Thermodynamics

Avogadro’s Law (V1/n1 = V2/n2)

Definition

Avogadro’s law states that at a fixed pressure and temperature the volume of a gas is proportional to the amount of gas in it, so V1/n1 = V2/n2. Double the moles and you double the volume. Equal volumes of any two gases under the same conditions hold equal numbers of molecules.

Push more air into a balloon and it answers in the only way it can: by getting bigger. The gas inside is no hotter than it was and no more tightly packed — the skin simply stretches until the pressure inside matches the room again.

That one-for-one bargain between how much gas you have and how much room it takes up is Avogadro’s law. It is why a lungful, a party balloon and a gas holder the size of a house all run on the same arithmetic — and why a litre of helium and a litre of carbon dioxide hold exactly the same number of molecules while weighing wildly different amounts.

What Is Avogadro’s Law?

Avogadro’s law says that the volume a gas occupies is proportional to the number of molecules in it, provided the pressure and the temperature are held still.

Everything hangs on that word proportional. Not “roughly tracks”, not “tends to increase with” — proportional, in the strict sense that plotting volume against amount gives a straight line through the origin. No molecules, no volume.

Three things are worth separating before the algebra starts.

  • What is held still: the pressure and the temperature. Both have to be the same before and after, or the comparison is meaningless.
  • What moves together: the amount of gas, measured in moles, and the volume it fills. Triple one and you triple the other.
  • What never enters the sum: which gas it is. Helium, air, carbon dioxide, a mixture you invented yourself — the volume per mole is identical.

That last point is the one students find hardest to believe, and it is the heart of the law. A carbon dioxide molecule is about eleven times heavier than a helium atom, yet at the same pressure and temperature the two gases demand exactly the same room per mole.

The idea started life as a hypothesis in the early nineteenth century: equal volumes of gases, compared at the same pressure and temperature, contain equal numbers of particles. Chemistry largely ignored it for roughly half a century. It is the reason a measured volume of gas can stand in for a count of molecules at all.

Here is what the proportionality looks like as a table of readings, for air at 100 kPa and 0 °C.

Amount of gas (mol) Volume (L) Mass of air (g) Molecules V/n (L/mol)
0.102.2712.8966.022e2222.711
0.5011.3614.483.011e2322.711
1.0022.7128.966.022e2322.711
2.0045.4257.921.204e2422.711
3.0068.1386.881.807e2422.711
5.00113.6144.83.011e2422.711

One column is doing the real work. Volume, mass and molecule count all climb in step, but V/n sits at 22.711 L/mol and refuses to move, and that frozen column is Avogadro’s law.

A caution about arithmetic on that table, since it catches people out. The volume, mass and molecule columns are each rounded to four figures for display, while V/n was worked out before any rounding, so dividing the two rounded columns can leave the final digit adrift — 45.42 divided by 2.00 gives 22.71, not 22.711.

The Avogadro’s Law Formula, and the One Thing That Has to Stay Constant

The formula is V1/n1 = V2/n2: the volume divided by the amount of gas is the same before and after, so long as the pressure and the temperature have not moved between the two readings.

V1 / n1 = V2 / n2
Symbol What it is SI unit Worth knowing
V1the volume the gas fills to begin withcubic metre, m3litres are the practical unit; the ratio is unit-free, so both volumes may be in millilitres if you prefer
n1the amount of gas sitting in that volumemole, mola molecule count in disguise: one mole is 6.02214076e23 of them
V2the volume once gas has been added or drawn offcubic metre, m3read at the same pressure and temperature as V1, or the law has nothing to say
n2the amount of gas afterwardsmole, mollarger than n1 when gas goes in, smaller when it comes out
Vmthe molar volume: the room one mole takes up under those conditionscubic metre per mole, m3/molit is R T / P, and it is identical for every gas at that pressure and temperature

There is a second way to write the same law, and it is often the more useful one. Multiply the molar volume by the amount of gas and you have the volume outright.

V = Vm · n, with Vm = R T / P

R is the molar gas constant, 8.31446261815324 J/(mol K), exact in the SI. T is the absolute temperature in kelvin and P the pressure in pascals, so a molar volume comes out in cubic metres per mole and is then usually quoted in litres per mole. This is the piece of the ideal gas law that Avogadro’s law is carved out of: fix P and T in PV = nRT and V is left proportional to n.

If you already have three of the four quantities, the arithmetic is one division and one multiplication, and our Avogadro’s law calculator will grind it out: type in the volume and amount you started with plus whichever of the other two you know, and it returns the missing one. It never asks for a pressure or a temperature, because it works on the ratio, where both cancel.

Now the condition that the formula hides, and the one most worth remembering. Avogadro’s law needs the pressure to be free to stay constant, which means the container must be able to change size — a piston, a balloon skin, the roof of a gas holder, a chest.

Pump gas into a rigid steel bottle and the volume cannot move at all, so the pressure climbs instead. That is PV = nRT at constant volume, a different question with a different answer, and no amount of Avogadro’s law will rescue it.

The lab below puts two such cylinders side by side, with the amount of gas in each, the pressure they work against and the temperature they share all under your control. The thing worth hunting for is the invariance: choosing a different gas moves the mass and the density and leaves every volume exactly where it was.

Avogadro's Law Lab

Frozen on the page, the same arrangement looks like the diagram below: two identical cylinders, the same pressure pushing down on both pistons, and twice as much gas on the right as on the left.

Two identical cylinders side by side on the same baseline, each closed by a free piston with a gold arrow above it labelled 100 kPa. The left cylinder holds 1.00 mol of air and its gas column is drawn 100 pixels tall, labelled 22.71 L with 12 molecules drawn. The right cylinder holds 2.00 mol and its column is drawn 200 pixels tall, exactly twice as high, labelled 45.42 L with 24 molecules drawn at the same spacing. A dashed guide line runs across at the height of the smaller column and is marked half height.
Two identical cylinders at 100 kPa and 0 °C, drawn to one shared scale. Doubling the amount of gas doubles the height of the column and nothing else: same pressure on both pistons, same spacing between the molecules.

Why a Mole Fills 22.4 Litres, and Why Our Tools Say 22.711

Both figures are correct, and they differ because they are quoted at different pressures. A molar volume is meaningless without the pressure and the temperature it belongs to.

22.414 L/mol is the volume one mole of an ideal gas occupies at 0 °C and 1 atm, which is 101.325 kPa. That is the older convention, and it is the number printed in most textbooks — usually rounded to the 22.4 L everybody remembers.

22.711 L/mol is the same temperature at IUPAC’s standard pressure of 100 kPa. Lower pressure, more room, bigger number. Nothing has been corrected and no one has made a mistake; the reference pressure simply changed underneath the tradition, and both conventions are still in daily use.

The gap is easy to pin down, because Vm = R T / P makes the molar volume inversely proportional to pressure. Drop the pressure by a factor of 1.01325 and the volume grows by the same factor, which is 1.325 % — the whole of the 22.4-against-22.711 puzzle, in one division. Squeezing a gas at fixed temperature is the province of Boyle’s law, and this is its fingerprint.

Temperature pulls the other way, and far harder. T is the absolute temperature in kelvin, which is why 0 °C is 273.15 K and not zero, and why warming a gas at constant pressure inflates it — a change governed by Charles’s law rather than this one. It is the temperature that matters here, not how much heat went in, and the difference between heat and temperature is worth having straight before you use these figures.

Condition Pressure Temperature Molar volume V/n (L/mol)
IUPAC standard conditions (STP)100 kPa0 °C (273.15 K)22.711
The older standard, at 1 atm101.325 kPa0 °C (273.15 K)22.414
Standard ambient (SATP)100 kPa25 °C (298.15 K)24.790
A warm room, at 1 atm101.325 kPa20 °C (293.15 K)24.055
Body temperature, at 1 atm101.325 kPa37 °C (310.15 K)25.450
Flue gas in a furnace100 kPa500 °C (773.15 K)64.283

Read the last row and the reason for all this fuss becomes obvious. A mole of flue gas takes up nearly three times the room a mole of the same gas takes at 0 °C, so anyone who quotes “22.4 L” for a hot stack is out by a factor that dwarfs the 1.325 % everybody argues about.

A graph of gas volume in litres against amount of gas in moles, with the amount running from 0 to 5 moles and the volume from 0 to 330 litres on one shared scale. Three straight lines all start at the origin. The steepest is 100 kPa at 500 degrees Celsius, slope 64.283 litres per mole. The middle line is 100 kPa at 0 degrees Celsius, slope 22.711 litres per mole, with ringed points marked at 1.00 mole and 22.71 litres and at 2.00 moles and 45.42 litres, each with dotted guides to both axes. The shallowest, almost flat on this scale, is 500 kPa at 0 degrees Celsius, slope 4.542 litres per mole.
Three conditions, three straight lines, all through the origin. The slope of each line is the molar volume at that pressure and temperature; the fact that every line is straight and starts at zero is Avogadro’s law.

Avogadro’s Law Is Not Avogadro’s Number

They are two different things that happen to share a name, and confusing them is the single commonest error on this topic.

The law is the proportionality above: at one pressure and temperature, volume follows amount. It is a statement about how gases behave, and you can apply it all day without knowing how many molecules are actually in there.

The number is 6.02214076e23 per mole — the count of entities in one mole, and nothing to do with volume at all. Since the redefinition of the mole it has been exact by definition rather than measured, so the SI definition of the mole now fixes the count and the CODATA entry for the Avogadro constant carries no uncertainty on it.

That has a neat consequence for any molecule count you work out. N = n × NA is exact arithmetic; the only uncertainty in “1.204e24 molecules” comes from how well you knew the 2.00 mol in the first place.

A quick test of whether the two are straight in your head: the law would still be true if a mole had been defined as some other number of particles, because the volume per mole would simply come out different. The proportionality is the physics. The number is bookkeeping.

Equal Volumes, Very Different Masses

Equal volumes of two gases at the same pressure and temperature hold equal numbers of molecules, not equal masses — and the difference between those two statements is where most marks are lost.

Take 2.00 mol of each of three gases at 100 kPa and 0 °C. Each one fills 45.42 L and each one contains 1.204e24 molecules. Put them on a balance, though, and helium weighs 8.006 g, air 57.92 g and carbon dioxide 88.02 g.

Three identical boxes drawn the same size, each labelled 45.42 litres and 1.204e24 molecules, and each containing the same 24 dots. The first is helium, the second air, the third carbon dioxide. Under each box a mass bar is drawn on one shared scale whose full length is 88.02 grams: helium's bar is a short stub at 8.006 grams, air's reaches about two thirds at 57.92 grams, and carbon dioxide's fills the scale at 88.02 grams.
The same volume and the same molecule count in all three boxes; the mass bars beneath them share one scale. Carbon dioxide outweighs helium by a factor of 10.9943, which is exactly the ratio of their molar masses.

Mass follows molar mass, and molar mass is the one property the volume side of the law never asks about. Air is worth a footnote here: it is a mixture, so its 28.96 g/mol is a conventional mean molar mass for dry air rather than the molar mass of any one substance.

Density falls straight out of the same idea, because one mole has mass M and volume Vm. Divide and you get the density: at 100 kPa and 0 °C that is 0.17626 g/L for helium and 1.27516 g/L for air, which is why a helium balloon rises and an air-filled one does not.

Common Misconceptions About Avogadro’s Law

Four beliefs account for most of the wrong answers on this topic, and every one of them can be tested in the lab above in under a minute.

“Equal volumes of gases have equal masses”

They have equal molecule counts, which is a different claim entirely. Swap helium for carbon dioxide at the same pressure, temperature and amount and the volume does not stir, while the mass climbs by a factor of 10.9943 — the whole law in one comparison.

“Avogadro’s law only works at standard temperature and pressure”

It works at any pressure and any temperature, so long as both states share them. Standard conditions are a convention for quoting molar volumes, not a restriction on the law. At 500 °C the volume still doubles when the amount doubles; each mole simply occupies 64.283 L there instead of 22.711 L.

“Pumping gas into a sealed bottle is Avogadro’s law in action”

It is not, and this is the most expensive misconception of the four. A rigid container fixes the volume by construction, so adding gas raises the pressure and leaves the volume untouched. A sealed fizzy drink, a compressed-gas cylinder and a car tyre all belong in this category: put more gas in and what moves is the gauge, not the walls.

“This is the gas law about heating things”

That one is Charles’s law. The four classical gas laws are easy to tell apart once you ask which quantity is being varied: the amount of gas here, the temperature in Charles’s law, the pressure in Boyle’s law, and pressure against temperature in a rigid vessel for Gay-Lussac’s.

Where You Meet Avogadro’s Law

Anywhere a container is free to change size while gas goes in or comes out, you are watching Avogadro’s law happen.

  • A breath. Your chest expands, more air molecules come in, and the lungful is a volume standing in for a number of molecules at body temperature and room pressure.
  • A balloon at a party. Helium goes in from a cylinder and the skin grows to match, holding the inside pressure at roughly the pressure of the room throughout.
  • Dough rising. Yeast turns sugar into carbon dioxide, so the number of gas molecules trapped in the dough goes up, and the dough swells against the air above it.
  • A gas syringe on a school bench. The plunger is a free piston, so the scale reading is a direct proxy for how much gas the reaction has made.
  • Lift. A balloon and the air it pushed aside hold the same number of molecules; the lift is the mass difference that follows from their molar masses.

In practice the first thing to check in any of these is whether the pressure genuinely held still. A syringe whose plunger sticks is no longer a constant-pressure device, and a balloon stretched hard enough to strain its skin is pushing its own pressure up as it grows.

Helium cylinder inflating a balloon, an everyday example of Avogadro's law: more gas molecules in, more volume out
Filling a balloon from a helium cylinder. The skin holds the pressure at whatever the room is at, so the volume simply follows the number of molecules going in.

Where Avogadro’s Law Breaks Down

The law is exact for an ideal gas and only approximate for a real one, and it fails outright in four recognisable situations.

  • Under hard compression. Real molecules take up room and pull on one another, neither of which the straight line through the origin knows about. The departure grows as a gas is squeezed, and it is a qualitative statement here because nothing on this page measures it.
  • Near condensation. Cooling a gas at constant pressure shrinks it in proportion to its absolute temperature, the gas relative of thermal expansion, until it stops being a gas at all. Carbon dioxide is already a solid near -78 °C at ordinary pressures.
  • When the two readings do not share conditions. A volume measured in a cold store and another measured in a warm laboratory cannot be compared, and neither can a raw reading set against one already reduced to standard conditions.
  • When the amount changes behind your back. Gas leaking through a balloon skin, dissolving into a liquid beneath it, or being consumed in a reaction all move n without anybody deciding to.

Treat any figure a gas-law tool gives you at an extreme setting as the volume an ideal gas would occupy there, rather than a prediction about a real sample. That distinction costs nothing to state and saves a great deal of embarrassment.

Worked Problems

Problem 1
A cylinder fitted with a free piston holds 1.00 mol of air at 100 kPa and 0 degrees Celsius. A second mole is pumped in at the same pressure and temperature. What volume does the gas fill before and after?
Show Solution
Solution: Step 1: Get into SI units. T = 0 + 273.15 = 273.15 K, and P = 100 kPa = 100000 Pa. The piston is free to move, so the pressure really can stay constant and the law applies. Step 2: Find the molar volume. Vm = R T / P = 8.31446261815324 × 273.15 / 100000 = 0.02271095 m3/mol, which is 22.71095 L/mol. Step 3: Multiply by each amount. V1 = 1.00 mol × 22.71095 L/mol = 22.71095 L, and V2 = 2.00 mol × 22.71095 L/mol = 45.42190 L. Step 4: Check the ratio. 45.42190 / 22.71095 = 2.000000, which is exactly n2/n1 = 2.00/1.00. Answer: 22.71 L before and 45.42 L after, to four significant figures. Twice the gas, twice the volume, and the molar volume never moved because the pressure and temperature never did.
Problem 2
A textbook says one mole of gas fills 22.4 L at standard temperature and pressure, while our tools report 22.711 L/mol at 0 degrees Celsius. Which one is right, and where does the difference come from?
Show Solution
Solution: Step 1: Both are right, because they are quoted at different pressures. Vm = R T / P, so a molar volume is only defined once the pressure and temperature are named. Step 2: Work out both. At 101.325 kPa and 273.15 K, Vm = 22.4140 L/mol. At 100 kPa and the same 273.15 K, Vm = 22.7110 L/mol. Step 3: Compare them. The difference is 22.7110 – 22.4140 = 0.2970 L/mol, and 0.2970 / 22.4140 = 0.01325, which is 1.325 %. Step 4: Trace where that came from. The molar volume goes as 1/P, so the two should stand in the reciprocal ratio of the pressures: 101.325 / 100 = 1.01325 exactly, and 22.7110 / 22.4140 = 1.01325 to six figures. They match. Answer: neither is wrong. 22.414 L/mol belongs to 0 degrees Celsius at 1 atm, 22.711 L/mol belongs to 0 degrees Celsius at IUPAC’s 100 kPa, and the 1.325 % between them is the pressure ratio and nothing else.
Problem 3
A balloon holds 0.50 mol of air at 101.325 kPa and 20 degrees Celsius. You blow in enough air to bring it up to 1.50 mol at the same pressure and temperature. Find the volume before and after, both the long way and the short way.
Show Solution
Solution: Step 1: Convert. T = 20 + 273.15 = 293.15 K and P = 101325 Pa. A balloon skin holds the gas at the pressure of the room, so the constant-pressure condition is met. Step 2: The long way, through the molar volume. Vm = R T / P = 8.31446261815324 × 293.15 / 101325 = 0.02405512 m3/mol = 24.05512 L/mol. Step 3: Multiply. V1 = 0.50 × 24.05512 = 12.02756 L, and V2 = 1.50 × 24.05512 = 36.08268 L. Step 4: The short way, straight from the ratio. n2/n1 = 1.50/0.50 = 3, so V2 = 12.02756 × 3 = 36.08268 L, with no molar volume needed at all. Answer: 12.03 L before and 36.08 L after. Three times the gas, three times the volume — and the second route is the one to reach for in an exam, because it never needs R, T or P.
Problem 4
Two identical containers each hold 2.00 mol of gas at 100 kPa and 0 degrees Celsius, one of helium and one of carbon dioxide. Compare their volumes, their molecule counts and their masses. Molar masses: helium 4.003 g/mol, carbon dioxide 44.01 g/mol.
Show Solution
Solution: Step 1: Volume first. Avogadro’s law never asks which gas it is, so both fill 2.00 mol × 22.71095 L/mol = 45.42190 L. Step 2: Molecules next. N = n × NA = 2.00 × 6.02214076e23 = 1.204428152e24, so 1.204e24 in each container. Step 3: Now the masses, where the gases finally part company. m = n × M gives 2.00 × 4.003 = 8.006 g of helium and 2.00 × 44.01 = 88.02 g of carbon dioxide. Step 4: Take the ratio. 88.02 / 8.006 = 10.9943, and the molar masses give 44.01 / 4.003 = 10.9943. Identical, as they must be. Answer: both fill 45.42 L and both hold 1.204e24 molecules, but they weigh 8.006 g and 88.02 g, a factor of 10.9943 apart. Equal volumes mean equal counts, never equal masses.
Problem 5
Using the molar volume at 100 kPa and 0 degrees Celsius, find the density of helium, air and carbon dioxide, then work out the gross lift of a 10.00 L helium balloon at the same conditions. Molar masses: helium 4.003, air 28.96, carbon dioxide 44.01 g/mol.
Show Solution
Solution: Step 1: Set up the relation. One mole has mass M and volume Vm, so the density is rho = M / Vm, with Vm = 22.71095 L/mol here. Step 2: Divide each molar mass by it. Air: 28.96 / 22.71095 = 1.27516 g/L. Helium: 4.003 / 22.71095 = 0.17626 g/L. Carbon dioxide: 44.01 / 22.71095 = 1.93783 g/L. Step 3: Lift is the mass of air displaced minus the mass of the helium in its place. Per litre that is 1.27516 – 0.17626 = 1.09890 g. Step 4: Scale up to the balloon. 1.09890 × 10.00 = 10.9890 g. Answer: about 10.99 g of gross lift. Gross is the operative word: the skin, the string and any payload have still to be subtracted. And this 10.99 g is a mass in grams, unrelated to the pure-number ratio 10.9943 in the previous problem — the leading digits agreeing is pure coincidence.
Problem 6
A small gas holder contains 3.00 mol of air at 100 kPa and 20 degrees Celsius. Two moles are drawn off at the same pressure and temperature. What volume is left, and what percentage of the original volume has gone?
Show Solution
Solution: Step 1: Find the molar volume at these conditions. T = 293.15 K, P = 100000 Pa, so Vm = 8.31446261815324 × 293.15 / 100000 = 0.02437385 m3/mol = 24.37385 L/mol. Step 2: Volumes before and after. V1 = 3.00 × 24.37385 = 73.12155 L, and with 1.00 mol left, V2 = 1.00 × 24.37385 = 24.37385 L. Step 3: The volume lost. 73.12155 – 24.37385 = 48.74770 L. Step 4: As a proportion of where it started. 48.74770 / 73.12155 = 0.6667, which is 66.67 %. Answer: 24.37 L remains, and 66.67 % of the volume has gone. Two of the three moles left, so two thirds of the volume left with them — the percentage was fixed the moment the amounts were.
Problem 7
How many moles of air fill 50.00 L at 100 kPa and 0 degrees Celsius? Then take the 1.00 mol to 2.00 mol cylinder from problem 1 and heat it to 500 degrees Celsius at the same 100 kPa. What happens to the two volumes, and to their ratio?
Show Solution
Solution: Step 1: Run the law backwards. n = V / Vm = 50.00 / 22.71095 = 2.20158 mol, which rounds to 2.202 mol at the four significant figures the 50.00 L carries. Step 2: Now heat the cylinder. T = 500 + 273.15 = 773.15 K, so Vm = 8.31446261815324 × 773.15 / 100000 = 0.06428327 m3/mol = 64.28327 L/mol. Step 3: Recompute both volumes. V1 = 1.00 × 64.28327 = 64.28327 L and V2 = 2.00 × 64.28327 = 128.56654 L, which round to 64.28 L and 128.6 L at four significant figures. Step 4: Check what survived. V2/V1 = 128.56654 / 64.28327 = 2.000000, unchanged. What did change is the molar volume: 64.28327 / 22.71095 = 2.83050, and 773.15 / 273.15 = 2.83050 — that factor is Charles’s law, not Avogadro’s. Answer: 50.00 L holds 2.202 mol; heating the cylinder to 500 degrees Celsius swells the pair to 64.28 L and 128.6 L while the ratio stays at exactly 2.000. Temperature moves the molar volume. Only the amount of gas moves the ratio.

Frequently Asked Questions

What is the formula for Avogadro's law?
The formula is V1/n1 = V2/n2, where V is the volume of the gas and n is the amount of gas in moles. The same law can be written V = Vm n, where Vm is the molar volume, the room one mole takes up at the pressure and temperature you are working at. Both forms say one thing: the volume divided by the amount does not budge.
Why is the molar volume 22.4 litres?
22.414 L/mol is the volume one mole of an ideal gas fills at 0 degrees Celsius and 1 atm, which is the older standard most textbooks still quote and usually round to 22.4 L. Move to IUPAC’s standard pressure of 100 kPa at the same temperature and it becomes 22.711 L/mol. Neither is the molar volume; quote one and you must quote its conditions.
Is Avogadro's law the same as Avogadro's number?
No, they are two different things that happen to share a name. The law is a statement about gases: at one pressure and temperature, volume follows the amount of gas. The number is 6.02214076e23 per mole, the count of entities in a mole, exact by definition since the mole was redefined. You can use the law from start to finish without ever mentioning the number.
What stays constant in Avogadro's law?
The pressure and the temperature stay constant, and with them the molar volume that follows from the two. What changes is the amount of gas and the volume it fills, always in step. The identity of the gas never enters the arithmetic, which is why helium and carbon dioxide occupy the same room for the same number of moles.
Does Avogadro's law work for mixtures like air?
Yes, and air is the standard example. The law counts molecules without asking what kind they are, so a mixture behaves exactly as a pure gas of the same total amount would. Dry air is normally handled with a mean molar mass of 28.96 g/mol, which is a convenient average for the mixture rather than the molar mass of any single substance.
Why do equal volumes of different gases weigh different amounts?
Because volume counts molecules while a balance weighs them. Two containers of the same size at the same pressure and temperature hold the same number of molecules, but a carbon dioxide molecule is far heavier than a helium atom, so the masses stand in the ratio of the molar masses. At 100 kPa and 0 degrees Celsius, 45.42 L holds 8.006 g of helium or 88.02 g of carbon dioxide.
When does Avogadro's law fail?
It fails whenever the pressure cannot stay constant, most obviously in a sealed rigid container, where added gas raises the pressure and leaves the volume alone. It also drifts as a gas is compressed hard or cooled towards condensation, because real molecules occupy space and attract one another. At the condensation point it stops applying altogether: liquids are not described by gas laws.
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