Combined Gas Law Calculator (P₁V₁/T₁ = P₂V₂/T₂)
The combined gas law, and the two things that make it go wrong
One relation contains Boyle, Charles and Gay-Lussac: P₁V₁/T₁ = P₂V₂/T₂. The two ways it fails are a temperature left in Celsius and a pressure left as gauge, so this page converts both, shows the wrong Celsius ratio beside the right kelvin one, and gives the amount of gas in moles from PV = nRT.
Combined gas law: the new volume
10 L of air at 150 kPa gauge and 27 °C, warmed to 47 °C with the gauge pressure held at 150 kPa
Three named laws are one relation with something held fixed
- T
- the absolute temperature in kelvin, and the one quantity here that must not be left in Celsius. It appears as a ratio, and the test for whether a formula needs an absolute scale is whether adding 273.15 to every temperature leaves it unchanged. Here it does not: 20 and 100 give 5, while 293.15 and 373.15 give 1.273
- P
- the absolute pressure, measured from a vacuum. Gauges read from atmospheric, so a gauge reading needs an atmosphere added. Only the ratio matters to the answer, so kPa, bar, atm or psi all work provided both pressures are in the same one
- Boyle, Charles, Gay-Lussac
- the three named laws, which are this one relation with T, P and V held fixed in turn: PV constant at fixed temperature, V/T constant at fixed pressure, P/T constant at fixed volume. Learning the combined form is less work than learning three, and it never leaves you wondering which one applies
- n
- the amount of gas in moles, which the combined law assumes is FIXED. If gas escapes or is added, the combined law does not apply and you need PV = nRT at both ends separately. NCERT’s own example of warming a room at constant pressure and volume is exactly that case: nothing about the room changes except that about 80 moles of air leave it
- R
- the molar gas constant, 8.314 462 618 153 24 J mol−1 K−1, which is exact: it is kNA, the product of two constants that have been exact by definition since 2019. NCERT uses 8.314 and 0.0821 L atm mol−1 K−1, both of which are fine to four figures
Worked example
10 L of air at 150 kPa gauge and 27 °C, warmed to 47 °C with the gauge pressure held at 150 kPa
MAKE THE PRESSURES ABSOLUTE. 150 kPa on a gauge is 150 + 101.325 = 251.325 kPa absolute, at both ends. Here the two happen to be equal, so the pressure ratio is 1 and the correction cancels — but do it anyway, because it does not cancel the moment the two pressures differ
MAKE THE TEMPERATURES ABSOLUTE. 27 °C = 300.15 K and 47 °C = 320.15 K. The ratio is 320.15/300.15 = 1.066 63. Compare it with 47/27 = 1.740 74, which is what the Celsius numbers give and is not the ratio of anything physical. NCERT's own module makes the same point with 20 and 100 °C: an apparent factor of 5 that is really 1.27
SOLVE. V2 = V1 × (P1/P2) × (T2/T1) = 10 × 1 × 1.066 63 = 10.6663 L. A gain of 6.7 per cent for a 20-degree warming, which is the useful rule of thumb hidden in Charles's law: near room temperature a gas gains roughly 1/300 of its volume per kelvin, about a third of a per cent
HOW MUCH GAS IS THERE? n = PV/RT = 251.325 × 10 / (8.314 463 × 300.15) = 1.0071 mol, using kPa with litres so that the product is in joules. That is 6.06 × 1023 molecules and a molar volume of 9.93 L/mol — well below the familiar 22.4 L/mol, because this gas is at two and a half atmospheres rather than one
AND THE TWO ERRORS, SIDE BY SIDE. Done in Celsius: 10 × 47/27 = 17.41 L, an overstatement of 63.2 per cent. Done with the gauge pressures left as absolute: 10.6663 L, which is exactly right by luck, because the two pressures are equal. Change the final pressure to 250 kPa gauge and the honest answer is 7.6302 L while the gauge-only version gives 6.3998 — a 16.1 per cent understatement that was invisible in the first version of the question
What the two mistakes cost, on this page’s own default change
| Done this way | Pressure used | Temperature ratio used | New volume | Error against the right answer |
|---|---|---|---|---|
| Correctly: absolute pressure, kelvin | 251.325 kPa both ends | 320.15 / 300.15 = 1.066 63 | 10.6663 L | — |
| Celsius temperatures, absolute pressure | 251.325 kPa both ends | 47 / 27 = 1.740 74 | 17.4074 L | +63.2% |
| Gauge pressure, kelvin | 150 kPa both ends | 1.066 63 | 10.6663 L | 0.0% here |
| Both mistakes together | 150 kPa both ends | 1.740 74 | 17.4074 L | +63.2% |
| 273 instead of 273.15, kelvin | 251.325 kPa both ends | 320 / 300 = 1.066 67 | 10.6667 L | +0.0031% |
One relation, two traps, and why Celsius is the one that does the damage
Boyle, Charles and Gay-Lussac are not three laws, they are one relation with a different quantity held fixed each time. P1V1/T1 = P2V2/T2 contains all three: hold T and you have Boyle, hold P and you have Charles, hold V and you have Gay-Lussac. Learning the combined form is less work than learning three and it never leaves you deciding which applies. It holds for a FIXED amount of an ideal gas, which is the assumption that quietly fails whenever gas escapes — and the moment it does, you need PV = nRT at each end instead.
The first trap is temperature, and it is the expensive one. Temperature enters these laws as a ratio, and a ratio of Celsius numbers means nothing because the Celsius zero is an arbitrary offset. NCERT’s own module makes the point with a gas warmed from 20 °C to 100 °C: the Celsius numbers suggest a factor of five, the kelvin ones give 373/293 = 1.27, and the volume grows by 27 per cent and not by 400. On this page’s default change the Celsius version overstates the answer by 63 per cent. The test for whether a formula needs an absolute scale is simple and worth applying: add 273.15 to every temperature in it and see whether the answer changes. In Q = mcΔT it does not, so Celsius is fine. Here it does.
The second trap is pressure, and what makes it dangerous is that it sometimes cancels. Tyre gauges, pressure cookers and workshop instruments read from atmospheric, so a gauge reading needs 101.325 kPa added before it goes anywhere near a gas law: a tyre at 200 kPa gauge is at 301 kPa absolute, half again as much. But when the two pressures are EQUAL, as in a constant-pressure problem, the ratio is 1 either way and the mistake costs nothing — so a student can practise on such a question, get it right, and carry the mistake into the next one, where it is worth tens of per cent. This page prints both answers so the difference is visible rather than lucky.
273 or 273.15? The SI Brochure defines the degree Celsius against the kelvin with t/°C = T/K − 273.15, so 273.15 is the correct offset and it is exact by definition. NCERT’s own module rounds it, converting 0 °C to 273 K, 27 °C to 300 K and 100 °C to 373 K. The page prints both answers and the difference between them: on the default change it is 31 parts per million, which is far below anything a school answer quotes. The point of the row is not that 273 is wrong but that it is a CHOICE, and a number quoted to five significant figures is quoting that choice in its last two.
And what the ideal gas equation is for. PV = nRT turns a pressure, a volume and a temperature into an amount of substance, which is the step that makes the gas laws useful rather than merely consistent: 10 litres at 251 kPa and 300 K is 1.007 moles, about 6 × 1023 molecules. The constant R = 8.314 462 618 153 24 J mol−1 K−1 is exact, because it is the product of the Boltzmann and Avogadro constants, both of which have been exact by definition since 2019 — which also means R carries no uncertainty at all, unlike almost every other constant in physics. Where the model fails is at low temperature and high pressure: below about 100 K and above about twenty atmospheres the molecules’ own size and their attraction stop being negligible, and the page says so rather than pretending otherwise.
Frequently asked questions
Why must the temperature be in kelvin when Q = mcΔT is happy in Celsius?
Because here the temperature appears as a RATIO and there it appears as a DIFFERENCE. A difference is the same number in both scales, since the degree Celsius is defined equal in magnitude to the kelvin. A ratio is not: 100 °C divided by 20 °C is 5, while 373.15 K divided by 293.15 K is 1.273, and only the second is the ratio of two physical temperatures. The quick test is to add 273.15 to every temperature in a formula and see whether the answer moves. If it does, you need an absolute scale.
What is the difference between gauge and absolute pressure?
Absolute pressure is measured from a perfect vacuum; gauge pressure is measured from whatever the atmosphere is doing, so a gauge reading of zero means atmospheric and not nothing. Absolute = gauge + atmospheric, and this page adds the standard atmosphere of 101.325 kPa. At everyday pressures the difference dominates: a tyre at 200 kPa gauge is at 301 kPa absolute, a factor of 1.5. Gas laws always want absolute, which is also why a “vacuum” of −50 kPa gauge is a perfectly ordinary 51 kPa absolute.
Does the combined gas law work if some gas escapes?
No. P1V1/T1 = P2V2/T2 is PV = nRT at two moments with n and R cancelled out, so it assumes the amount of gas is identical at both. For a leaking tyre or an open room you need PV = nRT separately at each end and the amounts will differ. NCERT’s module has a neat example of exactly this: a 100 m³ room at atmospheric pressure warmed by 6 K contains about 4060 moles before and 3980 after, because the pressure and volume are fixed by the building, so roughly 80 moles of air simply leave.
Why 22.4 litres per mole, and why doesn’t this page’s molar volume match it?
22.4 L/mol is the molar volume at 273 K and 1 atm specifically, which NCERT’s module derives as 0.0821 × 273 / 1. It is not a universal constant: it is RT/P with a particular T and P put in, so it changes with both. On this page’s default inputs the gas is at 251 kPa, about two and a half atmospheres, so the molar volume is about 9.9 L/mol. Note also that the older figure of 22.4 uses the old standard pressure of 1 atm; with 100 kPa as the reference instead, the molar volume at 273.15 K is about 22.7 L/mol, which is why two textbooks can print different “standard” molar volumes and both be right about different standards.
When does the ideal gas equation stop working?
At low temperature and at high pressure, and for the same underlying reason: the model assumes the molecules take up no space and do not attract one another. Both assumptions fail as the molecules get closer together, which is what cooling and compressing do. Near liquefaction the errors are large and the gas eventually condenses, at which point the equation describes nothing at all. Above about twenty atmospheres expect errors of a few per cent in the direction of the real gas being less compressible than predicted. School questions live comfortably in the middle, and that is where this page is honest; it flags the two edges rather than quietly computing through them.
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References
- NCERT / CIET, Physics-02 (Keph_201301): Behaviour of Perfect Gases and Kinetic Theory, Unit 9 Module 1, “Ideal Gas” (chapter 13, class XI), 2019, on the CIET e-content server (read 11 November 2026). The source of this page’s framing and of its two worked warnings. It gives Boyle’s law as PV constant, Charles’s as V/T constant, Gay-Lussac’s as P/T constant, and uses P1V1/T1 = P2V2/T2 in its examples without naming it the combined gas law. R is given as 8.314 J mol−1 K−1 and as 0.0821 L atm mol−1 K−1. It converts temperatures with 273 rather than 273.15 — 0 °C to 273 K, 27 to 300, 100 to 373 — and it is the source of the warning this page repeats, that a Celsius temperature ratio of 5 to 1 between 20 and 100 °C is really 1.27. Its tyre example works in gauge pressure and converts to absolute by adding an atmosphere, which is where this page’s pressure select comes from, and its room example shows the fixed-amount assumption failing: 4060 moles before a 6 K warming and about 3980 after.
- CODATA Recommended Values of the Fundamental Physical Constants: 2022, NIST SP 961 (the constants wall chart), May 2024 (read 11 November 2026). k = 1.380 649 × 10−23 J K−1 and NA = 6.022 140 76 × 1023 mol−1, both marked exact, both fixed by definition and carrying no uncertainty. The molar gas constant is printed as R = 8.314 462 618… with an ellipsis and no “exact” tag, which is a typographical truncation and not an uncertainty: R = kNA is the product of two exact decimals, so its digits terminate, and multiplying them out gives exactly R = 8.314 462 618 153 24 J mol−1 K−1. This batch did that multiplication rather than take the ellipsis on trust.
- The International System of Units (SI), 9th edition, BIPM, published May 2019; the version read for this batch is 4.01, revised June 2026 (read 11 November 2026). The source of the kelvin’s CURRENT definition: the kelvin is defined by fixing the Boltzmann constant at exactly k = 1.380 649 × 10−23 kg m2 s−2 K−1, which the brochure states as an exact relation. The degree Celsius is defined equal in magnitude to the kelvin, with T0 = 273.15 K and t/°C = T/K − 273.15. Most school textbooks still define the kelvin as 1/273.16 of the triple point of water. The brochure records that as the PREVIOUS definition, and says the triple-point temperature must now be determined experimentally — it stood at 273.16 K with a relative standard uncertainty of 3.7 × 10−7 at the redefinition, and is a measured quantity from then on.
- COULD NOT BE READ, and nothing was substituted for any of it. The NCERT textbook pages on ncert.nic.in, and four CIET modules in the same series as the ones cited here — Keph_201101 (temperature and its measurement), Keph_201102 (thermal expansion), Keph_201103 (specific heat capacity and calorimetry) and Keph_201105 (conduction and convection) — were all refused by the server’s robots policy on 11 November 2026. That is why this batch quotes no NCERT table number, and why the coefficients of expansion and the thermal conductivities are attributed to the reference actually read rather than to the textbook table you are more likely to be holding. If your table disagrees, use your table: these pages take the constant from you for exactly that reason.