Exploring Charles's Law: Real-World Applications

V1 / T1 = V2 / T2
Where:

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

Example 1: Inflating a Balloon

Imagine you have a balloon filled with air at a temperature of 20°C (293 K). The volume of the balloon is 2.0 liters. If you heat the balloon to 80°C (353 K), what will be the new volume?

1. Convert temperatures to Kelvin:

• T1 = 20 + 273 = 293 K

• T2 = 80 + 273 = 353 K

  2. Use Charles’s Law:

• V1 = 2.0 L

• V2 = ?
• V1 / T1 = V2 / T2

• 2.0 L / 293 K = V2 / 353 K

  3. Cross-multiply to solve for V2:

• V2 = (2.0 L * 353 K) / 293 K

• V2 ≈ 2.41 L
  Thus, the volume of the balloon increases to approximately 2.41 liters when heated to 80°C.

Example 2: Gas Expansion in a Car Tire

Consider a car tire that has a volume of 10.0 liters at an initial temperature of 25°C (298 K). If the tire heats up to 60°C (333 K) during driving, what happens to the volume?

1. Convert temperatures to Kelvin:

• T1 = 25 + 273 = 298 K

• T2 = 60 + 273 = 333 K

  2. Use Charles’s Law:

• V1 = 10.0 L

• V2 = ?
• V1 / T1 = V2 / T2

• 10.0 L / 298 K = V2 / 333 K

  3. Cross-multiply to solve for V2:

• V2 = (10.0 L * 333 K) / 298 K

• V2 ≈ 11.16 L
  As a result, the volume of the gas in the tire expands to approximately 11.16 liters when the temperature increases to 60°C.

Conclusion

Charles’s Law illustrates the relationship between temperature and volume for gases. These examples show how real-world scenarios, like inflating a balloon or gas expansion in a tire, are directly influenced by changes in temperature. Understanding this law not only helps in academic settings but also in everyday life, where gas behavior can impact various activities.