Charles Law Problems: Complete Worksheet Answers Explained
Table of Contents :
Charles's Law is a fundamental principle in gas physics that relates the volume of a gas to its temperature when pressure is held constant. This law states that the volume of a given mass of gas is directly proportional to its absolute temperature (measured in Kelvin) as long as the pressure remains constant. In this article, we will explore Charles's Law problems, complete worksheet answers, and provide explanations to help solidify your understanding of the concept. 🌡️
Understanding Charles's Law
Charles's Law can be expressed mathematically as:
[ \frac{V_1}{T_1} = \frac{V_2}{T_2} ]
where:
- (V_1) = initial volume
- (T_1) = initial temperature (in Kelvin)
- (V_2) = final volume
- (T_2) = final temperature (in Kelvin)
This equation implies that if the temperature of a gas increases, the volume also increases, provided the pressure is constant.
Converting Temperature to Kelvin
Before solving any problem using Charles's Law, it's essential to convert temperatures from Celsius (°C) or Fahrenheit (°F) to Kelvin (K). The conversion formula is:
- From Celsius: (K = °C + 273.15)
- From Fahrenheit: (K = (°F - 32) \times \frac{5}{9} + 273.15)
This conversion is crucial as Charles's Law requires temperature to be in Kelvin for accurate calculations.
Example Problems
Let’s dive into some common problems involving Charles's Law and explore the complete worksheet answers.
Problem 1: Volume Increase with Temperature
Question: A balloon has a volume of 2.0 L at 25°C. What will be its volume at 50°C, assuming pressure remains constant?
Solution Steps:
-
Convert temperatures to Kelvin:
- (T_1 = 25 + 273.15 = 298.15 K)
- (T_2 = 50 + 273.15 = 323.15 K)
-
Use Charles's Law: [ \frac{V_1}{T_1} = \frac{V_2}{T_2} ] Rearranging gives us: [ V_2 = V_1 \times \frac{T_2}{T_1} ]
-
Substitute the values: [ V_2 = 2.0 L \times \frac{323.15 K}{298.15 K} \approx 2.17 L ]
Answer: The volume of the balloon at 50°C will be approximately 2.17 L. 🎈
Problem 2: Volume Decrease with Temperature
Question: A gas occupies a volume of 10.0 L at a temperature of 100°C. What will be its volume at 0°C if the pressure is constant?
Solution Steps:
-
Convert temperatures to Kelvin:
- (T_1 = 100 + 273.15 = 373.15 K)
- (T_2 = 0 + 273.15 = 273.15 K)
-
Use Charles's Law: [ V_2 = V_1 \times \frac{T_2}{T_1} ]
-
Substitute the values: [ V_2 = 10.0 L \times \frac{273.15 K}{373.15 K} \approx 7.32 L ]
Answer: The volume of the gas at 0°C will be approximately 7.32 L. 📏
Problem 3: Finding Final Temperature
Question: A gas has a volume of 5.0 L at 20°C. If the gas is heated to a volume of 8.0 L, what will be the final temperature?
Solution Steps:
-
Convert the initial temperature to Kelvin:
- (T_1 = 20 + 273.15 = 293.15 K)
-
Use Charles's Law to find (T_2): [ \frac{V_1}{T_1} = \frac{V_2}{T_2} ] Rearranging gives: [ T_2 = T_1 \times \frac{V_2}{V_1} ]
-
Substitute the values: [ T_2 = 293.15 K \times \frac{8.0 L}{5.0 L} \approx 468.88 K ]
-
Convert back to Celsius:
- (T_2 = 468.88 - 273.15 \approx 195.73°C)
Answer: The final temperature will be approximately 195.73°C. 🔥
Summary of Key Points
| Variables | Description |
|---|---|
| (V) | Volume of the gas |
| (T) | Absolute temperature in Kelvin |
| (V_1, T_1) | Initial volume and temperature |
| (V_2, T_2) | Final volume and temperature |
Important Note: "Always remember to convert temperatures to Kelvin when applying Charles's Law!" 🔄
Conclusion
Understanding Charles's Law and how to apply it to various gas problems is crucial for students in chemistry and physics. By mastering the conversion of temperatures and the application of the law, you'll be well-prepared to tackle any related problems. Practice with real-world examples, and you'll find these concepts become increasingly intuitive.