Audited 29 Jul 2026·Last updated 28 Oct 2026·8 citations·Tier 1·0 uses

Combined Gas Law Calculator — P₁V₁/T₁ = P₂V₂/T₂

Solve P₁V₁/T₁ = P₂V₂/T₂ for any one of six values, using absolute pressure and kelvin, and see whether your change is Boyle's, Charles's or Gay-Lussac's law.

Combined Gas Law Calculator (Boyle, Charles & Gay-Lussac)

Solve for
ABSOLUTE pressure, not gauge. 1 atm = 101.325 kPa exactly; 1 bar = 100 kPa; 1 psi = 6.894757 kPa; 1 mmHg (Torr) = 0.1333224 kPa. A tyre gauge reading 220 kPa at sea level is 321.325 kPa absolute.
kPa
Any volume unit works as long as V₁ and V₂ use the same one — only their ratio enters the equation. Litres are used here: 1 m³ = 1000 L, 1 mL = 0.001 L, 1 ft³ = 28.3168 L.
L
ABSOLUTE temperature in kelvin — never Celsius or Fahrenheit. K = °C + 273.15. 20 °C = 293.15 K. Entering Celsius does not make the answer slightly wrong; it makes it meaningless, because the law is a proportionality measured from absolute zero.
K
Absolute pressure of the second state, same unit as P₁. Set P₂ = P₁ to model a constant-pressure (isobaric) change — that is Charles's law.
kPa
Volume of the second state, same unit as V₁. Set V₂ = V₁ to model a rigid sealed container (isochoric) — that is Gay-Lussac's law, the one behind aerosol-can and gas-cylinder warnings.
L
Absolute temperature of the second state, in kelvin. Set T₂ = T₁ to model a constant-temperature (isothermal) change — that is Boyle's law.
K
Answer
15.426
The quantity you chose to solve for, in the unit of its matching field: kilopascals for a pressure, litres for a volume, kelvin for a temperature.
Initial pressure P₁
101.325 kPa
Initial volume V₁
5 L
Initial temperature T₁
293.15 K
Final pressure P₂
25 kPa
Final volume V₂
15.426 L
Final temperature T₂
223.15 K
PV/T (constant for both states)
1.7282 kPa·L/K
Which gas law this is
Combined — pressure, volume and temperature all change

Background.

Fill a party balloon in a warm kitchen, carry it outside on a frosty night, and it goes soft. Seal a plastic food tub at room temperature, put it in the fridge, and the lid dimples inward. Drive for an hour on a hot motorway and your tyre pressures climb by a few kilopascals. Each is the same fixed parcel of gas moving from one set of conditions to another, and one equation links the before and after: P₁V₁/T₁ = P₂V₂/T₂.

The relation says that for a sealed amount of gas, pressure times volume divided by absolute temperature stays the same. Give it any five of the six quantities and it returns the sixth: the volume of a balloon at altitude, the pressure in an aerosol can left in a parked car, the temperature at which a cylinder reaches its limit.

Three laws students learn separately are this one with a quantity held still. Keep the temperature fixed and it becomes Boyle's law, P₁V₁ = P₂V₂, published in 1662. Keep the pressure fixed and it becomes Charles's law, V₁/T₁ = V₂/T₂, which Jacques Charles measured around 1787 and Joseph Louis Gay-Lussac published in 1802. Keep the volume fixed and it becomes the pressure–temperature law usually credited to Gay-Lussac, P₁/T₁ = P₂/T₂. Enter equal values for the fixed quantity and the result names which law you are using.

Two conventions decide whether the answer means anything. Pressure must be absolute, measured from a vacuum, so a tyre gauge's 220 kPa is really 321.325 kPa at sea level. Temperature must be in kelvin, measured from absolute zero, so 20 °C is 293.15 K. Get either wrong and the arithmetic still produces a number, just not one connected to the gas in front of you.

The amount of gas never has to be entered, because it cancels between the two states. That convenience carries a condition: nothing may leak in or out, condense, dissolve or react between the before and the after.

What is combined gas law calculator (boyle, charles & gay-lussac)?

The combined gas law states that for a fixed amount of an ideal gas, the product of absolute pressure and volume divided by absolute temperature has the same value in any two states. Subscript 1 marks the starting state and subscript 2 the finishing one; the law relates the endpoints and says nothing about the path between them. The shared value of PV/T is nR, the amount of gas in moles times the molar gas constant, R = 8.314462618 J mol⁻¹ K⁻¹, a value that has been exact since the 2019 redefinition of the SI. Because a fixed sample keeps the same n, and R never changes, both drop out when two states are compared, which is why the law can answer practical questions about a container whose contents you have never weighed.

How to use this calculator.

  1. Pick the one quantity you do not know under Solve for. The other five fields must all be filled; the field you are solving for is ignored.
  2. Enter both pressures as absolute values in kilopascals. Add local atmospheric pressure to any gauge reading: 101.325 kPa at sea level, about 89.9 kPa at 1,000 m in the standard atmosphere. 1 atm = 101.325 kPa, 1 bar = 100 kPa, 1 psi = 6.894757 kPa.
  3. Enter both temperatures in kelvin: K = °C + 273.15, or K = (°F + 459.67) × 5/9 from Fahrenheit.
  4. Enter both volumes in the same unit. Litres are labelled, but cubic feet or millilitres work as long as V₁ and V₂ agree, and the answer comes back in that unit.
  5. To study one variable at a time, type the same value in both states for the quantity being held constant, and check that the gas-law output names the case you intended.
  6. Before trusting the answer, ask whether any gas could have escaped, been added or condensed between the two states. If so, the combined law does not apply; use PV = nRT on each state with its own amount of gas.

The formula.

P₁V₁ / T₁ = P₂V₂ / T₂

Start with the ideal gas equation, PV = nRT, written once for each state of the same sample. Divide each by its temperature and both sides equal nR, so P₁V₁/T₁ = P₂V₂/T₂. Solving for each unknown in turn gives six rearrangements: V₂ = P₁V₁T₂ / (T₁P₂), P₂ = P₁V₁T₂ / (T₁V₂), T₂ = T₁P₂V₂ / (P₁V₁), and the mirror images V₁ = P₂V₂T₁ / (T₂P₁), P₁ = P₂V₂T₁ / (T₂V₁) and T₁ = T₂P₁V₁ / (P₂V₂). A helpful way to read any of them is as a starting value multiplied by ratios. The new volume is the old one scaled by P₁/P₂ and by T₂/T₁: halve the absolute pressure and the volume doubles, raise the absolute temperature by 10% and the volume rises by 10%, and both effects multiply. Because the ratios are taken from absolute zero, temperature changes that sound dramatic are often modest: going from 20 °C to −50 °C is a fall from 293.15 K to 223.15 K, a factor of 0.7612. Pressure ratios, by contrast, can be enormous, which is why pressure usually dominates when both change. The PV/T output is the left-hand side, P₁V₁/T₁, in kPa·L/K. Since 1 kPa × 1 L = 1 J exactly, that is J/K, and dividing it by R returns the amount of gas in moles. All six quantities must be greater than zero: each appears in a denominator somewhere, and a zero or negative absolute pressure or temperature has no physical meaning, so a negative entry usually signals a Celsius value or a gauge reading below atmospheric. The model behind all of this is the ideal gas, molecules with no size and no attraction for one another. For air, nitrogen, helium and similar gases near atmospheric pressure and well above their boiling points the error is around a tenth of a percent; it grows at high pressure, where the molecules' own volume matters, and close to condensation, where their attraction does.

A worked example.

Example

A sounding balloon holds 5 L of helium at launch, where the pressure is 101.325 kPa and the air is 20 °C, or 293.15 K. It climbs to a height where the outside pressure is 25 kPa and the temperature is −50 °C, or 223.15 K. The envelope is slack, so the helium is always at the outside pressure, and none escapes. Solving for the final volume uses V₂ = P₁V₁T₂ / (T₁P₂). The numerator is 101.325 × 5 × 223.15 = 113,053.36875; the denominator is 293.15 × 25 = 7,328.75. Dividing gives V₂ = 15.426 L, a little over three times the launch volume. Splitting it into ratios shows why. The pressure ratio 101.325 ÷ 25 = 4.053 would quadruple the volume on its own, while the temperature ratio 223.15 ÷ 293.15 = 0.7612 shrinks it by about a quarter. Together, 4.053 × 0.7612 = 3.0852, and 5 × 3.0852 = 15.426 L. The cold does matter: at constant pressure alone it would have shrunk the helium to 3.806 L. It simply loses to a fourfold pressure drop. The PV/T output checks the result. For the launch state it is 101.325 × 5 ÷ 293.15 = 1.7282 kPa·L/K, and for the high-altitude state 25 × 15.426 ÷ 223.15 gives the same 1.7282. Divided by R, that is 0.2079 mol of helium, about 0.83 g, a quantity the calculation never needed to know. The process label reads combined, because pressure, volume and temperature all changed.

initial Volume5
initial Temperature293.15
final Temperature223.15
final Volume15.426
initial Pressure101.325
final Pressure25
solve ForfinalVolume

Frequently asked questions.

Why does a weather balloon get bigger as it climbs into colder air?
Because pressure falls much faster, in proportion, than absolute temperature does. In the worked example the pressure drops by a factor of 4.053 while the temperature falls only to 0.7612 of its launch value on the kelvin scale, so the volume grows by about 3.09 times. Sounding balloons are launched only partly inflated so the envelope has room for this expansion before it bursts.
My tyre gauge says 220 kPa. What pressure goes into the equation?
The absolute pressure, gauge plus atmospheric: 220 + 101.325 = 321.325 kPa at sea level. Warm that tyre from 15 °C to 45 °C at constant volume and it reaches 354.78 kPa absolute, which your gauge would show as 253.45 kPa. Feeding the gauge figure straight in predicts 242.90 kPa instead, more than 10 kPa too low.
What goes wrong if I enter temperatures in Celsius?
The law is a proportion measured from absolute zero, so Celsius values give the wrong ratio. Warming from 10 °C to 20 °C looks like a doubling in Celsius, but it is 283.15 K to 293.15 K, an increase of only about 3.5%. Below 0 °C a Celsius value is negative, which is rejected outright; above 0 °C nothing stops you, so convert every temperature first.
How high can the pressure in an aerosol can go inside a hot car?
In a rigid can the volume stays put, so P₂ = P₁ × T₂/T₁. A can at 300 kPa absolute and 20 °C (293.15 K) that heats to 60 °C (333.15 K) reaches 300 × 333.15 ÷ 293.15 = 340.93 kPa, a rise of about 13.6%. Many aerosol cans also hold a liquefied propellant, whose vapour pressure rises faster with temperature than an ideal gas would, so the true rise can be larger.
Why does the lid of a sealed tub sink in the fridge?
The air sealed inside at 22 °C (295.15 K) cools to about 4 °C (277.15 K). With the volume nearly fixed, its pressure falls in proportion: 101.325 × 277.15 ÷ 295.15 = 95.15 kPa. The kitchen air outside still presses at 101.325 kPa, so about 6.18 kPa of difference pushes the flexible lid inward until the volume shrinks enough to narrow the gap.
How do I use it for Boyle's law?
Enter the same temperature in both states. Seal the tip of a 60 mL syringe at atmospheric pressure, 101.325 kPa, and push the plunger to 20 mL slowly enough that the air stays at room temperature: P₂ = 101.325 × 60 ÷ 20 = 303.975 kPa. A third of the volume gives three times the absolute pressure, and the result is labelled isothermal, Boyle's law.
When should I use the ideal gas law instead?
Whenever the amount of gas changes, or when you need the amount itself. Pumping up a tyre, a cylinder leaking, steam condensing on a cold wall or a reaction that produces gas all change n between the two states, so the combined law no longer holds. PV = nRT applied to each state with its own n handles those; the combined law is for one sealed sample.
Which standard temperature and pressure should I use?
It depends on whose standard. IUPAC has used 273.15 K and 100 kPa since 1982, giving an ideal molar volume of 22.711 L/mol; the older 273.15 K and 101.325 kPa gives the textbook 22.414 L/mol. The US EPA's emissions rules in 40 CFR 60.2 use 293 K and 101.3 kPa, which makes a gas volume about 7% larger than at 273.15 K and 101.325 kPa. Enter whichever your source specifies as state 2.
When is the ideal-gas answer not good enough?
At high pressure, near condensation and for strongly attracting molecules. Air in a scuba cylinder at 200 bar, steam close to its saturation line, refrigerants, and carbon dioxide near its critical temperature of about 31 °C all depart measurably from ideal behaviour. For those cases the NIST Chemistry WebBook provides reference-quality equations of state for individual fluids, and engineers use real-gas equations such as Peng-Robinson.
Can I enter pressure in atmospheres or volume in cubic feet?
Yes, as long as both states use the same unit, because only the ratios P₁/P₂ and V₁/V₂ enter the equation. The answer comes back in that unit. Temperature is the exception: its ratio means something only on an absolute scale, so use kelvin, or Rankine for US customary work. The PV/T output is in kPa·L/K only when you use kilopascals and litres.

References& sources.

  1. [1]International Union of Pure and Applied Chemistry. Compendium of Chemical Terminology (the 'Gold Book'), 5th edition, online version 5.0.0 (2025), entry 'standard conditions for gases' (S05910): 'Temperature, 273.15 K (0 °C) and pressure of 10⁵ pascals. IUPAC recommends that the former use of the pressure of 1 atm as standard pressure (equivalent to 1.01325 × 10⁵ Pa) should be discontinued.' Source document: Pure Appl. Chem. 1990, 62, 2167 (Glossary of atmospheric chemistry terms), p. 2216. Independent standards body; open access; retrieved 2026-07-29.
  2. [2]International Union of Pure and Applied Chemistry. Compendium of Chemical Terminology (the 'Gold Book'), 5th edition, online version 5.0.0 (2025), entry 'standard pressure' (S05921): 'Chosen value of pressure denoted by p°. In 1982 IUPAC recommended the value 10⁵ Pa, but prior to 1982 the value 101 325 Pa (= 1 atm) was usually used.' Source documents: Green Book, 2nd ed., p. 54; Pure Appl. Chem. 1994, 66, 533, p. 536. This is the citation behind the 22.414 L/mol versus 22.711 L/mol discrepancy discussed in the FAQ. Open access; retrieved 2026-07-29.
  3. [3]Tiesinga, E., Mohr, P. J., Newell, D. B. & Taylor, B. N. CODATA Recommended Values of the Fundamental Physical Constants: 2022. NIST Standard Reference Database 121, 'molar gas constant' — R = 8.314 462 618… J mol⁻¹ K⁻¹, listed with standard uncertainty '(exact)' following the 2019 SI redefinition. R is used on this page only to interpret the PV/T output as nR; it does not enter the combined-gas-law arithmetic. Independent national metrology institute; open access; retrieved 2026-07-29.
  4. [4]US Environmental Protection Agency. 40 CFR §60.2, Definitions (Standards of Performance for New Stationary Sources, subpart A): 'Standard conditions means a temperature of 293 K (68 °F) and a pressure of 101.3 kilopascals (29.92 in Hg).' Cited as the SECOND, independent authority on reference conditions — a US federal regulator whose standard temperature is 20 K above IUPAC's, which is why this page never assumes a standard state. Current as of the eCFR text retrieved 2026-07-29.
  5. [5]International Union of Pure and Applied Chemistry. Quantities, Units and Symbols in Physical Chemistry (the 'Green Book'), 3rd edition, 2nd printing 2012, §2.10 'Chemical thermodynamics', p. 48 — symbols and coherent SI units for pressure p (Pa), volume V (m³), thermodynamic temperature T (K), and the definition of partial pressure pB = yB p. Establishes the unit and symbol conventions used throughout this page. Open access PDF; retrieved 2026-07-29.
  6. [6]National Institute of Standards and Technology. NIST Chemistry WebBook, SRD 69 — Thermophysical Properties of Fluid Systems. Reference-quality equations of state (Span-Wagner for CO₂, IAPWS-95 for water, Lemmon et al. for air and nitrogen) that supersede the ideal-gas approximation at high pressure and near saturation. Consult it when you need to know how far the combined gas law is from the truth for a specific fluid at specific conditions. Independent national metrology institute; open access; retrieved 2026-07-29.
  7. [7]Boyle, R. A Defence of the Doctrine Touching the Spring and Weight of the Air, against the Objections of Franciscus Linus (London, 1662). The appendix 'A Table of the Condensation of the Air' is the first published statement of the inverse pressure–volume relation now called Boyle's law — the isothermal special case of the equation on this page. PRINT / BIBLIOGRAPHIC REFERENCE ONLY: no stable open URL is given here because the 1662 edition's digitised copies live behind institution-specific handles that change; consult a rare-books catalogue or Early English Books Online.
  8. [8]Gay-Lussac, J. L. 'Recherches sur la dilatation des gaz et des vapeurs.' Annales de Chimie 1802, 43, 137–175. The paper that published Jacques Charles's earlier unpublished constant-pressure measurements and established that all gases expand by the same fraction per degree — the isobaric special case. Gay-Lussac's own constant-volume result (pressure proportional to absolute temperature at fixed volume) followed in 1809. PRINT / BIBLIOGRAPHIC REFERENCE ONLY to a 19th-century journal; scanned volumes exist in several national-library collections under unstable identifiers.

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P₁V₁ / T₁ = P₂V₂ / T₂
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