A balloon containing a sample of helium gas is warmed in an oven at constant pressure. If the balloon measures 1.25 L at room temperature (20.0°C), what is its volume at 90.0°C? Enter your answer in the box provided.
A balloon containing a sample of helium gas is warmed in an oven at constant pressure. If the balloon measures 1.25 L at room temperature (20.0°C), what is its volume at 90.0°C? Enter your answer in the box provided.
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:**
A balloon containing a sample of helium gas is warmed in an oven at constant pressure. If the balloon measures 1.25 L at room temperature (20.0°C), what is its volume at 90.0°C?
**Answer:**
To solve this problem, use Charles's Law, which relates the volume and temperature of a gas at constant pressure:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where:
- \( V_1 = 1.25 \, \text{L} \) (initial volume)
- \( T_1 = 20.0°C = 293 \, \text{K} \)
- \( V_2 \) is the final volume
- \( T_2 = 90.0°C = 363 \, \text{K} \)
Rearranging for \( V_2 \), we get:
\[ V_2 = V_1 \times \frac{T_2}{T_1} \]
Substitute the values:
\[ V_2 = 1.25 \times \frac{363}{293} \, \text{L} \]
Calculate \( V_2 \) to find the final volume.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F073eb5ef-2617-4d6f-9bce-a81ca84c4945%2F6a74b398-b3c6-4978-88a6-63856b6d4cd1%2Fkcbruj.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem:**
A balloon containing a sample of helium gas is warmed in an oven at constant pressure. If the balloon measures 1.25 L at room temperature (20.0°C), what is its volume at 90.0°C?
**Answer:**
To solve this problem, use Charles's Law, which relates the volume and temperature of a gas at constant pressure:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
Where:
- \( V_1 = 1.25 \, \text{L} \) (initial volume)
- \( T_1 = 20.0°C = 293 \, \text{K} \)
- \( V_2 \) is the final volume
- \( T_2 = 90.0°C = 363 \, \text{K} \)
Rearranging for \( V_2 \), we get:
\[ V_2 = V_1 \times \frac{T_2}{T_1} \]
Substitute the values:
\[ V_2 = 1.25 \times \frac{363}{293} \, \text{L} \]
Calculate \( V_2 \) to find the final volume.
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