Gas laws — Boyle (PV const at T), Charles (V - T const at P), Gay-Lussac (P - T const at V): Visual walkthrough

Step 1 — What is a "wall push"? (pressure from one bounce)

WHAT. Imagine one tiny ball inside a box, flying to the right, hitting the right wall, and bouncing straight back to the left. That single bounce gives the wall a tiny shove.

WHY. Before we can say anything about pressure, we must know what a push even is. Pressure is nothing mystical — it is the total shove per second, spread over the wall's area. So we begin with exactly one shove.

PICTURE. Look at figure s01. The purple ball comes in with speed $v$ (the coral arrow) and leaves with speed $v$ the other way (an elastic bounce, by our rules). The wall feels a kick — the mint arrow.

Figure — Gas laws — Boyle (PV const at T), Charles (V - T const at P), Gay-Lussac (P - T const at V)

We have not written pressure as a formula yet — we only have "a kick, stronger when the ball is faster or heavier." Hold that thought.

Step 2 — What is pressure? (many balls, counted per area)

WHAT. Now fill the box with $N$ balls (that capital $N$ means the actual head-count of balls — one, two, a billion). Give the wall an area $A$ (the size of the flat face, in square metres $m^{2}$ ). Pressure $P$ is: (how many kicks per second) × (how hard each kick) ÷ (the wall area $A$ ).

WHY. A single kick is not something you can measure with a gauge. A gauge feels a steady push — the smeared-out average of millions of kicks. So we must go from "one kick" to "kicks per second per area." That is the definition of pressure.

PICTURE. Figure s02: a wall of area $A$ being peppered by many balls. Two things raise the push: more balls arriving (left panel) and faster balls hitting harder (right panel).

Intuition Why does

$v^{2}$ (speed squared) appear, not just $v$ ? A ball's speed enters the pressure twice over, for two separate reasons:

Harder hit. A faster ball carries more momentum $m v$ , so each kick is stronger — one factor of $v$ .
More frequent hits. A faster ball also races back to the wall sooner — it hits more often per second — a second factor of $v$ . Multiply the two together: pressure $\propto m v \times v = m v^{2}$ . That is where the square comes from. One more subtlety: a ball moves in three directions (left–right, up–down, front–back), but only the left–right part pushes on the right wall. On average the motion splits evenly three ways, so only one third of the total $m v^{2}$ pushes any one wall. Carrying that $\frac{1}{3}$ through, kinetic theory gives $P V = \frac{1}{3} N m ⟨ v^{2} ⟩$ , where $⟨ v^{2} ⟩$ means the average of $v^{2}$ over all balls. Full grind in Kinetic Theory of Gases.

So pressure goes up two ways: crowd the balls (density) or speed them up (temperature). Keep these two levers separate — they become Boyle and Charles.

Step 3 — What is temperature? (speed, made into a number)

WHAT. Temperature $T$ , measured in Kelvin, is a direct scorecard for the average kinetic energy — the average of "how much motion" — of the balls.

WHY the straight-line link " $T$ = motion"? We keep saying "faster," so we need a dial that means "faster." Here is the reason it comes out linear (a plain proportion, not a curve). Kinetic energy of one ball is $\frac{1}{2} m v^{2}$ — half its mass $m$ (defined in Step 1) times its speed squared; that is the standard "amount of motion" formula. Now think of temperature as the shared bank balance of motion: when two gases touch, motion flows from the busy one to the lazy one until the average energy per ball is equal — that settled-average is exactly what a thermometer reports. Since a thermometer is defined to read this average energy on an evenly-spaced scale (equal steps of energy = equal steps of degrees), "average $\frac{1}{2} m v^{2}$ " and " $T$ " march up together in lockstep — a straight line. We choose the zero of that line at the place where the average energy is zero, i.e. no motion at all. That place is $0 K$ .

PICTURE. Figure s03: two boxes side by side, cold (slow, short arrows) and hot (fast, long arrows). Below them a thermometer runs down to $0 K$ , where the arrows vanish, and a small straight-line plot shows average motion rising in step with $T$ .

Step 4 — From ball-count $N$ to moles $n$ (why bulk chemistry uses $n$ )

WHAT. A real box holds an absurd number of balls — trillions of trillions. Chemists refuse to write such numbers, so they bundle them: one mole is a fixed pack of $N_{A} = 6.022 \times 1 0^{23}$ balls. The bulk amount $n$ (in moles) is just the head-count $N$ divided by that pack-size: $n = \frac{N}{N _{A}}$

WHY. Everything so far was about the microscopic count $N$ — the thing that sets "kicks per second." But a chemist weighs gases on a balance and speaks in moles, never in individual molecules. So we need a clean bridge: $N$ (what the physics feels) $\leftrightarrow$ $n$ (what the chemist writes). Avogadro's number $N_{A}$ is that exchange rate — like "12 eggs per dozen," but the dozen is $6.022 \times 1 0^{23}$ . See Avogadro's Law.

PICTURE. Figure s04: a huge swarm of tiny balls on the left, an equals-sign, then a neat labelled crate " $1$ mol $= 6.022 \times 1 0^{23}$ balls" on the right.

Step 5 — Assemble the master equation $P V = n R T$

WHAT. Put the ideas together. Pressure $P$ times volume $V$ equals amount $n$ times a fixed number $R$ times temperature $T$ : $P V = n R T$

WHY (and where does $R$ come from?). From Steps 2–3, pressure grows for two independent reasons: crowding (more balls per volume, i.e. $\frac{n}{V}$ from Step 4) and speed ( $T$ from Step 3). Combining two independent proportionalities gives $P \propto \frac{n}{V} and P \propto T ⟹ P \propto \frac{n T}{V} .$ A proportionality sign " $\propto$ " is not yet an equation — it only says "these rise together." To turn " $\propto$ " into " $=$ " we must multiply by whatever fixed number converts the units of $\frac{n T}{V}$ into the units of $P$ . Experiment shows this conversion number is the same for every ideal gas — so we give it one name, $R$ , the gas constant, and write $P = R \frac{n T}{V} ⟹ P V = n R T .$ That last move just multiplies both sides by $V$ to clear the fraction. (The full kinetic-theory grind that predicts $R$ lives in Kinetic Theory of Gases; here we name it and read off the shape.)

PICTURE. Figure s05 labels every symbol right on the box.

Definition The gas constant

$R$ — its value, units, and where it comes from

$R = 8.314 J mol^{- 1} K^{- 1}$ ::: the SI value. (In pressure-volume units, $R = 0.0821 L atm mol^{- 1} K^{- 1}$ — handy when $P$ is in atm and $V$ in litres.)
Why these units? Rearranged, $R = \frac{P V}{n T}$ , so its units are $\frac{( Pa ) ( m ^{3} )}{( mol ) ( K )} = \frac{N m}{mol K} = \frac{J}{mol K}$ — energy per mole per Kelvin. Every unit cancels cleanly, confirming both sides of $P V = n R T$ are an energy ( $Pa \cdot m^{3} = N m = J$ ).
Link to Boltzmann. $R$ is just Avogadro's number times the Boltzmann constant $k_{B}$ : $R = k_{B} N_{A}$ , with $k_{B} = 1.381 \times 1 0^{- 23} J K^{- 1}$ . So $R$ is the per-mole version and $k_{B}$ is the per-molecule version of the very same conversion — which is why the microscopic law $P V = N k_{B} T$ and the bulk law $P V = n R T$ are the same equation ( $n R = \frac{N}{N _{A}} \cdot k_{B} N_{A} = N k_{B}$ ).

Recall Units check on the master equation

Question: in $P V = n R T$ , do both sides carry the same units? ::: Left: $Pa \times m^{3} = N m^{- 2} \times m^{3} = N m = J$ . Right: $mol \times (J mol^{- 1} K^{- 1}) \times K = J$ . ✔ Both are energy.

This one line is the whole family tree. See it in Ideal Gas Equation PV=nRT. Every child law below is this line with one knob frozen.

Step 6 — Freeze $T$ : Boyle appears (squeeze = higher push)

WHAT. Hold temperature $T$ fixed and the amount $n$ fixed. Then the right side $n R T$ is a single frozen number. So: $P V = frozen = const n R T \Rightarrow P_{1} V_{1} = P_{2} V_{2}$

WHY. We choose to freeze $T$ because we want to isolate the crowding lever alone. With $T$ and $n$ locked, only $P$ and $V$ can move — and their product can't change. That is Boyle's Law.

PICTURE. Figure s06: same balls, box squeezed to half the width. Same speed (same $T$ ), but now they reach the wall twice as often → push doubles.

Step 7 — Freeze $P$ : Charles appears (heat = grow)

WHAT. Hold pressure $P$ and amount $n$ fixed. Rearrange $P V = n R T$ into $V = \frac{n R}{P} T$ . Everything in $\frac{n R}{P}$ is frozen, so $V$ just tracks $T$ : $\frac{V}{T} = \frac{n R}{P} = const \Rightarrow \frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2}}$

WHY. This time we freeze $P$ to isolate the speed lever. If we heat the balls but insist the wall push stay the same (that is what "constant $P$ " means — think a piston free to slide), the box has no choice but to grow so the faster balls hit less often per area, keeping the push steady.

PICTURE. Figure s07: a cylinder with a free piston on top. Heat it → arrows lengthen → piston rises → volume grows in step with $T$ . The extrapolated line hits $V = 0$ at $- 273.1 5^{\circ}$ C.

Step 8 — Freeze $V$ : Gay-Lussac appears (locked can = higher push)

WHAT. Hold volume $V$ and amount $n$ fixed. Then $P = \frac{n R}{V} T$ , and $\frac{n R}{V}$ is frozen: $\frac{P}{T} = \frac{n R}{V} = const \Rightarrow \frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$

WHY. We freeze $V$ to see what heating does when the box cannot grow (a rigid, sealed can). The balls can't spread out, so all the extra speed goes straight into harder, more frequent kicks — the push climbs with $T$ .

PICTURE. Figure s08: a rigid steel can, walls fixed. Heat it → arrows lengthen → the pressure gauge needle swings up. Nothing moves outward; only the push rises.

Step 9 — Freeze $P$ and $T$ : Avogadro appears (more gas = bigger room)

WHAT. Now turn the fourth dial — the amount $n$ — while holding both pressure $P$ and temperature $T$ fixed. Rearrange $P V = n R T$ into $V = \frac{R T}{P} n$ ; everything in $\frac{R T}{P}$ is frozen, so $V$ tracks $n$ : $\frac{V}{n} = \frac{R T}{P} = const \Rightarrow \frac{V _{1}}{n _{1}} = \frac{V _{2}}{n _{2}}$

WHY. We froze $P$ and $T$ this time so that only crowding can change. Pump in more balls at the same speed ( $T$ fixed) but insist the wall push stay the same ( $P$ fixed): the only way the extra balls don't raise the push is if the room grows to give them the same elbow-room each. So volume is proportional to the number of balls — Avogadro's Law. This is why equal volumes of any gas at the same $P$ and $T$ hold equally many molecules. See Avogadro's Law.

PICTURE. Figure s09: same push and same speed, but doubling the balls doubles the box size.

Step 10 — The degenerate cases (what happens at the edges)

WHAT. Check the extreme and broken inputs, so no scenario surprises you.

WHY. A rule you trust must survive its limits. Let's push each dial to the boundary.

PICTURE. Figure s10 shows three edge panels: the box vanishing as $T \to 0$ , the Boyle hyperbola flattening as $V \to \infty$ , and the "wrong-units" trap of feeding Celsius into a ratio.

Definition The edge cases, one by one

$T \to 0 K$ (absolute zero). Balls stop. Charles says $V \to 0$ ; the box would shrink to nothing. Real gases liquefy first (our "no forces" rule fails when balls slow down) — this is why $0 K$ is unreachable, only a limit.
Negative Kelvin? Not allowed. Since $T$ measures average motion, $T$ can approach $0 K$ but never drop below it — every formula here assumes $T > 0$ .
Using $0^{\circ}$ C in Charles. $\frac{V}{0}$ ?? Only if you forgot Kelvin — $0^{\circ}$ C is $273 K$ , not zero. Never divide by a Celsius temperature.
$T$ changes but you use Boyle. $P_{1} V_{1} \neq = P_{2} V_{2}$ any more, because the frozen $n R T$ unfroze. You must use the combined law (the one-picture summary).
$V \to \infty$ (Boyle). As volume grows without bound, $P = \frac{const}{V} \to 0$ . Push fades to nothing — matches the hyperbola flattening onto the axis.
$n \to 0$ (Avogadro). Remove all the balls: $V = \frac{R T}{P} n \to 0$ at fixed $P, T$ . No gas, no room needed — consistent.

The one-picture summary

Figure s11 puts all four children on one diagram: the master box $P V = n R T$ in the centre, with four arrows peeling off — freeze $T$ → Boyle, freeze $P$ → Charles, freeze $V$ → Gay-Lussac, freeze $P, T$ → Avogadro — and the first three reunite in the combined law $\frac{P _{1} V _{1}}{T _{1}} = \frac{P _{2} V _{2}}{T _{2}}$ .

Recall Feynman retelling — the whole walkthrough in plain words

Picture a box of bouncing balls that never stick to each other and never lose speed when they hit a wall (our ideal rules). One ball of mass $m$ hits a wall of area $A$ and shoves it — that shove, smeared over millions of balls per second per area, is pressure. How fast the balls zoom is set by temperature (measured from true rest at $0 K$ , and never negative). The speed enters twice — harder hits and more frequent hits — which is why $v^{2}$ shows up. Because trillions of balls are awkward to write, we bundle $6.022 \times 1 0^{23}$ of them into one mole and call the count $n$ . Pack $n$ moles into a room of size $V$ and you get the master rule $P V = n R T$ , where $R = 8.314 J mol^{- 1} K^{- 1} = k_{B} N_{A}$ is just the fixed number that turns "rises together" into "equals," with both sides carrying units of energy. Now play with the dials.

Squeeze the room (freeze $T$ ): balls hit the wall more often → push doubles when room halves. That's Boyle.
Heat with a free lid (freeze $P$ ): balls zoom faster, so the lid rises to keep the push steady → volume grows with temperature. That's Charles.
Heat a locked can (freeze $V$ ): balls zoom faster but can't spread → they slam the walls harder → pressure climbs. That's Gay-Lussac.
Pump in more balls (freeze $P$ and $T$ ): to keep the push and speed the same, the room grows in step with the number of balls. That's Avogadro. Cool everything to $0 K$ and all motion stops — and always measure temperature from that true zero, never from Celsius zero.

Connections

Ideal Gas Equation PV=nRT — the master line all four fall out of.
Kinetic Theory of Gases — the full derivation of pressure from bounces (and of $R$ , and the $\frac{1}{3}$ factor).
Absolute Zero and Kelvin Scale — why Step 3 needs Kelvin and why $T > 0$ .
Combined Gas Law — the one-picture unifier.
Avogadro's Law — the $n$ (amount) dial, Step 4 and Step 9.
Dalton's Law of Partial Pressures — many gases sharing one box.

Flashcards

Why must temperature be in Kelvin for Charles & Gay-Lussac?

Because

V

and

P

are proportional to absolute motion; the ratio only makes sense measured from true zero motion (

0 K

), not from Celsius zero.

Why is average kinetic energy proportional to

T

(a straight line)?

A thermometer is scaled to read the average

\frac{1}{2} m ⟨ v^{2} ⟩

per ball on evenly spaced degrees, with zero set where motion stops; so energy and

T

rise in lockstep.

Why does pressure depend on

v^{2}

, not

v

Faster balls hit harder (one

v

, via momentum

m v

) and more often (a second

v

), so

P \propto m v^{2}

What are the four ideal-gas assumptions?

Point-sized molecules, no intermolecular forces between hits, perfectly elastic collisions, straight-line flight between hits.

What is the value and units of

R

R = 8.314 J mol^{- 1} K^{- 1}

(or

0.0821 L atm mol^{- 1} K^{- 1}

); both sides of

P V = n R T

then carry units of energy.

How does

R

relate to the Boltzmann constant?

R = k_{B} N_{A}

, with

k_{B} = 1.381 \times 1 0^{- 23} J K^{- 1}

;

R

is per-mole,

k_{B}

is per-molecule.

What is Avogadro's number and what does it bridge?

N_{A} = 6.022 \times 1 0^{23} mol^{- 1}

; it converts the microscopic head-count

N

into the bulk amount

n = N / N_{A}

Why does turning "

\propto

" into "

=

" need

R

A proportionality only says quantities rise together;

R

is the fixed conversion number that matches units so

P = R n T / V

becomes a true equality.

Can Kelvin temperature be negative?

No —

T

measures average motion, which bottoms out at zero, so

T > 0

always; only Celsius can be negative.

Freeze

T

P V = n R T

→ which law?

Boyle,

P_{1} V_{1} = P_{2} V_{2}

Freeze

P

P V = n R T

→ which law?

Charles,

V_{1} / T_{1} = V_{2} / T_{2}

Freeze

V

P V = n R T

→ which law?

Gay-Lussac,

P_{1} / T_{1} = P_{2} / T_{2}

Freeze

P

and

T

, vary

n

→ which law?

Avogadro,

V_{1} / n_{1} = V_{2} / n_{2}

What does pressure mean microscopically?

Total wall kicks per second per unit area

A

— more when balls are denser, heavier, or faster.