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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Question
![**Problem Statement:**
A 36.4-L volume of methane gas presents a change of temperature from 76°C to 97°C at constant pressure. What is the final volume of the gas?
**Solution:**
To solve this problem, we can use Charles's Law, which states that, at constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin.
1. Convert the temperatures from Celsius to Kelvin:
\[
T_1 = 76°C + 273.15 = 349.15 \, K
\]
\[
T_2 = 97°C + 273.15 = 370.15 \, K
\]
2. Use the formula derived from Charles's Law:
\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]
Given \(V_1 = 36.4 \, L\), solve for \(V_2\):
\[
V_2 = V_1 \times \frac{T_2}{T_1} = 36.4 \times \frac{370.15}{349.15}
\]
3. Calculate the final volume:
\[
V_2 \approx 36.4 \times 1.06008 \approx 38.6 \, L
\]
Therefore, the final volume of the gas is approximately 38.6 liters.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd9907483-a41f-491a-9684-8dee08aa60aa%2F48c2c3fa-54f4-4573-8e10-e6435b435031%2Fvzi8ds_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A 36.4-L volume of methane gas presents a change of temperature from 76°C to 97°C at constant pressure. What is the final volume of the gas?
**Solution:**
To solve this problem, we can use Charles's Law, which states that, at constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin.
1. Convert the temperatures from Celsius to Kelvin:
\[
T_1 = 76°C + 273.15 = 349.15 \, K
\]
\[
T_2 = 97°C + 273.15 = 370.15 \, K
\]
2. Use the formula derived from Charles's Law:
\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]
Given \(V_1 = 36.4 \, L\), solve for \(V_2\):
\[
V_2 = V_1 \times \frac{T_2}{T_1} = 36.4 \times \frac{370.15}{349.15}
\]
3. Calculate the final volume:
\[
V_2 \approx 36.4 \times 1.06008 \approx 38.6 \, L
\]
Therefore, the final volume of the gas is approximately 38.6 liters.
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