Charles's Law, a fundamental gas law, describes the relationship between the volume and temperature of a gas at constant pressure. Understanding this law is crucial for anyone studying chemistry or physics. This guide provides a thorough explanation of Charles's Law, along with worked-out example problems and their solutions, to help you master this important concept.
Understanding Charles's Law
Charles's Law states that the volume of a given amount of gas held at a constant pressure is directly proportional to its absolute temperature. This means that as the temperature increases, the volume increases proportionally, and vice versa. Mathematically, this relationship is represented as:
V₁/T₁ = V₂/T₂
Where:
- V₁ is the initial volume of the gas
- T₁ is the initial absolute temperature of the gas (in Kelvin)
- V₂ is the final volume of the gas
- T₂ is the final absolute temperature of the gas (in Kelvin)
Crucial Note: Always remember to convert Celsius temperatures to Kelvin before applying Charles's Law. The conversion is simple: K = °C + 273.15
Charles's Law Problems with Solutions
Let's tackle some example problems to solidify your understanding:
Problem 1: A balloon has a volume of 2.5 L at 25°C. What will be its volume if the temperature is increased to 50°C, assuming constant pressure?
Solution:
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Convert Celsius to Kelvin:
- T₁ = 25°C + 273.15 = 298.15 K
- T₂ = 50°C + 273.15 = 323.15 K
-
Apply Charles's Law:
- V₁/T₁ = V₂/T₂
- 2.5 L / 298.15 K = V₂ / 323.15 K
-
Solve for V₂:
- V₂ = (2.5 L * 323.15 K) / 298.15 K
- V₂ ≈ 2.72 L
Therefore, the balloon's volume will be approximately 2.72 L at 50°C.
Problem 2: A gas occupies 500 mL at 27°C. To what temperature (in °C) must the gas be cooled to reduce its volume to 400 mL at constant pressure?
Solution:
-
Convert Celsius to Kelvin:
- T₁ = 27°C + 273.15 = 300.15 K
-
Apply Charles's Law:
- V₁/T₁ = V₂/T₂
- 500 mL / 300.15 K = 400 mL / T₂
-
Solve for T₂:
- T₂ = (400 mL * 300.15 K) / 500 mL
- T₂ ≈ 240.12 K
-
Convert Kelvin back to Celsius:
- T₂ = 240.12 K - 273.15 = -33.03°C
Therefore, the gas must be cooled to approximately -33.03°C to reduce its volume to 400 mL.
Problem 3 (More Complex): A sample of gas at 100°C and a volume of 10 L is heated until its volume expands to 15L. Assuming constant pressure, what is the final temperature of the gas in Celsius?
Solution: This problem follows the exact same procedure as the previous ones. Remember to convert to Kelvin first, solve for the unknown temperature (T2), and then convert back to Celsius.
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Convert Celsius to Kelvin: T1 = 100°C + 273.15 = 373.15 K
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Apply Charles's Law: 10 L / 373.15 K = 15 L / T2
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Solve for T2: T2 = (15 L * 373.15 K) / 10 L = 559.725 K
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Convert Kelvin back to Celsius: T2 = 559.725 K - 273.15 = 286.575 °C
The final temperature is approximately 286.58°C
Further Practice
Practice is key to mastering Charles's Law. Try solving additional problems from your textbook or online resources. Remember to always double-check your units and conversions! Understanding Charles's Law is a stepping stone to understanding more complex gas laws and their applications in various scientific fields.
