Ever wondered why a balloon shrinks when you dive underwater, or why a tire feels softer on a hot day? Those everyday quirks are all about gas laws – the rules that tell us how pressure, volume, and temperature dance together.
Gas laws are simple recipes that link how tightly gas particles are packed (pressure), how much space they occupy (volume), and how fast they jiggle (temperature). Change one, and the others adjust automatically.
What are Gas Laws?
In chemistry, a gas law is a mathematical relationship that describes how a gas behaves when you tweak its pressure, volume, or temperature. Think of it like a rulebook for a game of marbles: if you squeeze the box (increase pressure), the marbles spread out (volume changes), and if you warm the box (raise temperature), the marbles bounce faster.
Boyle’s Law – Pressure‑Volume Relationship
Pressure is the force that gas particles exert on the walls of their container, like water pushing against a pipe. Volume is the space the gas occupies, similar to the amount of water in a tank. Boyle’s Law says that if the temperature stays the same, the product of pressure (P) and volume (V) is constant: P₁V₁ = P₂V₂.
Example: A syringe contains 30 mL of air at 1 atm. If you push the plunger until the volume becomes 10 mL, what is the new pressure?
Using P₁V₁ = P₂V₂ → (1 atm)(30 mL) = P₂(10 mL) → P₂ = 3 atm. So the pressure triples.
Charles’s Law – Temperature‑Volume Relationship
Temperature measures how fast gas particles move, like how quickly kids run around a playground. Charles’s Law holds pressure steady and links volume (V) with absolute temperature (T, measured in Kelvin): V₁/T₁ = V₂/T₂.
Example: A balloon holds 2 L of gas at 300 K. Heat it to 350 K without letting any gas escape. What is the new volume?
V₂ = V₁ × (T₂/T₁) = 2 L × (350/300) ≈ 2.33 L.
Gay‑Lussac’s Law – Pressure‑Temperature Relationship
When volume is locked, pressure (P) and temperature (T) move together: P₁/T₁ = P₂/T₂. Imagine a sealed soda can; heat it and the pressure inside climbs.
Example: A sealed container has a pressure of 1 atm at 280 K. If the temperature rises to 310 K, the pressure becomes?
P₂ = P₁ × (T₂/T₁) = 1 atm × (310/280) ≈ 1.11 atm.
Combined Gas Law
Combine the three previous relationships into one tidy equation: P₁V₁/T₁ = P₂V₂/T₂. This is handy when more than one variable changes.
Example: A 5 L gas sample at 1 atm and 300 K is compressed to 2 L and heated to 350 K. Find the final pressure.
P₂ = P₁V₁T₂ / (V₂T₁) = (1 atm × 5 L × 350 K) / (2 L × 300 K) ≈ 2.92 atm.
Ideal Gas Law
The ideal gas law ties all four variables together: PV = nRT. Here, n is the amount of gas in moles (think of it as a “dozen” for molecules), and R is the universal gas constant (≈0.0821 L·atm·K⁻¹·mol⁻¹). It works best for gases that behave “ideally,” i.e., they don’t stick together and occupy negligible space.
Example: How many moles of oxygen are in a 10 L container at 2 atm and 298 K?
n = PV / RT = (2 atm × 10 L) / (0.0821 L·atm·K⁻¹·mol⁻¹ × 298 K) ≈ 0.82 mol.
Quick Comparison of the Main Gas Laws
| Law | What Changes? | Formula | Constant(s) |
|---|---|---|---|
| Boyle’s | Pressure ↔ Volume | P₁V₁ = P₂V₂ | Temperature |
| Charles’s | Volume ↔ Temperature | V₁/T₁ = V₂/T₂ | Pressure |
| Gay‑Lussac’s | Pressure ↔ Temperature | P₁/T₁ = P₂/T₂ | Volume |
| Combined | All three | P₁V₁/T₁ = P₂V₂/T₂ | None (all vary) |
| Ideal Gas | Pressure, Volume, Temperature, Moles | PV = nRT | R (gas constant) |
Tips for Solving Gas‑Law Problems in ISC Exams
- Always convert temperature to Kelvin (K = °C + 273).
- Check units: pressure in atm, volume in litres, R = 0.0821 L·atm·K⁻¹·mol⁻¹.
- Identify which variables stay constant; pick the simplest law that fits.
- When more than one variable changes, fall back on the combined gas law or the ideal gas law.
- Write down what you need to find, then rearrange the formula before plugging numbers.
📝 Likely Exam Questions
- Question: A 0.5 mol sample of nitrogen gas occupies 12 L at 1 atm and 300 K. What volume will it occupy at 2 atm and 350 K?
Answer: Use combined gas law: (P₁V₁/T₁) = (P₂V₂/T₂) → V₂ = (P₁V₁T₂)/(P₂T₁) = (1×12×350)/(2×300) = 7 L. - Question: Explain why a hot air balloon rises using Charles’s law.
Answer: Heating the air inside increases its temperature, which (at constant pressure) expands the volume, making the air less dense than the cooler outside air, so buoyancy lifts the balloon. - Question: Calculate the final pressure when 25 mL of a gas at 1.2 atm is compressed to 10 mL at constant temperature.
Answer: Boyle’s law: P₂ = P₁V₁/V₂ = (1.2 atm × 25 mL) / 10 mL = 3 atm. - Question: A sealed container holds 3 mol of gas at 25 °C and 1 atm. Find the volume using the ideal gas law.
Answer: Convert temperature: 25 °C = 298 K. V = nRT/P = (3 mol × 0.0821 × 298) / 1 ≈ 73.5 L. - Question: State two assumptions behind the ideal gas law.
Answer: (i) Gas particles have no volume of their own; (ii) No attractive or repulsive forces act between particles.