The Ideal Gas Law
- State the ideal gas law and identify each variable and its units
- Solve for pressure, volume, temperature, or moles given the others
- Apply the combined gas law to before-and-after changes in a gas
Four quantities, one equation
For a gas dilute enough that its molecules rarely interact — an ideal gas — four macroscopic quantities are locked together: pressure P, volume V, amount n (in moles), and absolute temperature T. Squeeze the volume and the pressure climbs; heat the gas and it pushes outward. The ideal gas law captures all of these relationships at once.
The microscopic version
The same law can be written per molecule instead of per mole: P·V = N·k_B·T, where N is the number of molecules and k_B is Boltzmann’s constant. This form connects directly to kinetic theory — pressure is the collective drumming of molecules colliding with the container walls, and raising T makes each impact more energetic and more frequent.
A rigid tank holds 0.50 mol of an ideal gas in a volume of 0.020 m³ at 300 K. What is the pressure?
- 1.Solve the ideal gas law for pressure: P = nRT / V.
- 2.Substitute: P = (0.50 mol)(8.314 J·(mol·K)⁻¹)(300 K) / (0.020 m³).
- 3.Numerator: 0.50 × 8.314 × 300 = 1247 J.
- 4.Divide by the volume: 1247 / 0.020 = 62 355 Pa.
A sealed rigid container of gas is at 27 °C and 100 kPa. It is heated to 127 °C. What is the new pressure?
In the combined gas law, temperature must always be in kelvin, but pressure and volume can stay in any consistent units because they appear as ratios (P₂/P₁, V₂/V₁). Only the temperature has an offset zero, so only it must be converted.
A balloon holds 6.0 L of gas at 100 kPa. It is squeezed at constant temperature until the pressure reaches 150 kPa. What is the new volume?
When a problem gives “before” and “after” states, reach for P₁V₁/T₁ = P₂V₂/T₂ and cancel whatever is held constant. Constant V → P ∝ T; constant T → P ∝ 1/V; constant P → V ∝ T.
Answer the 2 checkpoints as you read.
Sign in to save your progress