What Celsius temperature, T₂, is required to change the volume of the gas sample in Part A (T₁ = 14 °C. V₁= 1.88×10³ 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.
What Celsius temperature, T₂, is required to change the volume of the gas sample in Part A (T₁ = 14 °C. V₁= 1.88×10³ 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.
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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What celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 14 Celsius, V1 = 1.88*10^3 L) to a volume of 3.76*10^3 L? Assume no change in pressure or the amount of gas in the balloon.
![**Temperature and Volume**
**Charles's Law** states that the volume of a gas is directly related to the absolute temperature when there is no change in the pressure or amount of gas.
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
**Figure Explanation:**
There is a visual representation of Charles's Law showing two pistons containing gas:
- **Left Piston:**
- It displays a lower temperature, indicated by fewer red-colored gas molecules and a shorter volume.
- The piston is closer to the base, reflecting a smaller volume.
- Caption: "Lower temperature, Smaller volume"
- **Right Piston:**
- It shows a higher temperature with more blue-colored gas molecules and an increased volume.
- The piston is pushed upwards, indicating a larger volume.
- Caption: "Higher temperature, Larger volume"
**Part B Question:**
What Celsius temperature, \( T_2 \), is required to change the volume of the gas sample in Part A (\( T_1 = 14^\circ C \), \( V_1 = 1.88 \times 10^3 \, L \)) to a volume of \( 3.76 \times 10^3 \, L \)? Assume no change in pressure or the amount of gas in the balloon.
Express your answer with the appropriate units.
\( T_2 = \)
**Answer Attempt:**
- Input: 289
- Unit: C
**Feedback:**
- Incorrect. Try Again; 2 attempts remaining.
- Enter your answer using units of temperature.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6375f9a1-49ba-4f50-83f6-5d24e4a38946%2Fa18089ac-50ac-45a6-b84e-5d5b776deb16%2F4gb9sxa_processed.png&w=3840&q=75)
Transcribed Image Text:**Temperature and Volume**
**Charles's Law** states that the volume of a gas is directly related to the absolute temperature when there is no change in the pressure or amount of gas.
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
**Figure Explanation:**
There is a visual representation of Charles's Law showing two pistons containing gas:
- **Left Piston:**
- It displays a lower temperature, indicated by fewer red-colored gas molecules and a shorter volume.
- The piston is closer to the base, reflecting a smaller volume.
- Caption: "Lower temperature, Smaller volume"
- **Right Piston:**
- It shows a higher temperature with more blue-colored gas molecules and an increased volume.
- The piston is pushed upwards, indicating a larger volume.
- Caption: "Higher temperature, Larger volume"
**Part B Question:**
What Celsius temperature, \( T_2 \), is required to change the volume of the gas sample in Part A (\( T_1 = 14^\circ C \), \( V_1 = 1.88 \times 10^3 \, L \)) to a volume of \( 3.76 \times 10^3 \, L \)? Assume no change in pressure or the amount of gas in the balloon.
Express your answer with the appropriate units.
\( T_2 = \)
**Answer Attempt:**
- Input: 289
- Unit: C
**Feedback:**
- Incorrect. Try Again; 2 attempts remaining.
- Enter your answer using units of temperature.
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