Ideal gas law calculator
PV = nRT: the pressure or the volume of a gas from the rest.
Fill in the fields and the result will appear here automatically.
Finds absolute pressure or volume of an ideal gas from amount of substance and temperature. Selected Pa, kPa or atm and m³ or litres are converted into compatible units; Celsius is converted to K automatically. Since 1 kPa·L = 1 Pa·m³ = 1 J, the same numerical R works for kPa with litres: the error comes from incompatible unit combinations, not from this pair. Positive temperature describes the gas model; 0 K is accepted only as a formal algebraic limit.
How it works
Formula and logic
PV = nRT, so P = nRT/V or V = nRT/P. The calculation uses R = 8.314462618 J/(mol·K), a rounded approximation of exact R = N_Ak = 8.31446261815324 J/(mol·K). 1 L = 0.001 m³ and 1 kPa = 1000 Pa, so 1 kPa·L = 1 J and the numerical R is the same in kPa·L/(mol·K). atm·L needs a different numerical R; the form performs that conversion. With positive n and V, T = 0 formally gives P = 0; with positive n and P, it gives V = 0. Neither is a physical gas state.
Example
2 mol, 300 K and 0.05 m³ (50 L) give P = 99,773.551416 Pa → 99,773.55 Pa. For 1 mol at 273.15 K, volume is 22.414 L at 1 atm = 101.325 kPa, but 22.711 L at 100 kPa: state the pressure explicitly.
Fields and units
- What to find — list option
- Amount of substance — mol
- Temperature unit — list option
- Temperature — K/°C
- Volume unit — list option
- Volume — m³/L
- Pressure unit — list option
- Pressure — Pa/kPa/atm
How to use
- — Choose pressure or volume, the two modes offered by this form.
- — Enter a positive amount of substance in moles and a temperature; choose K or °C.
- — For pressure, enter positive volume in m³ or litres; for volume, enter positive absolute pressure in Pa, kPa or atm.
- — Celsius is converted automatically: 0 °C = 273.15 K; a value below −273.15 °C is rejected.
- — Check the pressure reference: convert a gauge reading to absolute pressure using the actual ambient pressure first.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax Chemistry 2e §9.2: ideal and combined gas laws, units and model limits NIST CODATA 2022: exact gas constant and distinct standard molar-volume conditions
- Limitation
- Ideal-gas model, generally a better approximation for real gases at relatively low pressure and away from condensation. Pressure is absolute. 0 K is a formal algebraic limit only, not a real gas state. Substance properties and non-ideal behaviour are not modelled.
FAQ
Why are kPa and litres compatible with R = 8.314462618?
1000 Pa × 0.001 m³ = 1 J, so Pa·m³ and kPa·L have the same numerical R. Pa with litres without conversion differs by a factor of 1000; the form converts the selected units automatically.
Must I convert Celsius to kelvin myself in this form?
No. Select °C and enter that reading; the form adds 273.15. For example, 0 °C means 273.15 K, not zero pressure.
What does the PV = nRT result at 0 K mean?
Only the formal limit of the equation: at fixed positive volume, P = 0; at fixed positive pressure, V = 0. A real gas usually condenses before this point. The result does not describe an existing ideal gas at absolute zero.
Can gauge pressure be used in PV = nRT?
Pressure must be absolute. p_abs = p_gauge + p_ambient, using the ambient pressure for the actual measurement. The atm unit means a standard atmosphere, not a measurement of the local air.
Why is one mole not always 22.414 L?
Volume depends on temperature and absolute pressure. At 273.15 K and 101.325 kPa it is 22.414 L; at 100 kPa it is 22.711 L. A label such as “standard conditions” is insufficient by itself.
Does the ideal law give the same result for different gases?
For the same n, T and P, the ideal approximation gives the same volume. Real-gas deviations depend on the substance and conditions; high pressure or proximity to condensation requires other data and a different model.
Is the gas constant used here exact or rounded?
R = N_Ak itself is exact because both defining constants are exact. The calculator uses the rounded 8.314462618 rather than the full 8.31446261815324. Displayed decimal places do not promise the same accuracy in measured inputs.