In this lesson we’ll focus on gases and how their volume changes when we play with temperature and pressure. Think of a gas like a crowd of people in a room – the more people (molecules) or the bigger the room (temperature), the more space they need.
When you heat a gas, its molecules move faster and want more space. This is described by Charles’s Law:
\$V \propto T\$
or in equation form:
\$V = kT\$
🔍 Analogy: Imagine blowing up a balloon. The hotter the air inside, the bigger the balloon gets.
If you squeeze a gas, its molecules are forced closer together, reducing the space they occupy. Boyle’s Law states:
\$P \propto \dfrac{1}{V}\$
or:
\$PV = \text{constant}\$
💨 Analogy: Think of a spring-loaded air pump. Squeezing it (increasing pressure) makes the air inside compress (decrease volume).
When both temperature and pressure change, we combine the two laws:
\$\dfrac{P1 V1}{T1} = \dfrac{P2 V2}{T2}\$
📊 Example: A sealed balloon at 20 °C (293 K) and 1 atm has a volume of 2 L. If you heat it to 60 °C (333 K) while keeping pressure constant, the new volume is:
\$V2 = V1 \dfrac{T2}{T1} = 2\,\text{L} \times \dfrac{333}{293} \approx 2.27\,\text{L}\$
The most general equation for a gas is the Ideal Gas Law:
\$PV = nRT\$
Where:
🔬 Tip: Always convert temperature to Kelvin before using the equation.
💡 Practice Question: A gas occupies 5 L at 1 atm and 300 K. If the pressure is increased to 2 atm while keeping temperature constant, what is the new volume?
\$V2 = \dfrac{P1 V1}{P2} = \dfrac{1 \times 5}{2} = 2.5\,\text{L}\$
| Law | Relationship | Equation |
|---|---|---|
| Charles’s Law | \$V \uparrow \Leftrightarrow T \uparrow\$ | \$V = kT\$ |
| Boyle’s Law | \$P \uparrow \Leftrightarrow V \downarrow\$ | \$PV = \text{constant}\$ |
| Combined Gas Law | \$P,V,T\$ all vary | \$\dfrac{P1V1}{T1} = \dfrac{P2V2}{T2}\$ |
| Ideal Gas Law | All variables | \$PV = nRT\$ |