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

Below we build the whole alphabet of the topic, one symbol at a time. Each new one is allowed to use only the ones defined before it. If the parent page assumed you already knew it, we stop and draw it here.

1. The gas itself: $P$ , $V$ , $T$

Picture a box with a sliding lid (a piston). Inside are countless tiny gas molecules bouncing around.

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

Read the figure above: the yellow bar is the piston (it sets $V$ ); the blue dots with little arrows are molecules whose speed is $T$ and whose drumming on the walls is $P$ . This one box picture is the stage on which everything else happens — keep it in mind for the rest of the page.

Why the topic needs all three: an adiabatic curve is a path traced on a $P$ – $V$ diagram, and $T$ is what silently changes along it. You cannot state $P V^{γ} =$ const without first owning $P$ and $V$ ; you cannot explain "expansion cools" without $T$ .

2. Counting the gas: $n$ and $R$

Why the topic needs it: the whole derivation is a game of "I have three variables $P, V, T$ but I only want two." $P V = n R T$ is the tool that eliminates one variable at will — used to swap $P$ for $n R T / V$ , and later $V$ for $n R T / P$ .

3. Stored warmth: internal energy $U$

4. The grammar of change: $d$ , $δ$ , $Δ$

Before we can talk about energy flowing, we need the notation for "a tiny amount." These are not new physics — they are the grammar of change — so we meet them here, before first use.

5. Energy in transit: heat $Q$ and work $W$ (with sign conventions)

Energy can enter or leave the gas's bank ( $U$ ) in exactly two ways. Now that we own the $d / δ$ symbols, we can write them.

Read the figure above: the blue curve is one possible $P$ – $V$ path. The yellow strip has height $P$ and width $d V$ , so its area is exactly $δ W = P d V$ — the energy the gas spends widening by that whisker. Adding up all such strips along a path gives the total work; this is why work is "the area under the $P$ – $V$ curve."

6. The First Law — the energy balance in one line

Now we can state the law that ties $Q$ , $U$ , and $W$ together — the beating heart the whole derivation stands on.

Substituting the two pieces we already built ( $d U = n C_{V} d T$ from Section 3, and $δ W = P d V$ from Section 5) into the adiabatic first law $0 = d U + δ W$ gives the parent's starting equation: $0 = n C_{V} d T + P d V .$

7. The heat capacities: $C_{V}$ , $C_{P}$ , and Mayer's bridge

Why the topic needs this: Step 4 of the derivation replaces the ugly $R / C_{V}$ with a clean $γ - 1$ . That replacement is Mayer's relation in disguise.

8. The star of the show: $γ$

Read the figure above: through one shared yellow point we draw two curves. The green isotherm ( $P V =$ const, $T$ held fixed) falls gently; the red adiabat ( $P V^{γ} =$ const) falls steeper because $γ > 1$ . That extra steepness is the visual signature of the gas cooling as it expands — pressure drops faster when no heat refills the energy.

9. The undo-tool: the logarithm $ln$

10. Putting the pieces together — the derivation, step by step

Now every symbol is defined. Watch how the three tools ( $P V = n R T$ , Mayer's relation, $ln$ ) turn the first law into the adiabatic relations. Each step says what we do and why.

Start. Adiabatic first law from Section 6: $0 = n C_{V} d T + P d V .$ Why: $Q = 0$ removes income, so work is paid entirely from internal energy.

Step A — eliminate $P$ . We have three variables $P, V, T$ but want a relation in just $T$ and $V$ . The ideal gas law gives $P = \frac{n R T}{V}$ ; substitute: $0 = n C_{V} d T + \frac{n R T}{V} d V .$ Why: $P V = n R T$ is exactly the tool for trading one variable away; we keep $T, V$ because we're heading for the $T V$ relation.

Step B — separate variables. Divide every term by $n C_{V} T$ so each variable sits with its own differential: $\frac{d T}{T} + \frac{R}{C _{V}} \frac{d V}{V} = 0.$ Why: to integrate, each side must contain only one variable; this "separation" makes both terms integrable on their own.

Step C — inject $γ$ . Mayer's relation gives $\frac{R}{C _{V}} = \frac{C _{P} - C _{V}}{C _{V}} = γ - 1$ : $\frac{d T}{T} + (γ - 1) \frac{d V}{V} = 0.$ Why: this is the moment $γ$ — the gas's fingerprint — enters the maths.

Step D — integrate with $ln$ . Each $\frac{d x}{x}$ integrates to $ln x$ : $ln T + (γ - 1) ln V = const .$ Why: the logarithm is the natural antiderivative of $1/ x$ ; a sum of logs equal to a constant means a product is constant.

Step E — read off the result. $T V^{γ - 1} = const, then using T = \frac{P V}{n R} : P V^{γ} = const .$ Why: multiplying $T$ by $V^{γ - 1}$ adds the $V$ -powers, $V^{1 + (γ - 1)} = V^{γ}$ , and the constant $n R$ absorbs into the new constant.

This is the full skeleton the parent page fleshes out — every step now uses only symbols we defined from zero.

How it all feeds the topic

Read it top-down: the five gas symbols build the ideal gas law; $T$ and $C_{V}$ build $d U$ ; $Q = 0$ plus the work term build the first law; the ideal gas law eliminates $P$ ; Mayer's relation delivers $γ - 1$ ; and $ln$ finishes the job.

Equipment checklist

Test yourself — cover the right side of each line before revealing.

What does

P

physically measure?

Force per unit area from molecular collisions on the walls (in

Pa

atm

What does

T

measure, and which unit must it be in here?

Average molecular kinetic energy; always in kelvin (

K

State the ideal gas law and its job in the derivation.

P V = n R T

; it lets you eliminate one of

P, V, T

For an ideal gas,

U

depends on what alone?

On temperature

T

only.

Why can we write

d U = n C_{V} d T

even when volume changes?

Because

U = U (T)

for an ideal gas, so the

V

-subscript on

C_{V}

is only about how it was measured.

State the First Law and name each term's sign meaning.

δ Q = d U + δ W

;

Q

= heat added to the gas,

W

= work done by the gas.

What does "adiabatic" mean in one symbol, and what does the First Law become?

Q = 0

; the First Law becomes

0 = d U + δ W

Why write

δ Q

and

δ W

instead of

d Q

d W

Heat and work aren't stored quantities; they exist only in transit.

Write the small-expansion work and its sign for expansion.

δ W = P d V

; positive when the gas expands (

d V > 0

What is

C_{V}

C_{P}

Heat per mole per kelvin at constant volume vs constant pressure;

C_{P} > C_{V}

State Mayer's relation.

C_{P} - C_{V} = R

Define

γ

and give its range.

γ = C_{P} / C_{V}

, and

γ > 1

What does

R / C_{V}

simplify to?

γ - 1

What does

ln

do that we exploit when integrating?

Turns products into sums (and

\int d V / V = ln V

), so a sum of logs = const means a product = const.

Connections

Parent: the full derivation
First Law of Thermodynamics — supplies the energy balance $δ Q = d U + δ W$ .
Isothermal Process — the contrasting case where $T$ (and thus $U$ ) is held fixed.
Mayer's Relation — where $C_{P} - C_{V} = R$ comes from.
Degrees of Freedom and $\gamma$ — why $γ$ takes the values $5/3$ , $7/5$ , ...
Speed of Sound in Gases — uses the adiabatic bulk modulus $γ P$ .
Work Done in Thermodynamic Processes — the meaning of $δ W = P d V$ .