Ideal Gas Law
The central equation of gas chemistry: PV = nRT links pressure, volume, moles, and temperature in one statement.
What it means
The ideal gas law unites the three gas laws into one equation: any gas's pressure, volume, mole count, and absolute temperature are locked together by the constant R. It is the workhorse of gas stoichiometry — you can find any one of the four quantities from the other three, and the choice of R's units (8.314 with SI units, 0.08206 with liters and atmospheres) simply has to match the pressure and volume units you use. When to use it: for any calculation involving a gas whose P, V, n, and T are all known or findable, especially converting gas volumes into moles for stoichiometry.
Variables
| Symbol | Name | Unit | Meaning |
|---|---|---|---|
| Pressure | atm (or Pa, kPa) | Force per unit area exerted by gas collisions on the container walls. | |
| Volume | L (or m^3) | Space occupied by the gas — effectively the container's volume. | |
| Moles | mol | Amount of gas present. | |
| Gas constant | 0.08206 L·atm/(mol·K) or 8.314 J/(mol·K) | The gas constant; pick the value matching your P and V units. | |
| Temperature | K | Absolute temperature of the gas — always kelvin. |
Assumptions
- The gas behaves ideally: point-mass molecules, no attractive or repulsive forces.
- Temperature must be absolute (kelvin).
- Most reliable at high temperature and low pressure, where real gases approximate ideal behavior.
Worked example
How many moles of gas occupy 1.50 L at 2.00 atm and 300 K?
Given:
- P = 2.00 atm, V = 1.50 L, T = 300 K
- R = 0.08206 L·atm/(mol·K) — chosen so atm, L, and K cancel
Write the ideal gas law: PV = nRT.
Solve for n: n = PV / RT.
Substitute: n = (2.00 atm)(1.50 L) / (0.08206 L·atm/(mol·K) × 300 K).
Numerator: 2.00 × 1.50 = 3.00. Denominator: 0.08206 × 300 = 24.618.
Divide: 3.00 / 24.618 = 0.12186, which rounds to 0.122 mol.
Three atmospheres' worth of PV at room temperature corresponds to just over a tenth of a mole — a small, lab-scale amount of gas.