Ideal Gas Law Calculator

Ideal Gas Law is evaluated from Pressure, Volume and Moles of Gas. The calculation reports Pressure, Pressure and Pressure.

Results

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About the Ideal Gas Law Calculator

Ideal Gas Law is treated here as a quantitative relation between Pressure, Volume, Moles of Gas and Temperature and Pressure, Pressure, Pressure and Volume.

The calculator uses a multi formula configuration. Each reported value is read as a direct evaluation of the stored rules with the declared field formats and units.

Formula basis:
PV = nRT (R = 0.08206 L·atm/mol·K)
P = nRT/V (pressure)
V = nRT/P (volume)
n = PV/RT (moles)
T = PV/nR (temperature in K)
Always convert T to Kelvin: T(K) = T( degC) + 273.15

Interpret the outputs in the order shown by the result fields. Optional inputs affect only the outputs that depend on those variables.

Formula & How It Works

The calculation applies the following relations exactly as recorded in the metadata:

PV = nRT (R = 0.08206 L·atm/mol·K)
P = nRT/V (pressure)
V = nRT/P (volume)
n = PV/RT (moles)
T = PV/nR (temperature in K)
Always convert T to Kelvin: T(K) = T( degC) + 273.15

Each output field is produced by substituting the supplied inputs into the relevant relation and then applying the declared rounding or text format.

Worked Examples

Example 1: Car Tire Pressure (Summer vs. Winter)

Inputs

V: 14 n: 0.71 T_C: 35
Pressure: 1.2824 atm. Pressure: 129.939 kPa. Pressure: 18.846 psi. Volume: 14 L. Moles: 0.71 mol. Temperature: 35 degC. Temperature: 308.15 K

With Volume = 14, Moles of Gas = 0.71 and Temperature = 35 as the stated inputs, the result is Pressure = 1.2824 atm, Pressure = 129.939 kPa and Pressure = 18.846 psi. Each value corresponds to the declared output fields.

Example 2: Scuba Tank — Moles of Compressed Air

Inputs

P: 204 V: 12 T_C: 20
Pressure: 204 atm. Pressure: 20,670.3 kPa. Pressure: 2,997.964 psi. Volume: 12 L. Moles: 101.763 mol. Temperature: 20 degC. Temperature: 293.15 K

With Pressure = 204, Volume = 12 and Temperature = 20 as the stated inputs, the result is Pressure = 204 atm, Pressure = 20,670.3 kPa and Pressure = 2,997.964 psi. Each value corresponds to the declared output fields.

Example 3: Volume of CO₂ from Baking Soda

Inputs

P: 1 n: 0.011905 T_C: 180
Pressure: 1 atm. Pressure: 101.325 kPa. Pressure: 14.696 psi. Volume: 0.4427 L. Moles: 0.0119 mol. Temperature: 180 degC. Temperature: 453.15 K

With Pressure = 1, Moles of Gas = 0.011905 and Temperature = 180 as the stated inputs, the result is Pressure = 1 atm, Pressure = 101.325 kPa and Pressure = 14.696 psi. Each value corresponds to the declared output fields.

Example 4: Airbag Inflation Chemistry

Inputs

P: 1.5 V: 60 T_C: 300
Pressure: 1.5 atm. Pressure: 151.988 kPa. Pressure: 22.044 psi. Volume: 60 L. Moles: 1.9136 mol. Temperature: 300 degC. Temperature: 573.15 K

With Pressure = 1.5, Volume = 60 and Temperature = 300 as the stated inputs, the result is Pressure = 1.5 atm, Pressure = 151.988 kPa and Pressure = 22.044 psi. Each value corresponds to the declared output fields.

Common Use Cases

  • Calculate pressure of gas in a container at given temperature
  • Find volume of gas produced from a chemical reaction
  • Determine moles of gas from PVT measurements