power $()^{2/3}$ multiplies the exponents cleanly ( $3 \times \frac{2}{3} = 2$ ), turning an awkward root into an exact integer power — no calculator needed.
Read off the divergence law. In general $T \propto V^{- (γ - 1)}$ , so as $V \to 0$ , $T \to \infty$ like a power law with exponent $γ - 1$ . Pressure diverges even faster: $P \propto V^{- γ}$ . Why this step? Reading the functional form (not just the number) tells us the qualitative behaviour of the whole family of compressions: the $- (γ - 1)$ exponent is why $T$ blows up more slowly than $P$ , whose exponent is the full $- γ$ .

Verify: $30000$ K is enormous (hotter than the Sun's surface) — this is exactly why diesel engines ignite fuel by adiabatic compression alone. Sign ✔ (heating), magnitude physically sensible for a $1000 \times$ squeeze.

The figure below plots this heating law and contrasts it with the (wrong) intuition that $T$ should grow linearly with the compression ratio — the amber dot marks this very example.

Figure s01 — Adiabatic heating is a power law. Temperature $T$ versus compression ratio $V_{1} / V_{2}$ for a monatomic gas ( $γ = 5/3$ ). The cyan curve is the true law $T = T_{1} (V_{1} / V_{2})^{γ - 1}$ ; the dashed white line is the naive "linear" guess $T \propto V_{1} / V_{2}$ ; the amber dot marks Ex 6 at $(1000, 30000 K)$ .

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

Look at the cyan curve: it rises steeply but stays well below the dashed white "linear guess," because the exponent $γ - 1 = \frac{2}{3}$ is less than 1. The amber dot at $(1000, 30000)$ is exactly Ex 6.

Worked example Ex 7 — Cell G: real-world word problem (cloud formation)

A parcel of dry air ( $γ = 1.4$ ) at the ground has $T_{1} = 300$ K and pressure $P_{1} = 1.00$ atm. It rises until the surrounding pressure falls to $P_{2} = 0.70$ atm. Assuming it rises fast enough to be adiabatic, find its temperature $T_{2}$ .

Forecast: rising air expands (lower outside pressure) → it cools. This cooling is what makes water vapour condense into clouds.

Pick the tool. Given a pressure ratio, want temperature → $T^{γ} P^{1 - γ} = const$ , rearranged as in Ex 3: $\frac{T _{2}}{T _{1}} = (\frac{P _{2}}{P _{1}})^{\frac{γ - 1}{γ}} .$
Plug in. $\frac{P _{2}}{P _{1}} = 0.70$ , exponent $= \frac{0.4}{1.4} = 0.2857$ : $T_{2} = 300 \times 0.7 0^{0.2857} = 300 \times e^{0.2857 l n 0.70} = 300 \times e^{- 0.1019} = 300 \times 0.9031 \approx 271 K .$ Why this step? Because $P_{2} / P_{1} = 0.70 < 1$ and the exponent is positive, the factor $0.7 0^{0.2857} < 1$ , guaranteeing $T_{2} < T_{1}$ — the algebra automatically encodes "expansion cools," so the sign is a built-in physics check.

Verify: $271 < 300$ ✔ — the rising parcel cooled by about $29$ K, dropping below the typical dew point, so vapour condenses → cloud forms. This is the everyday face of $Q = 0$ .

Worked example Ex 8 — Cell H: exam twist, adiabat

then isotherm A gas ( $γ = 1.4$ ) starts at state 1: $P_{1} = 6$ atm, $V_{1} = 1$ L. Step I: adiabatic expansion to $V_{2} = 2$ L. Step II: isothermal compression back to $V_{3} = 1$ L. Find the final pressure $P_{3}$ .

Forecast: Step I drops pressure below the naive $6/2 = 3$ (adiabat is steeper). Step II compresses at fixed $T$ , raising it again — but does it return to 6 atm? Guess.

Step I — adiabatic, use $P V^{γ}$ . $P_{2} = P_{1} (\frac{V _{1}}{V _{2}})^{γ} = 6 \times (0.5)^{1.4} = 6 \times e^{1.4 l n 0.5} = 6 \times e^{- 0.9704} = 6 \times 0.3789 \approx 2.273 atm .$ Why this step? No heat during Step I → the $γ$ -exponent applies (Step I is reversible adiabatic, so $P V^{γ} = const$ holds).
Step II — isothermal, use $P V = const$ (i.e. exponent 1). $P_{3} = P_{2} (\frac{V _{2}}{V _{3}})^{1} = 2.273 \times \frac{2}{1} = 4.546 atm .$ Why this step? Step II holds $T$ fixed, so use the isotherm law (Isothermal Process), not $γ$ .
Compare to the start. $P_{3} \approx 4.55 atm < P_{1} = 6 atm$ , even though we returned to the same volume $V_{3} = V_{1}$ . Why this step? The two paths used different exponents ( $γ$ vs $1$ ), so the gas ends colder than it started — the cycle is not closed in temperature.

Verify: $P_{3} < P_{1}$ ✔ makes sense: Step I cooled the gas, and the isothermal return can't undo that cooling, so at equal volume the pressure is lower. The trap is applying $γ$ to Step II — that would wrongly give back $6$ atm.

Worked example Ex 9 — Cell I: inverse problem, find the volume ratio

A diatomic gas ( $γ = 1.4$ ) is compressed adiabatically and its temperature rises from $T_{1} = 300$ K to $T_{2} = 450$ K. By what factor was its volume reduced — i.e. find $V_{1} / V_{2}$ ?

Forecast: the gas got hotter → it was squeezed, so $V_{1} / V_{2} > 1$ (final volume smaller). This is the reverse of Ex 1: we know the temperatures and hunt for the volume ratio.

Pick the tool. We have a temperature ratio and want a volume ratio → still $T V^{γ - 1} = const$ , just solved the other way. Why this step? The same relation links $T$ and $V$ ; which quantity is "unknown" only changes what we isolate, not the tool.
Solve for the volume ratio. $T_{1} V_{1}^{γ - 1} = T_{2} V_{2}^{γ - 1} \Rightarrow (\frac{V _{1}}{V _{2}})^{γ - 1} = \frac{T _{2}}{T _{1}} \Rightarrow \frac{V _{1}}{V _{2}} = (\frac{T _{2}}{T _{1}})^{\frac{1}{γ - 1}} .$ Why this step? To free $V_{1} / V_{2}$ we raise both sides to the power $1/ (γ - 1)$ — the inverse exponent, which is the whole trick of reverse problems.
Plug in. $\frac{T _{2}}{T _{1}} = \frac{450}{300} = 1.5$ , $\frac{1}{γ - 1} = \frac{1}{0.4} = 2.5$ : $\frac{V _{1}}{V _{2}} = 1. 5^{2.5} = e^{2.5 l n 1.5} = e^{1.0136} = 2.755.$ Why this step? The exponent $1/ (γ - 1) = 2.5 > 1$ amplifies the modest temperature rise into a larger volume ratio — the mirror image of Ex 1, where the small exponent $γ - 1$ shrank a big volume ratio into a modest temperature rise.

Verify: $V_{1} / V_{2} \approx 2.76 > 1$ ✔ — the gas was compressed to about $1/2.76$ of its volume, consistent with heating. Sanity cross-check: feeding $V_{1} / V_{2} = 2.755$ back into Ex 1's formula gives $T_{2} = 300 \times 2.75 5^{0.4} = 450$ K ✔.

Which cell is this? (self-test)

Recall Match each question opener to its cell and tool

"Compressed to half, find $T$ " — which relation? ::: Cell A → $T V^{γ - 1} = const$ . "Expands, pressure doubles, find $T$ " — which relation? ::: Cell C → $T^{γ} P^{1 - γ} = const$ . "Cools from $T_{1}$ to $T_{2}$ , find work" — which relation? ::: Cell D → $W_{by} = n C_{V} (T_{1} - T_{2})$ . " $γ \to 1$ , what curve?" ::: Cell E → the adiabat becomes the isotherm $P V = const$ . "Volume shrinks $1000 \times$ , how does $T$ grow?" ::: Cell F → $T \propto V^{- (γ - 1)}$ , a power law. "Temperature ratio known, find $V$ -ratio" ::: Cell I → invert $T V^{γ - 1}$ : $V_{1} / V_{2} = (T_{2} / T_{1})^{1/ (γ - 1)}$ . "Why write $δ Q$ but $d U$ ?" ::: $U$ is a stored state function (exact $d$ ); $Q, W$ are path-dependent flows (inexact $δ$ ).

Connections

1.7.16 Adiabatic relations — PV^γ = const, TV^(γ−1) = const (derivation) (Hinglish) — the parent derivation these examples exercise.
First Law of Thermodynamics — powers the work example (Ex 4) and the $δ Q = d U + δ W$ sign convention.
Isothermal Process — the degenerate limit (Ex 5) and the second leg of Ex 8.
Mayer's Relation — why $γ > 1$ , so compression always heats.
Degrees of Freedom and $\gamma$ — where $γ = 5/3$ and $7/5$ come from.
Speed of Sound in Gases — the same $γ$ governs how sound compresses air adiabatically.
Work Done in Thermodynamic Processes — alternative $\int P d V$ route for Ex 4.