B3-2. The Empirical Gas Laws

1. Overview of the Gas Laws

The macroscopic behavior of a fixed mass of gas can be systematically analyzed by looking at how its pressure, volume, and absolute temperature interact. The three empirical gas laws each hold one of these variables constant to isolate the relationship between the remaining two.

2. Boyle's Law (Constant Temperature)

Boyle's Law states that for a fixed mass of gas kept at a constant temperature, the pressure is inversely proportional to its volume. As you compress a gas into a smaller volume, its particles collide with the container walls more frequently, increasing the overall pressure.

Boyle's Law Equation:

$$p \propto \dfrac{1}{V} \quad \Rightarrow \quad p_1 V_1 = p_2 V_2$$
p V T = Constant

Example 1: Applying Boyle's Law

Problem: A gas cylinder occupies a volume of $0.040$ m³ at a pressure of $2.0 \times 10^5$ Pa. If the gas is allowed to expand into a new container with a volume of $0.160$ m³ while maintaining a constant temperature, calculate its new pressure.

Solution:

  • Step 1: Identify the known variables.
    $p_1 = 2.0 \times 10^5$ Pa, $V_1 = 0.040$ m³, $V_2 = 0.160$ m³
  • Step 2: Rearrange Boyle's Law for $p_2$.
    $$p_1 V_1 = p_2 V_2 \implies p_2 = \dfrac{p_1 V_1}{V_2}$$
  • Step 3: Substitute and solve.
    $$p_2 = \dfrac{(2.0 \times 10^5) \times 0.040}{0.160} = \dfrac{8000}{0.160} = \mathbf{5.0 \times 10^4 \text{ Pa}}$$

3. Charles's Law (Constant Pressure)

Charles's Law states that for a fixed mass of gas kept at a constant pressure, the volume is directly proportional to its absolute temperature. Heating a gas causes its particles to move faster and hit the boundary walls with more force; to keep the pressure steady, the volume must expand.

Charles's Law Equation:

$$V \propto T \quad \Rightarrow \quad \dfrac{V_1}{T_1} = \dfrac{V_2}{T_2}$$
V T (K) p = Constant

Crucial Rule: The straight line points directly toward the origin ($0$ K), meaning you must always convert Celsius values to Kelvin ($T = \theta^\circ\text{C} + 273.15$) before computing!

Example 2: Applying Charles's Law

Problem: A flexible balloon contains $2.50$ m³ of air at a room temperature of $20^\circ\text{C}$. If the balloon is placed outside in a cold environment where the temperature drops to $-10^\circ\text{C}$ while keeping pressure constant, find its new volume.

Solution:

  • Step 1: Convert temperatures to Kelvin.
    $T_1 = 20 + 273.15 = 293.15$ K
    $T_2 = -10 + 273.15 = 263.15$ K
  • Step 2: Rearrange Charles's Law for $V_2$.
    $$\dfrac{V_1}{T_1} = \dfrac{V_2}{T_2} \implies V_2 = \dfrac{V_1 T_2}{T_1}$$
  • Step 3: Substitute and solve.
    $$V_2 = \dfrac{2.50 \times 263.15}{293.15} = \dfrac{657.875}{293.15} = \mathbf{2.24 \text{ m³}}$$

4. Gay-Lussac's / Pressure Law (Constant Volume)

Gay-Lussac's Law states that for a fixed mass of gas kept at a constant volume, the pressure is directly proportional to its absolute temperature. When a gas inside a rigid container is heated, the higher kinetic energy leads to more violent and frequent impacts per second against the fixed surface area.

Gay-Lussac's Law Equation:

$$p \propto T \quad \Rightarrow \quad \dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}$$
p T (K) V = Constant

Example 3: Applying the Pressure Law

Problem: A rigid steel canister holds an ideal gas at a pressure of $1.01 \times 10^5$ Pa when the temperature is $27^\circ\text{C}$. The canister possesses a safety valve that breaks if internal pressure reaches $1.50 \times 10^5$ Pa. Calculate the temperature at which the safety valve will fail.

Solution:

  • Step 1: Convert the initial temperature to Kelvin.
    $T_1 = 27 + 273.15 = 300.15$ K
  • Step 2: Rearrange Gay-Lussac's Law for the final temperature $T_2$.
    $$\dfrac{p_1}{T_1} = \dfrac{p_2}{T_2} \implies T_2 = \dfrac{p_2 T_1}{p_1}$$
  • Step 3: Substitute and solve.
    $$T_2 = \dfrac{(1.50 \times 10^5) \times 300.15}{1.01 \times 10^5} = \dfrac{4.50225 \times 10^7}{1.01 \times 10^5} \approx \mathbf{445.8 \text{ K}}$$
  • Converting back to Celsius (optional check): $\theta = 445.8 - 273.15 = \mathbf{172.65^\circ\text{C}}$