Chapter 11 Review Gases Answer Key

Chapter 11 Review: Gases – A Comprehensive Guide with Answers

This comprehensive guide delves into the key concepts covered in Chapter 11 of most general chemistry textbooks focusing on gases. We will explore the major gas laws, calculations, and problem-solving techniques. This detailed review is designed to help you master the material, prepare for exams, and build a strong foundation in gas chemistry. We will provide detailed explanations and answer keys to common chapter review questions. Remember, understanding the why behind the equations is just as important as knowing how to use them.

1. Fundamental Gas Laws: A Recap

Before tackling complex problems, let's revisit the fundamental gas laws that govern the behavior of gases. Understanding these laws is the cornerstone to tackling Chapter 11.

1.1 Boyle's Law: Pressure and Volume Relationship

Boyle's Law states that at a constant temperature, the volume of a gas is inversely proportional to its pressure. This means if you increase the pressure on a gas, its volume will decrease, and vice-versa. Mathematically, it's represented as:

P₁V₁ = P₂V₂

where:

• P₁ = initial pressure
• V₁ = initial volume
• P₂ = final pressure
• V₂ = final volume

Example: A gas occupies 5.0 L at a pressure of 1.0 atm. What will its volume be if the pressure is increased to 2.5 atm, assuming constant temperature?

Solution: Using Boyle's Law, we have:

(1.0 atm)(5.0 L) = (2.5 atm)(V₂)

V₂ = (1.0 atm * 5.0 L) / 2.5 atm = 2.0 L

Therefore, the new volume will be 2.0 L.

1.2 Charles's Law: Volume and Temperature Relationship

Charles's Law states that at a constant pressure, the volume of a gas is directly proportional to its absolute temperature (in Kelvin). This means if you increase the temperature, the volume will increase proportionally. The equation is:

V₁/T₁ = V₂/T₂

where:

• V₁ = initial volume
• T₁ = initial temperature (in Kelvin)
• V₂ = final volume
• T₂ = final temperature (in Kelvin)

Remember: Always convert Celsius temperatures to Kelvin (K = °C + 273.15) before using gas laws.

Example: A gas occupies 10.0 L at 25°C. What will its volume be at 50°C, assuming constant pressure?

Solution: First, convert Celsius to Kelvin:

T₁ = 25°C + 273.15 = 298.15 K T₂ = 50°C + 273.15 = 323.15 K

Now, use Charles's Law:

(10.0 L) / (298.15 K) = V₂ / (323.15 K)

V₂ = (10.0 L * 323.15 K) / 298.15 K ≈ 10.84 L

The new volume is approximately 10.84 L.

1.3 Gay-Lussac's Law: Pressure and Temperature Relationship

Gay-Lussac's Law states that at a constant volume, the pressure of a gas is directly proportional to its absolute temperature. This means if you increase the temperature, the pressure will increase proportionally. The equation is:

P₁/T₁ = P₂/T₂

where:

• P₁ = initial pressure
• T₁ = initial temperature (in Kelvin)
• P₂ = final pressure
• T₂ = final temperature (in Kelvin)

Example: A gas has a pressure of 1.5 atm at 20°C. What will its pressure be at 100°C, assuming constant volume?

Solution: Convert Celsius to Kelvin:

T₁ = 20°C + 273.15 = 293.15 K T₂ = 100°C + 273.15 = 373.15 K

Using Gay-Lussac's Law:

(1.5 atm) / (293.15 K) = P₂ / (373.15 K)

P₂ = (1.5 atm * 373.15 K) / 293.15 K ≈ 1.91 atm

The new pressure is approximately 1.91 atm.

2. The Ideal Gas Law: A Unified Approach

The Ideal Gas Law combines Boyle's, Charles's, and Gay-Lussac's Laws into a single, powerful equation:

PV = nRT

where:

• P = pressure
• V = volume
• n = number of moles of gas
• R = the ideal gas constant (0.0821 L·atm/mol·K or other units depending on the problem)
• T = temperature (in Kelvin)

This equation is incredibly versatile and allows you to solve a wide range of gas problems.

Example: What is the volume of 2.0 moles of an ideal gas at 27°C and 1.2 atm?

Solution:

First, convert Celsius to Kelvin: T = 27°C + 273.15 = 300.15 K

Then, use the Ideal Gas Law:

V = nRT/P = (2.0 mol)(0.0821 L·atm/mol·K)(300.15 K) / (1.2 atm) ≈ 41 L

The volume is approximately 41 liters.

3. Dalton's Law of Partial Pressures: Gas Mixtures

Dalton's Law states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the individual gases. The partial pressure of a gas is the pressure that gas would exert if it occupied the entire volume alone.

P_Total = P₁ + P₂ + P₃ + ...

where P₁, P₂, P₃, etc., are the partial pressures of each gas.

Example: A container holds a mixture of nitrogen (N₂), oxygen (O₂), and carbon dioxide (CO₂). The partial pressures are: P_N₂ = 0.7 atm, P_O₂ = 0.2 atm, and P_CO₂ = 0.1 atm. What is the total pressure in the container?

Solution:

P_Total = P_N₂ + P_O₂ + P_CO₂ = 0.7 atm + 0.2 atm + 0.1 atm = 1.0 atm

4. Gas Stoichiometry: Connecting Gases to Chemical Reactions

Gas stoichiometry involves using the Ideal Gas Law and stoichiometric principles to relate the amounts of gases involved in chemical reactions.

Example: Consider the reaction: 2H₂(g) + O₂(g) → 2H₂O(g). If 5.0 L of hydrogen gas reacts completely at STP (standard temperature and pressure: 0°C and 1 atm), what volume of oxygen gas is required?

Solution:

At STP, 1 mole of any gas occupies 22.4 L (molar volume).

From the balanced equation, 2 moles of H₂ react with 1 mole of O₂. Therefore, the mole ratio is 2:1.

Moles of H₂ = (5.0 L) / (22.4 L/mol) ≈ 0.223 mol

Moles of O₂ required = (0.223 mol H₂) * (1 mol O₂ / 2 mol H₂) ≈ 0.112 mol

Volume of O₂ required = (0.112 mol) * (22.4 L/mol) ≈ 2.5 L

5. Kinetic Molecular Theory: Understanding Gas Behavior at a Microscopic Level

The Kinetic Molecular Theory (KMT) provides a microscopic explanation for the macroscopic behavior of gases. Key postulates include:

• Gases are composed of tiny particles that are in constant, random motion.
• The volume of the gas particles themselves is negligible compared to the volume of the container.
• There are no significant attractive or repulsive forces between gas particles.
• Gas particles collide with each other and with the walls of the container, and these collisions are perfectly elastic (no energy is lost).
• The average kinetic energy of the gas particles is directly proportional to the absolute temperature of the gas.

Understanding KMT helps explain why gases behave as they do and provides a framework for understanding deviations from ideal gas behavior.

6. Deviations from Ideal Gas Behavior: Real Gases

Real gases deviate from the Ideal Gas Law, particularly at high pressures and low temperatures. These deviations occur because the KMT assumptions are not perfectly accurate. At high pressures, the volume of gas particles becomes significant, and at low temperatures, attractive forces between gas particles become more important.

The van der Waals equation is a more accurate model for real gases, taking into account these deviations:

(P + a(n/V)²)(V - nb) = nRT

where 'a' and 'b' are van der Waals constants specific to each gas.

7. Applications of Gas Laws: Real-World Examples

The gas laws have numerous applications in various fields, including:

• Meteorology: Understanding atmospheric pressure, temperature, and humidity.
• Medicine: Respiration and anesthesia.
• Industry: Chemical processes, storage and transportation of gases.
• Environmental Science: Air pollution and climate change.

8. Advanced Topics (Often Included in Chapter 11):

Some textbooks include more advanced topics within Chapter 11, such as:

• Graham's Law of Effusion: Describes the rate of effusion (escape of gas through a small hole) of different gases.
• Root-Mean-Square Speed (u_rms): A measure of the average speed of gas particles.
• Effusion and Diffusion: The processes by which gases mix and spread out.

9. Chapter 11 Review Questions and Answers (Example Set)

This section would typically include a series of practice problems mirroring those found in your textbook's Chapter 11 review. Unfortunately, without knowing the specific questions from your Chapter 11, I cannot provide tailored answers. However, the examples throughout this article illustrate the problem-solving techniques necessary to tackle most Chapter 11 review questions. Make sure to consult your textbook for practice problems and their corresponding solutions. Remember, consistent practice is key to mastering gas laws!

This detailed guide covers the essential concepts of a typical Chapter 11 on gases. By understanding these fundamentals, practicing example problems, and consulting your textbook, you’ll be well-prepared to confidently tackle any challenge related to gas behavior and calculations. Remember to focus on understanding the underlying principles rather than simply memorizing formulas. Good luck with your studies!