Charles Law Worksheet With Answers: Master The Concepts!
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Charles' Law is an essential principle in chemistry and physics, describing the relationship between the volume and temperature of a gas at constant pressure. Mastering Charles' Law is crucial for students who want to excel in their understanding of gas laws and thermodynamics. This article will not only delve into the concepts behind Charles' Law but also provide a worksheet with answers to help reinforce your knowledge and application of the law.
Understanding Charles' Law
Charles' Law states that the volume of a gas is directly proportional to its temperature when the pressure is held constant. This relationship can be expressed mathematically as:
[ \frac{V_1}{T_1} = \frac{V_2}{T_2} ]
Where:
- ( V_1 ) and ( V_2 ) are the initial and final volumes of the gas, respectively.
- ( T_1 ) and ( T_2 ) are the initial and final temperatures in Kelvin.
Important Note: Always convert temperatures to Kelvin when using Charles' Law. The formula is derived based on absolute temperature.
Key Concepts of Charles' Law
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Direct Proportionality:
- When the temperature increases, the volume of the gas increases as well, provided the pressure remains constant. Conversely, if the temperature decreases, the volume also decreases.
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Absolute Zero:
- Charles' Law implies that at absolute zero (0 Kelvin), the volume of a gas would theoretically reach zero. This provides an insight into the behavior of gases and the nature of temperature.
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Practical Applications:
- Charles' Law has practical applications in various fields including meteorology, cooking (baking), and understanding the behavior of gases in various environments.
Example Problem Using Charles' Law
Let's solve a problem to illustrate how to apply Charles' Law effectively.
Problem: A balloon has a volume of 2.0 liters at a temperature of 20°C. What will be the volume of the balloon when the temperature rises to 50°C, assuming constant pressure?
Solution:
-
Convert temperatures to Kelvin:
- ( T_1 = 20°C + 273.15 = 293.15 K )
- ( T_2 = 50°C + 273.15 = 323.15 K )
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Use Charles' Law: [ \frac{V_1}{T_1} = \frac{V_2}{T_2} ] [ \frac{2.0 L}{293.15 K} = \frac{V_2}{323.15 K} ]
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Solve for ( V_2 ): [ V_2 = \frac{2.0 L \times 323.15 K}{293.15 K} \approx 2.21 L ]
Charles' Law Worksheet
To enhance your understanding, here is a worksheet with a few problems related to Charles' Law. Try solving these before looking at the answers provided afterward.
Problems
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A gas occupies a volume of 4.5 liters at a temperature of 300 K. What will be the volume of the gas if the temperature is increased to 600 K?
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A gas has a volume of 1.2 liters at 25°C. What will be the temperature in Kelvin if the volume is decreased to 0.8 liters at constant pressure?
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A balloon filled with helium has a volume of 3.0 liters at a temperature of 15°C. If the temperature is lowered to 5°C, what will be the new volume?
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A sample of gas is kept in a rigid container at a temperature of 350 K and has a volume of 5.0 liters. If the temperature increases to 500 K, what is the new volume of the gas?
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If a gas occupies 10 liters at 400 K, what would be its volume at 200 K?
Answers to Worksheet Problems
Here are the solutions to the above problems:
-
Given:
- ( V_1 = 4.5 L, T_1 = 300 K, T_2 = 600 K )
- Find: ( V_2 ) [ V_2 = \frac{4.5 L \times 600 K}{300 K} = 9.0 L ]
-
Given:
- ( V_1 = 1.2 L, T_1 = 25°C = 298.15 K, V_2 = 0.8 L )
- Find: ( T_2 ) [ T_2 = \frac{0.8 L \times 298.15 K}{1.2 L} = 199.45 K ]
-
Given:
- ( V_1 = 3.0 L, T_1 = 15°C = 288.15 K, T_2 = 5°C = 278.15 K )
- Find: ( V_2 ) [ V_2 = \frac{3.0 L \times 278.15 K}{288.15 K} \approx 2.91 L ]
-
Given:
- ( T_1 = 350 K, V_1 = 5.0 L, T_2 = 500 K )
- Find: ( V_2 ) [ V_2 = \frac{5.0 L \times 500 K}{350 K} \approx 7.14 L ]
-
Given:
- ( V_1 = 10 L, T_1 = 400 K, T_2 = 200 K )
- Find: ( V_2 ) [ V_2 = \frac{10 L \times 200 K}{400 K} = 5.0 L ]
Conclusion
Understanding Charles' Law and its application is a fundamental part of studying gas laws. By practicing with problems and worksheets, students can solidify their grasp of how temperature and volume relate at constant pressure. Utilize the concepts and methods discussed in this article to master Charles' Law! Happy studying! 🎉📚