Avogadros Law Calculator
Hold the temperature and the pressure steady and the only thing left that moves a gas volume is how many particles you put in — twice the count, twice the space.
Avogadros Law Calculator
Results recalculate instantly on every keystroke. Nothing you type is transmitted.
What this result does not account for
- Ideal-gas law at fixed temperature and pressure
- Real gases drift near liquefaction or high pressure
In short: One mole of any ideal gas at 273.15 K and 1 atm fills 22.413970 L. Double the count at the same temperature and pressure and the volume doubles: 2 mol → 44.827940 L. Halve it and the flask needs only 11.206985 L. The quotient V/n agrees on both sides of the law — 22.413970 L/mol each time — and that quotient is the molar volume the ideal gas page derives from R·T/P.
Formula
V₁/n₁ = V₂/n₂ · V₂ = V₁·n₂/n₁ · Vₘ = R·T/P = 22.413970 L/mol at 273.15 K, 1 atm
Avogadro’s law freezes two of the four state variables: temperature and pressure stay put, so the volume is proportional to nothing but the particle count. The proportionality constant is the molar volume — about 22.414 litres per mole at the old STP convention — and it is the same number for helium, nitrogen or sulfur hexafluoride, because gases at the same conditions carry the same count per litre whatever the particles weigh.
Worked Example
- Type the reference volume and count.
- Fill in the new count to solve the new volume, or blank the new count and type the volume it must fill.
- Leave exactly one of the new-state boxes blank — that is the unknown the law solves.
- Read the solved state and the constant quotient.
Defaults: 22.413970 L, 1 mol → 2 mol gives 44.827940 L and the quotient pins at 22.413970 L/mol. Type 11.206985 into the new volume instead and the count solves to 0.5 mol.
Strengths & Limits Of This Model
Where this engine is strong
- Solves either new-state variable
- The molar volume quotient printed as a check
Where it stops
- Temperature and pressure frozen — not a state solver
- No real-gas correction
Practical Use Cases
Stoichiometry of gases
volume counts, moles price
Balloon and bag filling
count in, volume out
Teaching
the fourth gas law, isolated
Methodology & Editorial Standards
Computation runs in IEEE-754 double precision at full internal precision; rounding to two decimal places occurs strictly at the display layer, so no cumulative drift enters the result. All monetary outputs use accounting presentation — grouped thousands, two decimals, negatives in parentheses — so figures can be transcribed directly into a model or working paper. Division-by-zero and out-of-domain inputs return an em-dash rather than a misleading number.
This engine was reconciled against an independent reference implementation and hand-verified for the worked example above before release. Our full five-stage review process is published on the About Us page.
Disclaimer. This calculator is provided for informational and modelling purposes only and does not constitute financial, tax, legal, medical, or engineering advice. Verify all figures with a qualified professional before acting on them.
Avogadros Law Calculator — 8 Expert FAQs
8 analyst-written answers to the questions practitioners actually ask — optimised for voice and answer-engine retrieval.
Why does the identity of the gas not matter?
Because the law counts particles, not mass. A mole of hydrogen and a mole of sulfur hexafluoride hold the same number of molecules, and ideal-gas volume answers to the count. The molecules’ weight shows up in density — the gas density page weighs the difference — but never in the volume-to-count ratio.
How is this different from the ideal gas law page?
Scope. The ideal gas page solves the full state P·V = n·R·T — any one missing piece out of four. Avogadro’s law freezes temperature and pressure and tracks only how volume follows the count. Use it when the bath is thermostatted and the pressure is the room’s.
What happens at the count of zero?
An honest zero: no particles, no volume. The page prints 0.000000 L rather than an error, because an evacuated flask is a real object. A negative count is refused — that is not a state, it is a typo.
Why does the reference state refuse zero?
The reference anchors the ratio. V₁/n₁ with either entry at zero is either no gas at all or a flask with no room in it, and neither defines a proportionality to scale from. Type a real reference — 22.413970 L at 1 mol is the classic.
Is the molar volume exactly 22.414 litres?
At 273.15 K and exactly 1 atm, R·T/P computes 22.413970 L/mol from the exact gas constant. IUPAC’s current STP uses 100 kPa instead of 1 atm, which pushes the same quantity to 22.710955 L — about 1.3% apart. The ideal gas page splits the two conventions honestly; pick the one your table uses.
Why does the chip card keep quoting 22.413970?
Because the quotient V/n IS the molar volume, and at 273.15 K and 1 atm it computes to 22.413970 L/mol from the exact gas constant. The chips are set so the default drive is the classic one-mole flask: solve 2.000000 mol from it and the card shows 44.827940, exactly double, with the quotient unchanged at 22.413970 — the number agreeing with itself is the whole proof.
Can I run liquids through this law?
No, and the reason is worth stating: the law lives on compressibility. Gas particles are mostly empty space, so squeezing the count into a different volume costs almost nothing; liquids are already touch-tight, and their volume barely answers to the count at all. A mole of water is 18 millilitres whether you push or not — gases obey, liquids shrug.
Does the law survive at high pressure?
Less and less. Real molecules take up room and attract one another, so near liquefaction the volume-to-count ratio drifts from the ideal quotient. The law is exact in the ideal limit, good to percent-level for most permanent gases at a few atmospheres, and a lie near the boiling point.