Thermal Expansion of Gases

In short

Gases expand enormously when heated — far more than solids or liquids. At constant pressure, a fixed amount of an ideal gas obeys Charles's law:

\frac{V_1}{T_1} = \frac{V_2}{T_2} \qquad \text{or equivalently} \qquad V \propto T,

where T is the absolute (Kelvin) temperature. At constant volume, the same gas obeys Gay-Lussac's law:

\frac{P_1}{T_1} = \frac{P_2}{T_2} \qquad \text{or} \qquad P \propto T.

Both laws are only simple when T is measured from the correct zero. Plot V against Celsius temperature and you get a straight line whose backward extrapolation hits V = 0 at T_C = -273.15\,^\circC — regardless of which gas, regardless of pressure. That universal intercept is absolute zero. Shifting the temperature zero to this point (defining T_K = T_C + 273.15) turns the two laws into clean proportionalities. The Kelvin scale is not arbitrary; it is the scale a gas itself picks out.

The coefficient of volume expansion at constant pressure of an ideal gas is \gamma = 1/T (in kelvins), which at 0 °C gives \gamma \approx 1/273 per °C — an enormous number compared to the \sim 10^{-4} /K of a liquid and \sim 10^{-5} /K of a solid. Gases expand about 10,000 times more easily than solids, which is why they power everything from hot-air balloons to steam turbines.

At the Pushkar Camel Fair each November, hot-air balloons lift off into the dawn sky. A burner heats the air inside the envelope to about 100\,^\circC; the air outside is about 10\,^\circC. That temperature difference is the whole engine. The hot air inside the balloon is lighter than the cold air outside — because heating the trapped air has made it expand, spill some out through the open mouth, and left the remaining air less dense than the ambient atmosphere. Buoyancy does the rest, and a thousand kilograms of wicker basket, burner, fuel, and excitable tourists rise into the sky.

The physics powering that lift is the thermal expansion of a gas, and it is ferocious. A steel rail gets a few millimetres longer per metre when it is heated from a cool morning to a scorching afternoon — a barely visible change that only matters because of how long the rail is. A gas, heated through the same range at constant pressure, undergoes a relative expansion roughly ten thousand times larger. Air at 10 °C occupies a little less than 95% of the volume it occupies at 100 °C — a massive fractional change. That is the reason the balloon rises at all.

This article derives the gas-expansion laws that make this work — Charles's law for volume, Gay-Lussac's law for pressure — shows why they take a trivially simple form only when you use the Kelvin scale, and reveals the extrapolation argument that pinned down absolute zero before anyone had ever produced a temperature within 70 degrees of it.

Why gases expand more than solids or liquids

Before any derivation, get a feel for the scale of the effect. For a temperature change of \Delta T = 100 K:

Material	Coefficient of volume expansion \gamma (per K)	Relative volume change over \Delta T = 100 K
Iron (solid)	\sim 3.6 \times 10^{-5}	0.0036 (0.36%)
Water (liquid)	\sim 2.1 \times 10^{-4}	0.021 (2.1%)
Air (gas, at 0 °C, constant P)	\sim 3.66 \times 10^{-3}	0.366 (36.6%)

The gas expands by more than a third of its original volume over a 100 K heating. Iron expands by less than half a percent.

The reason is microscopic. In a solid, each atom is held tightly in place by bonds to its neighbours; heating increases the amplitude of its thermal vibration, slightly nudging the average spacing, but only by about one part in 10^5 per kelvin. In a liquid, molecules are close-packed but can rearrange; thermal motion loosens them a little, producing a larger expansion of about one part in 10^4 per kelvin. In a gas, molecules are far apart — the spacing between them is about ten times their diameter — and nothing holds them in place but the walls of the container. Heat the gas and they move faster; their collisions with the walls get more forceful and more frequent; if the walls can move (constant-pressure piston), the gas pushes them outward until equilibrium is re-established. There is no neighbour-bonding to resist. That is why gases are thousands of times more responsive to temperature than solids.

Charles's law — volume at constant pressure

Take a fixed amount of an ideal gas — say, a mole of dry nitrogen — trapped in a cylinder by a frictionless piston that can slide freely. The atmosphere pushes down on the piston from outside with constant pressure P_0; the gas pushes up from inside with the same pressure (the piston floats in equilibrium). Heat the gas. The piston rises, keeping the pressure constant at P_0. You measure the volume V of the gas at a range of temperatures.

The data fall on a straight line. If you extend the line far to the left — past any temperature at which the gas would actually liquefy — the line passes through the point where V = 0 at T = -273.15\,^\circC, the same absolute-zero intercept you first met in Temperature and Thermometers.

Volume of a fixed amount of ideal gas at constant pressure as a function of temperature. The measured points (circles) fall on a straight line. Extrapolating the line backwards (dashed) shows that it would intersect $V = 0$ at $T = -273.15\,^\circ$C — the same extrapolated zero for every gas, every pressure. Shifting the zero of the temperature axis to this point (the Kelvin axis below) turns the relation into $V \propto T$.

Deriving the law

The empirical statement is: at constant pressure, V is a linear function of the Celsius temperature.

V(T_C) = V_0 (1 + \gamma\, T_C), \tag{1}

where V_0 is the volume at 0\,^\circC and \gamma is the volume expansion coefficient at constant pressure — the fractional change in volume per degree of warming, measured from 0 °C.

Step 1. Find the extrapolated zero.

Set V(T_C) = 0 in equation (1):

0 = V_0 (1 + \gamma\, T_{C,0}) \quad \Longrightarrow \quad T_{C,0} = -\frac{1}{\gamma}.

Why: if the line that describes real data crosses zero volume at some temperature, algebra just hands you the intercept. Here \gamma is measured, not assumed, and the formula says a single number -1/\gamma locates the intercept.

Step 2. Read \gamma from experiment.

Careful measurements (going back to the work of Jacques Charles and Joseph Gay-Lussac around 1800, confirmed by every experiment since) give \gamma \approx 1/273.15 per °C — for every ideal gas, for every pressure (as long as the gas really is behaving ideally, i.e. not too cold or too dense). The universality is what makes the result physical and not merely a fact about some particular substance.

Step 3. Therefore T_{C,0} = -273.15\,^\circC.

This is the pristine experimental origin of absolute zero: the temperature where an ideal gas's volume would vanish by extrapolation.

Step 4. Introduce the absolute (Kelvin) temperature T_K = T_C + 273.15.

Substitute T_C = T_K - 273.15 into equation (1):

V = V_0 \left(1 + \frac{T_K - 273.15}{273.15}\right) = V_0 \left(\frac{273.15 + T_K - 273.15}{273.15}\right) = V_0 \cdot \frac{T_K}{273.15}.

Why: after replacing T_C and using \gamma = 1/273.15, the two constants collapse into a single ratio. The messy affine relation in Celsius becomes a clean proportionality in Kelvin.

Step 5. Write in ratio form.

Let V_0 / 273.15 = k, a constant that depends on the amount of gas and the pressure but not on the temperature. Then

\boxed{\; V = k\,T_K \;} \qquad \Longleftrightarrow \qquad V \propto T_K.

Equivalently, for any two states (V_1, T_{K,1}) and (V_2, T_{K,2}) at the same pressure:

\boxed{\; \frac{V_1}{T_{K,1}} = \frac{V_2}{T_{K,2}}. \;}

Why: if V/T_K = k always, then the two sides must be equal whenever you evaluate them at any two temperatures. Useful for problems: you do not need to know V_0 or P_0 explicitly, just the ratio of two states.

This is Charles's law. The clean form hides the experimental work that went into pinning down the number 273.15 — and the deeper fact that the number is universal across gases.

What \gamma = 1/T really says

At any temperature T_K, the coefficient of volume expansion of an ideal gas at constant pressure is

\gamma_{P}(T_K) = \frac{1}{V}\,\left(\frac{\partial V}{\partial T_K}\right)_{P} = \frac{1}{V}\,k = \frac{1}{V}\,\frac{V}{T_K} = \frac{1}{T_K}.

Why: for an ideal gas at constant pressure, V = kT_K means \partial V/\partial T_K = k and V = kT_K, and their ratio is 1/T_K. The coefficient depends on where on the temperature scale you are.

So at T_K = 273.15 K (the ice point), \gamma = 1/273.15 \approx 3.66 \times 10^{-3} /K. At T_K = 373 K (boiling water), \gamma = 1/373 \approx 2.68 \times 10^{-3} /K. The expansion coefficient is smaller at higher temperatures — another way of saying the V–T_K line has the same slope at every point (it is a straight line) but the fractional change per degree is smaller when the volume is already larger.

Gay-Lussac's law — pressure at constant volume

The companion experiment keeps volume fixed and watches pressure. Seal a gas in a rigid container (a strong bulb, or an LPG cylinder with the valve closed) and heat it. The molecules speed up, hit the walls harder and more often, and the pressure rises.

Over the same range of temperatures where Charles's law holds, a plot of P versus T_C is also a straight line, with the same extrapolated zero at T_C = -273.15\,^\circC. The same chain of reasoning as for volume gives:

P = k'\,T_K, \qquad \text{so} \qquad \boxed{\; \frac{P_1}{T_{K,1}} = \frac{P_2}{T_{K,2}} \;}

at constant V and constant amount of gas. This is Gay-Lussac's law (sometimes called Amontons's law).

The physics is the same as for Charles's law — an ideal gas's pressure is proportional to its absolute temperature — and the fact that the extrapolated zero is again -273.15\,^\circC is another universal fingerprint of the Kelvin scale.

Why this is why the LPG cylinder warnings exist

The LPG cylinder in your kitchen is filled to a pressure of around 17 atmospheres at room temperature T_1 = 300 K. What if it is left in direct afternoon sunlight and warms to T_2 = 330 K (57 °C, hot but not impossible in a Chennai summer)? Treating the cylinder walls as rigid so volume is constant, Gay-Lussac says:

P_2 = P_1 \cdot \frac{T_2}{T_1} = 17 \times \frac{330}{300} = 18.7 \text{ atm}.

A 10% temperature rise produces a 10% pressure rise. Cylinders are engineered with safety margins to handle this, but the cooking-gas safety pamphlets that ask you to keep cylinders away from direct sunlight and flames are there because the physics is exactly this: pressure is proportional to absolute temperature, and absolute temperature is not small — a fire licking a cylinder could take it from 300 K to 1000 K, tripling the pressure catastrophically.

At fixed volume, the pressure of an ideal gas is strictly proportional to the absolute temperature. The line passes through the origin — $P = 0$ at $T = 0$ K — because molecular motion ceases at absolute zero and there is nothing left to push on the walls. A 10% temperature rise produces a 10% pressure rise.

The absolute zero argument made precise

The most surprising feature of Charles's law and Gay-Lussac's law is that the extrapolated zero is the same -273.15\,^\circC for every gas. Here is the argument spelled out.

Fact 1 (experimental). For any gas at low enough density, V at constant P is a linear function of T_C, and P at constant V is a linear function of T_C. Both are of the form X = X_0(1 + \alpha T_C) with X_0 the value at 0\,^\circC and \alpha a coefficient. This is an observation, not a theory.

Fact 2 (experimental, universal). The coefficient \alpha has the same numerical value — 1/273.15 per °C — for every gas (hydrogen, nitrogen, oxygen, carbon dioxide, argon, …) and at every pressure, provided the gas is far enough from liquefaction to be well-described as "ideal."

Why this matters: \alpha being the same for all gases means the extrapolated zero T_{C,0} = -1/\alpha is also the same for all gases. If \alpha had depended on the gas, each gas would pick out a different extrapolated zero, and there would be no single "absolute zero" to speak of. Fact 2 says nature has done the picking for you.

Conclusion. There is a single temperature, the universal extrapolated zero, which every gas's expansion curve points to. Declare that temperature to be zero on a new scale (the Kelvin scale, with T_K = T_C + 273.15). On that scale, Charles's law is V \propto T_K and Gay-Lussac's law is P \propto T_K — no additive shifts, no arbitrary intercepts.

Fact 2 — universality of \alpha — is the physics content. The Kelvin scale is what it forces upon you.

Why ideal-gas liquefaction doesn't kill the argument

A real gas liquefies before it gets close to absolute zero — air liquefies at about 77 K. You cannot actually take a bulb of air down to -273\,^\circC; it would have condensed into a puddle long before. The extrapolation is therefore not a direct measurement of what happens at zero kelvin; it is an extension of the behaviour observed in the range where the gas is still a gas. But the extension is unambiguous — the line is straight, and straight lines have a unique intercept.

Does the ideal-gas law actually hold all the way down to absolute zero? No — real gases deviate from ideal behaviour at low temperatures, and quantum effects eventually dominate. The argument is logical, not experimental, in its final step: if the ideal-gas law held down to V = 0, the line would hit that point at -273.15\,^\circC. Absolute zero is defined as the temperature where this extrapolated V = 0 occurs.

The deeper argument from the kinetic theory of gases — that absolute temperature is proportional to the mean kinetic energy of molecules — eventually ties the absolute zero to something physical (no kinetic energy left), independent of any particular gas's behaviour. The extrapolation argument was the first historical route to the number; the kinetic-theory argument is the modern reason it has to be that number.

Worked examples

Example 1: Charles's law — the hot-air balloon

A hot-air balloon at Pushkar has an envelope volume of 2500\,\text{m}^3. The ambient air temperature at dawn is 10\,^\circC. The pilot fires the burner until the air inside the envelope reaches 100\,^\circC. The balloon's envelope is open at the bottom, so pressure inside equals atmospheric pressure throughout. If the envelope's walls held shape rigidly at volume 2500\,\text{m}^3, how much air (by volume, measured at ambient temperature and pressure) must be expelled through the opening as the balloon heats up?

Before: 2500 m³ of cool air (283 K) fills the envelope. After: the remaining air has been heated to 373 K, still at atmospheric pressure, and some has spilled out through the opening at the bottom. The fraction spilled is what makes the balloon buoyant.

Step 1. Convert temperatures to Kelvin.

T_1 = 10\,^\circ\text{C} = 283.15 K. T_2 = 100\,^\circ\text{C} = 373.15 K.

Why: Charles's law is V \propto T only when T is measured on the Kelvin scale. Plug in Celsius and you will get nonsense.

Step 2. Figure out what stays fixed and what changes.

Before heating, the envelope contains V_1 = 2500 m³ of air at T_1 = 283.15 K. Call the number of moles of gas in this starting state n_1. At the final state, after the burner has worked, the envelope still contains volume V_\text{env} = 2500 m³, but at the higher temperature T_2 = 373.15 K; it now holds n_2 < n_1 moles because the rest has been expelled through the bottom opening.

The question asks for the volume of the expelled air, measured at ambient conditions (283.15 K, atmospheric pressure).

Step 3. Use Charles's law to find what volume the remaining n_2 moles would occupy at ambient temperature.

If you cooled the remaining gas back down to T_1 = 283.15 K at constant pressure, its volume would shrink:

V_\text{remaining at } T_1 = V_\text{env} \cdot \frac{T_1}{T_2} = 2500 \cdot \frac{283.15}{373.15} = 2500 \cdot 0.7588 = 1897\text{ m}^3.

Why: the same parcel of air, at the same pressure, occupies volumes in the ratio T_1 : T_2 of the two temperatures. Going from hot (373 K) to cool (283 K) shrinks the volume by T_1/T_2.

Step 4. The expelled volume (at ambient temperature and pressure) is the difference.

V_\text{expelled} = V_1 - V_\text{remaining at } T_1 = 2500 - 1897 = 603\text{ m}^3.

Result: About 603 m³ of air — roughly a quarter of the original mass of air in the envelope — has been pushed out through the opening at the bottom. That is what the burner has done: not created heat for its own sake, but ejected air, leaving the envelope with less mass than an equal volume of ambient air. The buoyant force follows.

What this shows: A Charles's-law calculation gives you directly the volume-fraction of air expelled when a balloon is heated. The number is large — a quarter of the air leaves for a rise of 90 K — because gases expand so strongly. This is the quantitative payoff of \gamma_\text{gas} being so much larger than \gamma_\text{liquid} or \gamma_\text{solid}.

Example 2: Gay-Lussac — the LPG cylinder in summer sun

An LPG cylinder holds liquefied gas plus a column of vapour above the liquid. The vapour pressure is 8.0 \times 10^5 Pa at T_1 = 27\,^\circC. The cylinder is left on a Chennai rooftop and warms to T_2 = 57\,^\circC. Treating the vapour space as a fixed volume and approximating the vapour as an ideal gas (a rough model — real LPG vapour is close to, but not exactly, ideal), find the new pressure.

Vapour above the liquid LPG at two temperatures. The rigid cylinder fixes the volume; Gay-Lussac's law then predicts that pressure rises in proportion to absolute temperature.

Step 1. Convert to Kelvin.

T_1 = 27 + 273 = 300 K (I am using 273 K for round numbers; if you need more precision, use 273.15). T_2 = 57 + 273 = 330 K.

Step 2. Apply Gay-Lussac's law.

\frac{P_1}{T_1} = \frac{P_2}{T_2} \qquad \Longrightarrow \qquad P_2 = P_1 \cdot \frac{T_2}{T_1}.

Why: at fixed volume and fixed amount of gas, P is proportional to the absolute temperature. The ratio form lets you skip the intermediate constant and relate the two states directly.

Step 3. Substitute.

P_2 = 8.0 \times 10^5 \cdot \frac{330}{300} = 8.0 \times 10^5 \cdot 1.10 = 8.8 \times 10^5 \text{ Pa}.

Step 4. Interpret.

A 30 K rise — a perfectly routine thermal change for a Chennai rooftop in May — raises the pressure by 10%, a 0.8 \times 10^5 Pa (about 0.8 atmosphere) increase. The cylinder is engineered to withstand multiples of its rated pressure, so this change is safe. But the calculation shows why LPG cylinder safety rules are firm: leave a cylinder next to a fire, and "absolute temperature" suddenly becomes 1000 K or more, and the pressure triples or quadruples — easily enough to breach the cylinder.

Result: The vapour pressure rises from 8.0 \times 10^5 Pa to 8.8 \times 10^5 Pa over a heating of 30 K.

What this shows: Even a small-looking Celsius change (27 °C to 57 °C) corresponds to a meaningful fractional change in absolute temperature (300 K to 330 K is 10%), which Gay-Lussac's law translates directly into a 10% pressure rise. The factor T_2/T_1 in Kelvin is the whole story. In Celsius, 57/27 = 2.11 — a 111% "rise" — which is not the pressure change at all. The absolute scale is not just nicer; it is necessary.

Common confusions

"V \propto T is true in Celsius." No. Plug T_C = 0 into V = kT_C and you get V = 0 — but the gas does not have zero volume at the freezing point of water. The proportionality holds only when T is measured from absolute zero, i.e. in Kelvin. In Celsius the law becomes the affine form V = V_0(1 + T_C/273.15), which is not a simple proportionality.
"Different gases have different expansion coefficients." They do at high densities, where intermolecular forces matter and the gas is no longer ideal. But at low densities — conditions where every gas approaches the ideal-gas behaviour — the coefficient is universal, 1/273.15 per °C at 0 °C. This universality is exactly what makes the Kelvin scale a physical scale and not a convention.
"A hot-air balloon lifts because hot air is lighter." Yes, but spell out the mechanism. The envelope is open at the bottom, so its internal pressure equals atmospheric. Heating the gas at constant pressure forces some of it out through the opening (Example 1). The same volume of envelope now contains fewer moles of air, so its density is lower than the cold air outside. Buoyancy lifts it.
"Gay-Lussac's law and Charles's law are different laws." They are two faces of the same physics: an ideal gas's PV product is proportional to the absolute temperature. Holding P constant gives Charles's law (V \propto T); holding V constant gives Gay-Lussac's (P \propto T). You will see them combined into the full ideal gas law PV = nRT in the next chapter.
"The coefficient \gamma = 1/273 is per degree Celsius, not per kelvin." Both are right, because Celsius and Kelvin degrees are the same size — only their zeros differ. A temperature change of 1 °C and 1 K are identical, and any coefficient expressed as a fractional change per degree is the same number on both scales.
"Air does not obey the ideal gas law." Approximately it does, at everyday temperatures and pressures. At 27\,^\circC and 1 atm, the deviation from ideal behaviour is less than 0.1% — smaller than you can probably measure in a school laboratory. The ideal-gas approximation breaks down only when the gas is close to liquefying (low T, high P) or when intermolecular forces matter (polar molecules, very high densities). For the balloon, the cylinder, and almost every JEE problem, ideal is fine.

If you have followed Charles's law and Gay-Lussac's law and can use them in problems, you have the practical content. What follows is the deeper picture — where these laws come from at the molecular level, how to combine them with Boyle's law into the full ideal-gas equation, and what "absolute zero" means once you know what temperature really is.

The combined gas law

If the amount of gas is fixed but you vary all three state variables — P, V, and T — the combined law is

\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} = \text{constant (for a fixed amount of gas)}.

This is a single statement that contains Charles's (P fixed), Gay-Lussac's (V fixed), and Boyle's (T fixed) laws as special cases. The constant depends on the amount of gas — specifically, on the number of moles n — and an empirical constant R (the universal gas constant, R = 8.314 J/(mol·K)). The full equation is

\boxed{\; PV = nRT \;}

— the ideal gas equation of state, derived in full in Ideal Gas Laws. Every problem you will meet involving gases at school and college level reduces to this equation plus the substance-specific information (n, the initial conditions).

Why \gamma = 1/T from kinetic theory

The kinetic theory of gases gives a microscopic derivation of the ideal-gas law. The pressure P of a gas is the force per unit area with which its molecules bombard the walls of the container:

P = \tfrac{1}{3} n m \overline{v^2},

where n is the number density of molecules, m is the mass per molecule, and \overline{v^2} is the mean-squared speed. The deep result of kinetic theory is that \overline{v^2} is set by the absolute temperature:

\tfrac{1}{2} m \overline{v^2} = \tfrac{3}{2} k_B T,

where k_B = 1.38 \times 10^{-23} J/K is Boltzmann's constant. Combining these two gives P = n k_B T, which at a constant density (i.e., constant V and constant amount of gas) shows P \propto T — Gay-Lussac's law. At constant P, rearranging gives n = P/(k_B T), so the density falls as temperature rises, and for the same amount of gas the volume must grow in proportion to T — Charles's law.

From this microscopic view, absolute zero is the temperature at which \overline{v^2} = 0 — molecules have no kinetic energy, no motion, nothing to push on the walls. P \to 0 and (at constant P) V \to 0 follow automatically. The extrapolated intercept observed in the laboratory is exactly the temperature at which molecular motion ceases.

(Quantum mechanics modifies this: there is a residual "zero-point" motion even at T = 0 K, so molecules never quite stop moving. But the classical V \propto T line hits zero at exactly T = 0 K to within parts in 10^9 for any real gas, because the quantum correction is a small fraction of the classical thermal energy at any temperature where the gas is still a gas.)

Real-gas corrections

The ideal-gas laws assume molecules are point particles with no volume and no interactions. Real molecules have finite size (which lowers the volume available for motion, raising the pressure) and attract each other slightly at short range (which reduces the pressure by pulling molecules inward). The van der Waals equation makes both corrections:

\left(P + \frac{a n^2}{V^2}\right)(V - nb) = nRT,

where a and b are empirical constants specific to each gas. At low density (V \gg nb, (a n^2/V^2) \ll P), the van der Waals equation reduces to the ideal gas equation. At high density — close to liquefaction — the corrections become large, and Charles's and Gay-Lussac's laws fail.

Liquefaction itself is the point at which the gas ceases to obey even the van der Waals equation and instead undergoes a phase transition. For nitrogen this happens around 77 K at 1 atm. Below that, you cannot compress the substance as a gas; it turns into a liquid. This is why the V–T line in the Charles's law plot is drawn dashed below the liquefaction temperature: the gas isn't there to be measured.

Isobaric vs isothermal vs isochoric — the three named processes

In a pressure–volume diagram (P on y, V on x), the three basic gas-law processes look like:

Isobaric (constant P): horizontal line. V \propto T (Charles's law applies).
Isochoric (constant V): vertical line. P \propto T (Gay-Lussac's law applies).
Isothermal (constant T): a hyperbola PV = \text{const} (Boyle's law).

The combined gas law is the statement that all three are slices of a single surface PV/T = \text{const} in (P, V, T) space. Any real process can be decomposed as a sequence of these three (and a fourth — adiabatic — where heat does not enter or leave, which is the topic of a later article).

Rebuilding the Kelvin scale from scratch

Given the gas laws, you could construct the Kelvin scale without ever using the word "temperature" in the usual sense. The procedure:

Take a bulb of dilute gas at fixed volume and attach a pressure gauge.
Place the bulb in various reference systems (ice-water, boiling water, liquid nitrogen, a hot oven). Record the pressure at each.
Define temperature ratios by pressure ratios: T_2/T_1 \equiv P_2/P_1.
Pick any one fixed temperature (say, the triple point of water) and assign it a number (273.16).

This is the constant-volume gas thermometer realisation of the Kelvin scale. Historically it was the primary practical temperature standard until the ITS-90 (International Temperature Scale) took over in 1990. Even today, a constant-volume hydrogen thermometer is the gold-standard absolute thermometer against which everything else is calibrated. The whole scale is built on the universality of Gay-Lussac's law.

Where this leads next

Ideal Gas Laws — combining Charles's, Gay-Lussac's, and Boyle's laws into the single equation PV = nRT, and solving problems where all three variables change.
Kinetic Theory — Pressure and Temperature — the microscopic derivation of the ideal gas law from Newton's laws applied to molecules, identifying temperature with average kinetic energy.
First Law of Thermodynamics — when the gas does work by expanding, how much of the input heat goes to that work and how much becomes internal energy.
Thermal Expansion of Solids and Liquids — the gentler expansion of condensed matter, and how solids and gases are two ends of the same continuum.
Temperature and Thermometers — the operational meaning of temperature and the zeroth law that makes it a single number per body.