Gay Lussac's Law Worksheet Answers

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

Understanding gas laws is crucial in chemistry, and Gay-Lussac's Law is a fundamental concept. This 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. This article provides a detailed explanation of Gay-Lussac's Law, including its formula, applications, and a comprehensive worksheet with detailed answers to help you master this important concept.

Introduction to Gay-Lussac's Law

Gay-Lussac's Law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. In simpler terms, if you increase the temperature of a gas while keeping its volume the same, the pressure will increase proportionally. Conversely, decreasing the temperature will decrease the pressure. This relationship is crucial for understanding various phenomena, from inflating balloons to designing engines. The law is applicable to ideal gases, which are theoretical gases that obey certain assumptions. Real gases deviate slightly from this behavior under extreme conditions of temperature and pressure.

The Formula and its Components

Gay-Lussac's Law is mathematically expressed as:

P₁/T₁ = P₂/T₂

Where:

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

Remember: Temperature must be in Kelvin. To convert Celsius to Kelvin, use the formula: K = °C + 273.15

Understanding the Direct Proportionality

The equation highlights the direct proportionality between pressure and temperature. This means that if the temperature doubles (while volume remains constant), the pressure will also double. Similarly, if the temperature is halved, the pressure will be halved. This relationship is fundamental to understanding how gases behave under changing thermal conditions.

Applications of Gay-Lussac's Law

Gay-Lussac's Law has numerous practical applications in various fields:

• Aerosol Cans: The pressure inside an aerosol can increases significantly when exposed to heat. This is a direct application of Gay-Lussac's Law. Never expose aerosol cans to high temperatures, as this could lead to explosion.
• Pressure Cookers: Pressure cookers utilize Gay-Lussac's Law. By increasing the pressure inside a sealed container, the boiling point of water is elevated, allowing food to cook faster at higher temperatures.
• Tire Pressure: The pressure in car tires increases on hot days and decreases on cold days. This is due to the temperature change affecting the air pressure inside the tire (assuming the volume remains relatively constant).
• Engine Design: The principles of Gay-Lussac's Law are crucial in the design and operation of internal combustion engines. The pressure and temperature changes within the cylinders are directly related, and understanding this relationship is essential for efficient engine performance.
• Weather Balloons: Weather balloons expand as they ascend to higher altitudes because the atmospheric pressure decreases. This expansion is governed by the combined gas laws, including Gay-Lussac's Law.

Limitations of Gay-Lussac's Law

While Gay-Lussac's Law is a valuable tool, it's crucial to understand its limitations:

• Ideal Gas Assumption: The law is based on the ideal gas model. Real gases deviate from ideal behavior, particularly at high pressures and low temperatures. Intermolecular forces and the finite volume of gas molecules become significant under these conditions, affecting the accuracy of the law.
• Constant Volume: The law strictly applies only when the volume of the gas remains constant. Any change in volume will invalidate the application of the law. For situations with changing volumes, the combined gas law should be employed.

Gay-Lussac's Law Worksheet and Solutions

Now let's put your understanding of Gay-Lussac's Law to the test with a worksheet. Remember to always convert Celsius to Kelvin before applying the formula!

Worksheet Questions:

1. A gas in a rigid container has a pressure of 1.5 atm at a temperature of 25°C. What will the pressure be if the temperature is increased to 50°C?

2. A sealed container holds a gas at a pressure of 2.0 atm and a temperature of 300 K. If the pressure is increased to 3.0 atm, what is the new temperature?

3. A sample of gas at 100 kPa and 27°C is heated to 77°C. What is the new pressure if the volume remains constant?

4. A gas occupies a volume of 5.0 L at 20°C and 1.0 atm. If the temperature is raised to 40°C, what is the new pressure (assuming constant volume)?

5. A weather balloon is filled with helium at a pressure of 1.0 atm and a temperature of 25°C. As it rises, the pressure decreases to 0.5 atm. Assuming the volume remains constant, what is the new temperature of the helium in the balloon (in °C)?

Worksheet Solutions:

1.

• Step 1: Convert temperatures to Kelvin: T₁ = 25°C + 273.15 = 298.15 K; T₂ = 50°C + 273.15 = 323.15 K
• Step 2: Apply Gay-Lussac's Law: P₁/T₁ = P₂/T₂ => 1.5 atm / 298.15 K = P₂ / 323.15 K
• Step 3: Solve for P₂: P₂ = (1.5 atm * 323.15 K) / 298.15 K ≈ 1.63 atm

Answer: The pressure will be approximately 1.63 atm.

2.

• Step 1: Apply Gay-Lussac's Law: P₁/T₁ = P₂/T₂ => 2.0 atm / 300 K = 3.0 atm / T₂
• Step 2: Solve for T₂: T₂ = (3.0 atm * 300 K) / 2.0 atm = 450 K

Answer: The new temperature is 450 K (or 176.85°C).

3.

• Step 1: Convert temperatures to Kelvin: T₁ = 27°C + 273.15 = 300.15 K; T₂ = 77°C + 273.15 = 350.15 K
• Step 2: Apply Gay-Lussac's Law: P₁/T₁ = P₂/T₂ => 100 kPa / 300.15 K = P₂ / 350.15 K
• Step 3: Solve for P₂: P₂ = (100 kPa * 350.15 K) / 300.15 K ≈ 116.67 kPa

Answer: The new pressure is approximately 116.67 kPa.

4.

• Step 1: Convert temperatures to Kelvin: T₁ = 20°C + 273.15 = 293.15 K; T₂ = 40°C + 273.15 = 313.15 K
• Step 2: Apply Gay-Lussac's Law: P₁/T₁ = P₂/T₂ => 1.0 atm / 293.15 K = P₂ / 313.15 K
• Step 3: Solve for P₂: P₂ = (1.0 atm * 313.15 K) / 293.15 K ≈ 1.07 atm

Answer: The new pressure is approximately 1.07 atm.

5.

• Step 1: Convert initial temperature to Kelvin: T₁ = 25°C + 273.15 = 298.15 K
• Step 2: Apply Gay-Lussac's Law: P₁/T₁ = P₂/T₂ => 1.0 atm / 298.15 K = 0.5 atm / T₂
• Step 3: Solve for T₂: T₂ = (0.5 atm * 298.15 K) / 1.0 atm = 149.075 K
• Step 4: Convert T₂ back to Celsius: T₂ = 149.075 K - 273.15 = -124.075 °C

Answer: The new temperature of the helium is approximately -124.08°C.

Frequently Asked Questions (FAQs)

• Q: What happens if I use Celsius instead of Kelvin in Gay-Lussac's Law?

  • A: You will get an incorrect answer. Gay-Lussac's Law requires absolute temperature (Kelvin) because it describes a direct proportionality that only holds true with a zero point on the temperature scale.

• Q: Can Gay-Lussac's Law be used for mixtures of gases?

  • A: Yes, provided the mixture behaves ideally and the volume remains constant. The total pressure of the mixture will be directly proportional to the absolute temperature.

• Q: What if the volume is not constant? Which law should I use?

  • A: If the volume is not constant, you need to use the combined gas law, which incorporates the relationships between pressure, volume, and temperature.

• Q: Why is it important to understand Gay-Lussac's Law?

  • A: Understanding Gay-Lussac's Law is crucial for predicting and explaining pressure changes in gases due to temperature variations. This has significant applications in various fields, as mentioned earlier.

Conclusion

Gay-Lussac's Law is a fundamental principle in chemistry that governs the relationship between the pressure and temperature of a gas at constant volume. Understanding its formula, applications, and limitations is essential for mastering gas laws. By working through the worksheet and understanding the solutions, you've gained a solid foundation in applying Gay-Lussac's Law to real-world scenarios. Remember to always carefully consider the conditions and use the appropriate gas law for accurate calculations. This knowledge will serve as a strong base for more advanced studies in chemistry and related fields.