Charles Law Problems Worksheet Answers Explained
Table of Contents :
Charles's Law is a fundamental principle in chemistry that describes the relationship between the volume and temperature of a gas. Specifically, it states that the volume of a given mass of gas is directly proportional to its temperature (in Kelvin) when the pressure is held constant. This law is crucial for understanding gas behavior, and many problems associated with it can provide insights into real-world applications. In this blog post, we will delve into common problems related to Charles's Law, provide explanations, and clarify the answers step by step. Let's get started! 🌡️
Understanding Charles's Law
Charles's Law can be mathematically expressed 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)
Key Concepts
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Direct Proportionality: The volume increases as the temperature increases, provided the pressure remains constant. This is often demonstrated through various experiments and real-life scenarios.
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Absolute Temperature: Temperature must always be expressed in Kelvin when using Charles's Law. The conversion from Celsius to Kelvin can be done using the formula:
[ K = °C + 273.15 ]
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Applications: Understanding this law is essential in fields such as meteorology, engineering, and even cooking!
Common Charles's Law Problems
Let’s explore some typical problems and solutions related to Charles's Law, using our formula to derive the necessary answers.
Problem 1: Expanding Balloon
Problem Statement: A balloon has a volume of 2.0 L at a temperature of 20°C. What will its volume be if the temperature increases to 50°C?
Solution Steps:
-
Convert Temperatures to Kelvin:
- ( T_1 = 20°C = 20 + 273.15 = 293.15 K )
- ( T_2 = 50°C = 50 + 273.15 = 323.15 K )
-
Use Charles's Law:
- ( \frac{V_1}{T_1} = \frac{V_2}{T_2} )
- Substitute the known values: [ \frac{2.0 L}{293.15 K} = \frac{V_2}{323.15 K} ]
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Solve for ( V_2 ):
- Cross-multiply: [ V_2 = \frac{2.0 L \times 323.15 K}{293.15 K} \approx 2.21 L ]
Answer:
The volume of the balloon at 50°C is approximately 2.21 L. 🎈
Problem 2: Cooling Gas
Problem Statement: A gas occupies a volume of 5.0 L at a temperature of 300 K. If the temperature decreases to 250 K, what will be the new volume?
Solution Steps:
-
Known Values:
- ( V_1 = 5.0 L )
- ( T_1 = 300 K )
- ( T_2 = 250 K )
-
Use Charles's Law:
- ( \frac{V_1}{T_1} = \frac{V_2}{T_2} ) [ \frac{5.0 L}{300 K} = \frac{V_2}{250 K} ]
-
Solve for ( V_2 ):
- Cross-multiply: [ V_2 = \frac{5.0 L \times 250 K}{300 K} \approx 4.17 L ]
Answer:
The volume of the gas at 250 K is approximately 4.17 L. ❄️
Problem 3: Unknown Final Temperature
Problem Statement: A gas is contained in a cylinder with a volume of 3.0 L at 350 K. If the gas is allowed to expand to a volume of 4.5 L, what will be the new temperature?
Solution Steps:
-
Known Values:
- ( V_1 = 3.0 L )
- ( T_1 = 350 K )
- ( V_2 = 4.5 L )
-
Use Charles's Law:
- Rearranging the equation gives: [ T_2 = \frac{V_2 \times T_1}{V_1} ]
- Substitute the known values: [ T_2 = \frac{4.5 L \times 350 K}{3.0 L} \approx 525 K ]
Answer:
The new temperature after expansion is approximately 525 K. 🌞
Summary of Solutions
<table> <tr> <th>Problem</th> <th>Volume (L)</th> <th>Temperature (K)</th> </tr> <tr> <td>Balloon Expansion</td> <td>2.21 L</td> <td>323.15 K</td> </tr> <tr> <td>Cooling Gas</td> <td>4.17 L</td> <td>250 K</td> </tr> <tr> <td>Unknown Temperature</td> <td>4.5 L</td> <td>525 K</td> </tr> </table>
Important Notes
- Always remember to convert temperatures to Kelvin when working with gas laws.
- It is helpful to practice various problems to gain a better understanding of how temperature and volume relate in gases according to Charles's Law.
Understanding Charles's Law provides a foundation for more complex gas laws, including Boyle's Law and the Ideal Gas Law. Mastery of these concepts is crucial for success in chemistry and related fields, enhancing both academic knowledge and practical applications. With continued practice and exploration, you can confidently tackle a wide range of problems related to gas behavior!