Part B What Celsius temperature, T₂, is required to change the volume of the gas sample in Part A (T₁ = 14 °C, V₁= 1.88x103 L) to a volume of 3.76x103 L? Assume no change in pressure or the amount of gas in the balloon. Express your answer with the appropriate units. ▸ View Available Hint(s) μA ?
Part B What Celsius temperature, T₂, is required to change the volume of the gas sample in Part A (T₁ = 14 °C, V₁= 1.88x103 L) to a volume of 3.76x103 L? Assume no change in pressure or the amount of gas in the balloon. Express your answer with the appropriate units. ▸ View Available Hint(s) μA ?
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![**Educational Resource: Understanding Charles's Law**
---
### Topic: Temperature and Volume
**Charles's Law** states that the volume of a gas is directly related to its absolute temperature, provided there is no change in the pressure or amount of gas. This is expressed mathematically as:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
---
### Figure Description
Two diagrams illustrate the relationship between temperature and gas volume:
1. **Left Diagram:**
- A container with gas particles at a **lower temperature**, resulting in a **smaller volume**.
- The particles are shown in blue with less spacing between them.
- A flame is shown beneath the container indicating heat.
2. **Right Diagram:**
- A container with gas particles at a **higher temperature**, leading to a **larger volume**.
- The particles have more space between them, indicated by their dispersion.
- A larger flame is depicted, indicating increased heat input.
---
### Practice Problem
**Part B:**
Determine the Celsius temperature \( T_2 \) needed to alter the volume of a gas sample from a starting condition:
- Initial Temperature \( T_1 = 14^\circ C \)
- Initial Volume \( V_1 = 1.88 \times 10^3 \, L \)
To a final volume of:
- \( V_2 = 3.76 \times 10^3 \, L \)
Assume no change in pressure or gas amount. Use appropriate units for the answer.
**Input Attempt:**
- Entered Value: \( T_2 = 289^\circ C \)
- Note: The input was incorrect, indicated by a prompt to "Try Again," with 2 attempts remaining.
---
Please use this information to further understand the application of Charles's Law in thermodynamics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6375f9a1-49ba-4f50-83f6-5d24e4a38946%2Fe4ec0e20-53c9-46a1-8ef4-be92f589070e%2Fimztcz8_processed.png&w=3840&q=75)
Transcribed Image Text:**Educational Resource: Understanding Charles's Law**
---
### Topic: Temperature and Volume
**Charles's Law** states that the volume of a gas is directly related to its absolute temperature, provided there is no change in the pressure or amount of gas. This is expressed mathematically as:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
---
### Figure Description
Two diagrams illustrate the relationship between temperature and gas volume:
1. **Left Diagram:**
- A container with gas particles at a **lower temperature**, resulting in a **smaller volume**.
- The particles are shown in blue with less spacing between them.
- A flame is shown beneath the container indicating heat.
2. **Right Diagram:**
- A container with gas particles at a **higher temperature**, leading to a **larger volume**.
- The particles have more space between them, indicated by their dispersion.
- A larger flame is depicted, indicating increased heat input.
---
### Practice Problem
**Part B:**
Determine the Celsius temperature \( T_2 \) needed to alter the volume of a gas sample from a starting condition:
- Initial Temperature \( T_1 = 14^\circ C \)
- Initial Volume \( V_1 = 1.88 \times 10^3 \, L \)
To a final volume of:
- \( V_2 = 3.76 \times 10^3 \, L \)
Assume no change in pressure or gas amount. Use appropriate units for the answer.
**Input Attempt:**
- Entered Value: \( T_2 = 289^\circ C \)
- Note: The input was incorrect, indicated by a prompt to "Try Again," with 2 attempts remaining.
---
Please use this information to further understand the application of Charles's Law in thermodynamics.
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