When the pressure of 5 L of `N_(2)` is double and its temperature is raised from 300 K to 600 K, the final volume of the gas would be

Options

A. 10 L
B. 5 L
C. 15 L
D. 20 L

Explanation

To solve the problem, we will use the ideal gas law, which states that for a given amount of gas, the ratio of pressure (P), volume (V), and temperature (T) is constant. The relationship can be expressed as:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial volume, \( V_1 = 5 \, \text{L} \)
- Initial temperature, \( T_1 = 300 \, \text{K} \)
- Initial pressure, \( P_1 = x \) (let's assume it as \( x \))

2. **Identify Final Conditions**:
- Final temperature, \( T_2 = 600 \, \text{K} \)
- Final pressure, \( P_2 = 2x \) (since the pressure is doubled)

3. **Set Up the Equation**:
Using the ideal gas law, we can express the relationship between the initial and final states:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

4. **Rearrange to Solve for Final Volume \( V_2 \)**:
Rearranging the equation gives:
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

5. **Substitute Known Values**:
Substitute \( P_1 = x \), \( V_1 = 5 \, \text{L} \), \( T_2 = 600 \, \text{K} \), \( P_2 = 2x \), and \( T_1 = 300 \, \text{K} \):
\[ V_2 = \frac{x \cdot 5 \cdot 600}{2x \cdot 300} \]

6. **Simplify the Equation**:
The \( x \) terms cancel out:
\[ V_2 = \frac{5 \cdot 600}{2 \cdot 300} = \frac{3000}{600} = 5 \, \text{L} \]

7. **Final Answer**:
The final volume of the gas \( V_2 \) is \( 5 \, \text{L} \).