Gay Lussac Law Worksheet Answers

Gay-Lussac's Law: A Comprehensive Guide with Worksheet Answers

Gay-Lussac's Law, also known as the pressure-temperature law, describes the relationship between the pressure and temperature of a gas when the volume is held constant. Understanding this fundamental gas law is crucial in various scientific fields, from chemistry and physics to engineering and meteorology. This comprehensive guide will delve into the intricacies of Gay-Lussac's Law, providing a clear explanation, practical examples, and detailed solutions to common worksheet problems. We'll also explore the scientific principles behind the law and address frequently asked questions.

Understanding Gay-Lussac's Law: The Basics

Gay-Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to its absolute temperature. In simpler terms, if you increase the temperature of a gas in a sealed container, the pressure inside will increase proportionally. Conversely, if you decrease the temperature, the pressure will decrease proportionally. This relationship is mathematically represented as:

P₁/T₁ = P₂/T₂

Where:

• P₁ represents the initial pressure of the gas.
• T₁ represents the initial absolute temperature of the gas (always in Kelvin).
• P₂ represents the final pressure of the gas.
• T₂ represents the final absolute temperature of the gas (always in Kelvin).

The Importance of Absolute Temperature (Kelvin)

It's crucial to understand that Gay-Lussac's Law, and all gas laws, utilize the absolute temperature scale (Kelvin). This is because absolute zero (0 Kelvin or -273.15°C) represents the theoretical point where all molecular motion ceases, and thus, pressure would theoretically be zero. Using Celsius or Fahrenheit would lead to inaccurate results. To convert Celsius to Kelvin, use the formula:

K = °C + 273.15

Step-by-Step Guide to Solving Gay-Lussac's Law Problems

Let's break down the process of solving problems related to Gay-Lussac's Law:

1. Identify the known variables: Carefully read the problem statement and identify the values provided for pressure (P₁) and temperature (T₁), as well as either P₂ or T₂. Remember to convert Celsius temperatures to Kelvin.

2. Identify the unknown variable: Determine which variable you need to solve for (either P₂ or T₂).

3. Apply the formula: Substitute the known values into the Gay-Lussac's Law equation (P₁/T₁ = P₂/T₂).

4. Solve for the unknown: Rearrange the equation algebraically to isolate the unknown variable and solve for its value.

5. Check your units: Ensure that your final answer has the correct units (pressure in atmospheres, Pascals, etc., and temperature in Kelvin).

Example Problem 1:

A gas has a pressure of 2.5 atm at a temperature of 25°C. If the temperature is increased to 50°C while keeping the volume constant, what will the new pressure be?

Solution:

1. Known variables: P₁ = 2.5 atm, T₁ = 25°C + 273.15 = 298.15 K, T₂ = 50°C + 273.15 = 323.15 K

2. Unknown variable: P₂

3. Apply the formula: P₁/T₁ = P₂/T₂

4. Solve for the unknown: P₂ = (P₁ * T₂) / T₁ = (2.5 atm * 323.15 K) / 298.15 K = 2.71 atm (approximately)

5. Check units: The answer is in atmospheres, which is correct.

Example Problem 2:

A sample of gas in a rigid container exerts a pressure of 1.2 atm at a temperature of 273 K. To what temperature (in Celsius) must the gas be heated to increase its pressure to 1.8 atm?

Solution:

1. Known variables: P₁ = 1.2 atm, T₁ = 273 K, P₂ = 1.8 atm

2. Unknown variable: T₂

3. Apply the formula: P₁/T₁ = P₂/T₂

4. Solve for the unknown: T₂ = (P₂ * T₁) / P₁ = (1.8 atm * 273 K) / 1.2 atm = 409.5 K. To convert back to Celsius: 409.5 K - 273.15 = 136.35°C (approximately)

5. Check units: The final temperature is in Kelvin and correctly converted to Celsius.

Worksheet Problems with Detailed Answers

Here are some practice problems with step-by-step solutions to solidify your understanding of Gay-Lussac's Law:

Problem 1: A gas inside a steel container has a pressure of 100 kPa at 20°C. What will the pressure be if the temperature is increased to 100°C?

Answer:

1. Convert temperatures to Kelvin: T₁ = 20°C + 273.15 = 293.15 K; T₂ = 100°C + 273.15 = 373.15 K
2. Apply Gay-Lussac's Law: P₁/T₁ = P₂/T₂
3. Solve for P₂: P₂ = (P₁ * T₂) / T₁ = (100 kPa * 373.15 K) / 293.15 K ≈ 127.2 kPa

Problem 2: A balloon filled with helium has a pressure of 1 atm at 25°C. If the pressure increases to 1.2 atm while maintaining a constant volume, what will the new temperature be in Celsius?

Answer:

1. Convert the initial temperature to Kelvin: T₁ = 25°C + 273.15 = 298.15 K
2. Apply Gay-Lussac's Law: P₁/T₁ = P₂/T₂
3. Solve for T₂: T₂ = (P₂ * T₁) / P₁ = (1.2 atm * 298.15 K) / 1 atm = 357.78 K
4. Convert T₂ to Celsius: 357.78 K - 273.15 = 84.63°C

Problem 3: A gas cylinder containing oxygen has a pressure of 150 atm at 20°C. If the pressure drops to 120 atm, what will the new temperature be in Celsius? (Assume constant volume).

Answer:

1. Convert the initial temperature to Kelvin: T₁ = 20°C + 273.15 = 293.15 K
2. Apply Gay-Lussac's Law: P₁/T₁ = P₂/T₂
3. Solve for T₂: T₂ = (P₂ * T₁) / P₁ = (120 atm * 293.15 K) / 150 atm = 234.52 K
4. Convert T₂ to Celsius: 234.52 K - 273.15 = -38.63°C

Problem 4: A sealed container holds nitrogen gas at a pressure of 5 atm and a temperature of -10°C. If the pressure is reduced to 3 atm, what will the new temperature be in degrees Celsius?

Answer:

1. Convert the initial temperature to Kelvin: T₁ = -10°C + 273.15 = 263.15 K
2. Apply Gay-Lussac's Law: P₁/T₁ = P₂/T₂
3. Solve for T₂: T₂ = (P₂ * T₁) / P₁ = (3 atm * 263.15 K) / 5 atm = 157.89 K
4. Convert T₂ to Celsius: 157.89 K - 273.15 = -115.26°C

The Scientific Explanation Behind Gay-Lussac's Law

Gay-Lussac's Law is a direct consequence of the kinetic theory of gases. This theory postulates that gas particles are in constant, random motion and that the pressure exerted by a gas is due to the collisions of these particles with the walls of their container. Temperature is a measure of the average kinetic energy of these particles.

When the temperature of a gas increases, the gas particles gain kinetic energy and move faster. This leads to more frequent and forceful collisions with the container walls, resulting in a higher pressure. Conversely, when the temperature decreases, the particles slow down, leading to fewer and less forceful collisions and lower pressure. The constant volume ensures that the number of collisions per unit area remains directly proportional to the kinetic energy (and thus, temperature).

Frequently Asked Questions (FAQs)

Q1: What happens if the volume is not constant?

A1: If the volume is not constant, Gay-Lussac's Law is not applicable. You would need to use a more general gas law, such as the combined gas law or the ideal gas law, to describe the relationship between pressure, volume, and temperature.

Q2: Can I use Gay-Lussac's Law for all gases?

A2: Gay-Lussac's Law works best for ideal gases – gases that behave according to the ideal gas law. Real gases may deviate slightly from this law at high pressures or low temperatures.

Q3: Why is it important to use Kelvin in Gay-Lussac's Law calculations?

A3: Using Kelvin ensures accurate results because it's an absolute temperature scale. Celsius and Fahrenheit scales have arbitrary zero points, which would lead to incorrect calculations of the pressure-temperature relationship.

Q4: What are some real-world applications of Gay-Lussac's Law?

A4: Gay-Lussac's Law has numerous applications, including: pressure cookers (increased pressure at higher temperatures), weather forecasting (relationship between air pressure and temperature), and designing safety mechanisms in pressurized systems.

Conclusion

Gay-Lussac's Law is a fundamental principle in chemistry and physics, providing a crucial understanding of the relationship between the pressure and temperature of a gas at constant volume. Mastering this law requires a thorough understanding of its formula, the importance of using Kelvin, and the ability to apply it to various problem-solving scenarios. By practicing with numerous examples and worksheets, you can confidently apply this principle in various scientific and engineering contexts. Remember, the key is to carefully identify the knowns and unknowns, convert temperatures to Kelvin, and accurately apply the formula. With practice, you will become proficient in understanding and utilizing Gay-Lussac's Law.