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 Statement
A sample of nitrogen gas that contains 0.030 mole is collected at 2.8 atm and 12.3°C. What is the volume of the gas?
### Solution Approach
To find the volume of the gas, we'll use the Ideal Gas Law equation, which is:
\[ PV = nRT \]
Where:
- \( P \) is the pressure of the gas in atmospheres (atm)
- \( V \) is the volume of the gas in liters (L)
- \( n \) is the number of moles of the gas
- \( R \) is the ideal gas constant, which is \( 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)
- \( T \) is the temperature of the gas in Kelvin (K)
### Steps
1. **Convert the temperature from Celsius to Kelvin**:
\[ T = 12.3^\circ\text{C} + 273.15 = 285.45 \, K \]
2. **Substitute the known values into the ideal gas law equation**:
\[ P = 2.8 \, \text{atm} \]
\[ n = 0.030 \, \text{moles} \]
\[ R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \]
\[ T = 285.45 \, K \]
\[ (2.8 \, \text{atm}) \cdot V = (0.030 \, \text{moles}) \cdot (0.0821 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}) \cdot 285.45 \, K \]
3. **Solve for \( V \)**:
\[ V = \frac{(0.030 \, \text{moles}) \cdot (0.0821 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf4fd61a-46d9-45b2-8502-88fe4eac45cd%2F1457db50-56cb-4cf9-ba6c-25dca2864b0c%2Fgs1x8on_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement
A sample of nitrogen gas that contains 0.030 mole is collected at 2.8 atm and 12.3°C. What is the volume of the gas?
### Solution Approach
To find the volume of the gas, we'll use the Ideal Gas Law equation, which is:
\[ PV = nRT \]
Where:
- \( P \) is the pressure of the gas in atmospheres (atm)
- \( V \) is the volume of the gas in liters (L)
- \( n \) is the number of moles of the gas
- \( R \) is the ideal gas constant, which is \( 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \)
- \( T \) is the temperature of the gas in Kelvin (K)
### Steps
1. **Convert the temperature from Celsius to Kelvin**:
\[ T = 12.3^\circ\text{C} + 273.15 = 285.45 \, K \]
2. **Substitute the known values into the ideal gas law equation**:
\[ P = 2.8 \, \text{atm} \]
\[ n = 0.030 \, \text{moles} \]
\[ R = 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1} \]
\[ T = 285.45 \, K \]
\[ (2.8 \, \text{atm}) \cdot V = (0.030 \, \text{moles}) \cdot (0.0821 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol}^{-1}) \cdot 285.45 \, K \]
3. **Solve for \( V \)**:
\[ V = \frac{(0.030 \, \text{moles}) \cdot (0.0821 \, \text{L} \cdot \text{atm} \cdot \text{K}^{-1} \cdot \text{mol
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