Gas laws describe the relationships between pressure, volume, temperature, and amount of gas, and they form one of the most calculation-friendly topics in general chemistry. This guide covers the individual gas laws, the ideal gas law, kinetic molecular theory, and real gas behavior — with detailed worked examples to help you handle typical university assignments. These concepts also underpin the Kp calculations discussed in chemical equilibrium.
1. The Four States of Matter (Brief Overview)
- Solid: fixed shape and volume; particles vibrate in fixed positions.
- Liquid: fixed volume, variable shape; particles are close together but can flow past each other.
- Gas: no fixed shape or volume; particles are far apart and move independently, filling their container.
- Plasma: ionized gas with free electrons and ions; found in stars and specialized lab conditions.
This guide focuses primarily on the gas phase, where the mathematical relationships are most heavily tested.
2. The Individual Gas Laws
2.1 Boyle’s Law (Constant Temperature and Amount)
Pressure and volume are inversely proportional: P₁V₁ = P₂V₂
Worked Example: A gas occupies 4.00 L at 1.20 atm. What volume will it occupy at 2.40 atm (constant T)?
V₂ = P₁V₁/P₂ = (1.20)(4.00)/2.40 = 2.00 L
2.2 Charles’s Law (Constant Pressure and Amount)
Volume and absolute temperature (in Kelvin) are directly proportional: V₁/T₁ = V₂/T₂
Worked Example: A balloon has a volume of 2.50 L at 25°C. What is its volume at 60°C (constant P)?
Convert to Kelvin: T₁ = 298 K, T₂ = 333 K V₂ = V₁T₂/T₁ = (2.50)(333)/298 = 2.79 L
2.3 Gay-Lussac’s Law (Constant Volume and Amount)
Pressure and absolute temperature are directly proportional: P₁/T₁ = P₂/T₂
Worked Example: A sealed gas canister at 2.00 atm and 20°C (293 K) is heated to 100°C (373 K). Find the new pressure.
P₂ = P₁T₂/T₁ = (2.00)(373)/293 = 2.55 atm
2.4 Avogadro’s Law (Constant Temperature and Pressure)
Volume is directly proportional to the number of moles: V₁/n₁ = V₂/n₂. At STP (standard temperature and pressure, 0°C and 1 atm), one mole of any ideal gas occupies 22.4 L.
2.5 Combined Gas Law
When amount is constant but P, V, and T all change: P₁V₁/T₁ = P₂V₂/T₂
Worked Example: A gas at 1.50 atm, 3.00 L, and 300 K is compressed to 2.00 L and heated to 350 K. Find the new pressure.
P₂ = P₁V₁T₂/(T₁V₂) = (1.50)(3.00)(350)/[(300)(2.00)] = 1575/600 = 2.625 atm
3. The Ideal Gas Law
PV = nRT, where R = 0.08206 L·atm/(mol·K) (or 8.314 J/(mol·K) in SI units)
This single equation combines all the individual gas laws and is the most versatile tool for gas calculations.
Worked Example: How many moles of gas are contained in a 5.00 L tank at 3.00 atm and 27°C?
T = 300 K n = PV/RT = (3.00)(5.00)/[(0.08206)(300)] = 15.0/24.62 = 0.609 mol
3.1 Gas Density and Molar Mass
Rearranging PV = nRT with n = mass/M (M = molar mass):
M = mRT/(PV) = dRT/P (where d = density = mass/volume)
Worked Example: A gas has a density of 1.964 g/L at 1.00 atm and 273 K. Find its molar mass.
M = dRT/P = (1.964)(0.08206)(273)/1.00 = 44.0 g/mol (consistent with CO₂)
3.2 Dalton’s Law of Partial Pressures
For a mixture of gases: Ptotal = P₁ + P₂ + P₃ + …, and each gas’s partial pressure relates to its mole fraction: Pᵢ = χᵢ × Ptotal
Worked Example: A container has 2.0 mol N₂ and 3.0 mol O₂ at a total pressure of 5.0 atm. Find the partial pressure of O₂.
Mole fraction of O₂ = 3.0/(2.0+3.0) = 0.60 P(O₂) = 0.60 × 5.0 = 3.0 atm
4. Kinetic Molecular Theory (KMT)
KMT explains gas behavior at the molecular level with these key postulates:
- Gas particles are in constant, random, straight-line motion.
- The volume of individual gas particles is negligible compared to the total volume of the container.
- Gas particles do not attract or repel each other (no intermolecular forces).
- Collisions between particles (and with container walls) are perfectly elastic (no energy lost).
- The average kinetic energy of gas particles is directly proportional to the absolute temperature (in Kelvin), and is the same for all gases at a given temperature, regardless of molar mass.
4.1 Root-Mean-Square Speed
urms = √(3RT/M), where M must be in kg/mol and R = 8.314 J/(mol·K)
Worked Example: Calculate urms for O₂ gas (M = 0.0320 kg/mol) at 298 K.
urms = √[3(8.314)(298)/0.0320] = √(232,036) = ≈ 482 m/s
4.2 Graham’s Law of Effusion/Diffusion
Lighter gases move faster and effuse/diffuse more quickly than heavier gases:
rate₁/rate₂ = √(M₂/M₁)
Worked Example: Compare the effusion rates of H₂ (M = 2.02) and O₂ (M = 32.00).
rate(H₂)/rate(O₂) = √(32.00/2.02) = √15.84 = ≈ 3.98
Hydrogen effuses about 4 times faster than oxygen — a classic Graham’s Law calculation.
5. Real Gases: Deviations from Ideal Behavior
Real gases deviate from ideal behavior most significantly at high pressure (molecules are forced close together, so their volume is no longer negligible) and low temperature (molecules move slowly enough for intermolecular attractions to matter).
5.1 The Van der Waals Equation
(P + an²/V²)(V − nb) = nRT
- The a term corrects for intermolecular attractive forces (larger a = stronger attractions, common in polar or larger molecules).
- The b term corrects for the finite volume of gas molecules themselves (larger b = larger molecular size).
Worked Example (conceptual): Explain why CO₂ deviates more from ideal behavior than He at the same conditions.
CO₂ is a larger, more polarizable molecule with stronger intermolecular (dispersion) forces, giving it a larger a value, and also has a larger molecular volume, giving it a larger b value, compared to the very small, weakly-interacting He atom. Both factors make CO₂ deviate more from ideal gas behavior, especially at high pressure or low temperature (e.g., near its condensation point).
5.2 Compressibility Factor
Z = PV/nRT. For an ideal gas, Z = 1 exactly. Z < 1 indicates that attractive forces dominate (actual volume less than ideal prediction); Z > 1 indicates that repulsive/volume-exclusion effects dominate (typically at very high pressure).
6. Connecting Gas Laws to Stoichiometry
Gas law problems frequently combine with mole-ratio stoichiometry from reaction equations.
Worked Example: How many liters of O₂ (at STP) are needed to completely combust 10.0 g of propane, C₃H₈ + 5O₂ → 3CO₂ + 4H₂O?
Moles of C₃H₈ = 10.0 g / 44.1 g/mol = 0.2268 mol Moles of O₂ needed = 0.2268 × 5 = 1.134 mol Volume at STP = 1.134 mol × 22.4 L/mol = 25.4 L
7. Common Assignment Pitfalls
- Forgetting to convert temperature to Kelvin before using any gas law — this is the single most common error in gas law problems.
- Using inconsistent units for R (e.g., mixing atm with Pa, or L with m³) — always match units to the value of R being used.
- Forgetting that STP conventions can differ: the traditional STP (0°C, 1 atm, molar volume 22.4 L) is still widely taught, though IUPAC’s more recent standard uses 100 kPa (molar volume 22.7 L) — check which convention your course uses.
- Forgetting that Dalton’s Law partial pressures must be based on mole fraction, not mass fraction.
- Applying ideal gas assumptions at very high pressure or very low temperature, where van der Waals corrections are actually necessary.
Further Support With Gas Law Problems
Gas law problems often involve multiple steps, including unit conversions, selecting the appropriate equation, and interpreting the given information. For additional guidance with gas laws and related chemistry topics, see our Chemistry Assignment Help resource.
8. Full Worked Problem
Question: A 2.00 L flask contains a mixture of gases at 1.50 atm and 300 K. If 0.500 mol N₂ and an unknown amount of O₂ are present, and the total moles equal 0.122 mol… (Let’s instead pose a cleanly solvable version:)
Revised Question: A 10.0 L container holds N₂ and O₂ at a total pressure of 2.00 atm and 298 K. If the partial pressure of N₂ is 1.20 atm, find the moles of O₂ present.
Solution: P(O₂) = Ptotal − P(N₂) = 2.00 − 1.20 = 0.80 atm
Using PV = nRT for O₂ alone: n(O₂) = PV/RT = (0.80)(10.0)/[(0.08206)(298)] = 8.0/24.45 = 0.327 mol O₂
Gas laws provide essential quantitative tools that reappear throughout general chemistry, particularly in Kp calculations within chemical equilibrium and in reaction stoichiometry problems generally. Practice converting fluently between the individual gas laws and the combined/ideal gas law, and always double-check your units before plugging numbers into any equation.
