Equation of state pV = nRT · gas process

Leave one field empty — I’ll compute it. R = 8.314 J/(mol·K).
Temperature always in kelvin (K = °C + 273.15). Standard conditions: p = 101,325 Pa, T = 273.15 K → 1 mol occupies 22.4 dm³.
The ideal gas law (Clapeyron equation) relates pressure, volume, amount of substance and temperature:
where . Enter three of the four quantities ( in Pa, in m³, in moles, in kelvin) and the calculator finds the missing one.
The second mode solves a gas process for a fixed amount of gas:
From it come the special processes: isothermal ( = const → ), isobaric ( = const) and isochoric ( = const). Remember: temperature is always in kelvin.
Because the ideal gas law relies on absolute temperature. Using °C would give a wrong result — always convert: K = °C + 273.15.
22.4 dm³ (at p = 101,325 Pa and T = 273.15 K). This follows directly from the ideal gas law.
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