A sample of mole is collected at J.8 atm and 12.3°G, what is the volume cf the gas? nitrogon gas that conteins 0.030 .....

General Chemistry - Standalone book (MindTap Course List)
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter5: The Gaseous State
Section: Chapter Questions
Problem 5.128QP: Plot the data given in Table 5.3 for oxygen at 0C to obtain an accurate molar mass for O2. To do...
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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
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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