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Ideal Gas Law Chemistry

Solve for

Kelvin = °C + 273.15

Result —
Unit of the result
—
Pressure in kPa
—
Temperature in °C
—
Molar volume (L/mol)
—

The ideal gas law PV = nRT links pressure, volume, amount and temperature of a gas. Pick the unknown, enter the other three in atm, litres, moles and kelvin, and the result follows from R = 0.082057 L·atm/(mol·K). One mole at 273.15 K and 1 atm, for example, occupies 22.414 L.

About this tool

General chemistry problems keep returning to the same four quantities: how hard a gas pushes, how much space it takes, how many moles are present and how hot it is. This calculator rearranges PV = nRT for whichever one is missing, shows the substituted equation, and reports the pressure in kPa, the temperature in °C and the molar volume alongside the answer so you can sanity-check it against 22.4 L/mol. Inputs use the units most textbooks pair with R = 0.082057 L·atm/(mol·K): atmospheres, litres, moles and kelvin. The limit is the model itself: real gases at high pressure or near their condensation point stray from ideal behaviour, and the tool warns when your numbers drift into that region.

How to use it

  1. Choose the unknown

    Select pressure, volume, moles or temperature under Solve for; its input box disappears.

  2. Enter the known values

    Type pressure in atm, volume in litres, amount in moles and temperature in kelvin.

  3. Read the result and working

    The answer, its unit, kPa and °C conversions and the substituted equation update as you type.

Examples

1 mol at 0 °C and 1 atm

Solve for
Volume (V)
Pressure
1
Amount of gas
1
Temperature
273.15

Result Result: 22.414
Unit of the result: L
Pressure in kPa: 101.325
Temperature in °C: 0
Molar volume (L/mol): 22.414

  1. R = 0.082057366 L·atm/(mol·K)
  2. V = nRT / P = 1 × 0.082057366 × 273.15 / 1
  3. Volume = 22.414 L

The classic molar volume: one mole of an ideal gas at 273.15 K and 1 atm fills about 22.414 L.

10 L at 2 atm and 300 K

Solve for
Moles (n)
Pressure
2
Volume
10
Temperature
300

Result Result: 0.8124
Unit of the result: mol
Pressure in kPa: 202.65
Temperature in °C: 26.85
Molar volume (L/mol): 12.309

  1. R = 0.082057366 L·atm/(mol·K)
  2. n = PV / RT = 2 × 10 / (0.082057366 × 300)
  3. Amount of gas = 0.8124 mol

A 10 L vessel at 2 atm and 27 °C holds about 0.81 mol of gas, roughly 26 g of oxygen.

0.5 mol in 5 L at 25 °C

Solve for
Pressure (P)
Volume
5
Amount of gas
0.5
Temperature
298.15

Result Result: 2.4465
Unit of the result: atm
Pressure in kPa: 247.896
Temperature in °C: 25
Molar volume (L/mol): 10

  1. R = 0.082057366 L·atm/(mol·K)
  2. P = nRT / V = 0.5 × 0.082057366 × 298.15 / 5
  3. Pressure = 2.4465 atm

Half a mole squeezed into 5 L at room temperature exerts about 2.45 atm, or 247.9 kPa.

How it is calculated

PV = nRT, so P = nRT/V, V = nRT/P, n = PV/(RT), T = PV/(nR)

P
pressure in atmospheres
V
volume in litres
n
amount of gas in moles
R
gas constant, 0.082057366 L·atm/(mol·K)
T
absolute temperature in kelvin

The gas constant comes from the exact SI value 8.314462618 J/(mol·K) divided by 101.325 J per L·atm. The tool solves the rearranged equation directly, then converts pressure to kPa by multiplying by 101.325 and temperature to °C by subtracting 273.15. Molar volume is V divided by n, which equals 22.414 L/mol at 0 °C and 1 atm for any ideal gas.

When not to use it

  • Gases above roughly 10 atm, where the van der Waals equation fits better.
  • Vapours close to condensing, such as steam just above 100 °C.
  • Liquids, solids or solutions, which do not obey PV = nRT.

Common mistakes

  • Entering temperature in °C instead of kelvin, which can make T zero or negative.
  • Using millilitres for volume when R is in litre units.
  • Mixing R = 8.314 with atmospheres and litres instead of pascals and cubic metres.

Frequently asked questions

Which value of R does the calculation use?

It uses 0.082057366 L·atm/(mol·K), the exact SI gas constant 8.314462618 J/(mol·K) expressed in litre-atmospheres. That pairing matches the atm, L, mol and K inputs, so no conversion factor is needed.

Why must temperature be in kelvin?

The law depends on absolute temperature, which starts at zero where an ideal gas would have no volume. Celsius has its zero at the freezing point of water, so doubling a Celsius value does not double the volume. Add 273.15 to a Celsius reading before entering it.

How do I convert kPa or mmHg to atmospheres?

Divide kPa by 101.325, or divide mmHg (torr) by 760. A 750 mmHg reading is 0.9868 atm, and 200 kPa is 1.974 atm.

What is the molar volume figure in the results?

It is volume divided by moles. At 0 °C and 1 atm every ideal gas gives 22.414 L/mol; at 25 °C and 1 atm it rises to about 24.47 L/mol. A value far from these at ordinary conditions points to an input unit slip.

How do I find the mass of gas from the result?

Multiply moles by the molar mass. For 0.8124 mol of oxygen (32.00 g/mol) the mass is about 26.0 g; for the same amount of helium (4.003 g/mol) it is 3.25 g.

When does the ideal gas law stop working?

Deviations grow when molecules are pushed close together or move slowly: high pressure, low temperature, or gases with strong attractions such as ammonia and water vapour. At 1 atm and room temperature errors for nitrogen or oxygen are well below 1 %.

Cite this tool

Using this tool in an assignment, article or report? Copy a citation, or a link for your website.

APA

Tools Directory Editorial Team. (2026, October 27). Ideal Gas Law Chemistry. Tools Directory. Retrieved October 27, 2026, from https://tools.milunir.com/chemistry/gas-laws-and-thermodynamics/ideal-gas-law-chemistry/

MLA

Tools Directory Editorial Team. “Ideal Gas Law Chemistry.” Tools Directory, 27 Oct. 2026, https://tools.milunir.com/chemistry/gas-laws-and-thermodynamics/ideal-gas-law-chemistry/. Accessed 27 Oct. 2026.

Chicago

Tools Directory Editorial Team. “Ideal Gas Law Chemistry.” Tools Directory. Last modified October 27, 2026. Accessed October 27, 2026. https://tools.milunir.com/chemistry/gas-laws-and-thermodynamics/ideal-gas-law-chemistry/.

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