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?

Chemistry
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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
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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
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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