Temperature — thermal equilibrium, thermometers, scales: Visual walkthrough

We use only these ideas, each defined the moment it appears: a gas in a box, the push it makes on a wall (pressure), the temperature you read off a thermometer, and a straight line on a graph. Nothing else is assumed.

Step 1 — A box of gas that pushes on a lid

WHAT. Trap some gas — air, helium, anything thin — under a lid that can slide but is held in place so the volume never changes (constant volume). The gas drums on the lid from inside. That steady drumming, spread over the lid's area, is the pressure $P$ : how hard the gas pushes per unit of surface.

WHY. We need a thermometric property — a physical quantity that changes reliably when hotness changes. We pick pressure at fixed volume because (the parent told us) all dilute gases behave the same way here, unlike mercury or alcohol which each tell a slightly different story. Pressure is our honest witness.

PICTURE. The orange arrows are the gas particles hammering the walls; the teal number on the gauge is the pressure they produce. Squeeze in more warmth and the hammering gets fiercer.

Figure — Temperature — thermal equilibrium, thermometers, scales

Step 2 — Cool it down, watch the push weaken

WHAT. Lower the temperature $t_{C}$ (in Celsius) and record the pressure. Warmer gas → harder push; cooler gas → gentler push. Plot each measurement as a dot: temperature across the bottom, pressure up the side.

WHY. Before writing any formula, we look. We are asking the raw experimental question: is there a pattern? If the dots scatter randomly, there's no law to find. Nature is kinder than that.

PICTURE. As we sweep from warm (right) to cold (left), the pressure gauge falls and the dots march downward and to the left in a strikingly straight file.

Recall Why look before you leap?

Question ::: Why plot the data before assuming a formula? Answer ::: Because the shape of the data (a straight line) is what justifies the linear law. We are not guessing linearity — we are reading it off the graph.

Step 3 — The dots lie on a straight line

WHAT. Draw the best straight line through the dots. A straight line is captured by one equation: $P = P_{0} (1 + α t_{C})$

Let's read every symbol where it stands:

$P$ — the pressure we measure right now (the height of a dot).
$t_{C}$ — the temperature in Celsius we set (the horizontal position of that dot).
$P_{0}$ — the pressure at the ice point, $t_{C} = 0 ° C$ . It's where the line crosses the vertical axis. Call it the anchor height.
$α$ — the fractional slope: how much of $P_{0}$ the pressure gains for each degree of warming. Experiment gives $α \approx \frac{1}{273.15} ° C^{- 1}$ for every dilute gas.

WHY a straight line and not a curve? Because that's what the dots show, and it is the simplest law that fits — one intercept ( $P_{0}$ ) and one slope ( $α P_{0}$ ). A curve would need extra parameters the data don't demand.

PICTURE. The teal line hits the axis at $P_{0}$ when $t_{C} = 0$ ; the little step-triangle shows the rise $α P_{0}$ per $1 ° C$ .

Step 4 — Follow the line downward: where does the push vanish?

WHAT. The line keeps sloping down as we cool. Ask the sharp question: at what temperature would the pressure reach exactly zero? Set $P = 0$ and solve for $t_{C}$ .

$0 = P_{0} (1 + α t_{C})$

WHY set $P = 0$ ? Because Step 1 told us pressure can never go negative — it's a push, and a push can only weaken to nothing. So $P = 0$ is the natural floor. Finding the $t_{C}$ that produces it finds the coldest conceivable temperature.

Now solve, one honest move at a time:

$P_{0} \neq = 0$ (there's real gas in the box), so we may divide it out: $0 = 1 + α t_{C}$ .
Subtract $1$ : $α t_{C} = - 1$ .
Divide by $α$ : $t_{C} = - \frac{1}{α}$ .

Put in the measured slope $α = \frac{1}{273.15}$ : $t_{C} = - \frac{1}{1/273.15} = - 273.15 ° C$

PICTURE. The straight line is extended (dashed) past the data until it pierces the horizontal axis. That piercing point — where the push dies — sits at $- 273.15 ° C$ . That is absolute zero.

Step 5 — Edge case: does the answer depend on which gas?

WHAT. Run the experiment with helium, then nitrogen, then argon. Each has its own anchor height $P_{0}$ (different amount of gas, different starting push). So we get three different lines. Do they cross the axis at three different places?

WHY this matters. If absolute zero depended on the substance, it couldn't be a universal constant — it'd just be a quirk of helium. We must check the degenerate-looking possibility that the crossing wanders.

The algebra of the crossing. Every line, whatever its $P_{0}$ , obeys $0 = P_{0} (1 + α t_{C})$ . Since $P_{0}$ divides out, the crossing depends only on $α$ — and $α$ is the same for all dilute gases. So all three lines, however steep, pierce the axis at the same $- 273.15 ° C$ .

PICTURE. Three lines of different steepness (three gases) fan out from three different intercepts on the left — yet they all converge to one shared point on the temperature axis.

Step 6 — Slide the origin: build the Kelvin scale

WHAT. We have found a true zero of temperature. It is wasteful to keep it at the ugly number $- 273.15$ . So invent a new scale $T$ whose zero sits exactly at absolute zero, keeping each degree the same size as a Celsius degree: $T = t_{C} + 273.15$

Read the symbols:

$T$ — the new Kelvin temperature (unit: K, no degree sign).
$t_{C}$ — the same Celsius reading as before.
$+ 273.15$ — the shift that moves the origin from the ice point down to absolute zero.

WHY shift, not rescale? We love the size of the Celsius degree (100 steps between ice and steam), so we keep the slope. We only slide the origin so that the physically special point — where pressure vanishes — becomes the numerical zero.

The payoff. Put $t_{C} = - 273.15$ (absolute zero) into the shift: $T = - 273.15 + 273.15 = 0$ K. And rewrite the pressure law on the new scale: $P = P_{0} (1 + α t_{C}) = P_{0} α (t_{C} + 1/ α) = P_{0} α (t_{C} + 273.15) = (P_{0} α) T$ So on the Kelvin scale, pressure is simply proportional to temperature: $P \propto T$ . Clean, no offset.

PICTURE. Two rulers stacked: the Celsius ruler with its awkward $- 273.15$ , and the Kelvin ruler slid so its $0$ lands on absolute zero. Same tick spacing, different starting number.

Step 7 — Anchoring the scale with one fixed point

WHAT. A line through the origin needs just one known point to fully pin it. We use the triple point of water — the one exact temperature where ice, liquid, and vapour coexist — defined to be $T_{tr} = 273.16$ K. Measure the gas pressure there, $P_{tr}$ , and any other temperature follows by proportion: $T = T_{tr} \frac{P}{P _{tr}}$

$T$ — temperature we want.
$P$ — pressure of the gas in the unknown bath.
$P_{tr}$ — pressure of the same gas at the triple point.
$T_{tr} = 273.16$ K — the fixed anchor.

WHY only one point now? Because Step 6 gave us a line through the origin (zero is fixed by physics, not by choice). A line through the origin has one free number — its slope — so one measurement locks everything.

Degenerate check. If $P \to 0$ , then $T \to 0$ — consistent with Step 4. If $P = P_{tr}$ , then $T = T_{tr}$ — the anchor reproduces itself. Both sanity checks pass.

PICTURE. One line from the origin, one marked dot at $(P_{tr}, T_{tr})$ ; any pressure $P$ reads off a temperature by riding the ray.

The one-picture summary

Everything above, on one graph: the data dots, the fitted line, its dashed extension down to the axis at $- 273.15 ° C$ , and the second Kelvin axis slid so that piercing point reads $0$ K. Cooling the gas walks you left along the line; you run out of push exactly at absolute zero, and Kelvin is just Celsius with its zero moved there.

Recall Feynman retelling — the whole walk in plain words

Question ::: Retell the derivation of absolute zero to a curious kid. Answer ::: Put some air in a sealed jar so it can't grow or shrink, and stick a pressure gauge on it. The air bangs on the walls — that banging is pressure, its "push." Now chill the jar. The colder it gets, the lazier the banging, so the push drops. If you dot down every (temperature, push) pair, the dots fall on a perfectly straight line sloping down to the left. Grab a ruler and keep drawing that line past where you stopped measuring: it crosses the "zero push" mark at $- 273.15 ° C$ . But a gas can't push inward-less-than-nothing — zero push is the end of the road, so $- 273.15 ° C$ is the coldest temperature there can ever be. Try it with helium, nitrogen, argon: each starts at a different push, so each line is a different steepness — yet they all aim at that same crossing point, because a weaker gas also fades more gently. Since that point is so special, we build a new ruler (Kelvin) with its zero sitting right there, keeping the same degree size. On that ruler the push is just plain proportional to the temperature — double the Kelvin temperature, double the push. One good measurement at water's triple point ( $273.16$ K) sets the whole scale.

Prerequisites & neighbours: parent topic · Ideal Gas Law · Kinetic Theory of Gases · Thermal Expansion · Zeroth Law of Thermodynamics · Heat and Internal Energy