500 mL of nitrogen at `27^(@)C` is cooled to `-5^(@)C` at the same pressure. The new volume becomes

Options

A. 326.32 mL
B. 446.66 mL
C. 546.32 mL
D. 771.56 mL

Explanation

To find the new volume of nitrogen gas when it is cooled from 27°C to -5°C at constant pressure, we can use Charles's Law, which states that the volume of a gas is directly proportional to its absolute temperature when pressure is held constant.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin:**
- The initial temperature \( T_1 \) is given as 27°C. To convert this to Kelvin:
\[ T_1 = 27 + 273 = 300 \, \text{K} \]
- The final temperature \( T_2 \) is given as -5°C. To convert this to Kelvin:
\[ T_2 = -5 + 273 = 268 \, \text{K} \]

2. **Use Charles's Law:**
- According to Charles's Law:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
- Where:
- \( V_1 = 500 \, \text{mL} \) (initial volume)
- \( T_1 = 300 \, \text{K} \) (initial temperature)
- \( V_2 \) is the final volume we need to find.
- \( T_2 = 268 \, \text{K} \) (final temperature)

3. **Rearranging the Equation:**
- Rearranging the equation to solve for \( V_2 \):
\[ V_2 = V_1 \times \frac{T_2}{T_1} \]

4. **Substituting the Values:**
- Now substituting the known values into the equation:
\[ V_2 = 500 \, \text{mL} \times \frac{268 \, \text{K}}{300 \, \text{K}} \]

5. **Calculating \( V_2 \):**
- Performing the calculation:
\[ V_2 = 500 \times \frac{268}{300} = 500 \times 0.8933 \approx 446.66 \, \text{mL} \]

6. **Final Answer:**
- The new volume of nitrogen gas when cooled to -5°C is approximately:
\[ V_2 \approx 446.66 \, \text{mL} \]

### Summary:
The new volume of nitrogen gas after cooling from 27°C to -5°C at constant pressure is approximately **446.66 mL**.