If a balloon is initially at 25.0 °C with a volume of 2.80 L, what would the volume of the balloon be when the temperature rises to 125 °C? Assuming the pressure is constant.
If a balloon is initially at 25.0 °C with a volume of 2.80 L, what would the volume of the balloon be when the temperature rises to 125 °C? Assuming the pressure is constant.
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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![**Problem Description:**
If a balloon is initially at 25.0 °C with a volume of 2.80 L, what would the volume of the balloon be when the temperature rises to 125 °C? Assuming the pressure is constant.
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**Explanation and Solution:**
This problem can be solved using Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. The formula is:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where:
- \( V_1 \) is the initial volume (2.80 L),
- \( T_1 \) is the initial temperature in Kelvin,
- \( V_2 \) is the final volume,
- \( T_2 \) is the final temperature in Kelvin.
First, convert temperatures from Celsius to Kelvin:
\[ T_1 = 25.0 + 273.15 = 298.15 \, \text{K} \]
\[ T_2 = 125 + 273.15 = 398.15 \, \text{K} \]
Now, use Charles's Law to find the final volume \( V_2 \):
\[ \frac{2.80}{298.15} = \frac{V_2}{398.15} \]
Solve for \( V_2 \):
\[ V_2 = \frac{2.80 \times 398.15}{298.15} \]
\[ V_2 \approx 3.74 \, \text{L} \]
Therefore, when the temperature rises to 125 °C, the balloon will have a volume of approximately 3.74 L, assuming constant pressure.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F69d3c8b6-4012-43b0-80a6-d85ebf5fdc33%2Fb51a507a-51b5-4113-80b1-315f6900afd9%2F61hkue_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
If a balloon is initially at 25.0 °C with a volume of 2.80 L, what would the volume of the balloon be when the temperature rises to 125 °C? Assuming the pressure is constant.
---
**Explanation and Solution:**
This problem can be solved using Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. The formula is:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where:
- \( V_1 \) is the initial volume (2.80 L),
- \( T_1 \) is the initial temperature in Kelvin,
- \( V_2 \) is the final volume,
- \( T_2 \) is the final temperature in Kelvin.
First, convert temperatures from Celsius to Kelvin:
\[ T_1 = 25.0 + 273.15 = 298.15 \, \text{K} \]
\[ T_2 = 125 + 273.15 = 398.15 \, \text{K} \]
Now, use Charles's Law to find the final volume \( V_2 \):
\[ \frac{2.80}{298.15} = \frac{V_2}{398.15} \]
Solve for \( V_2 \):
\[ V_2 = \frac{2.80 \times 398.15}{298.15} \]
\[ V_2 \approx 3.74 \, \text{L} \]
Therefore, when the temperature rises to 125 °C, the balloon will have a volume of approximately 3.74 L, assuming constant pressure.
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