Thermodynamics

The Combined Gas Law: When Pressure and Temperature Both Move

Definition

The combined gas law states that for a fixed amount of gas, the pressure times the volume divided by the absolute temperature keeps the same value through any change: P1V1/T1 = P2V2/T2. Hand it any five of those six quantities and it returns the sixth. Temperatures go in kelvin, pressures are absolute.

A sealed bag of crisps carried up a mountain road arrives looking like a pillow. The air outside thinned as you climbed, the air trapped inside had nowhere to go, and the packet stretched to take up the slack. Meanwhile the temperature fell, quietly pulling the other way.

That tug-of-war is the whole subject. Boyle’s law handles a squeeze at a steady temperature, Charles’s law a warming at a steady pressure — but a balloon, a tyre and a crisp packet rarely hold anything steady. One equation covers the case where two things move at once.

What Is the Combined Gas Law?

The combined gas law says that for a sealed sample of gas, the group PV/T keeps the same value no matter how the three quantities inside it change. Move two of them and the third has to shift so the group stays put.

Think of it as a budget. The gas is handed a fixed allowance at the start; spend some of it on a higher pressure and something has to give, either the volume or the temperature. Nothing is created and nothing is lost, because the allowance was set by how much gas is in there.

Fixed amount is the condition that does all the work. The same molecules have to be present at the end as at the beginning — no leak, no gas dissolving into a liquid, no reaction quietly eating or making them. Break that and the law has nothing to say.

The name is literal. Three older relations each pinned one quantity and let the other two move; this one refuses to pin anything, which is why it is the version that survives contact with a real balloon.

The Combined Gas Law Formula

Written for two states of the same sample, the law is a single equality between two groups.

P1V1 / T1 = P2V2 / T2

Subscript 1 is the state before, subscript 2 the state after. Rearranged for the quantity people usually want, the volume at the end, it becomes one multiplication over another.

V2 = P1V1T2 / (T1P2)

Any of the six can be the unknown, so there are six rearrangements and no one of them is more fundamental than the others. If you would rather not do the algebra by hand, the combined gas law calculator takes any five values, in kilopascals, litres, kelvin, Celsius or Fahrenheit, and returns the sixth with the working shown.

  • P2 = P1V1T2 / (T1V2)
  • T2 = T1P2V2 / (P1V1)
  • V1 = P2V2T1 / (T2P1)
  • P1 = P2V2T1 / (T2V1)
  • T1 = T2P1V1 / (P2V2)
SymbolWhat it isSI unitWorth knowing
P1pressure of the gas before the changePaabsolute pressure, never a gauge reading; kilopascals, atmospheres, bars and psi all work provided P2 uses the same one
V1volume it occupies beforem3litres are the practical unit here; the volume unit cancels, so it only has to match V2
T1temperature beforeKkelvin only, because the law divides by it — NIST puts it plainly that zero degrees Celsius is 273.15 K
P2pressure afterPathe second state has to be a settled one, not a reading taken mid-squeeze
V2volume afterm3the quantity asked for in most exam questions, and the one the simulator below computes for you
T2temperature afterKa value of exactly 0 K sends the volume to zero, which is why a good tool refuses it rather than printing it

Notice what is missing from that table: the amount of gas, and any constant of nature. Both cancel when you divide one state by the other, which is exactly why the units of pressure and volume are free and the unit of temperature is not.

The simulator below puts the equation on screen. Five sliders set the pressure, volume and temperature before, and the pressure and temperature after; the volume after is not a slider at all, because that one belongs to the physics. Drag any slider and watch which way the second cylinder moves.

Combined Gas Law Lab

Where the Combined Gas Law Comes From

It is not a new law. Start from the equation of state for an ideal gas and hold one thing fixed, and this drops out in a line.

PV = nRT

Divide both sides by T and you get PV/T = nR. For a sealed sample the number of moles n never changes, and R is the same number for every gas, so the whole right-hand side is a fixed quantity for that particular sample. Whatever PV/T was before, it is that again afterwards.

That single step is worth more than it looks, because it tells you precisely what breaks the law: anything that changes n. Our guide to the ideal gas law carries the full derivation and the value of R; this page takes the finished result and uses it.

One warning about the word constant. PV/T is constant for one sample of gas, not across samples — double the amount of gas and the constant doubles with it. It is a bookkeeping value for the gas in front of you, not a property of the universe.

The Three Gas Laws Hiding Inside It

Set any one of the three quantities equal on both sides and a familiar law falls out of the algebra. That is the sense in which the combined law contains the others rather than competing with them.

Hold this equalWhat is leftWhose law it isRead more
T2 = T1P1V1 = P2V2Boyle’s lawsqueeze it at a steady temperature
P2 = P1V1/T1 = V2/T2Charles’s lawwarm it at a steady pressure
V2 = V1P1/T1 = P2/T2Gay-Lussac’s lawheat it inside something rigid

Read the two halves separately and you get the two shapes below. Hold the pressure after fixed and sweep the temperature after, and the volume lands on a straight line through the origin; the graph is drawn from the exact values, not from readouts.

A line graph of the volume after in litres against the temperature after in kelvin, for a fixed sample that starts at 100 kilopascals, 2.00 litres and 300 kelvin and finishes at 200 kilopascals. A single gold straight line rises from the origin of the axes at zero kelvin and zero litres to 2.0000 litres at 600 kelvin, with marked points at 100 kelvin and 0.3333 litres, 200 kelvin and 0.6667 litres, 300 kelvin and 1.0000 litres, 400 kelvin and 1.3333 litres, 500 kelvin and 1.6667 litres, and 600 kelvin and 2.0000 litres. The vertical axis is volume after in litres, ticked 0, 0.5, 1.0, 1.5 and 2.0; the horizontal axis is temperature after in kelvin, ticked 0, 100, 200, 300, 400, 500 and 600. Doubling the kelvin temperature from 300 to 600 kelvin doubles the volume from 1.0000 to 2.0000 litres.
Volume after against temperature after, with 100 kPa, 2.00 L, 300 K becoming 200 kPa. The line passes through the origin of the graph, which is a property of the equation on an absolute scale — not a prediction that a real gas shrinks away to nothing.

Hold the temperature after fixed instead and sweep the pressure after, and the same equation bends into a hyperbola. Doubling the pressure after from 100 kPa to 200 kPa takes the volume from 3.0000 L to exactly 1.5000 L, and the curve flattens without ever touching the axis.

A curve of the volume after in litres against the pressure after in kilopascals, for a fixed sample that starts at 100 kilopascals, 2.00 litres and 300 kelvin and finishes at 450 kelvin. A single gold curve falls steeply from 15.0000 litres at 20 kilopascals, through 6.0000 litres at 50 kilopascals, 3.0000 litres at 100 kilopascals, 1.5000 litres at 200 kilopascals, 1.0000 litres at 300 kilopascals and 0.7500 litres at 400 kilopascals, to 0.6000 litres at 500 kilopascals, flattening as it goes but never reaching the horizontal axis. The vertical axis is volume after in litres, ticked 0, 4, 8, 12 and 16; the horizontal axis is pressure after in kilopascals, ticked 0, 100, 200, 300, 400 and 500. Doubling the pressure from 100 to 200 kilopascals exactly halves the volume from 3.0000 to 1.5000 litres.
Volume after against pressure after, with 100 kPa, 2.00 L, 300 K becoming 450 K. Every doubling of the pressure halves the volume, so the curve falls steeply at low pressure and barely moves at high pressure.

There is a trap in the third row of that table, and it is worth flagging here. Imposing V2 = V1 as a condition — a gas sealed in a rigid steel cylinder — genuinely gives you Gay-Lussac’s law. A volume that merely happens to come back to its starting figure is a different thing entirely, and the misconceptions section deals with it.

Two Routes, One Answer: The Check That Can Actually Fail

Here is the uncomfortable part of every gas-law page, stated plainly. When a tool computes V2 from P1V1/T1 = P2V2/T2 and then shows you that PV/T is the same before and after, it has proved nothing at all. The two are equal because the answer was chosen to make them equal.

So what would count as evidence? Doing the change a different way and seeing whether you arrive at the same place. Squeeze first and then warm, or warm first and then squeeze — different arithmetic, different numbers in the middle, and no guarantee in advance that the two agree.

Worked problem 6 below puts numbers on it. A sample sitting at 120 kPa and 5.00 L at 250 K is taken to 300 kPa and 400 K, and the change is made three different ways.

A line graph of the volume of a fixed sample of gas in litres against the stage of a change, for 120 kilopascals, 5.00 litres and 250 kelvin becoming 300 kilopascals and 400 kelvin. Three paths start together at 5.00 litres on the left. The gold path, Boyle first then Charles, drops to a midway volume of 2.00 litres and then rises to 3.20 litres. The wine path, Charles first then Boyle, rises to a midway volume of 8.00 litres and then falls to 3.20 litres. The dark dashed path is the one combined step and goes straight from 5.00 litres to 3.20 litres with no midway point. The vertical axis is volume in litres, ticked 0, 2, 4, 6 and 8, with a dashed guide line at 3.20 litres where all three paths finish. The horizontal axis has three positions labelled Before, Midway and After and is the stage of the two-step recipe, not a time axis.
Squeezing first drops the gas to 2.00 L before the warming lifts it back; warming first pushes it out to 8.00 L before the squeeze brings it down. Both finish on 3.20 L, and so does the single combined step — its dashed line is only a connector, because a one-step change has no midway volume to plot. The horizontal axis is the stage of the recipe, not the passage of time.

The gap between those two midpoints is the point of the picture. One route passes through 2.00 L, the other through 8.00 L, and they are four times apart — yet the finish is identical to the last digit either way.

This was checked over a grid of 10,584 starting and finishing states while the cluster was being built. The worst disagreement between a two-step route and the single step was 4.0645e-16, which is the size of a rounding error in double-precision arithmetic. In 2,142 of those 10,584 states the two-step answer was not even bit-identical to the one-shot answer, because the multiplications happen in a different order — so the agreement is physics, not the same expression written twice.

One caution before you try it on paper. Carry the midpoint at full precision; rounding it to two decimals and then continuing changes the printed answer in about one state in five. Taking 50 kPa, 0.5 L and 150 K to 90 kPa and 330 K, the true midpoint of 0.277778 L shown as 0.28 L and carried on gives 0.62 L when the honest answer is 0.61 L.

Do not read that as a warning against the two-step method. Four times out of five the rounded midpoint lands on the same printed answer anyway; the method is sound and it is the rounding that is fragile.

You can watch both routes at once. Their two volumes are printed on screen already, in the simulator above and on the combined gas law simulator page; pressing Show both routes draws them as two dashed levels across the cylinders. Drag the pressure after from there and route A’s middle slides while route B’s stays exactly where it was — and both routes still land on whatever the one-step answer has become.

The Two Mistakes That Wreck a Combined Gas Law Answer

Almost every wrong answer on this equation comes from what went into it, not from the arithmetic. Two inputs do nearly all the damage, and neither announces itself: both produce a number that looks perfectly reasonable.

Celsius fed straight into the ratio

The law divides by temperature, and the Celsius scale starts wherever water happens to freeze. That zero is a convenience, not a physical floor, so ratios of Celsius numbers mean nothing. Add 273.15 to every reading first — our article on absolute zero covers why the kelvin scale stops where it does.

The table below starts a 2.00 L sample at 25 °C, holds the pressure steady and warms it. The middle column uses the honest kelvin ratio; the next column uses the Celsius numbers as they stand.

Warmed toCorrect volumeVolume from the Celsius ratioHow far out
50 °C2.1677 L4.0000 L84.5 % too big
100 °C2.5031 L8.0000 L219.6 % too big
150 °C2.8385 L12.0000 L322.8 % too big
200 °C3.1739 L16.0000 L404.1 % too big
250 °C3.5093 L20.0000 L469.9 % too big
300 °C3.8447 L24.0000 L524.2 % too big

Two things make it worse. A bigger change widens the gap, as every row of the table shows, and a start nearer 0 °C collapses the Celsius denominator — at exactly 0 °C the shortcut divides by zero and returns nothing at all.

Is it ever near enough? Sweeping every whole-degree change between 1 °C and 400 °C, the smallest error the shortcut can make is 0.1016 %, going from 400 °C to 399 °C. So the honest statement is that it is always wrong when the temperature moves, not usually wrong.

Gauge pressure fed straight into the ratio

A gauge on a tyre or a cylinder reads the excess over the air around it, so its zero is today’s weather rather than a vacuum. The law wants absolute pressure. Add the atmosphere before the number goes anywhere near the equation.

At sea level that addition is 101.325 kPa, the standard atmosphere — a defined figure, fixed by agreement rather than measured on the day. Worked problem 5 puts the cost of skipping it at 3.78 %: 11.659427 L from the gauge readings against 12.117974 L done properly.

Three point eight per cent sounds survivable until you notice that nothing in the output flags it. The wrong answer has the right units, the right order of magnitude and the right sign.

Common Misconceptions About the Combined Gas Law

Two of the five below carry figures measured against this cluster’s own simulator while it was being built, by sweeping a systematic sample of the states its sliders can reach. The other three are points of logic, not guesses about what students find hard.

  • “The two PV/T readouts agreeing proves the law.” They agree because the answer was computed to make them agree. Treat them as the law being displayed, never as a result you have confirmed — the two-route check above is the one that could come out wrong and does not.
  • “You can check it by multiplying the numbers on screen.” Over 406,016 states sampled across the simulator’s slider ranges, recomputing PV/T from the rounded volume on the display disagrees with the printed value in 331,281 of them, or 81.59 %. Your arithmetic is fine; the second decimal place is not there.
  • “When the volume ends where it started, the volume was held.” Nothing held it. The pressure ratio happened to match the temperature ratio and the two effects cancelled. Gay-Lussac’s law applies when a rigid container forces the volume to stay put, which is a condition, not a coincidence.
  • “The answers come out to round numbers.” Of 67,865 sampled simulator states, only 1,543 — 2.27 % — give a volume that lands exactly on two decimal places. Textbook problems are chosen to be tidy; reality is not.
  • “It describes what happens on the way.” It relates one settled state to another settled state and says nothing about the middle. That silence is precisely why two different routes arriving at the same answer is interesting rather than obvious.

There is a smaller display effect worth knowing about too. Even the two PV/T readouts can disagree with each other at four significant figures, in 791 of those 406,016 sampled states — 0.19 %, and purely a rounding artefact rather than a crack in the physics.

The last misconception is about scope rather than arithmetic. This is an ideal-gas result, and real gases drift from it when the molecules are crowded enough for their own size and their attraction to each other to matter — near a boiling point, or at pressures where the gas is close to liquefying.

Where You Meet the Combined Gas Law in Real Life

The cases that need this equation rather than one of its three special cases share a signature: something sealed travels between two places whose pressure and temperature both differ. That is more common than the textbook suggests.

A weather balloon at launch, the everyday case the combined gas law is built for
Thin, cold air is two changes at once, which is exactly the case one law alone cannot handle.
  • Weather balloons. Worked problem 7 is the standard one: 8.00 L at the ground becomes 49.9 L high in the atmosphere, a factor of 6.24. Getting that figure right needs both effects in the sum, because the thinning air on its own would have predicted 67.55 L and the cold pulls the honest answer well below it. An envelope is designed around the real number.
  • Tyres that barely flex. Heat from a motorway run has almost nowhere to go, so the pressure takes nearly all of it. That is why a recommended pressure is always a cold one, and why worked problem 5 is about the gauge rather than the volume.
  • Compressed gas let out into the open. The whole reason a diving cylinder is worth its weight is the volume factor when the contents reach ordinary pressure and a different temperature. The two changes fight, and the pressure wins by a very wide margin.
  • Anything sealed that travels. Crisp packets over a mountain pass, shampoo bottles in an unpressurised hold, instrument housings in a cargo bay. The number that matters is usually the volume factor, because that is what the seam has to hold.
  • The gas-syringe practical. Trap air, change the pressure and the temperature, predict the reading, then measure it. A gap between prediction and measurement is nearly always a leak or a thermometer that has not settled — which is itself a lesson about the fixed-amount condition.
  • Volumes measured on different days. Two gas volumes recorded at different laboratory pressures and temperatures cannot be compared until both are brought to the same conditions. Doing that conversion is what the phrase “corrected to standard conditions” means on a data sheet.

How the Combined Gas Law Fits the Rest of Thermodynamics

It occupies one narrow slot, and knowing the edges of that slot saves a lot of wasted effort. The combined gas law is the two-state reading of PV = nRT with the amount of gas held fixed. Everything it cannot do follows from that sentence.

When the amount of gas is the thing that changes — pumping more air in, or a reaction producing a gas — the relation you want is Avogadro’s law, which ties volume to the number of moles instead. When you have a single state and want to know how much gas is in it, you need the mole count and the gas constant back, so it is the ideal gas law.

And when the question is about energy, this equation is silent. It will tell you the volume a gas ends up at; it will not tell you how much heat went in, how much work the gas did on its surroundings, or how long any of it took. Those belong to the first law and its neighbours.

Which is a fair summary of the whole page: six quantities, one relation, two states, and a very clear boundary around what it promises.

Worked Problems

Problem 1
A sealed syringe holds 2.00 L of air at 100 kPa and 300 K. It is squeezed to 200 kPa and warmed to 450 K. What volume does the air end up at?
Show Solution
Solution: Step 1: The amount of gas is fixed, so P1V1/T1 = P2V2/T2. Rearranged for the volume after, V2 = P1V1T2 / (T1P2). Step 2: Substitute, carrying units. V2 = (100 kPa × 2.00 L × 450 K) / (300 K × 200 kPa). The kilopascals cancel, the kelvin cancel, and litres are left. Step 3: Top: 100 × 2.00 × 450 = 90000. Bottom: 300 × 200 = 60000. So V2 = 90000 / 60000 = 1.50. Answer: V2 = 1.50 L Sanity check: the doubled pressure halves 2.00 L to 1.00 L, and the 1.5-fold rise in kelvin temperature lifts that to 1.50 L. Three-quarters of where it started, from two changes pulling opposite ways.
Problem 2
A gas occupies 4.00 L at 150 kPa and 27.0 °C. It is compressed to 300 kPa and heated to 127.0 °C. Find its new volume.
Show Solution
Solution: Step 1: Convert both temperatures before anything else. T1 = 27.0 + 273.15 = 300.15 K, and T2 = 127.0 + 273.15 = 400.15 K. Step 2: V2 = P1V1T2 / (T1P2) = (150 × 4.00 × 400.15) / (300.15 × 300). Top: 150 × 4.00 = 600, and 600 × 400.15 = 240090. Bottom: 300.15 × 300 = 90045. Step 3: V2 = 240090 / 90045 = 2.66633 L, which is 2.67 L to three significant figures. Answer: V2 = 2.67 L Two warnings live in this one problem. Putting 127 over 27 into the ratio instead gives (150 × 4.00 × 127) / (27 × 300) = 76200 / 8100 = 9.4074 L, which is 252.8 % too big. And rounding the kelvin values to 300 and 400 gives 240000 / 90000 = 2.6667 L, which already disagrees in the fourth figure — carry the 0.15.
Problem 3
A 2.50 L sample of gas at 101.325 kPa and 293.15 K is compressed to 1.75 L and warmed to 313.15 K. What is its new pressure?
Show Solution
Solution: Step 1: The unknown is a pressure this time, so rearrange for it: P2 = P1V1T2 / (T1V2). Step 2: P2 = (101.325 × 2.50 × 313.15) / (293.15 × 1.75). Top: 101.325 × 2.50 = 253.3125, and 253.3125 × 313.15 = 79324.809375. Bottom: 293.15 × 1.75 = 513.0125. Step 3: P2 = 79324.809375 / 513.0125 = 154.62549 kPa, or 154.6 kPa to four significant figures. Answer: P2 = 154.6 kPa Worth a glance for plausibility: the squeeze on its own, at a steady temperature, would have given 101.325 × 2.50 / 1.75 = 144.75 kPa, and the 20 K of warming supplies the rest.
Problem 4
A 3.00 L sample at 200 kPa and 288.15 K is squeezed to 1.00 L at 500 kPa. What temperature has it reached, in kelvin and in Celsius?
Show Solution
Solution: Step 1: Rearrange for the temperature after: T2 = T1P2V2 / (P1V1). Step 2: T2 = (288.15 × 500 × 1.00) / (200 × 3.00). Top: 288.15 × 500 = 144075, and × 1.00 leaves 144075. Bottom: 200 × 3.00 = 600. Step 3: T2 = 144075 / 600 = 240.125 K. Converting, 240.125 – 273.15 = -33.025, so -33.0 °C. Answer: T2 = 240.125 K, which is -33.0 °C A negative Celsius answer is not a mistake; it is an ordinary kelvin temperature seen through a shifted scale. What you must never do is feed that -33.0 back into a ratio in a later part of the question.
Problem 5
A tyre gauge reads 220 kPa on a cold morning at 15.0 °C and 250 kPa after a motorway run at 45.0 °C. The 12.0 L of air inside has bulged the casing slightly. What volume does it now occupy, and what does using the gauge readings cost?
Show Solution
Solution: Step 1: Convert both pressures to absolute by adding the standard atmosphere. P1 = 220 + 101.325 = 321.325 kPa and P2 = 250 + 101.325 = 351.325 kPa. The temperatures become T1 = 288.15 K and T2 = 318.15 K. Step 2: V2 = (321.325 × 12.0 × 318.15) / (288.15 × 351.325). Top: 321.325 × 12.0 = 3855.9, and 3855.9 × 318.15 = 1226754.585. Bottom: 288.15 × 351.325 = 101234.29875. Step 3: V2 = 1226754.585 / 101234.29875 = 12.117974 L, or 12.12 L. Answer: V2 = 12.12 L Now the same sum on the gauge readings: (220 × 12.0 × 318.15) / (288.15 × 250) = 839916 / 72037.5 = 11.659427 L. That is 3.78 % low, and nothing about it looks wrong.
Problem 6
A gas at 120 kPa occupies 5.00 L at 250 K, and is taken to 300 kPa and 400 K. Work out the final volume in one step, then in two steps in both possible orders.
Show Solution
Solution: Step 1: One combined step. V2 = (120 × 5.00 × 400) / (250 × 300). Top: 120 × 5.00 = 600, and 600 × 400 = 240000. Bottom: 250 × 300 = 75000. So V2 = 240000 / 75000 = 3.20 L. Step 2: Squeeze first, at a steady 250 K, which is Boyle’s law: V = 5.00 × 120 / 300 = 5.00 × 0.400 = 2.00 L. Then warm at a steady 300 kPa, which is Charles’s law: V2 = 2.00 × 400 / 250 = 2.00 × 1.600 = 3.20 L. Step 3: Warm first, at a steady 120 kPa: V = 5.00 × 400 / 250 = 5.00 × 1.600 = 8.00 L. Then squeeze at a steady 400 K: V2 = 8.00 × 120 / 300 = 8.00 × 0.400 = 3.20 L. Answer: V2 = 3.20 L by all three methods The two routes pass through 2.00 L and 8.00 L on the way — a factor of four apart — and still finish on the same figure. That agreement is the closest thing to evidence this equation offers.
Problem 7
A weather balloon holds 8.00 L of gas at 101.325 kPa and 293.15 K at launch. High in the atmosphere the pressure is 12.00 kPa and the temperature 216.65 K. What volume does the gas occupy there?
Show Solution
Solution: Step 1: V2 = P1V1T2 / (T1P2), with both temperatures already in kelvin. Step 2: V2 = (101.325 × 8.00 × 216.65) / (293.15 × 12.00). Top: 101.325 × 8.00 = 810.6, and 810.6 × 216.65 = 175616.49. Bottom: 293.15 × 12.00 = 3517.8. Step 3: V2 = 175616.49 / 3517.8 = 49.92225 L, or 49.9 L to three significant figures. Answer: V2 = 49.9 L, which is 6.24 times the launch volume The pressure drop on its own would have given 8.00 × 101.325 / 12.00 = 67.55 L. Cooling from 293.15 K to 216.65 K then multiplies that by 216.65 / 293.15 and brings it back to 49.9 L, which is why a balloon designer cannot use the pressure ratio alone.
Problem 8
10.0 L of gas at 100 kPa and 300 K is compressed to 400 kPa at a steady 300 K, then heated to 600 K at a steady 400 kPa. Find the volume after each stage, then check it with one combined step.
Show Solution
Solution: Step 1: Stage one holds the temperature, so it is Boyle’s law: V = 10.0 × 100 / 400 = 10.0 × 0.250 = 2.50 L. Step 2: Stage two holds the pressure, so it is Charles’s law: V2 = 2.50 × 600 / 300 = 2.50 × 2.000 = 5.00 L. Step 3: Now ignore the stages and go straight there. V2 = (100 × 10.0 × 600) / (300 × 400). Top: 100 × 10.0 = 1000, and 1000 × 600 = 600000. Bottom: 300 × 400 = 120000. So V2 = 600000 / 120000 = 5.00 L. Answer: 2.50 L after the compression, then 5.00 L — exactly half the original 10.0 L Naming the stages is the useful part here. Each half of the change is a law with a name, and putting them end to end reproduces the combined result exactly.

Frequently Asked Questions

What is the combined gas law formula?
The formula is P1V1/T1 = P2V2/T2, with the pressure, volume and absolute temperature of one state on the left and the same three quantities for a second state on the right. It holds for a fixed amount of gas that neither leaks nor reacts. Rearranged for the volume at the end, it reads V2 = P1V1T2 divided by T1P2.
Is the combined gas law the same as the ideal gas law?
No. The ideal gas law, PV = nRT, describes a single state and carries the amount of gas in moles along with the gas constant. The combined gas law compares two states of one fixed sample, so the moles and the constant cancel out and nothing is left but the six measured quantities.
Do I always have to convert Celsius to kelvin?
Yes, whenever the temperature changes. The law divides by temperature, and the Celsius scale is measured from an arbitrary zero, so ratios of Celsius readings carry no meaning. Add 273.15 to every Celsius figure first. A change from 25 to 150 degrees Celsius is a factor of 1.4193, not a factor of six.
What units should the pressures and volumes be in?
Any units you like, provided each quantity uses the same one on both sides. Pressure appears once above and once below, and so does volume, so kilopascals, atmospheres, bars and psi all cancel, as do litres, millilitres and cubic metres. Temperature is the fussy one: it has to be kelvin.
Can I do the change in two steps instead of one?
Yes, and either order gives the same result. Change the pressure first at the starting temperature, then the temperature at the new pressure, or reverse the two. The orders pass through different intermediate volumes, 2.00 L against 8.00 L in worked problem 6, and still land on the same 3.20 L.
When does the combined gas law stop working?
It fails the moment the amount of gas changes, so a leak, a gas dissolving into a liquid or a reaction that makes or consumes molecules all break it. It also drifts from reality near a boiling point and at high pressures, where the ideal-gas picture behind it gives way. Both states must be settled ones.
How do I rearrange the combined gas law for the temperature before?
Cross-multiply, then divide. Multiplying both sides by T1 and then by T2 gives T2P1V1 = T1P2V2, and dividing by P2V2 leaves T1 = T2P1V1 divided by P2V2. Every rearrangement is built that way, and there are six of them because any one of the six quantities can be the unknown. The answer arrives in kelvin.
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