Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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![### Problem Statement
15.0 L of an ideal gas at 298 K and 3.36 bar are heated to 350 K with a new pressure of 5.35 bar. What is the new volume in litres?
### Explanation
This question involves an ideal gas undergoing changes in temperature and pressure. To find the new volume, we can use the combined gas law:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Where:
- \( P_1 \): Initial pressure = 3.36 bar
- \( V_1 \): Initial volume = 15.0 L
- \( T_1 \): Initial temperature = 298 K
- \( P_2 \): Final pressure = 5.35 bar
- \( T_2 \): Final temperature = 350 K
- \( V_2 \): Final volume (unknown)
### Steps to Solve:
1. **Rearrange the combined gas law to solve for the final volume (\( V_2 \))**:
\[
V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}
\]
2. **Substitute the given values into the equation**:
\[
V_2 = \frac{(3.36 \, \text{bar}) \times (15.0 \, \text{L}) \times (350 \, \text{K})}{(5.35 \, \text{bar}) \times (298 \, \text{K})}
\]
### Mathematical Solution:
\[
V_2 = \frac{(3.36) \times (15.0) \times (350)}{(5.35) \times (298)}
\]
Calculate step by step:
1. **Numerator Calculation**:
\[
3.36 \times 15.0 = 50.4
\]
\[
50.4 \times 350 = 17640
\]
2. **Denominator Calculation**:
\[
5.35 \times 298 = 1594.3
\]
3. **Final Volume Calculation**:
\[
V_2 = \frac{17640}{1594.3} \approx 11.06 \, \text{L}
\]
### Final](https://content.bartleby.com/qna-images/question/bc3b72c7-eb99-406f-83fe-82fc41b2a8b5/67d3c769-1286-4194-8426-9f5ff1c578b9/vglapkn_processed.jpeg)
Transcribed Image Text:### Problem Statement
15.0 L of an ideal gas at 298 K and 3.36 bar are heated to 350 K with a new pressure of 5.35 bar. What is the new volume in litres?
### Explanation
This question involves an ideal gas undergoing changes in temperature and pressure. To find the new volume, we can use the combined gas law:
\[
\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}
\]
Where:
- \( P_1 \): Initial pressure = 3.36 bar
- \( V_1 \): Initial volume = 15.0 L
- \( T_1 \): Initial temperature = 298 K
- \( P_2 \): Final pressure = 5.35 bar
- \( T_2 \): Final temperature = 350 K
- \( V_2 \): Final volume (unknown)
### Steps to Solve:
1. **Rearrange the combined gas law to solve for the final volume (\( V_2 \))**:
\[
V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}
\]
2. **Substitute the given values into the equation**:
\[
V_2 = \frac{(3.36 \, \text{bar}) \times (15.0 \, \text{L}) \times (350 \, \text{K})}{(5.35 \, \text{bar}) \times (298 \, \text{K})}
\]
### Mathematical Solution:
\[
V_2 = \frac{(3.36) \times (15.0) \times (350)}{(5.35) \times (298)}
\]
Calculate step by step:
1. **Numerator Calculation**:
\[
3.36 \times 15.0 = 50.4
\]
\[
50.4 \times 350 = 17640
\]
2. **Denominator Calculation**:
\[
5.35 \times 298 = 1594.3
\]
3. **Final Volume Calculation**:
\[
V_2 = \frac{17640}{1594.3} \approx 11.06 \, \text{L}
\]
### Final
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