Boyle’s Law, Charles’s Law, Avogadro’s Law

Introduction

Key Concepts

Boyle’s Law

Boyle’s Law describes the inverse relationship between the pressure and volume of a gas at constant temperature. Formulated by Robert Boyle in the 17th century, this law states that for a fixed amount of gas at a constant temperature, the pressure of the gas is inversely proportional to its volume.

The mathematical expression of Boyle’s Law is:

$$P \cdot V = k$$

where:

• P is the pressure of the gas.
• V is the volume of the gas.
• k is a constant for a given amount of gas at constant temperature.

Alternatively, Boyle’s Law can be expressed as:

$$P_1 V_1 = P_2 V_2$$

where:

• P₁ and V₁ are the initial pressure and volume.
• P₂ and V₂ are the final pressure and volume.

Example:
If a gas occupies 2 liters at a pressure of 1 atmosphere, what volume will it occupy if the pressure increases to 3 atmospheres, assuming temperature remains constant?
Using Boyle’s Law:

$$P_1 V_1 = P_2 V_2$$

$$1 \, \text{atm} \times 2 \, \text{L} = 3 \, \text{atm} \times V_2$$

Solving for V₂:

$$V_2 = \frac{1 \times 2}{3} = \frac{2}{3} \, \text{L}$$

The volume decreases to approximately 0.667 liters.

Charles’s Law

Charles’s Law establishes a direct relationship between the volume and temperature of a gas at constant pressure. Formulated by Jacques Charles, this law indicates that the volume of a gas is directly proportional to its absolute temperature (measured in Kelvin) when pressure remains constant.

The mathematical expression of Charles’s Law is:

$$\frac{V}{T} = k$$

where:

• V is the volume of the gas.
• T is the absolute temperature of the gas.
• k is a constant for a given amount of gas at constant pressure.

Alternatively, it can be expressed as:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

where:

• V₁ and T₁ are the initial volume and temperature.
• V₂ and T₂ are the final volume and temperature.

Example:
A balloon has a volume of 3 liters at a temperature of 300 K. If the temperature increases to 450 K while maintaining constant pressure, what is the new volume of the balloon?
Using Charles’s Law:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

$$\frac{3 \, \text{L}}{300 \, \text{K}} = \frac{V_2}{450 \, \text{K}}$$

Solving for V₂:

$$V_2 = \frac{3 \times 450}{300} = \frac{1350}{300} = 4.5 \, \text{L}$$

The volume increases to 4.5 liters.

Avogadro’s Law

Avogadro’s Law posits that the volume of a gas is directly proportional to the number of moles of gas present at constant temperature and pressure. Proposed by Amedeo Avogadro, this law implies that equal volumes of gases, at the same temperature and pressure, contain an equal number of molecules.

The mathematical expression of Avogadro’s Law is:

$$\frac{V}{n} = k$$

where:

• V is the volume of the gas.
• n is the number of moles of the gas.
• k is a constant for a given amount of gas at constant temperature and pressure.

Alternatively, it can be expressed as:

$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$

where:

• V₁ and n₁ are the initial volume and number of moles.
• V₂ and n₂ are the final volume and number of moles.

Example:
If 2 moles of gas occupy 5 liters, how volume will 5 moles occupy under the same conditions?
Using Avogadro’s Law:

$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$

$$\frac{5 \, \text{L}}{2 \, \text{mol}} = \frac{V_2}{5 \, \text{mol}}$$

Solving for V₂:

$$V_2 = \frac{5 \times 5}{2} = \frac{25}{2} = 12.5 \, \text{L}$$

The volume increases to 12.5 liters.

Advanced Concepts

In-depth Theoretical Explanations

The gas laws—Boyle’s, Charles’s, and Avogadro’s Laws—are specific cases of the Ideal Gas Law, which combines these individual laws into a single equation:

$$PV = nRT$$

where:

• P is the pressure of the gas.
• V is the volume of the gas.
• n is the number of moles of the gas.
• R is the universal gas constant ($R = 8.314 \, \text{J mol}^{-1} \text{K}^{-1}$).
• T is the absolute temperature of the gas.

Each of the three gas laws can be derived from the Ideal Gas Law by holding two of the variables constant:

• Boyle’s Law: Hold temperature and number of moles constant.
• Charles’s Law: Hold pressure and number of moles constant.
• Avogadro’s Law: Hold pressure and temperature constant.

The Ideal Gas Law assumes that gas molecules do not interact and occupy no volume, which holds true under low pressure and high temperature conditions. Deviations from the Ideal Gas Law are accounted for by more complex equations like the Van der Waals equation.

Mathematical Derivations and Proofs

Deriving Boyle’s, Charles’s, and Avogadro’s Laws from kinetic molecular theory provides deeper insights into gas behavior.

From kinetic molecular theory, the pressure exerted by a gas arises from collisions of gas molecules with the container walls. Pressure (P) is given by:

$$P = \frac{n}{V}RT$$

Rearranging, we obtain the Ideal Gas Law:

$$PV = nRT$$

By holding different variables constant, we derive the individual gas laws:

• At constant T and n:

  $$P \propto \frac{1}{V} \Rightarrow PV = k$$

  This is Boyle’s Law.
• At constant P and n:

  $$V \propto T \Rightarrow \frac{V}{T} = k$$

  This is Charles’s Law.
• At constant P and T:

  $$V \propto n \Rightarrow \frac{V}{n} = k$$

  This is Avogadro’s Law.

Complex Problem-Solving

Problem 1:
A sample of gas initially at a pressure of 2 atm and volume of 4 liters is compressed isothermally to a new volume of 1 liter. Calculate the final pressure.
Using Boyle’s Law:

$$P_1 V_1 = P_2 V_2$$

$$2 \times 4 = P_2 \times 1$$

$$P_2 = 8 \, \text{atm}$$

Problem 2:
If 3 moles of an ideal gas occupy 24 liters at 300 K, what volume would 5 moles occupy at 300 K and the same pressure?
Using Avogadro’s Law:

$$\frac{V_1}{n_1} = \frac{V_2}{n_2}$$

$$\frac{24}{3} = \frac{V_2}{5}$$

$$V_2 = \frac{24 \times 5}{3} = 40 \, \text{liters}$$

Problem 3:
A gas sample at 1.5 atm and 30°C has a volume of 10 liters. What will be its volume at 1.5 atm and 60°C?
First, convert temperatures to Kelvin:
$$T_1 = 30 + 273.15 = 303.15 \, \text{K}$$
$$T_2 = 60 + 273.15 = 333.15 \, \text{K}$$
Using Charles’s Law:

$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$

$$\frac{10}{303.15} = \frac{V_2}{333.15}$$

$$V_2 = \frac{10 \times 333.15}{303.15} \approx 11 \, \text{liters}$$

The volume increases to approximately 11 liters.

Interdisciplinary Connections

Gas laws are integral to various scientific and engineering fields. For instance:

• Aerospace Engineering: Understanding gas behavior under different pressures and temperatures is essential for designing propulsion systems and life-support systems.
• Chemistry: Gas laws are fundamental in reactions involving gases, stoichiometry, and determining reaction yields.
• Medicine: Boyle’s Law is applied in respiratory therapy and hyperbaric medicine, where understanding pressure-volume relationships can be critical.
• Environmental Science: Gas laws help in modeling atmospheric phenomena and understanding greenhouse gas behaviors.

Comparison Table

Summary and Key Takeaways