charles law problems answer key

Lemon

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04-02-2025

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Charles's Law, a fundamental gas law, describes the relationship between the volume and temperature of a gas at constant pressure. Understanding and applying Charles's Law is crucial for students of chemistry and physics. This guide provides a comprehensive overview of Charles's Law, along with worked examples and solutions to common problems.

Understanding Charles's Law

Charles's Law states that at constant pressure, the volume of a given mass of an ideal gas is directly proportional to its absolute temperature. Mathematically, this relationship is expressed as:

V₁/T₁ = V₂/T₂

Where:

• V₁ is the initial volume
• T₁ is the initial absolute temperature (always in Kelvin!)
• V₂ is the final volume
• T₂ is the final absolute temperature (always in Kelvin!)

Remember: To convert Celsius to Kelvin, add 273.15 (K = °C + 273.15). Failure to use Kelvin will result in incorrect answers.

Common Charles's Law Problems and Solutions

Let's work through some example problems to solidify your understanding.

Problem 1: A balloon filled with helium gas occupies 2.50 L at 25°C. What volume will the balloon occupy at 50°C, assuming the pressure remains constant?

Solution:

1. Convert Celsius to Kelvin:

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

2. Apply Charles's Law:

   • V₁/T₁ = V₂/T₂
   • 2.50 L / 298.15 K = V₂ / 323.15 K

3. Solve for V₂:

   • V₂ = (2.50 L * 323.15 K) / 298.15 K
   • V₂ ≈ 2.71 L

Therefore, the balloon will occupy approximately 2.71 L at 50°C.

Problem 2: A sample of gas has a volume of 4.00 L at 27°C. To what temperature (in °C) must the gas be cooled to reduce its volume to 2.00 L at constant pressure?

Solution:

1. Convert Celsius to Kelvin:

   • T₁ = 27°C + 273.15 = 300.15 K

2. Apply Charles's Law:

   • V₁/T₁ = V₂/T₂
   • 4.00 L / 300.15 K = 2.00 L / T₂

3. Solve for T₂:

   • T₂ = (2.00 L * 300.15 K) / 4.00 L
   • T₂ = 150.08 K

4. Convert Kelvin back to Celsius:

   • T₂ = 150.08 K - 273.15 = -123.07 °C

Therefore, the gas must be cooled to approximately -123.07 °C to reduce its volume to 2.00 L.

Problem 3 (More Challenging): A gas occupies 10.0 L at 20°C and 1.0 atm. If the temperature is increased to 40°C while maintaining the same pressure, what is the new volume?

Solution: This problem reinforces that only temperature and volume change according to Charles's Law; pressure remaining constant is already stated.

1. Convert Celsius to Kelvin:

   • T₁ = 20°C + 273.15 = 293.15 K
   • T₂ = 40°C + 273.15 = 313.15 K

2. Apply Charles's Law:

   • V₁/T₁ = V₂/T₂
   • 10.0 L / 293.15 K = V₂ / 313.15 K

3. Solve for V₂:

   • V₂ = (10.0 L * 313.15 K) / 293.15 K
   • V₂ ≈ 10.68 L

Therefore, the new volume is approximately 10.68 L.

Key Takeaways and Further Exploration

Remember always to use Kelvin for temperature in Charles's Law calculations. Practice solving various problems to build your confidence and understanding. Explore more complex scenarios involving combined gas laws, which incorporate Charles's Law alongside Boyle's Law and Avogadro's Law. This will provide a deeper understanding of gas behavior under different conditions.