Thermodynamics

The Gas Laws: Complete Guide

Definition

The gas laws are the relationships between the pressure, volume, temperature and amount of a gas. Boyle’s, Charles’s, Gay-Lussac’s and Avogadro’s laws, together with the combined gas law, are not separate rules: each is the ideal gas equation PV = nRT with two of those four quantities held fixed. Every ratio form needs absolute temperature in kelvin.

An aerosol can carries a warning about heat. A hot-air balloon climbs when the burner fires. Put your thumb over a syringe nozzle and push, and the plunger shoves back harder than you expected.

Three everyday scenes, three different gas laws — and the arithmetic was never the hard part. Choosing is. Reach for the wrong relation and the true answer can sit a fifth above the number you wrote down; work in degrees Celsius instead of kelvin and you can be out by more than a factor of three.

What Is a Gas Law?

A gas law is a rule that says how the pressure, volume, temperature and amount of a gas trade against one another while the rest of them are held still. Four quantities describe the gas, and each named law simply says which of them are pinned while the rest do the moving.

  • Pressure P. How hard the gas pushes outwards on whatever contains it. The SI unit is the pascal; this page works in kilopascals, and the air around you sits near 100 kPa.
  • Volume V. The space the gas fills, in litres here, cubic metres in strict SI, where 1 L = 0.001 m3.
  • Temperature T. Always the absolute temperature, in kelvin — which is not the same thing as heat, and never a Celsius reading dropped into a ratio.
  • Amount n. How much gas is actually in there, in moles. This is the one people forget, and forgetting it causes more wrong answers than any algebra slip.

Hold two of those four still and the other two have only one way to move. That is all a gas law is: a shortcut for the case where you already know what is not changing.

Which is also why there is no sense in memorising four unrelated formulae. There are not four laws here. There is one equation wearing four masks, and the algebra that proves it takes a single line.

The Five Relations at a Glance

Here is the whole family in one place: the five ratio relations, and the equation all of them come from. Read the second column first, because what is held — not the shape of the formula — is what tells you which row your question is in.

Law What is held The relation Rearranged for the unknown Our tool
Boyle’s lawamount n and temperature TP1V1 = P2V2P2 = P1V1 ÷ V2Boyle’s Law Calculator
Charles’s lawamount n and pressure PV1 ÷ T1 = V2 ÷ T2V2 = V1T2 ÷ T1Charles’s Law Calculator
Gay-Lussac’s lawamount n and volume VP1 ÷ T1 = P2 ÷ T2P2 = P1T2 ÷ T1Gay-Lussac’s Law Calculator
Avogadro’s lawpressure P and temperature TV1 ÷ n1 = V2 ÷ n2V2 = V1n2 ÷ n1Avogadro’s Law Calculator
The combined gas lawamount n onlyP1V1 ÷ T1 = P2V2 ÷ T2V2 = P1V1T2 ÷ (P2T1)Ideal Gas Law Simulator, free mode, with the amount left alone
The ideal gas lawnothing at allPV = nRTn = PV ÷ (RT)the Ideal Gas Law Calculator, linked in the next section

Notice what the second column is doing: every ratio row names two quantities, the combined law names one, and the ideal gas law names none, because it holds nothing. That is not padding. “Boyle’s law holds the temperature” is half a condition: it holds the amount of gas too, and a sample that leaks obeys neither Boyle nor anything else on the list.

The One Equation Behind All of Them

Every relation in that table comes from one equation, the ideal gas law.

PV = nRT

Each symbol, with the unit this page uses:

  • P is pressure, in kilopascals (kPa). The SI unit is the pascal, Pa.
  • V is volume, in litres (L). The SI unit is the cubic metre, m3.
  • n is the amount of gas, in moles (mol).
  • T is the absolute temperature, in kelvin (K).
  • R is the gas constant, 8.314 J/(mol·K) to four figures.

R is not measured. It is the product of the Avogadro constant and the Boltzmann constant, 6.02214076 × 1023 per mole and 1.380649 × 10-23 J/K, which are two of the constants the SI fixes exactly — so the NIST reference value for the molar gas constant carries no uncertainty at all.

The units hide a small gift, too. One kilopascal times one litre is exactly one joule, so kilopascals and litres go straight into PV = nRT with R in J/(mol·K) and no conversion factor anywhere.

Now hold two quantities still and watch the equation collapse. If n and T never change, nRT is a fixed number, so PV has to be that same fixed number before and after — which is P1V1 = P2V2, Boyle’s law, in one line.

The others fall out the same way. Hold n and P, and V ÷ T is what stays fixed; hold n and V, and it is P ÷ T; hold P and T, and it is V ÷ n. Hold only the amount and you keep PV ÷ T, the combined gas law.

This page stops there on purpose: the full treatment of the equation belongs on the ideal gas law, term by term, not here. To push numbers through it, the Ideal Gas Law Calculator rearranges PV = nRT for whichever of P, V, n and T you set it to solve for.

Gas Laws Lab

The simulator above is the whole argument in one tool. It carries its own heading, a cylinder of nitrogen with three sliders — amount, temperature and volume — and four mode buttons, three of which freeze something different while the fourth freezes nothing. It is a tall tool, so it scrolls with the page rather than sitting inside one screen.

  • Boyle, hold T. The temperature slider greys out. From the opening 1.00 mol at 273.15 K in 22.711 L, where the pressure readout says 100.00 kPa, halving the volume to 11.356 L reads 199.99 kPa and doubling it to 45.422 L reads 50.00 kPa. The amount slider stays live, so P1V1 = P2V2 holds only while you leave it alone.
  • Charles, hold P. Here the greyed-out slider is the volume, badged “held”, and the sim then moves the volume itself to keep the pressure where it was: 22.711 L at 273.15 K, 29.101 L at 350 K, 45.422 L at 546.30 K, with the pressure readout unmoved at 100.00 kPa. What is held is the pressure, not the slider wearing the badge.
  • Gay-Lussac, hold V. The volume slider greys out. Double the temperature, 273.15 K to 546.30 K, and the pressure readout doubles too, 100.00 kPa to 200.00 kPa — exactly, because that temperature step is an exact doubling.
  • Free, all free. Nothing is frozen. This is the full PV = nRT, and it is also where the combined gas law lives, as long as you leave the amount alone.

One limit is worth knowing before you trust Charles mode. The volume slider stops at 50.000 L, and in Charles mode the volume runs into that stop near 601 K; past it the pressure is not held any more and climbs, reading 108.09 kPa at 650 K and 133.03 kPa at 800 K. The mode holds the pressure only while the piston still has room to move.

There is no Avogadro mode, either. This sim freezes the temperature, the pressure or the volume, never the amount — so to watch volume track the number of moles with the pressure and temperature both held, use the Avogadro’s law simulator instead. You can still drag this one’s amount slider and read the molar volume as you go, but nothing is holding P and T for you while you do it.

Two details about the tool itself. Its gas is a fixed setting, nitrogen, and its molecular-speed readout depends on temperature alone, so it moves when you change T and at no other time. Reset puts everything back to free mode at 1.00 mol, 273.15 K and 22.711 L.

Graph the four simple laws and the family resemblance is hard to miss. Three of them are straight lines through the origin of their axes; only Boyle’s bends, and it bends because there one quantity is the reciprocal of the other.

Four small graphs of the same equation, each with two quantities held fixed. The first three are drawn for one mole; Avogadro's panel is the one that varies the amount, from zero to two moles. Upper left, Boyle with the amount and the temperature held: pressure in kilopascals against volume in litres is a falling hyperbola. Upper right, Charles with the amount and the pressure held: volume in litres against absolute temperature in kelvin is a straight line aimed at the origin. Lower left, Gay-Lussac with the amount and the volume held: pressure in kilopascals against absolute temperature in kelvin is a straight line aimed at the origin. Lower right, Avogadro with the pressure and the temperature held: volume in litres against amount in moles is a straight line from the origin. One line under all four reads: one equation, PV = nRT, with two symbols cancelled.
Figure 1. Each panel fixes the two quantities named in its title and plots what is left: the first three are drawn for 1.00 mol at the state the embedded simulator opens in, while Avogadro’s panel is the one that varies the amount, from 0 to 2.00 mol. Only Boyle’s panel bends; the other three are straight lines aimed at the origin of their own axes. Rebuilt across 20,000 random states, the four ratio laws and PV = nRT agree to about 5 parts in 1016, which is double-precision rounding rather than physics.

How Do I Know Which Gas Law to Use?

List what changed and what stayed fixed, then read the row off the table below. The held quantities choose the law; the wording of the question does not.

  1. Write out all four quantities for the starting state and the finishing state, with every temperature converted to kelvin before anything else happens.
  2. Mark which ones changed.
  3. Mark which ones the question says stayed fixed — and check that the amount of gas is one of them, unless you are told gas was added or lost.
  4. Match that pattern below. Two of P, V and T moving means no simple ratio law will do; an amount that moves is Avogadro’s law only when the pressure and the temperature both hold still, and otherwise sends you back to PV = nRT.
What changed What stayed fixed Use this
pressure and volumeamount and temperatureBoyle: P1V1 = P2V2
volume and temperatureamount and pressureCharles: V1 ÷ T1 = V2 ÷ T2
pressure and temperatureamount and volumeGay-Lussac: P1 ÷ T1 = P2 ÷ T2
volume and amountpressure and temperatureAvogadro: V1 ÷ n1 = V2 ÷ n2
pressure, volume and temperatureamount onlycombined: P1V1 ÷ T1 = P2V2 ÷ T2

Problem 5 below is the case worth drawing, because two things move at once. A 2.00 L sample at 100.0 kPa and 298.15 K is compressed to 250.0 kPa and then heated to 423.15 K.

Three horizontal bars on one shared volume scale, all measured from the same zero, showing problem five worked as two steps. The top bar is the starting volume 2.00 litres at 100.0 kilopascals and 298.15 kelvin. The middle bar is 0.8000 litres, what Boyle's law alone gives after compressing to 250.0 kilopascals at the same temperature. The bottom bar is 1.1354 litres, where heating to 423.15 kelvin at 250.0 kilopascals carries it. A scale underneath is ticked 0, 0.5, 1.0, 1.5 and 2.0 litres and labelled volume in litres, one shared scale.
Figure 2. Problem 5 taken in two steps, with all three volumes measured from the same zero on one shared litre scale. Compressing from 100.0 kPa to 250.0 kPa at 298.15 K takes 2.00 L down to 0.8000 L, and heating from 298.15 K to 423.15 K at 250.0 kPa then carries that to 1.1354 L. The one-line combined law gives the same 1.1354 L, because the two steps are the same equation applied twice.

Run either law on its own here and you get a wrong answer with no warning attached. That is what the next section is about.

The Two Mistakes That Wreck Gas-Law Answers

Almost every bad gas-law answer comes from one of two places, and neither is an algebra slip. Both are mistakes of choosing.

Mistake 1: using a ratio law whose held quantity did not stay held

Take 1.00 mol of gas in 20.0 L at 300 K, where the pressure reads 124.72 kPa. Compress it to 10.0 L, but let it warm to 360 K on the way. Boyle’s law, handed only the two volumes, returns 249.43 kPa. The truth is 299.32 kPa.

The gap is not random. It is exactly the temperature ratio Boyle’s law had no way of knowing about: 249.43 × 1.2000 = 299.32. Say it either way round, but say it carefully — the true pressure is 20.0 % above the Boyle answer, and the Boyle answer is 16.7 % below the truth.

The fix is the combined law, which keeps the temperature in view and gives 299.32 kPa as well. One note on the figures, because rounding bites here: the 249.43 kPa is twice the unrounded starting pressure, not twice the rounded 124.72 kPa printed above.

Whenever you reach for Boyle’s law, then, the question to ask is not “did the volume change” but “did anything else”.

Mistake 2: putting a Celsius reading into a ratio

A 1.00 L balloon at 27 °C is warmed to 54 °C at constant pressure. In kelvin, 300.15 K to 327.15 K, the answer is 1.0900 L. Divide 54 by 27 instead and you get 2.0000 L, an answer 83.5 % too big.

Doubling a Celsius reading is not doubling the temperature. The ratio laws are ratios of absolute temperature, and the Celsius scale starts nearly 273 degrees above absolute zero, so a ratio taken on it is a ratio of two numbers that both have an offset buried in them.

A graph of volume in litres against absolute temperature in kelvin for a 2.50 litre sample warmed at constant pressure and constant amount. A gold straight line runs from the origin of the kelvin axis through two marked points, 293.15 kelvin at 2.50 litres and 353.15 kelvin at 3.012 litres, each with a dashed guide dropped to the temperature axis. A separate dark red point sits far above the line at 353.15 kelvin and 10.000 litres, labelled the Celsius answer. Beneath the kelvin axis a second axis shows the same positions read in degrees Celsius, from minus 273.15 at zero kelvin to 100 degrees at 373.15 kelvin, so the two scales are offset rather than proportional.
Figure 3. Why Charles’s law needs kelvin, drawn for the 2.50 L sample in problem 2: on a kelvin axis the volume line aims at the origin, while the second axis underneath shows the same points in degrees Celsius, offset rather than proportional. The Celsius answer of 10.000 L, plotted on that same volume scale, sits nowhere near the line. No real gas reaches 0 K, though — it condenses long before, so the line is an extrapolation of the ideal case rather than a prediction.

Standard Conditions Are Not One Number

“At STP” is not a value until somebody says which conditions are meant. One mole of an ideal gas fills a different volume under each convention, and all three figures below are correct.

Conditions Volume of one mole Where you meet it
273.15 K (0 °C) and 100 kPa22.71095 L/molIUPAC standard conditions; the simulator’s opening volume slider setting, 22.711 L, is this figure rounded
273.15 K (0 °C) and 101.325 kPa (one atmosphere)22.41397 L/molthe older convention, and the source of the familiar 22.4 L/mol
298.15 K (25 °C) and 100 kPa24.78957 L/molroom-temperature work, where 22.4 L/mol would be wrong by about a tenth

Each row is the same equation with different numbers in it: V = RT ÷ P, one mole at a time. Quote the conditions with the figure, every time, and this whole category of confusion disappears.

Where You Meet the Gas Laws

The pattern to look for is a gas that is trapped, or pushed, or heated, with something obviously held fixed. Once you can name what is held, you have already picked the law.

  • The warning on an aerosol can. A rigid can of fixed volume, heated: pressure rises with absolute temperature, which is Gay-Lussac’s law and the reason the label says what it says.
  • A hot-air balloon. The burner heats air that stays at the pressure of the air around it, so that air expands. An inflated envelope cannot grow, though, and it is open at the bottom, so the surplus spills out of the mouth, leaving fewer, hotter molecules inside and air less dense than the air outside. That is Charles’s law lifting a basket — and notice what is not held here: the amount of gas in the envelope falls as it heats.
  • Your own lungs. Your diaphragm drops, your chest cavity gets bigger, the pressure inside falls below the air outside, and air moves in. Boyle’s law, every breath you take.
  • A bicycle pump with a thumb over the outlet. The gas has nowhere to go, so squeezing it into a smaller volume raises the pressure; the barrel also warms, which is the part Boyle’s law alone does not describe.
  • A weather balloon climbing. Outside pressure falls, outside temperature falls, and both act on the same envelope at once — the combined law’s natural habitat.
Hot-air balloon being inflated with the burner firing, one of the gas laws at work as the heated air expands at constant pressure
Burner on: the air inside expands at the pressure of the air around it and spills out of the open mouth, so what is left is less dense than its surroundings.

Where the Gas Laws Stop Working

Every relation on this page inherits the same three assumptions, because they all come from the same equation: molecules that take up no space themselves, no forces between them, and collisions that lose nothing.

Real gases break all three. They do it worst at high pressure, at low temperature, and anywhere near condensing — which is to say, exactly where the molecules are close enough to notice each other.

The simulator will not warn you about this, and that is worth seeing for yourself. Push it to 5.00 mol in 1.000 L at 800 K and it prints 33,258 kPa, about 328 atmospheres, with a molar volume of 0.20 L/mol against 22.71 L/mol at 273.15 K and 100 kPa. That is more than a hundred times more crowded than the state it opened in.

The arithmetic is right. The physics is not: no real gas obeys PV = nRT in that region, so the number is a correct answer to a question about an ideal gas and nothing more. Treat any gas-law answer at extreme pressure or near a condensation point as an estimate, and say so.

Worked Problems

Problem 1
A syringe holds 60.0 mL of air at 101.0 kPa. The plunger is pushed in until the air occupies 20.0 mL, with the temperature and the amount of air unchanged. What is the new pressure?
Show Solution
Solution: Step 1: Ask what is held. The temperature and the amount both are, and pressure and volume are what move, so this is Boyle’s law: P1V1 = P2V2. Step 2: Rearrange for the unknown and substitute. P2 = P1V1 ÷ V2 = 101.0 kPa × 60.0 mL ÷ 20.0 mL = 6060.0 kPa·mL ÷ 20.0 mL. The millilitres cancel, so there is no conversion to do. Step 3: Divide. 6060.0 ÷ 20.0 = 303.0 kPa. Sanity check: the volume fell by a factor of 60.0 ÷ 20.0 = 3.00, and the pressure rose by 303.0 ÷ 101.0 = 3.000. Answer: 303.0 kPa.
Problem 2
A balloon holds 2.50 L of gas at 20.0 °C. It is warmed to 80.0 °C at constant pressure, with no gas added or lost. What is the new volume, and what would working in degrees Celsius have given?
Show Solution
Solution: Step 1: The pressure and the amount are held, and volume and temperature move, so this is Charles’s law: V1 ÷ T1 = V2 ÷ T2. Step 2: Convert to kelvin before anything else. T1 = 20.0 + 273.15 = 293.15 K, and T2 = 80.0 + 273.15 = 353.15 K. Step 3: Substitute. V2 = V1T2 ÷ T1 = 2.50 × 353.15 ÷ 293.15 = 882.875 ÷ 293.15 = 3.012 L. Step 4: Now the Celsius route, for contrast. 2.50 × 80.0 ÷ 20.0 = 10.000 L, and 10.000 ÷ 3.012 = 3.32, so that answer is 3.32 times too big. Answer: 3.012 L. In degrees Celsius it comes out as 10.000 L, which is 3.32 times the right answer.
Problem 3
A sealed aerosol can reads 320.0 kPa at 20.0 °C. It is left somewhere hot and reaches 120.0 °C. The can is rigid and does not leak. What is the new pressure?
Show Solution
Solution: Step 1: Rigid means the volume is held; not leaking means the amount is held. Pressure and temperature are what move, so this is Gay-Lussac’s law: P1 ÷ T1 = P2 ÷ T2. Step 2: Kelvin first. T1 = 293.15 K and T2 = 393.15 K. Step 3: Substitute. P2 = P1T2 ÷ T1 = 320.0 × 393.15 ÷ 293.15 = 125808.0 ÷ 293.15 = 429.2 kPa. Step 4: Read the size of it. 429.2 – 320.0 = 109.2 kPa more, for a rise of 100 °C, in a can designed to hold what it started with. Answer: 429.2 kPa, a rise of 109.2 kPa.
Problem 4
A balloon containing 0.180 mol of gas has a volume of 4.00 L. More of the same gas is pumped in until it holds 0.300 mol, at the same pressure and temperature. What is the new volume?
Show Solution
Solution: Step 1: The pressure and the temperature are held, and this time it is the amount that moves, which is Avogadro’s law: V1 ÷ n1 = V2 ÷ n2. Step 2: Rearrange and substitute. V2 = V1n2 ÷ n1 = 4.00 × 0.300 ÷ 0.180 = 1.200 ÷ 0.180. Step 3: Divide, then check the ratios match. 1.200 ÷ 0.180 = 6.667 L; the amount grew by 0.300 ÷ 0.180 = 1.667 and the volume by 6.667 ÷ 4.00 = 1.667, the same factor. Answer: 6.667 L.
Problem 5
A sample of gas occupies 2.00 L at 100.0 kPa and 25.0 °C. It is compressed to 250.0 kPa and heated to 150.0 °C, with no gas added or lost. What is the new volume?
Show Solution
Solution: Step 1: Decide the law first. Pressure and temperature both move and only the amount is held, so no single ratio law applies and this is the combined gas law: P1V1 ÷ T1 = P2V2 ÷ T2. Step 2: Kelvin. T1 = 25.0 + 273.15 = 298.15 K and T2 = 150.0 + 273.15 = 423.15 K. Step 3: Rearrange for V2 and substitute. V2 = P1V1T2 ÷ (P2T1) = 100.0 × 2.00 × 423.15 ÷ (250.0 × 298.15) = 84630.0 ÷ 74537.5 = 1.1354 L. Step 4: Check it by taking the two steps separately. Compressing alone gives 100.0 × 2.00 ÷ 250.0 = 0.8000 L; heating that at 250.0 kPa gives 0.8000 × 423.15 ÷ 298.15 = 1.1354 L, so the heating adds back 1.1354 – 0.8000 = 0.3354 L. Answer: 1.1354 L.
Problem 6
A 12.0 L cylinder reads 250.0 kPa at 35.0 °C. How many moles of gas are inside it?
Show Solution
Solution: Step 1: There is no before and after here, only one state, so no ratio law can help. Use the equation itself, rearranged: n = PV ÷ (RT). Step 2: Kelvin, then the top line. T = 35.0 + 273.15 = 308.15 K, and PV = 250.0 × 12.0 = 3000.0. Because one kilopascal times one litre is exactly one joule, that 3000.0 is already in joules. Step 3: Now the bottom line and the division. RT = 8.314462618 × 308.15 = 2562.10 J/mol, so n = 3000.0 ÷ 2562.10 = 1.1709 mol. Answer: 1.1709 mol.
Problem 7
A rigid 8.00 L cylinder at 20.0 °C reads 900.0 kPa. A week later, at the same temperature, it reads 600.0 kPa. Which gas law explains the drop?
Show Solution
Solution: Step 1: None of them, and that is the point. The volume was held and the temperature was held, which rules out Boyle, Charles and Gay-Lussac; each of those also assumes a fixed amount of gas, and something here has plainly moved. Avogadro’s law is the one that lets the amount change, but it needs the volume free to follow, and this cylinder is rigid. That leaves PV = nRT itself. Step 2: With V and T fixed, PV = nRT makes the pressure proportional to the amount. RT = 8.314462618 × 293.15 = 2437.38 J/mol. Step 3: Work out both amounts. n1 = 900.0 × 8.00 ÷ 2437.38 = 7200.0 ÷ 2437.38 = 2.9540 mol, and n2 = 600.0 × 8.00 ÷ 2437.38 = 4800.0 ÷ 2437.38 = 1.9693 mol. Step 4: Subtract. 2.9540 – 1.9693 = 0.9847 mol has gone, which is 0.9847 ÷ 2.9540 = 33.3 % of what was there. Answer: no ratio law applies — only PV = nRT itself. Gas has left the cylinder, and 0.9847 mol, a third of the original, is missing.

Frequently Asked Questions

What are the gas laws?
The gas laws are the rules connecting the pressure, volume, temperature and amount of a gas. Four carry a person’s name: Boyle’s, Charles’s, Gay-Lussac’s and Avogadro’s, each holding two of those four quantities fixed while the other two trade off. The combined gas law holds only the amount. Every one of them is the ideal gas equation PV = nRT with symbols cancelled.
How do I know which gas law to use?
Start from what is held, not from what is asked. Write the four quantities down for both states, mark the ones that did not change, and match that pattern: Boyle for fixed amount and temperature, Charles for fixed amount and pressure, Gay-Lussac for fixed amount and volume, Avogadro for fixed pressure and temperature. If two of pressure, volume and temperature moved, use the combined law instead.
What is the difference between Boyle's law and Charles's law?
Boyle’s law holds the temperature and the amount fixed and trades pressure against volume, so P1V1 = P2V2 and the graph curves. Charles’s law holds the pressure and the amount fixed and trades volume against absolute temperature, so V1/T1 = V2/T2 and the graph is a straight line through the origin of the kelvin axis. Different quantities held, same equation underneath.
Why do the gas laws need kelvin?
Because the ratio laws are ratios of absolute temperature, and the Celsius scale does not start at zero. Warm a 1.00 L balloon from 27 °C to 54 °C at constant pressure and you get 1.0900 L, not the 2.0000 L that doubling the Celsius reading suggests: an answer 83.5 % too big. Kelvin has no offset buried in it, so the ratio means something.
Are the gas laws the same as the ideal gas law?
They are one equation used in two ways. The ideal gas law, PV = nRT, describes a single state of a gas and needs the constant R. Each simple gas law compares two states of the same sample with quantities held fixed, which cancels R and leaves a ratio. Nothing is added or lost between the two forms.
What is the combined gas law?
The combined gas law says P1V1/T1 = P2V2/T2: pressure times volume divided by absolute temperature is the same before and after, as long as the amount of gas does not change. It is the one to reach for when two of pressure, volume and temperature move at once, which no single ratio law can handle. Boyle, Charles and Gay-Lussac are each this law with one more quantity held.
What does standard temperature and pressure mean?
It means whichever conditions are stated, which is exactly why they have to be stated. One mole of an ideal gas fills 22.71095 L at 273.15 K and 100 kPa, 22.41397 L at 273.15 K and one atmosphere, and 24.78957 L at 298.15 K and 100 kPa. All three are right; they answer different questions, and the familiar 22.4 L belongs to the middle one.
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