Thermodynamics

Gay-Lussac’s Law (P1/T1 = P2/T2)

Definition

Gay-Lussac’s law states that the pressure of a fixed quantity of gas in a rigid sealed container rises in direct proportion to its absolute temperature, so P1/T1 = P2/T2 with every temperature in kelvin. Heating the gas gives it no room to expand into, so the whole effect appears as pressure.

Every aerosol can carries a line telling you not to leave it in the sun. It is not decoration. The can is a sealed steel box that cannot grow, so warming the propellant inside has only one outlet: the pressure pushing outwards on the walls.

The same physics shows up in a duller place — the tyre pressure gauge on the forecourt. Check your tyres after half an hour on the motorway and every one of them reads high, not because air has been added, but because the air inside is hotter than it was on the driveway. One short equation covers both cases.

What Is Gay-Lussac’s Law?

Gay-Lussac’s law is the rule that the pressure of a fixed amount of gas held at constant volume is directly proportional to its absolute temperature. Heat the gas by a certain fraction on the kelvin scale and the pressure rises by exactly that fraction.

Three conditions have to hold, and each one earns its place. The container must keep the same volume, so there is nowhere for the expansion to go. The amount of gas must stay the same, which rules out leaks. And the temperature must be measured on an absolute scale, in kelvin.

Put like that, the law is a statement about a ratio. Divide the pressure by the absolute temperature at any moment and you get the same number every time, however hot or cold the gas becomes. That number is fixed by how much gas you sealed in and how big the container is.

You will meet the same law under two other names. Some textbooks call it the pressure-temperature law, which says what it does; others call it Amontons’s law, which is closer to who found it. All three names describe one behaviour: a sealed, rigid container of gas whose pressure tracks its absolute temperature.

The Gay-Lussac’s Law Formula

Written for two states of the same sealed gas — before and after a temperature change — the law takes this form:

P1 / T1 = P2 / T2
Symbol Quantity SI unit Notes
P1absolute pressure in the starting statePameasured from a vacuum, not from the surrounding air; kPa, bar, atm and psi all work provided both pressures use the same unit
T1absolute temperature in the starting stateKkelvin only; T/K = temperature in °C + 273.15
P2absolute pressure in the final statePathe quantity you usually want
T2absolute temperature in the final stateKkelvin only, again
V, nvolume and amount of gasm3, moldo not appear in the formula because both are held fixed

Rearranged for the pressure you are usually chasing, it reads:

P2 = P1 × T2 / T1

So the working is short: convert both temperatures to kelvin, form the ratio T2/T1, and multiply the starting pressure by it. If you would rather have the arithmetic and the unit conversions done for you — including solving backwards for a temperature — the Gay-Lussac’s law calculator takes pressures in Pa, kPa, atm, bar or psi and temperatures in K, °C or °F.

One habit will save you marks. Keep the temperature ratio in full during the calculation and round only the final answer, because rounding the ratio first can shift the result by a whole digit. Problem 1 below shows exactly where that goes wrong.

The lab underneath lets you build the experiment instead of imagining it. You choose the pressure and the temperature at which the vessel was sealed, then move the current temperature and watch what happens.

Two things are worth staring at. The vessel outline never changes size at any setting, because the volume is fixed by the walls, and the pressure-temperature graph beside it stays a single straight line that runs back to zero pressure at -273.15 °C, whichever starting state you choose.

Gay-Lussac's Law Lab

Why the Temperature Must Be in Kelvin

Temperatures must be in kelvin because the law is a statement about ratios, and a ratio is only meaningful on a scale whose zero is the real zero. The Celsius zero is the melting point of ice — a useful landmark, but an arbitrary one as far as a gas is concerned.

Watch what the choice does to a number. Take gas sealed at 100.0 kPa at 27 °C and heat it to 327 °C. In kelvin the temperatures are 300.15 K and 600.15 K, the ratio is 1.9995002, and the pressure ends at 199.950 kPa. Work in Celsius and you would compute 327/27 = 12.111111, predicting 1211.11 kPa — too high by a factor of 6.057.

The kelvin scale exists precisely because gases behave this way. Measure the pressure of a sealed sample at several temperatures, plot the points and extend the straight line backwards, and it always reaches zero pressure at the same place: -273.15 °C. As NIST’s introduction to the kelvin puts it, the scale is absolute in the sense that it starts at absolute zero, and 0 K is exactly -273.15 °C.

Modern metrology no longer puts a gas in the definition, though gas thermometry is still one of the ways the kelvin is realised in the laboratory. The BIPM’s definition of the kelvin fixes the Boltzmann constant at 1.380 649 × 10-23 J/K, which pins the size of one kelvin to an energy. The gas experiments came first, though, and they are still the clearest way to see where the zero comes from.

Graph of absolute pressure against temperature in degrees Celsius for one sealed rigid flask. The horizontal axis runs from minus 300 to plus 320 degrees Celsius and the vertical axis from 0 to 220 kilopascals. A single straight gold line rises from left to right. It is dashed from the point where it meets zero pressure at minus 273.15 degrees Celsius, which is 0 kelvin, up to the sealed state at 20 degrees Celsius and 101.3 kilopascals; solid between that point and the heated state at 100 degrees Celsius and 128.9 kilopascals; and dashed again beyond. A dashed horizontal line marks atmospheric pressure at 101.325 kilopascals, which the sealed state sits on.
Figure 1. Absolute pressure against temperature for one sealed rigid vessel holding gas at 101.3 kPa and 20 °C. The solid section spans the two states used in Problem 1; the dashed line is extrapolation. Note that the sealed point sits on the atmospheric line, because the vessel was closed at ordinary air pressure, and that the shaded band marks temperatures at which ordinary air has condensed and the law no longer applies.

The table reads that same line off at nine temperatures. Every row belongs to one sealed vessel, and the last column is the giveaway: divide the pressure by the absolute temperature and you get 0.3456 kPa per kelvin every time.

Temperature (°C) Absolute temperature (K) Pressure (kPa) P/T (kPa per K)
-20073.1525.280.3456
-78195.1567.440.3456
-30243.1584.020.3456
0273.1594.390.3456
20 (sealed here)293.15101.30.3456
50323.15111.70.3456
100373.15128.90.3456
200473.15163.50.3456
300573.15198.10.3456

The coldest row carries a warning. Ordinary air condenses near -190 °C at atmospheric pressure, and once the gas has turned to liquid the law has nothing left to describe, so the -200 °C row says where the straight line goes, not where a flask of air would follow it.

What Happens Inside the Gas When You Heat It

Pressure is the drumming of molecules on the walls, and temperature sets how hard they drum. Raise the absolute temperature and each molecule carries more momentum and arrives at the wall more often — two effects, both growing with the molecular speed.

Two drawings of the same sealed rigid vessel side by side. On the left the vessel is sealed at 20 degrees Celsius and its gauge needle points to 101.3 kilopascals on a dial marked from 0 to 200; on the right the same vessel is heated to 100 degrees Celsius and the needle has moved to 128.9 kilopascals. The vessel outline, the gas volume and the sixteen molecules are drawn in identical positions in both panels, with longer motion streaks on the hot side. A dimension line under each vessel marks the same width, showing that the volume has not changed.
Figure 2. The same sealed rigid vessel at 20 °C and at 100 °C. The outline, the gas volume and the sixteen molecules are drawn in identical positions in both panels; only the motion streaks and the gauge needle differ, with the needle moving from 101.3 kPa to 128.9 kPa.

Here is where the square root earns its keep. Mean molecular speed grows as the square root of the absolute temperature, so warming a gas from 20 °C to 100 °C multiplies the speed by sqrt(373.15/293.15) = 1.1282277. The molecules therefore hit about 1.128 times as often and about 1.128 times as hard.

Multiply those two factors together and the square root undoes itself: 1.1282277 × 1.1282277 = 1.2728977, which is the temperature ratio 373.15/293.15 = 1.2728978 back again, to within the rounding of the square root. Pressure ends up proportional to the absolute temperature, not to its square root, and not to anything on the Celsius scale.

This is also the cleanest way to see why the law needs an absolute scale. Temperature in kelvin is a direct measure of the average kinetic energy of the molecules, which is why it is not the same quantity as heat: heat is energy in transit, temperature is the vigour of the jostling already there.

Gauge Pressure or Absolute Pressure?

Gay-Lussac’s law needs absolute pressure, measured from a vacuum, not the gauge reading you get from a tyre gauge or a workshop manometer. Those instruments read the amount by which the pressure inside beats the air outside, which is why a flat tyre reads zero rather than 101 kPa.

Two horizontal stacked bars comparing a car tyre before and after a fast drive. Each bar is measured on a scale from 0 to 400 kilopascals starting at a vacuum. The cold tyre at 10 degrees Celsius has a mist-coloured section of 101.325 kilopascals for the atmosphere plus a wine section of 220.0 kilopascals of gauge pressure, giving 321.3 kilopascals absolute. After the drive at 45 degrees Celsius the atmosphere section is unchanged and the gauge section has grown to 259.7 kilopascals, giving 361.0 kilopascals absolute. A dashed vertical line marks where a tyre gauge's own scale begins.
Figure 3. The tyre of Problem 2, before and after a fast drive. Each bar runs from a true vacuum, so the gauge reading is only the upper section and the atmosphere makes up the rest. Adding 101.325 kPa turns a gauge reading into the absolute pressure the law needs; subtracting it again turns the answer back into something a gauge would show.

The round trip has three steps. Add the atmosphere to the gauge reading, apply the law to the absolute pressures, then subtract the atmosphere from the answer. Skip the first step and the law is fed a number that is about 101 kPa too small, and every proportional change you compute comes out too small with it.

For the tyre in Problem 2 the honest answer is a gauge reading of 259.7 kPa; feeding the gauge figure straight into the law gives 247.194 kPa, which is 12.525 kPa short. A common student slip is to notice that the units match and conclude that the numbers must be interchangeable.

One practical caveat: 101.325 kPa is the standard atmosphere, an agreed reference value. The actual air pressure where you are standing departs from it with the weather, and departs much further with altitude, so treat the conversion as a working approximation rather than as an exact accounting. If the difference matters for your answer, measure the local pressure instead of assuming it.

Gay-Lussac’s Law, Boyle’s Law and Charles’s Law

The three simple gas laws are one experiment run three ways: hold one of the three state variables fixed and watch how the other two trade against each other. Gay-Lussac’s law is the member of that family whose walls cannot move.

Law Held constant What it links Relation
Boyle’s lawtemperature, amountpressure and volumeP1V1 = P2V2
Charles’s lawpressure, amountvolume and temperatureV1/T1 = V2/T2
Gay-Lussac’s lawvolume, amountpressure and temperatureP1/T1 = P2/T2
Combined gas lawamount onlyall three at onceP1V1/T1 = P2V2/T2
Ideal gas lawnothingpressure, volume, amount, temperaturePV = nRT

Reading down the table, the family resemblance is obvious. A syringe with its nozzle blocked, squeezed at room temperature, is Boyle’s law at work; a balloon shrinking in a freezer, free to change size at constant air pressure, is Charles’s law; a sealed steel cylinder in the same freezer is this one.

All three are special cases of the same statement. Fix any one variable in PV = nRT and the corresponding simple law drops out, which is why the ideal gas law is worth learning once you are comfortable with the three simple ones.

The practical question is which member of the family your container belongs to. If the walls cannot move, use Gay-Lussac’s law; if the container is free to change size while the outside pressure holds steady, you are in Charles’s territory instead.

Everyday Examples of Gay-Lussac’s Law

Sealed rigid containers are everywhere once you start looking, and the law explains something you can check in all of them.

  • An aerosol can in a hot car. A can at 300.0 kPa absolute on a 20 °C morning reaches 340.9 kPa if the cabin gets to 60 °C — a rise of nearly 41 kPa with nobody touching it. That is why the labels warn against heat and against throwing empties on a fire.
  • Tyres, checked cold. Manufacturers quote cold pressures because a warm tyre reads high. The same tyre at 220.0 kPa gauge on a 10 °C morning reads about 203 kPa gauge after a night at -5 °C, which is how a tyre can look soft in winter without having lost any air.
  • Constant-volume gas thermometers. Turn the law around and the pressure becomes the thermometer. A rigid bulb of gas of known pressure at the ice point measures temperature directly from the pressure it reaches, which is the instrument that mapped out the absolute scale in the first place.
  • Gas cylinders left outside. A cylinder filled to 250.0 kPa absolute in a 25 °C workshop falls to about 203.9 kPa at -30 °C. Nothing has escaped; the gauge simply shows the lower absolute temperature.
  • Sealed electronics and dry packaging. A sealed sensor housing or a dry sealed package pushes harder on its seams as it warms, and that load grows in proportion to the absolute temperature of the gas inside it. Put liquid in the container and this law no longer governs the total: the liquid’s own vapour adds a pressure that climbs far faster than the temperature does.
Warning label on an aerosol can telling the user not to expose it to heat, an everyday example of Gay-Lussac's law
The warning exists because the can cannot expand: heating it raises the pressure in proportion to the absolute temperature.

A word on those warnings. Aerosol labels commonly tell you not to expose the can to temperatures above 50 °C, and that is a manufacturer’s instruction rather than a number this law produces. The law tells you how the pressure grows; what a given can can withstand is a question for its maker.

Common Mistakes and Where Gay-Lussac’s Law Breaks Down

Most lost marks come from four habits, and the first two are just bookkeeping.

  • Using Celsius in the ratio. The commonest error of all, and the most expensive: in Problem 5 it inflates the predicted pressure more than sixfold.
  • Feeding a gauge reading into the law. Add the atmosphere first, then take it off the answer.
  • Rounding the temperature ratio early. Carry it in full to the end; Problem 1 shows how a fourth-figure rounding moves the printed answer.
  • Assuming any closed container is rigid. A glass flask or a steel cylinder is close to rigid. A car tyre is not.

That last one deserves a sentence of its own, because it is physics rather than arithmetic. A tyre’s sidewalls flex, so a warm tyre grows very slightly, and the extra volume takes the edge off the pressure rise — the real increase is a little smaller than the law predicts, though it is close enough for everyday checks.

The law is an ideal-gas result, and real gases drift away from it. At ordinary pressures the agreement is excellent; near liquefaction, and at pressures of many atmospheres, the departure becomes measurable, and a sealed cylinder at hundreds of bar is not a place to trust a one-line proportion.

Cooling has a harder limit. The straight line can be extended all the way to -273.15 °C on paper, but ordinary air has condensed near -190 °C long before it gets there — nitrogen boils at about -196 °C and oxygen at about -183 °C — and a liquid obeys no gas law. The extrapolation is how the absolute scale was inferred; it is not a prediction about a real flask.

Two more quiet assumptions are worth naming. The amount of gas must be constant, so a slow leak or gas dissolving into a liquid inside the container breaks the law before any physics does. And the container itself changes size a little when heated, since thermal expansion applies to steel and glass as much as to the gas inside.

Who Actually Discovered Gay-Lussac’s Law?

The name is a historical accident, and the honest answer is that Guillaume Amontons got there first, around 1700, when he found that the pressure of a fixed volume of air tracks its temperature. That is why many textbooks, chemistry ones especially, call the pressure-temperature relation Amontons’s law.

Gay-Lussac’s own paper of 1802 was about something adjacent: how gases expand when heated at constant pressure. In it he credited unpublished work by Jacques Charles, which is why that relation now usually carries Charles’s name — and why NASA’s Beginner’s Guide to Aeronautics still files the volume-temperature result under Charles and Gay-Lussac’s Law. The labels have been shuffled more than once.

There is a second Gay-Lussac’s law to keep clear of this one. In chemistry, the law of combining volumes says that gases react in simple whole-number ratios by volume, which is a statement about reactions, not about heating a sealed can. If you arrived here looking for that law, it is a different result with the same surname attached.

Worked Problems

Problem 1
A sealed rigid flask holds gas at 101.3 kPa at 20 C. It is heated to 100 C. What is the pressure inside?
Show Solution
Solution: Step 1: The volume and the amount of gas are fixed, so P2 = P1 × T2/T1 with both temperatures in kelvin. Step 2: T1 = 20 + 273.15 = 293.15 K and T2 = 100 + 273.15 = 373.15 K, so the ratio is 373.15/293.15 = 1.2728978. Step 3: P2 = 101.3 kPa × 1.2728978 = 128.945 kPa, a rise of 128.945 – 101.3 = 27.645 kPa. Answer: 128.9 kPa (about 27.6 kPa above the sealed pressure). Watch the rounding here: if you shorten the ratio to 1.273 first, 101.3 × 1.273 = 128.955 kPa, which prints as 129.0 kPa. Carry the full ratio and round once, at the end.
Problem 2
A tyre gauge reads 220.0 kPa on a cold morning at 10 C. After a fast drive the air inside is at 45 C. What will the gauge read now? Take atmospheric pressure as 101.325 kPa.
Show Solution
Solution: Step 1: The law needs absolute pressure, so convert the gauge reading first: P1 = 220.0 + 101.325 = 321.325 kPa. Step 2: T1 = 283.15 K and T2 = 318.15 K, so T2/T1 = 318.15/283.15 = 1.1236094, and P2 = 321.325 kPa × 1.1236094 = 361.044 kPa absolute. Step 3: Convert back for the gauge: 361.044 – 101.325 = 259.719 kPa. Answer: about 259.7 kPa on the gauge, or 37.67 psi, up from 31.91 psi. Had you put the gauge reading straight into the law, you would have got 220.0 × 1.1236094 = 247.194 kPa; that is 259.719 – 247.194 = 12.525 kPa short of the real reading.
Problem 3
An aerosol can is filled to 300.0 kPa absolute at 20 C and left in a car that reaches 60 C. By how much does its pressure rise, in kPa and as a percentage?
Show Solution
Solution: Step 1: The can is rigid and sealed, so P2 = P1 × T2/T1. Step 2: T1 = 293.15 K and T2 = 333.15 K, giving 333.15/293.15 = 1.1364489, so P2 = 300.0 kPa × 1.1364489 = 340.935 kPa. Step 3: The rise is 340.935 – 300.0 = 40.935 kPa, and the ratio gives the percentage directly: 1.1364489 – 1 = 0.1364489. Answer: the pressure rises by about 40.9 kPa to 340.9 kPa, an increase of 13.64 %. The percentage depends only on the two absolute temperatures, so every sealed can in that car gains the same 13.64 %, whatever it started at.
Problem 4
A constant-volume gas thermometer reads 100.0 kPa when its bulb is in melting ice. Placed in a warm bath, it reads 136.0 kPa. What is the bath temperature?
Show Solution
Solution: Step 1: Rearranged for the unknown temperature, the law reads T2 = T1 × P2/P1. Step 2: The ice point is T1 = 273.15 K, and P2/P1 = 136.0/100.0 = 1.36. Step 3: T2 = 273.15 K × 1.36 = 371.484 K, which is 371.484 – 273.15 = 98.334 °C. Answer: about 98.33 °C. As a sanity check, at exactly 100 °C this thermometer would read 136.610 kPa, so a reading of 136.0 kPa should indeed fall a little short of boiling.
Problem 5
Gas at 100.0 kPa and 27 C is heated to 327 C. Work out the new pressure, then work out what a student gets by using Celsius in the ratio, and compare.
Show Solution
Solution: Step 1: Convert: T1 = 27 + 273.15 = 300.15 K and T2 = 327 + 273.15 = 600.15 K. Step 2: The correct ratio is 600.15/300.15 = 1.9995002, so P2 = 100.0 kPa × 1.9995002 = 199.950 kPa. Step 3: In Celsius the student would form 327/27 = 12.111111 and predict 100.0 kPa × 12.111111 = 1211.11 kPa, which is 1211.11/199.950 = 6.057 times too big. Answer: 199.95 kPa, not 1211 kPa. Note also that this famous “doubling” is not quite exact: 27 °C is 300.15 K, not 300.00 K, so the true ratio is 1.9995 rather than 2.
Problem 6
The same flask from Problem 1 (101.3 kPa at 20 C) is cooled to -200 C. What pressure does the law predict, and how far can you trust it?
Show Solution
Solution: Step 1: P2 = P1 × T2/T1, as before. Step 2: T2 = -200 + 273.15 = 73.15 K, so T2/T1 = 73.15/293.15 = 0.24953096. Step 3: P2 = 101.3 kPa × 0.24953096 = 25.277 kPa. Answer: 25.28 kPa on paper. In practice you would never measure it with air: the sample condenses near -190 °C, so the last stretch of that line, including the point where it reaches zero pressure at -273.15 °C, is extrapolation rather than experiment.
Problem 7
A different rigid gas cylinder reads 205.0 kPa absolute at -30 C. It was filled to 250.0 kPa absolute. At what temperature was it filled, assuming nothing has leaked?
Show Solution
Solution: Step 1: Here the unknown is the starting temperature, so rearrange to T1 = T2 × P1/P2. Step 2: T2 = -30 + 273.15 = 243.15 K, and P1/P2 = 250.0/205.0 = 1.2195122. Step 3: T1 = 243.15 K × 1.2195122 = 296.524 K, which is 296.524 – 273.15 = 23.374 °C. Answer: about 23.4 °C, a normal indoor temperature. This is not the workshop cylinder from the examples list above, which was filled at 25 °C and reads 203.9 kPa at this same -30 °C. Get a filling temperature back that is absurd, and the assumption to question is “nothing has leaked” rather than the arithmetic.

Frequently Asked Questions

What is Gay-Lussac's law in simple terms?
Gay-Lussac’s law says that a gas sealed in a container that cannot change size pushes harder on the walls as it gets hotter, in exact proportion to its absolute temperature. Raise the kelvin temperature by 10 per cent and the pressure rises by 10 per cent. Cool the gas and the pressure drops by the same fraction.
Why must the temperature be in kelvin?
Temperature must be in kelvin because the law compares two temperatures as a ratio, and only a scale that starts at absolute zero makes such a ratio meaningful. Celsius starts at the melting point of ice, an arbitrary landmark as far as a gas is concerned. Heating from 27 to 327 degrees Celsius multiplies the kelvin temperature by 1.9995, not by 12.1.
What happens to the pressure if the temperature doubles?
The pressure doubles, provided the volume and the amount of gas are unchanged and the doubling is measured in kelvin. Going from 300 K to 600 K takes 100 kPa to 200 kPa. Celsius cannot express such a ratio at all: the gas thermometer of Problem 4 climbs from 100.0 kPa to 136.0 kPa, a rise of 36 per cent, while its Celsius reading goes from 0 to 98.33 and 98.33 divided by zero means nothing.
Can I put a tyre gauge reading straight into Gay-Lussac's law?
No, because a tyre gauge measures the pressure above the surrounding air rather than from a vacuum. Add about 101.3 kPa to the reading, apply the law, then subtract the same amount from the answer. A tyre at 220.0 kPa gauge warmed from 10 to 45 degrees Celsius really reaches 259.7 kPa gauge, not the 247.2 kPa the shortcut predicts.
Is Gay-Lussac's law the same as Amontons's law?
Yes, they are two names for the same relation between pressure and absolute temperature at constant volume. Guillaume Amontons measured it around 1700, roughly a century before Gay-Lussac’s work on gases, so chemists often prefer his name. Physics textbooks tend to say Gay-Lussac’s law or, avoiding the argument entirely, the pressure-temperature law.
Is this the same as Gay-Lussac's law of combining volumes?
No. The law of combining volumes is a chemistry result about reacting gases, stating that the volumes they consume and produce stand in simple whole-number ratios at the same temperature and pressure. It has nothing to do with heating a sealed container. The two share a name because the same scientist worked on both.
Does Gay-Lussac's law work for real gases?
It works well for real gases at everyday pressures, which is why tyre and aerosol estimates come out right. The law is derived for an ideal gas, so departures grow as the gas approaches liquefaction or is squeezed to many atmospheres. Once the gas condenses the law stops applying altogether, which is why very cold predictions are extrapolation.
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