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)
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.
- 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.
- 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.
- Enter both temperatures in kelvin: K = °C + 273.15, or K = (°F + 459.67) × 5/9 from Fahrenheit.
- 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.
- 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.
- 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.
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.
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.
Frequently asked questions.
Why does a weather balloon get bigger as it climbs into colder air?
My tyre gauge says 220 kPa. What pressure goes into the equation?
What goes wrong if I enter temperatures in Celsius?
How high can the pressure in an aerosol can go inside a hot car?
Why does the lid of a sealed tub sink in the fridge?
How do I use it for Boyle's law?
When should I use the ideal gas law instead?
Which standard temperature and pressure should I use?
When is the ideal-gas answer not good enough?
Can I enter pressure in atmospheres or volume in cubic feet?
References& sources.
- [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]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]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]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]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]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]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]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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- Method
- P₁V₁ / T₁ = P₂V₂ / T₂
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