Charles law, properly written Charles’s law, states that the volume of a fixed mass of gas held at constant pressure is directly proportional to its absolute temperature. Double the temperature in kelvin and the volume doubles. In symbols, V₁ divided by T₁ equals V₂ divided by T₂, with T always in kelvin.
Leave a balloon in a hot car and it strains at the knot, taut and glossy. Bring it out into a cold night and within minutes it sags, wrinkled, looking half-empty.
Nothing leaked. The same air sits inside, pressed by the same atmosphere — only the temperature moved, and the volume obediently followed. That is Charles’s law, doing its work in a car park.
What Is Charles’s Law?
Ask a class what heat does to a gas and someone will say “it makes it expand.” True. But that is a description, not a law — and a law has to tell you exactly how much.
Charles’s law supplies the number. Hold the pressure steady, keep the same quantity of gas, and the volume tracks the absolute temperature in strict proportion. Ten per cent hotter in kelvin means ten per cent bigger, every time.
That proportionality only works if you measure temperature from the right zero. Start counting at absolute zero, in kelvin, and the relationship is a clean straight line through the origin. Start at the freezing point of water, in Celsius, and it falls apart completely.
Two conditions do the heavy lifting here, and both are easy to lose sight of:
- Constant pressure. The gas must be free to push its container outward — a balloon, a piston, a flexible bag. A sealed steel cylinder does not qualify.
- Fixed amount of gas. No molecules added, none allowed to escape. The number of moles stays put.
Who Actually Discovered It?
Jacques Charles, a French physicist and balloonist, ran the experiments around 1787 and never wrote them up. Joseph Louis Gay-Lussac published the volume–temperature relationship in 1802 and pointedly credited Charles’s unpublished work — which is why the law bears the name of the man who did not publish it.
John Dalton reached much the same conclusion independently in 1801. NASA’s Glenn Research Center still teaches it as Charles and Gay-Lussac’s law, which is the fairer title, if a clumsier one.
The Charles’s Law Formula
Charles’s law wears two faces. The proportional form tells you what the physics is. The two-state form is what you actually calculate with.
Both say the same thing: the ratio V/T is a constant. Heat the gas, and the volume climbs by whatever factor keeps that ratio fixed.
- V₁ — initial volume, in cubic metres (m³). Litres are fine, provided V₂ uses the same unit.
- T₁ — initial absolute temperature, in kelvin (K). Never Celsius.
- V₂ — final volume, in the same unit as V₁.
- T₂ — final absolute temperature, in kelvin (K).
- k — the constant of proportionality, k = V/T, in cubic metres per kelvin (m³/K). Its value depends on the pressure and on how much gas you have.
Because both volumes carry the same unit, they cancel in the ratio. Temperature enjoys no such freedom — kelvin, or nothing.
Solving a Charles’s Law Problem in 5 Steps
- Convert both temperatures to kelvin. Add 273.15 to each Celsius value. Do this before you do anything else.
- Check the conditions. Is the pressure constant? Is the amount of gas fixed? If either answer is no, you need a different gas law.
- Write V₁/T₁ = V₂/T₂ and label the knowns. Three of the four quantities will be given.
- Rearrange for the unknown first, then substitute. Solving for volume gives V₂ = V₁ × T₂/T₁. Solving for temperature gives T₂ = T₁ × V₂/V₁.
- Sanity-check the direction. Hotter must mean bigger; colder must mean smaller. If your answer disagrees, you have inverted the ratio.
Step 5 catches more errors than the other four combined. In practice, a student who inverts T₂/T₁ produces an answer that is wrong by a plausible-looking factor, and nothing on the page objects — except the physics.
Prefer to check your working, or skip the arithmetic altogether? Our Charles’s Law Calculator solves for any of the four quantities — enter the three you know and it rearranges the ratio and converts your temperatures to kelvin automatically.
The Ratio, Checked Numerically
Take 2.00 L of gas at 300 K and warm it at constant pressure. Watch what stays fixed.
| Temperature T (K) | Temperature (°C) | Volume V (L) | V / T (L/K) |
|---|---|---|---|
| 300 | 26.85 | 2.00 | 0.00667 |
| 400 | 126.85 | 2.67 | 0.00667 |
| 500 | 226.85 | 3.33 | 0.00667 |
| 600 | 326.85 | 4.00 | 0.00667 |
V/T holds at 0.00667 L/K throughout (3 s.f.). Notice that the Celsius column does no such thing.
How Charles’s Law Works: Why Heating a Gas Pushes the Piston Out
Zoom in far enough and a gas is just molecules, flying, colliding, rebounding off the walls. Temperature is a measure of how fast they are going — specifically, of their average kinetic energy.
Heat the gas and every molecule speeds up. Each wall collision now lands harder, and collisions arrive more often. Pressure is force per unit area, so if the container could not move, the pressure would climb.
But under Charles’s law the container can move. The piston is loaded with a fixed weight, so the outside pressure never changes. The faster molecules shove the piston outward, the gas volume grows, and each square centimetre of wall now receives fewer hits per second because the molecules have farther to travel between them.
The piston stops when the hit rate per unit area has fallen back to exactly what it was before. That happens when the volume has grown in the same ratio as the absolute temperature. Which is the law.
Eight molecules on the left, eight on the right — the amount of gas never changes. Doubling the kelvin temperature doubles the volume.
Push the sliders below and watch the ratio hold. Set T₂ to twice T₁ and the volume should land on exactly twice V₁ — no more, no less.
Why Charles’s Law Only Works in Kelvin
Here is the single most expensive mistake in this topic, and it costs marks in every exam season.
Celsius has an arbitrary zero. It was pinned to the freezing point of water because water was convenient, not because anything special happens to molecular motion there. A gas at 0 °C has plenty of energy left in it.
Kelvin has an absolute zero. According to NIST’s definition of the kelvin, 0 K is absolute zero and 0 °C sits at 273.15 K. Only from that origin does “twice the temperature” mean “twice the molecular kinetic energy” — and therefore twice the volume.
Try it in Celsius and the absurdity is immediate. Going from 1 °C to 2 °C would double the volume. Going from 0 °C to 10 °C would multiply it by infinity. Going from −10 °C to −20 °C would somehow produce a positive volume from a negative one.
Plot volume against absolute temperature and you get a straight line aimed squarely at the origin. That is the whole law in one picture.
Extend the measured line backwards and it hits zero volume at 0 K. No gas ever gets there — it liquefies first — but the extrapolation is how absolute zero was first pinned down.
Real-World Examples of Charles’s Law
1. Hot-air balloons. A burner heats the air inside the envelope from roughly 15 °C to about 100 °C. At constant pressure the air cannot expand sideways — the envelope is already full — so it expands out of the mouth, leaving fewer molecules inside. The trapped air is now about 23% less dense than the air outside, and the balloon rises.
2. A balloon dunked in liquid nitrogen. The classic lecture demo. Drop an inflated balloon into nitrogen at 77 K and it shrivels to a fraction of its size within seconds. Lift it out and it reinflates as the air warms back to room temperature. Nothing escaped; volume simply tracked T.
3. Rescuing a dented ping-pong ball. Roll it into hot water and the dent pops out. The air inside warms, tries to occupy more volume, and the thin celluloid gives way at the weakest point — the dent.
4. Bread and cakes rising in the oven. Yeast and baking powder make the bubbles; the oven makes them big. Every trapped pocket of gas expands as it heats, and the dough sets around the enlarged bubbles. This is why an oven door opened too early gives you a sunken cake.
5. Thermals and rising warm air. At constant pressure a gas’s density is inversely proportional to its absolute temperature — the same proportionality, written the other way up. Sun-warmed air over a ploughed field is less dense than the air above it, so it rises. Gliders and buzzards both make a living from this.
Charles’s Law vs the Other Gas Laws
Four simple gas laws exist, and each one freezes two variables to watch the other two dance. Students mix them up constantly. The fix is to ask one question first: what is being held constant?
| Gas law | Relationship | Held constant | Two-state equation | Graph shape |
|---|---|---|---|---|
| Boyle’s law | P ∝ 1/V | T, n | P₁V₁ = P₂V₂ | Hyperbola (P against V) |
| Charles’s law | V ∝ T | P, n | V₁/T₁ = V₂/T₂ | Straight line through the origin (V against T) |
| Gay-Lussac’s law (pressure law) | P ∝ T | V, n | P₁/T₁ = P₂/T₂ | Straight line through the origin (P against T) |
| Avogadro’s law | V ∝ n | P, T | V₁/n₁ = V₂/n₂ | Straight line through the origin (V against n) |
| Combined gas law | PV/T is constant | n only | P₁V₁/T₁ = P₂V₂/T₂ | Surface, not a curve |
| Ideal gas law | PV = nRT | R = 8.314 J/(mol·K) | PV = nRT | Surface, not a curve |
One warning worth carrying into an exam. Some textbooks call the pressure–temperature relationship “Gay-Lussac’s law”, others call it “Amontons’s law”, and a handful use “Gay-Lussac’s law” to mean Charles’s law itself. Read the equation, not the name.
Common Misconceptions About Charles’s Law
Misconception 1: Doubling the Celsius temperature doubles the volume
It does not, and the error is enormous. Heating a gas from 20 °C to 40 °C looks like a doubling, but in kelvin it is 293.15 K to 313.15 K — a rise of just 6.8%.
The volume grows by 6.8%, not by 100%. A student who forgets to convert overstates the answer by nearly fifteen-fold.
Misconception 2: Charles’s law applies to any sealed container
Only if the container can change volume. Heat a sealed steel gas cylinder and its volume barely budges; the pressure rockets instead. That is the pressure law, not Charles’s law.
Ask yourself whether the walls can move. Balloon, piston, syringe, flexible bag — Charles’s law. Rigid tin, tyre, aerosol can — not Charles’s law.
Misconception 3: Gas volume really reaches zero at absolute zero
The straight line says so; nature does not oblige. Every real gas condenses into a liquid long before it gets near 0 K — nitrogen at 77 K, helium at 4.2 K.
Below that point there is no gas left to obey the law. The extrapolation is a mathematical device, and a spectacularly useful one: it is how absolute zero was located in the first place.
Misconception 4: Heating a balloon works by raising the pressure inside
Barely. A rubber balloon holds its internal pressure at atmospheric plus a small excess from the stretched skin. Heat it and that pressure stays almost unchanged.
What changes is the volume. The gas has an escape route — the balloon simply grows — so pressure never gets the chance to build. Take that escape route away and you have a different law entirely.
How Charles’s Law Relates to the Ideal Gas Law and Thermodynamics
Charles’s law is not a standalone rule. It is a special case, quietly hiding inside the ideal gas law.
Rearrange for V/T and you get V/T = nR/P. Hold n and P fixed and the entire right-hand side is a constant — which is precisely the statement V₁/T₁ = V₂/T₂. Charles’s law falls out in one line.
Boyle’s law and Avogadro’s law drop out the same way, each by freezing a different pair of variables. This is worth internalising, because it means you only ever need to remember one equation.
Underneath all of it sits kinetic theory, where the absolute temperature of a gas is directly proportional to the average kinetic energy of its molecules. That single link explains why the kelvin scale is the only one that works: it counts energy from zero, not from the freezing point of a puddle.
The distinction between energy transferred and temperature reached matters here too. If the difference between heat and temperature feels blurry, that is worth clearing up before you tackle gas laws in earnest.
How much heat you need to raise the temperature depends on the material’s specific heat capacity. And the constant-pressure expansion described by Charles’s law is exactly the process in which a gas does work on its surroundings — a first-law idea explored in the laws of thermodynamics.
Worked Problems
Show Solution
Solution:
Step 1: Charles’s law applies (constant pressure, sealed balloon). V₁/T₁ = V₂/T₂, so V₂ = V₁ × T₂/T₁.
Step 2: Convert to kelvin. T₁ = 27 + 273.15 = 300.15 K. T₂ = 87 + 273.15 = 360.15 K.
Step 3: V₂ = 2.50 L × (360.15 K / 300.15 K) = 2.50 L × 1.1999 = 3.00 L.
Answer: V₂ = 3.00 L
Show Solution
Solution:
Step 1: V₂ = V₁ × T₂/T₁. Both temperatures are already in kelvin.
Step 2: V₂ = 1.20 m³ × (280 K / 350 K).
Step 3: V₂ = 1.20 m³ × 0.800 = 0.960 m³.
Sanity check: the gas got colder, so the volume shrank. Correct direction.
Answer: V₂ = 0.960 m³
Show Solution
Solution:
Step 1: Solving for temperature, T₂ = T₁ × V₂/V₁.
Step 2: T₁ = 20.0 + 273.15 = 293.15 K. V₂/V₁ = 66.0 mL / 60.0 mL = 1.100.
Step 3: T₂ = 293.15 K × 1.100 = 322.47 K.
Step 4: Convert back. θ₂ = 322.47 − 273.15 = 49.3 °C.
Answer: T₂ = 322.5 K, or 49.3 °C
Show Solution
Solution:
Step 1: The claim uses Celsius. Convert. T₁ = 298.15 K, T₂ = 323.15 K.
Step 2: V₂/V₁ = T₂/T₁ = 323.15 / 298.15 = 1.0839.
Step 3: Percentage increase = (1.0839 − 1) × 100% = 8.39%.
The volume grows by about 8.4%, not 100%. Doubling the Celsius reading is not doubling the temperature.
Answer: about 8.4%
Show Solution
Solution:
Step 1: Rearrange V₁/T₁ = V₂/T₂ for T₁, giving T₁ = T₂ × V₁/V₂.
Step 2: T₁ = 250 K × (4.50 L / 3.75 L).
Step 3: T₁ = 250 K × 1.20 = 300 K.
Sanity check: the gas was hotter to begin with, because it started bigger. 300 K > 250 K. Correct.
Answer: T₁ = 300 K (26.85 °C)
Show Solution
Solution:
Step 1: Density is mass per unit volume, ρ = m/V. With mass fixed and V ∝ T, density is inversely proportional to T: ρ₂ = ρ₁ × T₁/T₂.
Step 2: T₁ = 273.15 K, T₂ = 373.15 K.
Step 3: ρ₂ = 1.29 kg/m³ × (273.15 K / 373.15 K) = 1.29 × 0.7320 = 0.944 kg/m³.
In practice this is why a chimney draws: hot flue gas is roughly 27% lighter than the cold air around it.
Answer: ρ₂ = 0.944 kg/m³
Show Solution
Solution:
Step 1: Find the density inside using ρ₂ = ρ₁ × T₁/T₂, with T₁ = 288.15 K and T₂ = 373.15 K.
Step 2: ρ_inside = 1.225 kg/m³ × (288.15 / 373.15) = 1.225 × 0.7722 = 0.946 kg/m³.
Step 3: Density difference Δρ = 1.225 − 0.946 = 0.279 kg/m³.
Step 4: Lifted mass = Δρ × V = 0.279 kg/m³ × 2800 m³ = 781 kg.
Step 5: Lift force = 781 kg × 9.81 m/s² = 7.66 × 10³ N.
Sanity check: a real envelope, basket, burner and fuel weigh several hundred kilograms, which is why balloons this size carry only a handful of passengers.
Answer: 781 kg of lift, or 7.66 kN
Show Solution
Solution:
Step 1: Volume varies linearly with Celsius temperature: V = V₀ + mθ, where V₀ = 25.00 mL is the volume at 0.00 °C.
Step 2: Find the gradient. m = (34.15 − 25.00) mL / (100.0 − 0.00) °C = 9.15 / 100.0 = 0.0915 mL/°C.
Step 3: Set V = 0 and solve for θ. 0 = 25.00 + 0.0915θ, so θ = −25.00 / 0.0915.
Step 4: θ = −273.2 °C.
This is how absolute zero was first estimated — not by reaching it, but by drawing a line towards it. The accepted value is −273.15 °C.
Answer: absolute zero ≈ −273.2 °C