Quick Answer
One mole of an ideal gas at 1 atm and 273.15 K (STP) occupies 22.41 L. At 2 atm and 300 K, 0.5 moles of gas occupies 6.15 L.
Common Examples
| Input | Result |
|---|---|
| n = 1 mol, T = 273.15 K, P = 1 atm | V = 22.41 L |
| P = 2 atm, n = 0.5 mol, T = 300 K | V = 6.15 L |
| V = 10 L, n = 0.5 mol, T = 350 K | P = 1.44 atm |
| P = 1 atm, V = 5 L, T = 298 K | n = 0.2044 mol |
| P = 3 atm, V = 8.2 L, n = 1 mol | T = 299.73 K |
How It Works
The Ideal Gas Law
PV = nRT
Where:
- P = pressure in atmospheres (atm)
- V = volume in liters (L)
- n = amount of substance in moles (mol)
- R = ideal gas constant = 0.08206 L atm / (mol K)
- T = absolute temperature in Kelvin (K)
Rearranged forms:
- Pressure: P = nRT / V
- Volume: V = nRT / P
- Moles: n = PV / (RT)
- Temperature: T = PV / (nR)
The Gas Constant
The value of R depends on the units used. Common values:
- R = 0.08206 L atm / (mol K) (when using atm and liters)
- R = 8.314 J / (mol K) (when using pascals and cubic meters)
- R = 62.36 L mmHg / (mol K) (when using mmHg and liters)
This calculator uses R = 0.08206 L atm / (mol K).
Standard Temperature and Pressure (STP)
At STP (0 degrees C = 273.15 K, 1 atm), one mole of any ideal gas occupies 22.414 liters. This value is called the molar volume at STP and is useful for quick conversions between moles and volume.
Converting to Kelvin
Temperature must be in Kelvin. To convert: K = C + 273.15. For Fahrenheit: K = (F - 32) x 5/9 + 273.15. Common reference points: water freezes at 273.15 K (0 C), room temperature is approximately 293 to 298 K (20 to 25 C), and water boils at 373.15 K (100 C).
Limitations
The ideal gas law assumes gas molecules have no volume and no intermolecular forces. It is most accurate at low pressures and high temperatures. For real gases at high pressures or low temperatures, the van der Waals equation or other corrections provide better results.
Worked Example
A sealed container holds 2.0 moles of nitrogen gas at 25 C (298.15 K) and 1.5 atm. What is the volume?
V = nRT / P = (2.0 x 0.08206 x 298.15) / 1.5 = 48.91 / 1.5 = 32.61 L
The gas occupies 32.61 liters under these conditions.
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